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CBSE Class 12 Maths Chapter 11 Three-Dimensional Geometry Important Questions

Question 1. The coordinates of the foot of the perpendicular drawn from the point (2, -3. 4) on the y-axis is?

1. (2, 3, 4)
2. (-2, -3, -4)
3. (0,-3,0)
4. (2,0,4)

Solution: 3. (0,-3,0)

The coordinates of the foot perpendicular on the y-axis from the points (2, -3, 4) are (0, -3, 0)

Question 2. Find the direction cosines of a line which makes equal angles with the coordinate axes.

Or,

A line passes through the point with position vector \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathrm{k}}\) and is in the direction of the vector \(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\). Find the equation of the line in Cartesian form.

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Solution:

Let the line make angles α, β, and γ with OX, OY, and OZ axes with its dc’s l, m, and n respectively

Given, \(\alpha=\beta=\gamma\)

∴ \(\ell^2+m^2+n^2=1 \Rightarrow \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\)

⇒ \(3 \cos ^2 \alpha=1\)

(because \(\alpha=\beta=\gamma\))

or \(\cos ^2 \alpha=\frac{1}{3} \Rightarrow \cos \alpha= \pm \frac{1}{\sqrt{3}}\)

∴ Direction cosines are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\) or \(\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\)

Or,

Since the line passes through the point (2,-1,4) and its direction ratios are 1,1,-2;

⇒ Equation of a line in cartesian form is : \(\frac{x-2}{1}=\frac{y+1}{1}=\frac{z-4}{-2}\)

Question 3. Show that the lines \(\frac{1-x}{2}=\frac{y-3}{4}=\frac{z}{-1} \text { and } \frac{x-4}{3}=\frac{2 y-2}{-4}=z-1\) are coplanar.
Solution:

Given equation of lines are \(\frac{1-x}{2}=\frac{y-3}{4}=\frac{z}{-1}\) and \(\frac{x-4}{3}=\frac{2 y-2}{-4}=z-1\)

⇒ \(\frac{x-1}{-2}=\frac{y-3}{4}=\frac{z-0}{-1} \text { and } \frac{x-4}{3}=\frac{y-1}{-2}=\frac{z-1}{1}\)

In vector form, both the lines can be expressed as : \(\vec{r}=(\hat{i}+3 \hat{j})+\lambda(-2 \hat{i}+4 \hat{j}-\hat{k}) \text { and } \vec{r}=(4 \hat{i}+\hat{j}+\hat{k})+\mu(3 \hat{i}-2 \hat{j}+\hat{k})\)

On comparing it with \(\vec{\mathrm{r}}=\vec{\mathrm{a}}_1+\lambda \vec{\mathrm{b}}_1 \text { and } \vec{\mathrm{r}}=\vec{\mathrm{a}}_2+\mu \vec{\mathrm{b}}_2 \text {; }\)

⇒ \(\vec{a}_1=\hat{i}+3 \hat{j}, \vec{b}_1=-2 \hat{i}+4 \hat{j}-\hat{k}\),

⇒ \(\vec{a}_2=4 \hat{i}+\hat{j}+\hat{k} \text { and } \vec{b}_2=3 \hat{i}-2 \hat{j}+\hat{k}\)

The given lines are coplanar if \(\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=0\)

Now; \(\vec{a}_2-\vec{a}_1=(4 \hat{i}+\hat{j}+\hat{k})-(\hat{i}+3 \hat{j})=3 \hat{i}-2 \hat{j}+\hat{k}\)

and \(\vec{b}_1 \times \vec{b}_2\)

= \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & 4 & -1 \\ 3 & -2 & 1\end{array}\right|\)=\(\hat{i}(4-2)-\hat{j}(-2+3)+\hat{k}(4-12)=2 \hat{i}-\hat{j}-8 \hat{k}\)

Now, \(\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=(3 \hat{i}-2 \hat{j}+\hat{k}) \cdot(2 \hat{i}-\hat{j}-8 \hat{k})\)=6+2-8=0

Hence, both lines are coplanar.

Question 4. Find the shortest distance between the following lines: r = \(3 \hat{i}+5 \hat{j}+7 \hat{k}+\lambda(\hat{i}-2 \hat{j}+\hat{k}) \text { and } \vec{r}=(-\hat{i}-\hat{j}-\hat{k})+\mu(7 \hat{i}-6 \hat{j}+\hat{k})\).
Solution:

Given lines are: \(\vec{r}=3 \hat{i}+5 \hat{j}+7 \hat{k}+\lambda(\hat{i}-2 \hat{j}+\hat{k})\)

and \(\vec{r}=(-\hat{i}-\hat{j}-\hat{k})+\mu(7 \hat{i}-6 \hat{j}+\hat{k})\)

⇒ \(\vec{a}_1=3 \hat{i}+5 \hat{j}+7 \hat{k}, \vec{b}_1=\hat{i}-2 \hat{j}+\hat{k}\)

⇒ \(\vec{a}_2=-\hat{i}-\hat{j}-\hat{k}, \vec{b}_2=7 \hat{i}-6 \hat{j}+\hat{k}\)

Now; \(\vec{a}_2-\vec{a}_1=-4 \hat{i}-6 \hat{j}-8 \hat{k}, \vec{b}_1 \times \vec{b}_2\)

= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 1 \\
7 & -6 & 1
\end{array}\right|=4 \hat{i}+6 \hat{j}+8 \hat{k}\)

∴ distance between lines is given by d = \(\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|\)

⇒ d = \(\left|\frac{(-4 \hat{i}-6 \hat{j}-8 \hat{k}) \cdot(4 \hat{i}+6 \hat{j}+8 \hat{k})}{|4 \hat{i}+6 \hat{j}+8 \hat{k}|}\right|=\left|\frac{-16-36-64}{\sqrt{16+36+64}}\right|\)

or d = \(\frac{116}{\sqrt{116}}=\sqrt{116}\) units

Question 5. Find the shortest distance between the lines: \(\vec{i}=2 \hat{i}-\hat{j}+\hat{k}+\lambda(3 \hat{i}-2 \hat{j}+5 \hat{k}) \text { and } \vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\mu(4 \hat{i}-\hat{j}+3 \hat{k})\)
Solution:

Given lines are: \(\vec{i}=2 \hat{i}-\hat{j}+\hat{k}+\lambda(3 \hat{i}-2 \hat{j}+5 \hat{k}) \text { and } \vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\mu(4 \hat{i}-\hat{j}+3 \hat{k})\)

Here; \(\vec{a}_1=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}_1=3 \hat{i}-2 \hat{j}+5 \hat{k}, \vec{a}_2=3 \hat{i}+2 \hat{j}-4 \hat{k}, \vec{b}_2=4 \hat{i}-\hat{j}+3 \hat{k}\)

⇒ \(\left(\vec{a}_2-\vec{a}_1\right)=\hat{i}+3 \hat{j}-5 \hat{k} \text { and } \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -2 & 5 \\
4 & -1 & 3
\end{array}\right|\)

⇒ \(\vec{b}_1 \times \vec{b}_2=-\hat{i}+11 \hat{j}+5 \hat{k}\)

∴ \(\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{1+121+25}=\sqrt{147}\)

⇒ The shortest distance b/w given lines is:

S.D. = \(\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|\)

= \(\left|\frac{(\hat{i}+3 \hat{j}-5 \hat{k}) \cdot(-\hat{i}+11 \hat{j}+5 \hat{k})}{\sqrt{147}}\right|\)

= \(\left|\frac{-1+33-25}{\sqrt{147}}\right|=\frac{7}{\sqrt{147}}\) units

Question 6. Find the shortest distance between the lines \(\vec{\mathrm{r}}=(4 \hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})\) and \(\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k})\).
Solution:

Equation of lines are Equation of lines are \(\vec{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})\) and \(\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k})\)

where \(\vec{a}_1=4 \hat{i}-\hat{j}, \vec{b}_1=\hat{i}+2 \hat{j}-3 \hat{k}\), \(\vec{a}_2=\hat{i}-\hat{j}+2 \hat{k}, \vec{b}_2=2 \hat{i}+4 \hat{j}-5 \hat{k}\)

⇒ \(\vec{a}_2-\vec{a}_1=-3 \hat{i}+2 \hat{k} \text { and } \vec{b}_1 \times \vec{b}_2\)

= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -3 \\
2 & 4 & -5
\end{array}\right|\)

∴ \(\vec{b}_1 \times \vec{b}_2=2 \hat{i}-\hat{j} \Rightarrow\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{5}\)

⇒ Required S.D. = \(\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\left|\frac{(-3 \hat{i}+2 k) \cdot(2 \hat{i}-\hat{j})}{\sqrt{5}}\right|=\left|\frac{-6}{\sqrt{5}}\right|=\frac{6}{\sqrt{5}}\) units

Question 7. Find the shortest distance between the following lines: \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Solution:

The given lines are \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\)….(1)

and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)….(2)

Line (1) passes through points (-1, -1, -1) with its Dr’s 7, -6, 1, and line (2) passes through points (3, 5, 7) with its Dr’s 1,-2, 1

So, vector equation of lines (1) and (2) are: \(\vec{r}_1=-\hat{i}-\hat{j}-\hat{k}+\lambda \cdot(7 \hat{i}-6 \hat{j}+\hat{k}) \text { and } \vec{r}_2=3 \hat{i}+5 \hat{j}+7 \hat{k}+\mu(\hat{i}-2 \hat{j}+\hat{k})\)

which are of the form: \(\vec{r}_1=\vec{a}_1+\lambda \vec{b}_1 \text { and } \vec{r}_2=\vec{a}_2+\mu \vec{b}_2\)

Here; \(\vec{a}_1=-\hat{i}-\hat{j}-\hat{k}, \vec{b}_1=7 \hat{i}-6 \hat{j}+\hat{k} \text {, }\)

⇒ \(\vec{\mathrm{a}}_2=3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}, \quad \vec{\mathrm{b}}_2=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)

⇒ \(\vec{\mathrm{a}}_2-\vec{\mathrm{a}}_1=4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}\)

and \(\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1\end{array}\right|=-4 \hat{i}-6 \hat{j}-8 \hat{k}\)

⇒ \(\left|\vec{\mathrm{b}}_1 \times \vec{\mathrm{b}}_2\right|=\sqrt{(-4)^2+(-6)^2+(-8)^2}=2 \sqrt{29}\)

∴ The shortest distance between given lines is:

S.D. = \(\left|\frac{\left(\vec{b}_1 \times \vec{b}_2\right) \cdot\left(\vec{a}_2-\vec{a}_1\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\left|\frac{(-4 \hat{i}-6 \hat{j}-8 \hat{k}) \cdot(4 \hat{i}+6 \hat{j}+8 \hat{k})}{2 \sqrt{29}}\right|\)

= \(\frac{116}{2 \sqrt{29}}=\frac{58}{\sqrt{29}}=2 \sqrt{29}\) units

CBSE Class 12 Maths Multiple Choice Questions And Answers

Question 1. If \(\mathrm{A}=\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\), then A is equal to

1. \(\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]\)
2. \(\left[\begin{array}{cc}0 & 2023 \\ 0 & 0\end{array}\right]\)
3. \(\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)
4. \(\left[\begin{array}{cc}2023 & 0 \\ 0 & 2023\end{array}\right]\)

Solution: 3. \(\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)

⇒ \(\mathrm{A}=\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\)

⇒ \(\mathrm{A}^2=\mathrm{A} \times \mathrm{A}=\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\)

= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

⇒ \(\mathrm{A}^{2023} = [latex]\mathrm{A}^2 \times \mathrm{A}^{2021}=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \cdot \mathrm{A}^{2021}=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Question 2. If \(\left[\begin{array}{ll}
2 & 0 \\
5 & 4
\end{array}\right]\)= P + Q, where P is symmetric and Q is a skew-symmetric matrix, then Q is equal to

1. \(\left[\begin{array}{cc}2 & 5 / 2 \\ 5 / 2 & 4\end{array}\right]\)
2. \(\left[\begin{array}{cc}0 & -5 / 2 \\ 5 / 2 & 0\end{array}\right]\)
3. \(\left[\begin{array}{cc}0 & 5 / 2 \\ -5 / 2 & 0\end{array}\right]\)
4. \(\left[\begin{array}{cc}2 & -5 / 2 \\ 5 / 2 & 4\end{array}\right]\)

Solution: 2. \(\left[\begin{array}{cc}0 & -5 / 2 \\ 5 / 2 & 0\end{array}\right]\)

⇒ \(\left[\begin{array}{ll}
2 & 0 \\
5 & 4
\end{array}\right]=\mathrm{P}+\mathrm{Q}\)

Let \(\mathrm{A}=\left[\begin{array}{ll}
2 & 0 \\
5 & 4
\end{array}\right]\)

A = \(\mathrm{P}+\mathrm{Q} \Rightarrow \mathrm{A}=\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)+\frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right)\)

∴ Q = \(\frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right)\)

= \(\frac{1}{2}\left\{\left[\begin{array}{ll}
2 & 0 \\
5 & 4
\end{array}\right]-\left[\begin{array}{ll}
2 & 5 \\
0 & 4
\end{array}\right]\right\}=\frac{1}{2}\left[\begin{array}{cc}
0 & -5 \\
5 & 0
\end{array}\right]=\left[\begin{array}{cc}
0 & -5 / 2 \\
5 / 2 & 0
\end{array}\right]\)

Question 3. If \(\left[\begin{array}{lll}
1 & 2 & 1 \\
2 & 3 & 1 \\
3 & a & 1
\end{array}\right]\) is a non-singular matrix and a ∈ A, then the set A is

1. R
2. {0}
3. {4}
4. R-{4}

Solution: 4. R-{4}

For non-singular matrix Δ ≠ 0

⇒ \(\left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 3 & 1 \\
3 & a & 1
\end{array}\right| \neq 0 \Rightarrow a-4 \neq 0 \Rightarrow a \neq 4 \Rightarrow A=R-\{4\}\) (because a ∈ A)

Question 4. If |A| = |k A|, where A is a square matrix of order 2, then the sum of all possible values of k is

1. 1
2. -1
3. 2
4. 0

Solution: 4. 0

|A|=|k A|

|A| = \(k^2|A|\)

⇒ \(k^2=1 \Rightarrow k= \pm 1\)

The sum of all possible values of k is 1+(-1)=0

Question 5. If \(\frac{d}{dx}\) [f(0)] = 0, then f(x) is equal to

1. \(a+b\)
2. \(\frac{a x^2}{2}+b x\)
3. \(\frac{a x^2}{2}+b x+c\)
4. \(b\)

Solution: 2. \(\frac{a x^2}{2}+b x\)

⇒ \(\frac{d}{d x}[f(x)]=a x+b\)

f(x) = \(\int(a x+b) d x\)

f(x) = \(\frac{a x^2}{2}+b x+C\)

f(0)=C=0

⇒ f(x) = \(\frac{a x^2}{2}+b x\)

Question 6. Degree of the differential equation \(\sin x+\cos \left(\frac{d y}{d x}\right)=y^2\) is

1. 2
2. 1
3. Not Defined
4. 0

Solution: 2. 1

⇒ \(\sin x+\cos \left(\frac{d y}{d x}\right)=y^2 \Rightarrow \cos \left(\frac{d y}{d x}\right)=y^2-\sin x \Rightarrow \frac{d y}{d x}\)

= \(\cos ^{-1}\left(y^2-\sin x\right)\)

∴ Degree =1

Question 7. The integrating factor of the differential equation \(\left(1-y^2\right) \frac{d x}{d y}+y x=a y \cdot(-1<y<1)\) is

1. \(\frac{1}{y^2-1}\)
2. \(\frac{1}{\sqrt{y^2-1}}\)
3. \(\frac{1}{1-y^2}\)
4. \(\frac{1}{\sqrt{1-y^2}}\)

Solution: 4. \(\frac{1}{\sqrt{1-y^2}}\)

⇒ \(\left(1-y^2\right) \frac{d x}{d y}+y x=a y,-1<y<1\)

⇒ \(\frac{d x}{d y}+x \cdot \frac{y}{1-y^2}=\frac{\text { ay }}{1-y^2}\)

This is a linear differential equation of the form, \(\frac{d x}{d y}+P x=Q\), where \(P=\frac{y}{1-y^2}\) and \(Q=\frac{a y}{1-y^2}\)

I.F. =\(\mathrm{e}^{\int \mathrm{p} \cdot \mathrm{dy}}=\mathrm{e}^{\int \frac{\mathrm{y}}{1-y^2} \mathrm{dy}}\)

Put \(1-\mathrm{y}^2=\mathrm{t} \Rightarrow-2 \mathrm{dy}=\mathrm{dt} \Rightarrow \mathrm{ydy}=-\frac{1}{2} \mathrm{dt}\)

I.F = \(\mathrm{e}^{-\frac{1}{2} \int \frac{\mathrm{d}}{1}}=\mathrm{e}^{-\frac{1}{2} \log \mathrm{t}}=\mathrm{e}^{\log \left(\mathrm{t}^{-122}\right.}\)

∴ I.F. = \(\frac{1}{\sqrt{t}}=\frac{1}{\sqrt{1-y^2}}\)

Question 8. Unit vector along \(\overline{\mathrm{PQ}}\), where coordinates of P and Q respectively are (2, 1,-1) and (4, 4, -7), is

1. \(2 \hat{i}+3 \hat{j}-6 \hat{k}\)
2. \(-2 \hat{i}-3 \hat{j}+6 \hat{k}\)
3. \(\frac{-2 \hat{i}}{7}-\frac{3 \hat{j}}{7}+\frac{6 \hat{k}}{7}\)
4. \(\frac{2 \hat{\mathrm{i}}}{7}+\frac{3 \hat{\mathrm{j}}}{7}-\frac{6 \hat{\mathrm{k}}}{7}\)

Solution: 4. \(\frac{2 \hat{\mathrm{i}}}{7}+\frac{3 \hat{\mathrm{j}}}{7}-\frac{6 \hat{\mathrm{k}}}{7}\)

⇒ PQ = \(\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}\)

⇒ PQ = \((4 \hat{i}+4 \hat{j}-7 \hat{k})-(2 \hat{i}+\hat{j}-\hat{k}) \Rightarrow P Q=2 \hat{i}+3 \hat{j}-6 \hat{k}\)

Unit vector along \(\overrightarrow{P Q}=\frac{\overrightarrow{P Q}}{\overrightarrow{|P Q|}}=\frac{2 \hat{i}+3 \hat{j}-6 \hat{k}}{\sqrt{49}}=\frac{2 \hat{i}}{7}+\frac{3 \hat{j}}{7}-\frac{6 \hat{k}}{7}\)

Question 9. The position vector of the mid-point of line segment AB is \(3 \hat{i}+2 \hat{j}-3 \hat{k}\). If the position vector of point A is \(2 \hat{i}+3 \hat{j}-4 \hat{k}\). then the position vector of the point B is

1. \(\frac{5 \hat{\mathrm{i}}}{2}+\frac{5 \hat{\mathrm{j}}}{2}-\frac{7 \hat{\mathrm{k}}}{2}\)
2. \(4 \hat{i}+\hat{j}-2 \hat{k}\)
3. \(5 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}\)
4. \(\frac{\hat{\mathrm{i}}}{2}-\frac{\hat{\mathrm{j}}}{2}+\frac{\hat{\mathrm{k}}}{2}\)

Solution: 2. \(4 \hat{i}+\hat{j}-2 \hat{k}\)

⇒ \(\overrightarrow{\mathrm{OC}}=\frac{\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}}{2}\)

⇒ 2 \(\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}} \)

⇒ \(\overrightarrow{\mathrm{OB}}=2 \overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} \)

= \(2(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})-(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})\)

= \(4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)

Question 10. Projection of vector \(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\) on the vector \(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}\) is

1. 0
2. 12
3. \(\frac{12}{\sqrt{13}}\)
4. \(\frac{-12}{\sqrt{13}}\)

Solution: 1. 0

Let \(\vec{a}=2 \hat{i}+3 \hat{j}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}\)

Projection \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}=\frac{6-6}{\sqrt{9+4}}=\frac{0}{\sqrt{13}}=0\)

Question 11. The equation of a line passing through the point (1, 1, 1) and parallel to the z-axis is

1. \(\frac{x}{1}=\frac{y}{1}=\frac{z}{1}\)
2. \(\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{1}\)
3. \(\frac{x}{0}=\frac{y}{0}=\frac{z-1}{1}\)
4. \(\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{1}\)

Solution: 4. \(\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{1}\)

Equation of the line passing through (1, 1, 1) and parallel to the z-axis is \(\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{1}\) (Dr’s of line parallel to z-axis are 0, 0, 1)

Question 12. If the sum of numbers obtained on throwing a pair of dice is 9, then the probability that the number obtained on one of the dice is 4, is

1. 1/9
2. 4/9
3. 1/18
4. 1/2

Solution: 4. 1/2

Let E: The number obtained on one of the dice is 4.

E = {(1,4), (2, 4), (3, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}

F: Sum of numbers a pair of dice is 9 F: {(6, 3), (3, 6), (5, 4), (4, 5)}

E∩F = {(5, 4). (4, 5)}

P(E/F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{2}{36}}{\frac{4}{36}}=\frac{1}{2}\)

Question 13. Anti-derivative of \(\frac{\tan x-1}{\tan x+1}\) with respect to x is

1. \(\sec ^2\left(\frac{\pi}{4}-x\right)+c\)
2. \(-\sec ^2\left(\frac{\pi}{4}-x\right)+c\)
3. \(\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c\)
4. \(-\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c\)

Solution: 3. \(\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c\)

Let I = \(\int \frac{\tan x-1}{\tan x+1} d x=-\int \frac{1-\tan x}{1+\tan x} d x=-\int \tan \left(\frac{\pi}{4}-x\right) d x\)

= \(\frac{-\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|}{-1}+C=\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+C\)

Question 14. If (a, b), (c, d), and (e, f) are the vertices of ΔABC and Δ denotes the area of ΔABC, then \(\left|\begin{array}{lll}
\mathrm{a} & \mathrm{c} & \mathrm{e} \\
\mathrm{b} & \mathrm{d} & \mathrm{f} \\
\mathrm{l} & \mathrm{l} & 1
\end{array}\right|\) is equal to

1. 2Δ²
2. 4Δ²
3. 2Δ
4. 4Δ

Solution: 2. 4Δ²

⇒ \(\Delta=\frac{1}{2}\left|\begin{array}{lll}
\mathrm{a} & \mathrm{b} & 1 \\
\mathrm{c} & \mathrm{d} & 1 \\
\mathrm{e} & \mathrm{f} & 1
\end{array}\right|\)

⇒ \(\left|\begin{array}{lll}
\mathrm{a} & \mathrm{b} & 1 \\
\mathrm{c} & \mathrm{d} & 1 \\
\mathrm{c} & \mathrm{f} & 1
\end{array}\right|=2 \Delta\)

⇒ \(\left|\begin{array}{lll}
\mathrm{a} & \mathrm{c} & \mathrm{c} \\
\mathrm{b} & \mathrm{d} & \mathrm{f} \\
1 & 1 & 1
\end{array}\right|=2 \Delta\)

[because |\(\mathrm{A}^{\prime}\)|= |A|]

⇒ \(\left|\begin{array}{lll}
\mathrm{a} & \mathrm{c} & \mathrm{e} \\
\mathrm{b} & \mathrm{d} & \mathrm{f} \\
\mathrm{l} & \mathrm{l} & 1
\end{array}\right|=4 \Delta^2\)

Question 15. The function f(x) = x |x| is

1. Continuous and differentiable at x = 0.
2. Continuous but not differentiable at x = 0.
3. Differentiable but not continuous at x = 0.
4. Neither differentiable nor continuous at x = 0.

Solution: 1. Continuous and differentiable at x = 0.

f(x) = \(x|x|=\left\{\begin{array}{cc}
x^2 ; x \geq 0 \\
-x^2 ; x<0
\end{array}\right.\)

⇒ \(f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h^2-0}{h}=0 \in R\)

⇒ \(f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{-h^2-0}{-h}=0 \in R\)

∴ \(f^{\prime}\left(0^{+}\right)=f^{\prime}\left(0^{-}\right)\)

So f(x) is differentiable at x=0.

Also \(\mathrm{f}(\mathrm{x})\) is continuous at x=0

Question 16. If \(\tan \left(\frac{x+y}{x-y}\right)=k\), then \(\frac{d y}{d x}\) is equal to

1. \(\frac{-y}{x}\)
2. \(\frac{y}{x}\)
3. \(\sec ^2\left(\frac{y}{x}\right)\)
4. \(-\sec ^2\left(\frac{y}{x}\right)\)

Solution: 2. \(\frac{y}{x}\)

⇒ \(\tan \left(\frac{x+y}{x-y}\right)=k \Rightarrow \frac{x+y}{x-y}=\tan ^{-1} k\)

Differentiate with respect to x, (x-y)\(\left[1+\frac{d y}{d x}\right]-(x+y)\left[1-\frac{d y}{d x}\right]=0\)

⇒ \(\frac{d y}{d x}[(x-y)+(x+y)]=(x+y)-(x-y) \Rightarrow \frac{d y}{d x}=\frac{2 y}{2 x}=\frac{y}{x}\)

Question 17. The objective function Z = ax + by of an LPP has a maximum value of 42 at (4, 6) and a minimum value of 19 at (3, 2). Which of the following is true?

1. a = 9, b = 1
2. a = 5, b = 2
3. a = 3, b = 5
4. a = 5, b = 3

Solution: 3. a = 3, b = 5

Z=ax+by

Let A(4,6), B(3,2)

∵ \(Z_A=42, Z_1=19\)

⇒ 4 a+6 b=42….(1)

and 3 a+2 b=19……(2)

from Equation (1) and (2)

∴ a=3, b=5

Question 18. The corner points of the feasible region of a linear programming problem are (0, 4), (8, 0) and \(\left(\frac{20}{3}, \frac{4}{3}\right)\). If Z = 30x + 24y is the objective function, then (maximum value of Z – minimum value of Z) is equal to

1. 40
2. 96
3. 120
4. 136

Solution: Bonus

Z_max – Z_min = 240- 96 = 144

Statement of Assertion (A) is followed by a statement of Reason (R).

1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. (A) is false but (R) is true.

Question 19. Assertion (A): Maximum value of (cos^-1 x)² is π².

Reason (R): Range of the principal value branch of \(\cos ^{-1} \mathrm{x} \text { is }\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text {. }\)

Solution: 3. (A) is true but (R) is false.

Assertion : \(\cos ^{-1} x \in[0, \pi], \forall x \in[-1,1]\)

max. of \(\cos ^{-1} x\) is \(\pi\)

∴ max of \(v\left(\cos ^{-1} x\right)^2=\pi^2\)

Hence, Assertion is true.

Reason: The range of the principal value branch of \(\cos ^{-1} \mathrm{x}\) is \([0, \pi]\).

The reason is false.

Question 20. Assertion (A): If a line makes angle α, β, γ with the positive direction of the coordinate axes, then sin²α + sin²β + sin²β = 2

Reason (R): The sum of squares of the direction cosines of a line is 1.

Solution: 1. Both (A) and (R) are true and (R) is the correct explanation of (A).

Assertion: \(l=\cos \alpha \cdot m=\cos \beta \cdot n=\cos \gamma\)

∵ \(l^2+m^2+n^2=1\)

∴ \(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\)

⇒ \(\left(1-\sin ^2 \alpha\right)+\left(1-\sin ^2 \beta\right)+\left(1-\sin ^2 \gamma\right)=1\)

⇒ \(\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=2\)

The assertion is true.

Reason: Reason is also true as \(l^2+m^2+n^2=1\).

Both (A) and (R) are true and R is the correct explanation of A.

Very Short Answer Type Questions And Answers

Question 1. Evaluate \(\sin ^{-1}\left(\sin \frac{3 \pi}{4}\right)+\cos ^{-1}(\cos \pi)+\tan ^{-1}(1) \text {. }\)

Or,

Draw the graph of cos x, where x [0, 1]. Also, write its range.

Solution:

sin \(^{-1}\left(\sin \frac{3 \pi}{4}\right)+\cos ^{-1}(\cos \pi)+\tan ^{-1}(1)\)

= \(\sin \left(\sin \left(\pi-\frac{\pi}{4}\right)\right)+\cos ^{-1}(\cos \pi)+\tan ^{-1}(1)\) (because Range of \(\sin ^{-1} x\) is \(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)

= \(\sin \left(\sin \frac{\pi}{4}\right)+\cos ^{-1}(\cos \pi)+\tan ^{-1}(1)=\frac{\pi}{4}+\pi+\frac{\pi}{4}=\frac{3 \pi}{2}\)

Or,

Let f(x) = cos^-1 x, where x ∈ [-1, 0]

Range of f(x) is (\(\frac{\pi}{\pi}\))

Question 2. A particle moves along the curve 3y = ax’ + 1 such that at a point with x-coordinate 1, the y-coordinate is changing twice as fast at x-coordinate. Find the value of a.
Solution:

Given cure, \(3 y=a x^3+1, x=1\) and \(\frac{d y}{d t}=2 \frac{d x}{d t} \Rightarrow \frac{d y}{d x}=2\)

Now, differentiating \(3 \mathrm{y}=\mathrm{ax}^3+1\) with respect to x \(\frac{3 \mathrm{dy}}{\mathrm{dx}}=3 \mathrm{ax}^2\)

3 x 2= \(3 \mathrm{a}(1)^2\) (because x=1)

6 = 3a

∴ a = 2

Question 3. If \(\vec{a}, \vec{b}, \vec{c}\) are three non-zero unequal vectors such that \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}\), yhen find the angle between \(\vec{a}\) and \(\vec{b}\) – \(\vec{c}\).
Solution:

Given that, \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}\)….(1)

Let angle between \(\vec{a}\) and \(\vec{b}-\vec{c}\) be \(\theta\) then \(\cos \theta=\frac{\vec{a} \cdot(\vec{b}-\vec{c})}{|\vec{a}||\vec{b}-\vec{c}|}\)

⇒ \(\cos \theta=\frac{\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{b}-\vec{c}|}\)

∵ \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} \neq \overrightarrow{0} \text { and } \overrightarrow{\mathrm{b}} \neq \overrightarrow{\mathrm{c}}\)

⇒ \(\cos \theta=0=\cos \frac{\pi}{2}\) (from equation (1))

⇒ \(\theta=\frac{\pi}{2}\)

Aliter: Given that, \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}\)

⇒ \(\vec{a} \cdot(\vec{b}-\vec{c})=0\)

Since \(\vec{a}, \vec{b}, \vec{c}\) are non-zero unequal vectors

∴ \(\vec{a} \perp(\vec{b}-\vec{c})\)

Hence angle between \(\vec{a}\) and \((\vec{b}-\vec{c})\) is \(\frac{\pi}{2}\).

Question 4. Find the coordinates of points on line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z+1}{2}\) which are at a distance of √11 units from the origin.
Solution:

Given line is \(\frac{x}{1}=\frac{y-1}{2}=\frac{z+1}{2}=\lambda\) (let)

Let arbitrary point on line is \(\mathrm{P}(\lambda, 2 \lambda+1,2 \lambda-1)\)

According to question \(\mathrm{OP}=\sqrt{11}\)

⇒ \(\sqrt{(\lambda-0)^2+(2 \lambda+1-0)^2+(2 \lambda-1-0)^2}=\sqrt{11}\)

⇒ \(\lambda^2+4 \lambda^2+4 \lambda+1+4 \lambda^2-4 \lambda+1=11 \Rightarrow 9 \lambda^2=9 \Rightarrow \lambda= \pm 1\)

Hence required points are (1,3,1)and (-1,-1,-3)

Question 5. If \(y=\sqrt{a x+b} \text {, prove that } y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2=0\)

Or,

If \(f(x)=\left\{\begin{array}{l}a x+b: 0<x \leq 1 \\ 2 x^2-x: 1<x<2\end{array}\right.\) is a differentiable function in (0, 2) then find the value of a and b.

Solution:

Given that, \(y=\sqrt{a x+b}\)….(1)

Differentiating with respect to x, \(\frac{d y}{d x}=\frac{1}{2 \sqrt{a x+b}} \times a=\frac{a}{2 y}\) [From eq.(1)]

⇒ \(2 y \frac{d y}{d x}=a \Rightarrow 2\left(\frac{d y}{d x}\right)^2+2 y \frac{d^2 y}{d x^2}=0\) [Again differentiating with respect to x]

⇒ \(y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0\)

Or,

Method-1: Given that. \(f(x)=\left\{\begin{array}{l}a x+b: 0<x \leq 1 \\ 2 x^2-x: 1<x<2\end{array}\right.\)

Every differentiable function is always continuous.

If f(x) is continuous at \(\mathrm{x}=1\) then \(\mathrm{RHL}=\mathrm{LHL}=\mathrm{f}(1)\)

R.H.L. \(\lim _{x \rightarrow 1} 2 x^2-x=1\)

f(1)=a+b

∴ a+b=1….(1)

Now, f(x) is differentiable at x=1

So, RHD = LHD

⇒ \(\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \mathrm{x}^2-\mathrm{x}\right)\right]_{\mathrm{x}=1}=\left[\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{ax}+\mathrm{b})\right]_{\mathrm{x}=1}\)

⇒ \((4 \mathrm{x}-1)_{\mathrm{x}=1}=\mathrm{a} \Rightarrow 4-1=\mathrm{a} \Rightarrow \mathrm{a}=3\)

3 + b = 1 [from eq.(1)]

∴ b = -2

Method-2: Given that, \(f(x)= \begin{cases}a x+b & ; 0<x \leq 1 \\ 2 x^2-x: & 1<x<2\end{cases}\)

Every differentiable function is always continuous. If f(x) is continuous at x=1 then RHL = LHL = f(1)

R.H.L. \(\lim _{x \rightarrow 1} 2 x^2-x=1\)

f(1)=a+b

∴ a+b=1….(1)

Now, f(x) is differentiable at x=1

∴ \(f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{-}\right)\)

⇒ \(\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}\)

⇒ \(\lim _{h \rightarrow 0} \frac{\left\{2(1+h)^2-(1+h)\right\}-(a+b)}{h}=\lim _{h \rightarrow 0} \frac{a(1-h)+b-(a+b)}{-h}\)

⇒ \(\lim _{h \rightarrow 0} \frac{2 h^2+3 h+(1-a-b)}{h}=\lim _{h \rightarrow 0} \frac{-a h}{-h}\)

⇒ \(\lim _{h \rightarrow 0} \frac{h(2 h+3)+(1-a-b)}{h}=a \Rightarrow \lim _{h \rightarrow 0}(2 h+3)+\lim _{h \rightarrow 0} \frac{1-a-b}{h}=a\)

⇒ \(\lim _{h \rightarrow 0}(2 h+3)+\lim _{h \rightarrow 0} \frac{1-1}{h}=a\) [From eq.(1)]

⇒ 3 + 0 = a

⇒ a = 3 and b = -2 [From eq.(1)]

Short Answer Type Questions And Answers

Question 1. Evaluate \(\int_0^{\pi / 4}\) log(1+tan x) dx

Or,

Find \(\int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}}\)

Solution:

Let I \(=\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x\)….(1)

⇒ I = \(\int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x \quad\left(\int_0^a f(x) d x=\int_0^a f(a-x) d x\right)\)

⇒ I = \(\int_0^{\frac{\pi}{4}} \log \left\{1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}\right\} d x\)

⇒ I = \(\int_0^{\frac{\pi}{4}} \log \left\{1+\frac{1-\tan x}{1+\tan x}\right\} d x\)

⇒ I = \(\int_0^{\frac{\pi}{1}}\left\{\log \frac{2}{(1+\tan x)}\right\} d x\)

⇒ I \(=\int_0^{\frac{\pi}{4}} \log 2 d x-\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x\)

⇒ \(\mathrm{I}=\int_0^{\frac{\pi}{4}} \log 2 \mathrm{dx}-\mathrm{I}\) (From (1))

⇒ \(2 \mathrm{I}=[\mathrm{x} \log 2]_9^{\frac{\pi}{4}}\)

⇒ \(2 \mathrm{I}=\frac{\pi}{4} \log 2 \Rightarrow \mathrm{I}=\frac{\pi}{8} \log 2\)

Or,

Let \(I=\int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}} \Rightarrow I=\int \frac{d x}{\sqrt{\sin ^3 x(\cos x \cdot \cos \alpha+\sin x \cdot \sin \alpha)}}\)

⇒ I = \(\int \frac{d x}{\sqrt{\sin ^4 x(\cot x \cdot \cos \alpha+\sin \alpha)}} \Rightarrow I=\int \frac{\mathrm{cosec}^2 x d x}{\sqrt{\cot x \cdot \cos \alpha+\sin \alpha}}\)

Put \(\cot x \cdot \cos \alpha+\sin \alpha=t^2 \Rightarrow-\mathrm{cosec}^2 x \cdot \cos \alpha d x=2 t d t \Rightarrow \mathrm{cosec}^2 x d x=\frac{-2 t d t}{\cos \alpha}\)

∴ I = \(-\frac{1}{\cos \alpha} \int \frac{2 \mathrm{t} d \mathrm{t}}{\sqrt{\mathrm{t}^2}} \Rightarrow \mathrm{I}=-\frac{2}{\cos \alpha} \int \mathrm{dt}=-\frac{2 \mathrm{t}}{\cos \alpha}+\mathrm{c}\)

I = \(-\frac{2}{\cos \alpha} \sqrt{\cot x \cos \alpha+\sin \alpha}+\mathrm{c} \Rightarrow \mathrm{I}=-\frac{2}{\cos \alpha} \sqrt{\frac{\cos x \cos \alpha+\sin \mathrm{x} \sin \alpha}{\sin x}}+\mathrm{c}\)

I = \(-\frac{2}{\cos \alpha} \sqrt{\frac{\cos (\mathrm{x}-\alpha)}{\sin x}}+\mathrm{c}\)

Question 2. Find \(\int e^{\cot ^{-1} x}\left(\frac{1-x+x^2}{1+x^2}\right) d x\)
Solution:

I = \(\int_{\log \sqrt{2}}^{\log \sqrt{5}} \frac{1}{\left(e^x+e^{-x}\right)\left(e^x-e^{-x}\right)} d x\)

I = \(\int_{\log \sqrt{2}}^{\log \sqrt{3}} \frac{e^{2 x}}{\left(e^{2 x}+1\right)\left(e^{2 x}-1\right)} d x\)

I = \(\int_{\log \sqrt{2}}^{\log \sqrt{3}} \frac{e^{2 x}}{\left(e^{2 x}\right)^2-1} d x\)

Put \(\mathrm{e}^{2 \mathrm{x}}=1 \Rightarrow \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}=\frac{1}{2} \mathrm{dt}\); when \(\mathrm{x}=\log \sqrt{3}\) then t =3; when \(\mathrm{x}=\log \sqrt{2}\) then t=2

Now, \(I=\frac{1}{2} \int_2^3 \frac{d t}{t^2-1}\)

Question 4. Find the general solution of the differential equation (xy-x²)dy = y² dx

Or,

Find the general solution of the differential equation \(\left(x^2+1\right) \frac{d y}{d x}+2 x y=\sqrt{x^2+4}\)

Solution:

Given (xy-x²)dy = y² dx

⇒ \(\frac{d y}{d x}=\frac{y^2}{x y-x^2}\)

Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

⇒ v+x \(\frac{d v}{d x}=\frac{v^2 x^2}{v x^2-x^2}\)

⇒ \(v+x \frac{d v}{d x}=\frac{v^2}{v-1}\)

⇒ \(x \frac{d v}{d x}=\frac{v^2}{v-1}-v\)

⇒ \(x \frac{d v}{d x}=\frac{v^2-v^2+v}{v-1}\)

⇒ \(x \frac{d v}{d x}=\frac{v}{v-1} \Rightarrow \frac{v-1}{v} d v=\frac{d x}{x}\)

On integrals both sides,

⇒ \(\int \frac{v-1}{v} d v=\int \frac{d x}{x} \Rightarrow \int\left(1-\frac{1}{v}\right) d v=\int \frac{d x}{x}\)

⇒ \(v-\log v=\log x+\log C\)

⇒ \(\frac{y}{x}-\log \frac{y}{x}=\log x+\log C\) (because y = vx)

⇒ \(\frac{y}{x}=\log \frac{y}{x}+\log x+\log C \Rightarrow \frac{y}{x}=\log C y\)

⇒ \(C y=e^{y / x} \Rightarrow y=C^{\prime} e^{y / x}\) (because \(\frac{1}{C}=C^{\prime}\))

Or,

Given differential equation is \(\left(x^2+1\right) \frac{d y}{d x}+2 x y=\sqrt{x^2+4}\)

⇒ \(\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{\sqrt{x^2+4}}{1+x^2}\)

This is a linear differential equation of the form \(\frac{d y}{d x}+P y=Q\).

where P = \(\frac{2 x}{1+x^2}\) and \(Q=\frac{\sqrt{x^2+4}}{1+x^2}\)

Now. I.F. = \(\mathrm{e}^{\int \mid {p dx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\left.\log x 1+\mathrm{x}^2\right)}\)

I.F. =1+x²

The solution of a given differential equation is given by

y (I.F.) = \(\int Q \cdot(\text { I.F. }) d x \Rightarrow y\left(1+x^2\right)=\int \frac{\sqrt{x^2+4}}{1+x^2}\left(1+x^2\right) d x\)

⇒ \(y\left(1+x^2\right)=\int \sqrt{x^2+4} d x\)

⇒ \(y\left(1+x^2\right)=\frac{x}{2} \int \sqrt{x^2+4}+2 \log \left|x+\sqrt{x^2+4}\right|+C\)

Question 5. Two balls are drawn at random one by one with replacements from an urn containing an equal number of red balls and green balls. Find the probability distribution of a number of red balls. Also, find the mean of the random variable.

Or,

A and B throw a die alternately till one of them gets a ‘6’ and wins the game. Find their respective probabilities of winning, if A starts the game first.

Solution:

Let X = number of reed balls, X = 0, 1, or 2

Mean = E(X) = ∑(X) P(X) = 1/2 + 1/2 = 1

Or,

Let S denote the success (getting a ‘6’) and F denote the failure (not getting a’6′)

∴ P(S) = 1/6, P(F) = 5/6

P(A wins in the first throw) = P(S) = 1/6

A gets the third throw, when the first throw by A and the second thrown by B result in failures.

i.e., P(A wins in third throw) = P(FFS) = P(F).P(F).P(S) = \(\frac{5}{6} \times \frac{5}{6}=\frac{1}{6}=\left(\frac{5}{6}\right)^2 \times \frac{1}{6}\)

Similarly P(A wins in the fifth throw) = P(FFFFS) = P(F) P(F) P(F) P(F) P(S)

= \(\left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)\) and so on

P(A wins) = \(\frac{1}{6}+\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)+\ldots \ldots . .=\frac{\frac{1}{6}}{1-\left(\frac{5}{6}\right)^{-2}}=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{\frac{1}{6}}{\frac{11}{36}}=\frac{6}{11}\)

P(B wins) = 1-P(A wins) = \(1-\frac{6}{11}=\frac{5}{11}\)

Question 6. Solve the following linear programming problem graphically. Minimize: Z = 5x + 10y

Subject to constraints:

• x + 2y ≤ 120
• x + y ≥ 60, x – 2y ≥ 0
• x ≥ 0, y ≥ 0

Solution:

Minimize Z = 5x + 10y

Subject to constraints x + 2y ≤120, x + y ≥ 60, x 0, x -2y ≥ 0

So, the feasible region lies in the first quadrant.

∴ The feasible region is ABCDA.

On solving equations x-2y = 0andx + y = 60. we get D(40,20)

And on solving equations x-2y = 0 and x + 2y = 120, we get C(60,30)

The corner points of the feasible region are. A(60,0), E(120,0) ,C(60,30)and D(40,20).

The values of Z at these points are as follows;

The minimum value of Z is 300 at A(60,0)

Long Answer Type Questions And Answers

Question 1. If \(A=\left[\begin{array}{ccc}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right], B=\left[\begin{array}{ccc}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]\), then find AB and use it to solve the following system of equations: x – 2y = 3, 2x – y – z = 2, and -2y + z = 3

Or,

If \(f(\alpha)=\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\), prove that f(α), f(-β) = f(α-β).

Solution:

Given, \(A=\left[\begin{array}{ccc}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right], B=\left[\begin{array}{ccc}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right]\)

AB = \(\left[\begin{array}{ccc}
-3 & -2 & -4 \\
2 & 1 & 2 \\
2 & 1 & 3
\end{array}\right]\left[\begin{array}{ccc}
1 & 2 & 0 \\
-2 & -1 & -2 \\
0 & -1 & 1
\end{array}\right]\)

= \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)

⇒ AB = \(\mathrm{I}\)

i.e. \(\mathrm{A}=\mathrm{B^-1}\)….(1)

Now, given system of equations, is

x-2y=3

2x-y-z=2

-2y+z=3

This system can be written in matrix equation form as

⇒ \({\left[\begin{array}{ccc}
1 & -2 & 0 \\
2 & -1 & -1 \\
0 & -2 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
3 \\
2 \\
3
\end{array}\right] }\)

B^T X=C

⇒ X = (B^T)^-1 C = (B^-1)^T C

[(A^T)^-1 = (A^-1)^T]

⇒ X = A^T C

⇒ \({\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{lll}
-3 & 2 & 2 \\
-2 & 1 & 1 \\
-4 & 2 & 3
\end{array}\right]\left[\begin{array}{l}
3 \\
2 \\
3
\end{array}\right] \Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
1 \\
-1 \\
1
\end{array}\right] }\)

∴ x=1, y=-1, z=1

Or,

Given, \(f(\alpha)=\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\)

f\((-\beta)=\left[\begin{array}{ccc}
\cos (-\beta) & -\sin (-\beta) & 0 \\
\sin (-\beta) & \cos (-\beta) & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\cos \beta & \sin \beta & 0 \\
-\sin \beta & \cos \beta & 0 \\
0 & 0 & 1
\end{array}\right]\)

∴ \(f(\alpha) \cdot f(-\beta)=\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{ccc}
\cos \beta & \sin \beta & 0 \\
-\sin \beta & \cos \beta & 0 \\
0 & 0 & 1
\end{array}\right]\)

⇒ f\((\alpha) \cdot f(-\beta)=\left[\begin{array}{ccc}
\cos \alpha \cos \beta+\sin \alpha \sin \beta & \cos \alpha \sin \beta-\sin \alpha \cos \beta & 0 \\
\sin \alpha \cos \beta-\cos \alpha \sin \beta & \sin \alpha \sin \beta+\cos \alpha \cos \beta & 0 \\
0 & 0 & 1
\end{array}\right]\)

⇒ f\((\alpha) \cdot f(-\beta)=\left[\begin{array}{ccc}
\cos (\alpha-\beta) & -\sin (\alpha-\beta) & 0 \\
\sin (\alpha-\beta) & \cos (\alpha-\beta) & 0 \\
0 & 0 & 1
\end{array}\right]\)

∴ \(f(\alpha), f(-\beta)=f(\alpha-\beta)\)

Question 2. Find the equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, -6), Q(5, -3,1), R(12, 4, 5) and S(11, 9, -2). Use these equations to find the point of intersection of diagonals.

Or,

A-line l passes through point (-1, 3, -2) and is perpendicular to both the lines \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) and \(\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}\). Find the vector equation of the line l. Hence obtain its distance from the origin.

Solution:

Given, the vertices of parallelogram PQRS are P(4, 2, -6), Q(5, -3,1), R(12, 4, 5) and S(11,9, -2).

Equation of diagonal PR is \(\frac{x-4}{12-4}=\frac{y-2}{4-2}=\frac{z+6}{5+6}\) i.e. \(\frac{x-4}{8}=\frac{y-2}{2}=\frac{z+6}{11}\)

Equation of diagonal QS is \(\frac{x-5}{11-5}=\frac{y+3}{9+3}=\frac{z-1}{-2-1}\) i.e. \(\frac{x-5}{6}=\frac{y+3}{12}=\frac{z-1}{-3}\)

Now, the coordinates of any point on diagonal PR is T(8λ + 4, 2λ+ 2, 11λ,- 6)

if point T also lies on the diagonal QS. then \(\frac{(8 \lambda+4)-5}{6}=\frac{(2 \lambda+2)+3}{12}=\frac{(11 \lambda-6)-1}{-3} \Rightarrow \frac{8 \lambda-1}{6}=\frac{2 \lambda+5}{12}=\frac{11 \lambda-7}{-3}\)

On solving, we get: λ = 1/2

Hence, the point of intersection of diagonals PR and QR is \(\mathrm{T}\left(8 \times \frac{1}{2}+4,2 \times \frac{1}{2}+2,11 \times \frac{1}{2}-6\right) \text { i.e. T }\left(8,3, \frac{-1}{2}\right)\)

Or,

Equation of a line l passes through point (-1, 3, 2 ) is \(\frac{x+1}{a}=\frac{y-3}{b}=\frac{z+2}{c}\)….(1)

If line l is perpendicular to lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}\) both, then a + 2b + 3c = 0 ……(2)

and -3a + 2b + 5c = 0 ….(2)

From equation (2) and (3) \(\frac{a}{10-6}=\frac{b}{-9-5}=\frac{c}{2+6} \text { i.c. } \frac{a}{4}=\frac{b}{-14}=\frac{c}{8}\)

So, Dr’s offline l are 4, – 14, 8 i.e. 2, -7, 4

Equation of line l in cartesian does is \(\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\)

Equation of line l in vector form is \(\vec{r}=(-\hat{i}+3 \hat{j}-2 \hat{k})+\lambda(2 \hat{i}-7 \hat{j}+4 \hat{k})\)

The coordinates of any point on the line l are P(2λ, -1, + 3, 4λ – 2)

dr’s of line OP are 2λ-1, -7λ + 3, 4λ -2

Since. OP ⊥ l

2(2λ – 1) -7 (-7λ + 3) + 4(4λ – 2) = 0

⇒ 69λ = 31

∴ λ = 31/69

So, the coordinates of point P are P\(\left(\frac{-7}{69}, \frac{-10}{69} \cdot \frac{-14}{69}\right)\)

Now, the distance of line l from the origin is

OP = \(\sqrt{\left(\frac{-7}{69}\right)^2+\left(\frac{-10}{69}\right)^2+\left(\frac{-14}{69}\right)^2}\)

= \(\sqrt{\frac{49+100+196}{(69)^2}}=\sqrt{\frac{345}{(69)^2}}=\sqrt{\frac{5}{69}} \text { units }\)

Aliter:

Distance of line (l) from the origin (0, 0, 0) is D = \(\frac{\left|\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)\right|}{|\vec{b}|}\)

⇒ D = \(\frac{|(2 \hat{i}-7 \hat{j}+4 \hat{k}) \times[(-\hat{i}+3 \hat{j}-2 \hat{k})-(0 \hat{i}+0 \hat{j}+0 \hat{k})]|}{|2 \hat{i}-7 \hat{j}+4 \hat{k}|}\)

⇒ D = \(\frac{|-2 \hat{i}+\hat{k}|}{|2 \hat{i}-7 \hat{j}+4 \hat{k}|}=\frac{\sqrt{5}}{\sqrt{69}} \text { units }\)

Question 3. Using integration, find the area of the region bounded by line y =√3x, the curve \(y=\sqrt{4-x^2}\), and the y-axis in the first quadrant.
Solution:

Given, line y = A√3X…..(1)

curve y = \(y=\sqrt{4-x^2}\)….(2)

from equation (1) and (2) \(y=\sqrt{4-x^2}\) = 3x

⇒ 4 -X²= 3X²

⇒ x² = 1

⇒ x = 1

from equation (1) when x = 1 ⇒ y = √3

The point of intersection of line (1) and curve (2) is B( 1, √3 )

Required Area = Area of the region APBOA

= Area of the region CBOC + area of the region APBCA

= \(\int_0^{\sqrt{3}} \frac{y}{\sqrt{3}} d y+\int_{\sqrt{3}}^2 \sqrt{4-y^2} d y=\left[\frac{y^2}{2 \sqrt{3}}\right]_0^{\sqrt{3}}+\left[\frac{y}{2} \sqrt{4-y^2}+\frac{4}{2} \sin ^{-1} \frac{y}{2}\right]_{\sqrt{3}}^2\)

= \(\left(\frac{\sqrt{3}}{2}-0\right)+\left[2 \cdot \frac{\pi}{2}-\left\{\frac{\sqrt{3}}{2}+2 \cdot \frac{\pi}{3}\right\}\right]=\frac{\sqrt{3}}{2}+\left[\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\right]=\frac{\pi}{3} \text { sq. units }\)

Question 4. A function f : [- 4, 4] → [O. 4] is given by f(x) = \(\sqrt{16-x^2}\) . Show that f is an onto function but not a one-one function. Further, find all possible values of ‘a’ for which f(a) =√7
Solution:

Given, function f:[-4,4] → [0,4] is given by \(f(x)=\sqrt{16-x^2}\)

One-one: f(-4) = \(\sqrt{16-(-4)^2}=0\)

f(4) = \(\sqrt{16-(4)^2}=0\)

⇒ \(f(-4)=f(4) \text { but }-4 \neq 4\)

Since different elements of the domain have the same image in the codomain.

So function f is not a one-to-one function.

Onto: Let \(y \in\lceil 0,4]\) such that \(f(x)=y=\sqrt{16-x^2} ; y \geq 0\)

⇒ x = \(\pm \sqrt{16-\mathrm{y}^2}\)

Here, \(16-y^2 \geq 0\)

⇒ (4-y)(4+y) \(\geq 0\)

⇒  (y-4)(y+4) \(\leq 0\)

⇒  \(-4 \leq y \leq 4 \text { and } y \geq 0\)

So, \(y \in[0,4]\)

Therefore the range of f=[0,4]= codomain of f.

So, function f is onto function.

if f(x)=\(\sqrt{7}\)

⇒ \(\sqrt{16-a^2}=\sqrt{7} \Rightarrow 16-a^2=7 \Rightarrow a^2=9 \Rightarrow a \pm 3\)

Assessment Questions And Answers

Question 1. Engine displacement is the measure of the cylinder volume swept by all the pistons of a piston engine. The piston moves inside the cylinder bore

The cylinder bore in the form of a circular cylinder open at the top is to be made from a metal sheet of area 75 π cm².

Based on the above information, answer the following questions:

1. If the radius of the cylinder is r cm and the height is h cm. then write the volume V of the cylinder in terms of radius r.
2. Find \(\frac{dV}{dr}\)
3. Find the radius of the cylinder when its volume is maximum.

Or,

For maximum volume, h > r. State true or false and justify.

Solution:

Given \(\mathrm{r}\) and \(\mathrm{h}\) be the radius and height of the cylinder bore open at the top.

Then, \(\pi \mathrm{r}^2+2 \pi \mathrm{rh}=75 \pi \mathrm{cm}^2\) (given)

⇒ \(\mathrm{r}^2+2 \mathrm{rh}=75 \Rightarrow \mathrm{h}=\frac{75-\mathrm{r}^2}{2 \mathrm{r}}\)

(1) \(\mathrm{V}=\pi \mathrm{r}^2 \mathrm{~h}\)

V = \(\pi r^2\left(\frac{75-r^2}{2 r}\right)=\frac{\pi}{2}\left[75 r-r^3\right]\) [from equation (1)]

(2) Differential equation (2) with respect to \(\frac{\mathrm{dV}}{\mathrm{dr}}=\frac{\pi}{2}\left[75-3 \mathrm{r}^2\right]\)

(3) For maximum volume \(\frac{\mathrm{dV}}{\mathrm{dr}}=0 \Rightarrow \frac{\pi}{2}\left[75-3 \mathrm{r}^2\right]=0 \Rightarrow \mathrm{r}^2=25 \Rightarrow \mathrm{r}=5\)

Now, \(\frac{d^2 V}{d r^2}=\frac{\pi}{2}[-6 r] \Rightarrow \frac{d^2 V}{d r^2})_{r=5}=-\frac{\pi}{2} \times 30<0\)

Hence, volume is maximum at r=5

Question 37. Recent studies suggest that roughly 12% of the world’s population is left-handed.

Depending on the parents, the chances of having a left-handed child are as follows :

• A: When both father and mother are left-handed: The chances of a left-handed child is 24%.
• B: When the father is right-handed and the mother is left-handed: The chances of a left-handed child is 22%.
• C: When the father is left-handed and the mother is right-handed: The chances of a left-handed child is 17%.
• D: When both father and mother are right-handed: The chances of a left-handed child is 9%.

Assuming that P(A) = P(B) = P(C) = P(D) = 1/4 and L denotes the event that the child is left-handed.

Based on the above information, answer the following questions:

1. Find P(L/C)
2. Find P(L/A)
3. Find P(A/L)

Or,

Find the probability that a randomly selected child is left-handed given that exactly one of the parents is left-handed.

Solution:

Given events are

• A: When both father and mother are left-handed: The chances of a left-handed child is 24%.
• B: When the father is right-handed and the mother is left-handed: The chances of left-handed children is 22%.
• C: When the father is left-handed and the mother is right-handed: The chances of a left-handed child is 17%.
• D: When both father and mother are right-handed: The chances of a left-handed child is 9%.

P(A) = P(B) = P(C) = P(D) = 1/4 and L: Child is left-handed.

Now, \(\mathrm{P}(\mathrm{L} / \mathrm{A})=\frac{24}{100}, \mathrm{P}(\mathrm{L} / \mathrm{B})=\frac{22}{100}, \mathrm{P}(\mathrm{L} / \mathrm{C})=\frac{17}{100}, \mathrm{P}(\mathrm{L} / \mathrm{D})=\frac{9}{100}\)

(1) \(\mathrm{P}(\mathrm{L} / \mathrm{C})=17 \%=\frac{17}{100}\)

(2) \(\mathrm{P}(\overline{\mathrm{L}} / \Lambda)=1-\mathrm{P}(\mathrm{L} / \mathrm{\Lambda})=1-\frac{24}{100}=\frac{76}{100}\)

(3) P(A/L)= \(\frac{P(A) \cdot P(L / A)}{P(A) \cdot P(L / A)+P(B) \cdot P(L / B)+P(C) \cdot P(L / C)+P(D) \cdot P(L / D)}\)

⇒ \(P(A / L)=\frac{\frac{1}{4} \times \frac{24}{100}}{\frac{1}{4} \times \frac{24}{100}+\frac{1}{4} \times \frac{22}{100}+\frac{1}{4} \times \frac{17}{100}+\frac{1}{4} \times \frac{9}{100}}=\frac{24}{24+22+17+9}=\frac{24}{72}=\frac{1}{3}\)

Or

Required probability = \(\mathrm{P}(\mathrm{L} / \mathrm{B} \cup \mathrm{C})=\frac{22}{100}+\frac{17}{100}=\frac{39}{100}\)

Question 3. The use of electric vehicles will curb air pollution in the long run.

The use of electric vehicles is increasing every year and the estimated electric vehicles in use at any time t is given by the function V: V(t) = 1/5 t³ — 5/2 t² + 25t-2 where t represents the time and t = 1, 2, 3 …. corresponds to years 2001, 2002, and 2003 respectively.

Based on the above information, answer the following questions

1. Can the above function be used to estimate the number of vehicles in the year 2000? Justify.
2. Prove that the function V(t) is an increasing function.

Solution:

Given that, \(V(t)=\frac{1}{5} t^3-\frac{5}{2} t^2+25 t-2\)….(1)

where t represents the time and t=1,2,3 \(\ldots\) corresponds to years 2001, 2002, 2003 ….respectively.

1. When \(\mathrm{t}=0\), then \(\mathrm{V}(0)=-2\) that means there is no vehicles are used in the year 2000.
2. Differentiating equation (1) concerning \(V^{\prime}(t)=\frac{3}{5} t^2-5 t+25=\frac{3}{5}\left(t^2-\frac{25}{3} t+\frac{125}{3}\right)\)

⇒ \(V^{\prime}(t)=\frac{3}{5}\left(t^2-\frac{25}{3} t+\left(\frac{25}{6}\right)^2-\left(\frac{25}{6}\right)^2+\frac{125}{3}\right)=\frac{3}{5}\left[\left(t-\frac{25}{6}\right)^2+\frac{875}{36}\right]\)

⇒ \(V^{\prime}(t)>0 \forall t \in 1,2,3 \ldots \ldots .\)

Hence, V(t) is increasing function.

CBSE Class 12 Maths Practical Notes Activity 1 Relation And Functions

Objective:  To verify that the relation R in the set L of all lines in a plane, defined by R = {(l,m):  \(\perp\)} is symmetric but neither reflexive nor transitive.

Material Required:

A piece of plywood, some pieces of wires (8). nails, white paper, glue, etc.

Method Of Construction:

Take a piece of plywood and paste a white paper on it. Fix the wires randomly on the plywood with the help of nails such that some of them are parallel, some are perpendicular to each other and some are inclined.

Demonstration:

1. Let the wires represent the lines \(l_1, l_2, \ldots, l_8\).

2. \(l_1\) is perpendicular to each of the lines \(l_2, l_3, l_4\).

3. \(l_6\) is perpendicular to \(l_7\).

4. \(l_2\) is parallel to \(l_3 \cdot l_7\) is parallel to \(l_4\) and \(l_5\) is parallel to \(l_8\).

5. \(\left(l_1, l_2\right),\left(l_1, l_3\right),\left(l_1, l_4\right),\left(l_6, l_7\right) \in \mathrm{R}\)

Observation :

1. In Given Above, no line is perpendicular to itself, so the relation R = {( l m): l \(\perp\) m) is not
reflexive (is/is not).

2. Given Above, \(l_1 \perp l_2.\mathrm{Is} l_2 \perp l_1\)? Yes (Yes/No)

⇒ \(\left(l_1, l_2\right) \in \mathrm{R} \Rightarrow\left(l_2, l_1\right) \in \mathrm{R}(\varepsilon / \in)\)

Similarly, \(l_1 \perp l_1,\) Is \(l_1 \perp l_2\)?  yes (Yes \No)

⇒ \(\left(l_3, l_1\right) \in \mathrm{R} \Rightarrow\left(l_1, l_1\right) \quad \mathrm{R}(\in / \in) \)

Also, \(l_6 \perp t_4\) Is \(l_7 \perp l_6 \text { ? yes (Yes/No) } \)

⇒ \(\left(l_0, l_0\right) \in \mathrm{R} \Rightarrow\left(l_3, l_4\right) \in \mathrm{R}(\epsilon / \in)\)

The relation R is symmetric (is/is not)

3. In , 1, \(l_1 \perp l_1\) and \(l_1 \perp l_3\). Is \(l_7 \perp l_3\) ? No. (Yes /No) i.e., \(\left(l_2, l_1\right) \in \mathrm{R}\) and \(\left(l_1, l_2\right) \in \mathrm{R} \Rightarrow\left(l_2, l_3\right) \notin \mathrm{R}(\notin f \in)\)

The relation R is not transitive (is/is not).

Result:

R = \(\{(l, \mathrm{~m}): l \perp \mathrm{m}\}\) is neither reflexive nor transitive but it is symmetric.

Application:

This activity can be used to check whether a given relation is an equivalence relation or not.

Viva Voice:

Question 1. Let R = {(a, b): a, b ∈ A} where A = (1. 2, 3. 4} if R is reflexive, write R in tabular form.
Answer:

R= {(1. 1). (2. 2), (3. 3), (4, 4)}.

Question 2. When does a relation R in set A called symmetric?
Answer:

If (a, b) ∈ R ⇒ (b, a) ∈ R for every a. b. c A, then the relation is called symmetric.

Question 3. When is a relation R in set A called reflexive?
Answer:

If (a, a) ∈ R. for every a ∈ A, it is called a reflexive relation.

Question 4. When is a relation R in set A called a transitive relation?
Answer:

If (a, b) ∈ R. (b. c) ∈ R ⇒ (a, c) ∈ R for every a. b, c ∈ A, then the relation is called transitive.

Question 5. If R – {(T₁, T₂): T₁ and T₂ are congruent triangles}, does R is reflexive?
Answer:

Yes. R is reflexive because each triangle is congruent to itself.

CBSE Class 12 Maths Practical Notes Activity 2 Relations And Functions

Objective: To verify whether the relation R in the set L of all lines in a plane, defined by R = {(l. m): l || m is an equivalence relation or not.

Material Required:

A piece of plywood, some pieces of wire (8). plywood, nails, white paper, glue.

Method Of Construction:

Take a piece of plywood of convenient size and paste a white paper on it. f ix the wares randomly on the plywood with the help of nails such that some of them are parallel, some are perpendicular to each other and some are inclined as shown below

Demonstration:

let the wires represent the lines \(l_1, l_7, \ldots, l_{\mathrm{x}}\).

⇒ \(l_1\) is perpendicular to each of the lines \(l_2, l_2, l_4\).

⇒ \(l_6\) is perpendicular to \(l_7\).

⇒ \(l_2\) is parallel to \(l_3, l_1\) is parallel to \(l_4\) and \(l_5\) is parallel to \(l_8\).

⇒ \(\left(I_2, l_3\right),\left(l_3, l_4\right),\left(l_5, l_8\right), \in \mathrm{R}\)

Observation:

1. In the above, every line is parallel to itself. So the relation R = {( l, m) : l || m } is reflexive relation (is/is not)

In The Above, observe that \(l_2 \| l_3\). Is \(l_3 \ldots l_2\)?

So,\(\left(l_3, l_3\right) \in \mathrm{R} \Rightarrow\left(l_3, l_2\right), \ldots \mathrm{R}(\mathbb{e} / \mathrm{e})\)

Similarly, \(l_3 \| l_4. Is l_4 \| l_3\) ?

So, \(\left(l_4, l_4\right) \in \mathrm{R} \Rightarrow\left(l_4, l_3\right) \in \mathrm{R}(\notin / \in)\)

and \(\left(l_3, l_8\right) \in \mathrm{R} \Rightarrow\left(l_8, l_3\right) \in \mathrm{R}(\notin / \in)\)

The relation R is symmetric (is/is not)

In the givenobserve that \(l_2 \| l_4\) and \(l_3 \| l_4\). Is \(l_2 \ldots l_4\)?

So, \(\left(l_2, l_2\right) \in \mathrm{R} and \left(l_3, l_4\right) \in \mathrm{R} \Rightarrow\left(l_2, l_4\right), \ldots \mathrm{R}(\in / \notin)\)

Similarly. \(l_3 \| l_4\) and \(l_4 \| l_2. Is l_3 \| l_2\) ?

So, \(\left(l_3, l_4\right) \in \mathrm{R},\left(l_4, l_2\right) \in \mathrm{R} \rightarrow\left(l_3, l_2\right) \in \mathrm{R}(\in, \notin)\)

Thus, the relation R is transitive (is/is not)

Hence, the relation R is reflexive and symmetric. and transitive. So, R is an equivalence relation.

Result:

The set of all lines in the plane that are parallel to each other defined as R = {(l, m): l|| m} is an equivalence relation.

Application:

This activity is useful in understanding the concept of an equivalence relation.

Viva Voice:

Question 1. If a relation is reflexive, symmetric, and transitive, then it is known as:
Answer:

An equivalence relation.

Question 2. If A = {1,2), B = {a, b}, then what is B x A?
Answer:

B x A = {(a. 1) (a. 2) (b, 1) (b, 2)}.

Question 3. What do you mean by an empty relation?
Answer:

A relation R in a set A is called an empty relation, if no element of A is related to any element of A i.e., \(\phi \subset A \times A\).

Question 4. Which methods are used to represent relation?
Answer:

There are two methods:

(1) Roster method (2) Set-builder method

Question 5. If R₁ and R₂ are two equivalence relations in set A. then R \(\cap R\). is equivalence or not.
Answer:

Yes R \(\cap\) R, will be an equivalence relation because both are reflexive, symmetric, and transitive therefore their intersection will be reflexive, symmetric, and transitive i.e. equivalence.

CBSE Class 12 Maths Practical Notes Activity 3 Relations And Functions

Objective: To demonstrate a function that is not one-one but is onto.

Material Required: Cardboard, nails, strings, adhesive, and plastic strips.

Method Of Construction:

1. Paste a plastic strip on the left-hand side of the cardboard and fix three nails on it as shown below. Name the nails as 1.2 and 3.

2. Paste another strip on the right-hand side of the cardboard and fix two nails on it as shown in Below. Name the nails as a and b.

3. Join nails on the left strip to the nails on the right strip as shown in the given below.

 

Demonstration:

1. Take the set X = {1,2.3}

2. Take the set Y = {a, b}

3. Join (correspondence) elements of X to the elements of Y as Shown Above

Observation:

1. The image of the element 1 of X in Y is a.

The image of the element 2 of X in Y is b.

The image of the element 3 of X in Y is b So, Given Above represents a Function.

2. Two elements in X have the Same image in Y. So. the function is not one-one (one-one/not one-one).

3. The pre-image of each element of Y in X exists (exists/does not exist). So. the function is onto (onto/not onto)

Result:

Here, the given function is not one-one but is onto.

Application:

This activity can be used to demonstrate the concept of one-one and function.

CBSE Class 12 Maths Practical Notes Activity  4 Relations And Functions

Objective:

To demonstrate a function that is one-one but not onto.

Material Required:

Cardboard, nails, strings, adhesive and plastic strips.

Method Of Construction:

1. Paste a plastic strip on the left-hand side of the cardboard and fix two nails in it as shown in the Given below. Name the nails as a and b.
2. Paste another strip on the right-hand side of the cardboard and fix three nails on it as shown in the given below. Name the nails on the right strip as 1.2 and 3.
3. Join nails on the left strip to the nails on the right strip as shown in the Given Below

Demonstration:

1. Take the set X = {a. b}

2. Take the set Y = {1,2.3}

3. Join (correspondence) elements of X to the elements of Y as shown in Above

Observation:

1. The image of the element a of X in Y is 2

The image of the clement b of X in Y is 3

So, The given Above represents a Function.

3. The pre-image of each element 1 of Y in X does not exist (exists/does not exist). So, the function is not onto (onto/not onto).

4. Thus, figure 4.3 represents a function that is one-one but not onto.

Result:

Here, The given function is one-on-one but not one-on-one.

Application:

This activity can be used to demonstrate the concept of one-one and onto function.

Viva-Voice:

Question 1. What is the domain of f(x) = \(\frac{1}{x-5}\) ?
Answer:

The domain of function f(x) is R – {5} because f(x) is not defined at x = 5.

Question 2. What are the domain and range of the function f = {(1,2). (4. 5). (6. 8)}?
Answer:

Domain is {1.4. 6). Range is {2. 5. 8}.

Question 3. Is every relation R a function?
Answer:

No: every relation is not necessarily a function.

Question 4. Is every function a relation?
Answer:

Yes. every function is a relation.

Question 5. Is {(1.2). (3. 4). (5. 6)} one-one and onto?
Answer:

yes, it is one-one and onto.

CBSE Class 12 Maths Practical Notes Activity 5 Relations And Functions

Objective:

To sketch the graphs of \(a^x\) and \(\log _3 x, a>0, a \neq 1\) and to examine that they are mirror images of each other.

Material Required:

Drawing board, geometrical instruments, drawing pins, thin wires, sketch pens, thick white paper, adhesive, pencil, eraser, a plane mirror, and squared paper.

Method Of Construction:

1. On the drawing board, fix a thick paper sheet of convenient size 20 cm x 20 cm (say) with adhesive.
2. On the sheet, take two perpendicular lines XOX’ and YOY’, depicting coordinate axes.
3. Mark graduations on the two axes as shown in the given
4. Find some ordered pairs satisfying y = ax and y logax. Plot these points corresponding to the ordered pairs and join them by free-hand curves in both cases.
5. Fix thin wires along these curves using drawing pins.
6. Draw the graph of y = x. and fix a wire along the graph, using drawing pins.

Demonstration:

1. For \(a^x\) take a = 2 (say), find ordered pairs satisfying it plot these ordered pairs on the squared paper, and fix a drawing pin at each point.

2. Join the bases of drawing pins with a thin wire. This will represent the graph of \(2^x\).

3. \(\log _2\) x=y = y gives x = \(2^x\), Some ordered pairs satisfying it are:

Plot these ordered pairs on the squared paper (graph paper) and fix a drawing pin at each plotted point.

Join the bases of the drawing pins with a thin wire. This will represent the graph of Logix.

4. Draw the graph of line y = x on the sheet.

5. Place a mirror along the wire representing y = x. It can be seen that the two graphs of the given functions are mirror images of each other in the line y = x.

Observation:

The image of the ordered pair (1. 2) on the graph of y = 2X in y = x is (2,1). It lies on the graph of y = \(\log _2 x\).

2. Image of the point (4. 2) on the graph y = log2X in y = x is (2,4) which lies on the graph of y= \(2^x\)

Repeat this process for some more points lying on the two graphs.

Result:

Hence; The graphs of ax and log x are mirror images of each other.

Application:

This activity is useful in understanding the concept of (exponential and logarithmic functions) which are mirror images of each other in y = x,

Viva Voice:

Question 1. What is the domain of the logarithmic function?
Answer:

The domain of the logarithmic function is (0. \(\infty\)) i.e., all positive numbers.

Question 2. What is the range of logarithmic function?
Answer:

All real numbers.

Question 3. Are logarithmic functions defined for negative values?
Answer:

No. The log can be found only of positive numbers.

Question 4. Which formulae are used in the calculation of the logarithm?
Answer:

⇒ \(\log _a(m \times n)=\log _a m+\log _a n\)

⇒ \(\log _a(m)^n=n \log _x m\)

⇒ \(\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n\)

Question 5. What is the use of logarithms?
Answer:

It is used to simplify the calculations in the field of Chemistry. Physics and Engineering.

CBSE Class 12 Maths Practical Notes Activity 6 Inverse Trigonometric Function

Objective:

To draw the graph of sin-1 x, use the graph of sin x and demonstrate the concept of mirror reflection (about the line y = x).

Material Required:

Cardboard, white chart paper, ruler, colored pens, adhesive, pencil, eraser, cutter, nails, and thin wires.

Method Of Construction:

1. Take cardboard of suitable dimensions, say, 30 cm x 30 cm.
2. On the cardboard, paste a white chart paper of size 25 cm x 25 cm (say).
3. On the paper, draw two lines, perpendicular to each other and name them XOX’ and YOY’ as rectangular axes.
4. Graduate the axes approximately as shown. by taking the unit on the X-axis = 1.25 times the unit on the Y-axis.
5. Mark approximately the points \((\frac{\pi}{6}, \sin \frac{\pi}{6}) \cdot(\frac{\pi}{4}, \sin \frac{\pi}{4}t) \ldots(\frac{\pi}{2}, \sin \frac{\pi}{2}\) in the coordinate plane and at each point fix a nail.
6. Repeat the above process on the other side of the x-axis, marking the points \(\left(\frac{-\pi}{6}, \sin \frac{-\pi}{6}\right) \cdot\left(\frac{-\pi}{4}, \sin \frac{-\pi}{4}\right), \ldots,\left(\frac{-\pi}{2}, \sin \frac{-\pi}{2}\right)\) approximately and fix nails on these points as \(\mathrm{N}_3^{\prime}, \mathrm{N}_2^{\prime}, \mathrm{N}_3^{+}, \mathrm{N}_4^{\prime}\), Also fix a nail at O.
7. Join the nails with the help of a tight wire on both sides of the x-axis to get the graph of sin x from \(\frac{-\pi}{2} \text { to } \frac{\pi}{2}\)
8. Draw the graph of the line y = x (by plotting the points (1, 1 ), (2, 2), (3, 3), … etc., and fixing a wire on these points).
9. From the nails \(\mathrm{N}_1, \mathrm{~N}_2, \mathrm{~N}_3, \mathrm{~N}_4\) draw perpendicular on the line y = x and produce these lines such that length of perpendicular on both sides of the line y – x are equal. At these points fix nails, \(\mathrm{I}_1, \mathrm{I}_2, \mathrm{I}_3, \mathrm{I}_4^{\prime}\)
10. Repeat the above activity on the other side of the X-axis and fix nails at \(\mathrm{I}_1, \mathrm{I}_2, \mathrm{I}_3, \mathrm{I}_4^{\prime}\)
11. Join the nails on both sides of the line y = x by a tight wire that will show the graph of y = sin’ x.

Demonstration:

Put a mirror on the line y = x. The image of the graph of sin x in the tile mirror will represent the graph of sin x showing that sin 1 x is a mirror reflection of sin x and vice versa.

Observation:

1. The image of point \(N_1\) in the mirror (the line y=x ) is  \(\mathbf{I}_1\).
2. The image of point \(\mathrm{N}_2\) in the mirror (the line \(\mathrm{y}=\mathrm{x} )\) is \(\mathrm{I}_2\).
3. The image of point N, in the mirror (the line y=x ) is \(1_3\).
4. The image of point \(\mathrm{N}_4\) in the mirror (the line \(\mathrm{y}=\mathrm{x} )\) is \(\mathbf{1}_4\).
5. The image of point \(\mathrm{N}_1^{+}\) in the mirror (the line \(\mathrm{y}=\mathrm{x} )\) is \(\mathrm{I}_1{ }^{\prime}\).
6. The image point of \(\mathrm{N}_2^{\prime}\) in the mirror (the line \(\mathrm{y}=\mathrm{x}) is \mathbf{I}_2{ }^{\prime}\).
7. The image point of \(\mathrm{N}_3^{\prime}\) in the mirror (the line \(\mathrm{y}=\mathrm{x} )\) is \(\mathbf{1}^{\prime}\).
8. The image point of \(\mathrm{N}_4^{+}\)in the mirror (the line \(\mathrm{y}=\mathrm{x} )\) is \(\mathbf{I}_{+}{ }^{\prime}\).
9. The image of the graph of sin x in y=x is the graph of {Sin}^{-1} x and the image of the graph of \(\sin ^{\prime} x\) in y=x is the graph of Sin x.

Result:

Hence; we have drawn the graph of Sin’ X using the graph of sin x and demonstrated the concept of mirror reflection (about the line y = x).

Application:

A similar activity can be performed for drawing the graphs of \(\cos ^{-1} x, \tan ^{-1} x\) etc.

Viva Voice:

Question 1. If We put the mirror on line v :x. then what will be the reflection of \(\sin ^{-1}x\).

Answer: sin x

Question 2. What is the value of \(\sin ^{-1} (sin x)\).
Answer:

⇒ \(\sin ^{-1}(\sin x)\)=x , where x \(\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

Question 3. What is \(\sin ^{-1}\left(\frac{1}{x}\right)\) ?
Answer:

⇒ \({cosec}^{-1} x\)

Question 4. How can you explain \(\sin ^{-1}\)x?
Answer:

⇒ \(\sin ^{-1}\)x is an angle, the value of whose sine is x.

Question 5. Whenever no branch, of an inverse trigonometric function is given then we consider which branch?
Answer:

Principal Value Branch

CBSE Class 12 Maths Practical Notes Activity 7 Continuity And Differentiability

Objective:

To find analytically the limit of a function f (x) at x = c and also to check the continuity of the function at that point.

Material Required: Paper, pencil, calculator.

Method Of Construction:

Consider the function given by f(x)=\(\{\begin{array}{cc}
\frac{x^2-16}{x-4} & x \neq 4 \\
10, & x=4
\end{array}\).

Take some points on the left and some points on the right side of c (= 4). which are very near to c.

Find the corresponding values of f (x) for each of the points considered in step 2 above.

Record the values of points on the left and right side of c as x and the corresponding values of f(x ) in the form of a table.

Demonstration:

The values of x and f (x) are recorded as follows:

Table 1: For points on the left of c (= 4).

Table 2: For points on the right of c (= 4).

Observation:

The value of f (x) is approaching 8 as x → 4 from the left.

The value off (x) is approaching to 8 , as x → 4 from the right

Result:

If lim \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow c} f(x)\). then f(x) is said to be continuous at x = c otherwise it is discontinuous.

Application:

This activity is useful in understanding the concept of limit and continuity of a function at a point.

Viva Voice:

Question 1. Is logarithmic function continuous everywhere?
Answer:

No. The logarithmic function is continuous only in its domain.

Question 2. What are the conditions for a function to be continuous at a point of its domain?
Answer:

Function f is said to be continuous at a point of its domain if the left-hand limit, right-hand limit, and value of the function at that particular point are equal.

Question 3. Are sine and cosine functions continuous everywhere?
Answer:

Yes. sine and cosine functions are continuous everywhere.

Question 4. Is the function defined by f(x) = |xj. a continuous function?
Answer:

Yes, the modulus function f(x) = |x| is continuous at all points.

Question 5. If f and g are two continuous functions at x = a, then what is the algebra of a continuous function?
Answer:

1. f + g is continuous at x = a.
2. f -g is continuous at x -a.
3. f.g is continuous at x = a.
4. \(\frac{\mathrm{f}}{\mathrm{g}}\) is continuous at x = a, provided g(a)\(\neq  0\)

CBSE Class 12 Maths Practical Notes Activity 8 Continuity And Differentiability

Objective:

To verify that for a function f to be continuous at a given point \(x_0 \cdot \Delta y=\left|f\left(x_0+\Delta x\right)-f\left(x_0\right)\right|\) is arbitrarily small provided. Ax is sufficiently small.

Material Required:

Hardboard, white sheets, pencil, scale, calculator, adhesive.

Method Of Construction :

1. Paste a white sheet on the hardboard.
2. Draw the curve of the given continuous function as represented in the Given Below.
3. Take any point A (\(x_0\). 0) on the positive side of the x-axis and corresponding to this point, mark the point P \(\left(x_0, y_0\right)\) on the curve.

Demonstration:

1. Take one more point \(M_1\left(x_0+\Delta x_1, 0\right)\) to the right of A, where \(\Delta x_1\) is an increment in x.
2. Draw the perpendicular from \(\mathrm{M}_1\), to meet the curve at \(\mathrm{N}_1\). Let the coordinates of \(\mathrm{N}_1 be\left(x_b+\Delta x_1 \cdot y_1+\Delta y_1\right)\)
3. Draw a perpendicular from the point \(\mathrm{P}\left(\mathrm{x}_{10}, \mathrm{y}_0\right)\) to meet \(\mathrm{N}_1 \mathrm{M}_1 at \mathrm{T}_1\).
4. Now measure \(\mathrm{AM}_1=\mathrm{Ax}_1\) (say) and record it and also measure \(\mathrm{N}_1 \mathrm{~T}_1=\Delta y_1\), and record it.
5. Reduce the increment in x to \(\Delta \mathrm{x}_2 (i.e.. \left.\Delta \mathrm{x}_2<\Delta \mathrm{x}_1\right)\) to get another point \(\mathrm{M}_2\left(\mathrm{x}_0+\Delta \mathrm{x}_2, 0\right)\). Get the corresponding point \mathrm{N}_2 on the curve
6. Let the perpendicular PT, intersects \(\mathrm{N}_2 \mathrm{M}_2 at \mathrm{T}_2\).
7. Again measure A \(M_2=\Delta x_2\) and record it. Measure \(N_2 T_2=\Delta y_2\) and record it.
8. Repeat the above steps for some more points so that \(\Delta\) x becomes smaller and smaller.

Observation:

So, \(\Delta\)y becomes Smaller when \(\Delta\)x becomes smaller.

Thus \(\lim _{x \rightarrow 0} \Delta\) y=0 for a continuous function.

Result:

Hence; we have verified that for a function f to be continuous at a given point \(x_0 \cdot \Delta y=\left|f\left(x_0+\Delta x\right)-f\left(x_0\right)\right|\) is arbitrarily small provided; \(\Delta\)x is sufficiently small.

Application:

This activity helps explain the concept of derivative (left hand or right hand) at any point on the curve corresponding to a function.

Viva Voice:

Question 1. A function f(x) is said to be continuous in an interval if it is continuous at every point in the domain of f(x). Is it true?

Answer: Yes

Question 2. Let f(x)=\(\begin{cases}{ll}\frac{|y|}{y}, & y \neq 0 \\ 0, & y=0\end{cases}\)., Is f continuous at y=0?
Answer:

No. \((\lim _{y \rightarrow 0}\). Does not exist as RHL =1, LHL=-1)

Question 3. Find the domain for the function \(f(x)=\frac{1}{x^2-7 x+12}\) where it represents a continuous function.
Answer:

⇒ \(\mathrm{R}-\{3,4\}\)

Question 4. Is the function defined by F(x)= \(\begin{cases}2 x-1, & x<0 \\ 2 x+1, & x \geq 0\end{cases}\) a continuous function at x=0
Answer:

Clearly. f(0)-2(0)+1-0+1=1

LHL =\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}[2(0-h)-1]\)

=\(\lim _{h \rightarrow 0}[-2 h-1]=0-1=-1\)

RHL =\(\lim _{x \rightarrow p^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0}[2(0+h)+1]=\lim _{h \rightarrow 0}[2 h+1]=1\)

CBSE Class 12 Maths Practical Notes Activity 9 Continuity And Differentiability

Objective: To verify Rolle’s Theorem.

Material Required:

A piece of plywood, wires of different lengths, white paper, sketch pen.

Method Of Construction:

1. Take a cardboard of a convenient size and paste a white paper on it.
2. Take two wires of convenient size and fix them on the white paper pasted on the plywood to represent the x-axis and y-axis
3. Take a piece of wire 15 cm in length bend it in the shape of a curve and fix it on the plywood as shown in the Given Below.
4. Take two straight wires of the same length and fix them in such a way that they are perpendicular to the x-axis at points A and B and meet the curve at points C and 1).

Demonstration:

1. In the figure, let the curve represent the function y = f(x). Let OA = a units and OB = b units.
2. The coordinates of the points A and B are (a. 0) and (b, 0), respectively.
3. There is no break in the curve in the interval [a, b]. So, the function f is continuous on [a, b],
4. The curve is smooth between x = a and x = b which means that at each point, a tangent can be drawn which in turn gives that the function f is differentiable in the interval (a, b).
5. As the wires at A and B are of equal lengths, i.e., AC = BD, so f (a) = f (b).
6. Because of steps (3), (4) and (5). conditions of Rolle’s theorem are satisfied. From the given above, we observe that tangents at P as well as Q are parallel to the x-axis. therefore f(x) at P and also at Q are zero.
7. Thus, there exists at least one value c of x in (a. b) such that f(c) = 0.
8. Hence, the Rolle’s theorem is verified.

Observation:

Let f(x)=\(x^2+2 x-8: x \in[-4,2]\)

a = -4 , b = 2

f(a)=0 , f(b)=0 Is f(a)=f(b) ?(Yes / No)

The slope of the tangent at P =0. So, \(\mathrm{f}(\mathrm{x})(\text { at } \mathrm{P})\)=0.

Result:

Thus, Rolle’s Theorem is verified.

Application:

This theorem may be used to find the roots of an equation.

Viva Voice

Question 1. What is Rolle’s Theorem?
Answer:

Let f be a real-valued function defined on [a, b] such that

f is contimuous on a, b

f is differentiable on (a, b)

f( a )=f( b )

Then. there exists at least a point c \(\in\)(a, b) such that \(f^{\prime}(\mathrm{c})\)=0

Question 2. If \(\left(\frac{d y}{d x}\right)_{x=0}\)=0 for a function, what is its meaning?
Answer:

Tangent to the curve at point x = a is parallel to the x-axis.

CBSE Class 12 Maths Practical Notes Activity 10 Continuity And Differentiability

Objective:

To verify Lagrange’s Mean Value Theorem.

Material Required:

A piece of plywood, wires, white paper, sketch pens, wires.

Method Of Instruction:

1. Take a piece of plywood and paste a white paper on it.
2. Take two wires of convenient size and fix them on the white paper pasted on the plywood to represent the x-axis and y-axis
3. Take a piece of wire about 10 cm in length and bend it in the shape of a curve. Fix this curved wire on the white paper pasted on the plywood.
4. Take two straight wires of lengths 10 cm and 13 cm and fix
   them at two different points of the curve parallel to the y-axis and their feet touching the x-axis.
5. Join the two points, where the two vertical wires meet the curve, using another wire.
6. Take one more wire of a suitable length and fix it in such a way that it is tangential to the curve and is parallel to the wire joining the two points on the curve.

Demonstration:

1. Let the curve represent the function y- f (x). In the figure, let OA = a units and OB = b units.
2. The coordinates of A and B are (a, 0) and (b, 0), respectively.
3. MN is a chord joining the points M (a. f (a) and N (b, f (b)).
4. PQ represents a tangent to the curve at the point. R (c, f (c)). in the interval (a, b).
5. f(c) is the slope of the tangent PQ at x = c.
6. \(\frac{f(b)-f(a)}{b-a}\) is the slope of the chord MN.
7. MN is parallel to PQ, therefore. f(c) = \(\frac{f(b)-f(a)}{b-a}\) Thus, Lagrange’s Mean Value Theorem is verified.

Observation:

1. Let f(x)=\(x^2-4 x-3 ; x \in[1,4]\)

a = 1 , b = 4 .

f(a)=-6 , f(b)=-3

2. f(b)-f(a)=3

b-a=3

3. \(\frac{f(b)-f(a)}{b-a}=1\) = Slope of MN.

4. Since \(\mathrm{PQ} \| \mathrm{MN} \Rightarrow\) Slope of \(\mathrm{PQ}=\mathrm{f}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}\)

2c-4 = 1 or c=2.5 \(\in\)[1.4]

Result :

Hence, we have verified Lagrange’s Mean Value Theorem.

Application:

Lagrange’s Mean Value Theorem has significant applications in calculus. For example, this theorem is used to explain the concavity of the graph.

CBSE Class 12 Maths Practical Notes Activity 11 Application Of Derivatives

Objective:

To understand the concepts of decreasing and increasing functions.

Material Required:

Pieces of wire of different lengths, pieces of plywood of suitable size, white paper, adhesive, geometry box, trigonometric tables.

Method Of Construction:

1. Take a piece of plywood of a convenient size and paste a white paper on it.
2. Take two pieces of wires of length say 20 cm each and fix them on the white paper to represent the x-axis and y-axis.
3. Take two more pieces of wire each of suitable length and bend them in the shape of curves representing two functions and fix them on the paper as shown.
4. Take two straight wires each of suitable length to show tangents to the curves at different points on them.

 

 

 

Demonstration:

1. Take one straight wire and place it on the curve (on the left) such that it is tangent to the curve at the point say \(P_1\), and make an angle \(\alpha_1\), with the positive direction of the x-axis.
2. \(\alpha_1\), is an obtuse angle, so tana\(\alpha_1\) is negative, i.e., the slope of the tangent at \(P_1 \)(a derivative of the function at P_1) is negative
3. Take another two points say \(P_2\) and\(P_3\) on the same curve* and make tangents using the same wire at \(P_2\) and\(P_3\) making \(\alpha_2\) And \(\alpha_3\), respectively with the positive direction of x-axis.
4. Here again \(\alpha_2\) And \(\alpha_3\). are obtuse angles and therefore slopes of the tangents tan \(\alpha_2\) And tan\(\alpha_3\) on are both negative, be, derivatives of the function at \(P_2\) and\(P_3\) are negative.
5. The function given by the curve (on the left) is a decreasing function,
6. On the curve (on the right), take three points \(Q_1, Q_2, Q_3 \)t and using the other straight wires, form tangents at each of these points making angles \(\beta_j, \beta_2, \beta_3\), respectively with the positive direction of the x-axis, as shown. \(\beta_j, \beta_2, \beta_3\),are all acute angles.
7. So, the derivatives of the function at these points are positive. Thus, the function given by this curve (on the right) is an increasing function.

Observation:

⇒ \(\alpha_1=\underline{120^{\circ}}>90^{\circ} \cdot \alpha_2-135^{\circ}>90^{\circ} \cdot \alpha_2=150^{\circ}>90^{\circ}, \tan \alpha_1\)

=\(-\sqrt{3}, (negative), \tan a_2=-1, (negative) \tan a_1-\frac{1}{\sqrt{3}}, (negative\)). Thus the function decreasing

⇒ \(\beta_1=\underline{30^{\circ}}<90^{\circ} \cdot \beta_2=45^{\circ},<90^{\circ}, \beta_3=60^{\circ},<90^{\circ}, \tan \beta_1\)

=\(\underline{1 / \sqrt{3}}. (positive), \tan \beta_2-1 (positive), \tan \beta_3-\underline{\sqrt{3}}(positive)\). Thus, the function is increasing.

Result:

1. The function is an increasing function at the point if the slope of the tangent is positive at that point.
2. The function decreases at the point if the slope of the tangent is negative at that point.
3. If a function is increasing in one interval and decreasing in another interval or vice-versa then for the entire interval, it is neither increasing nor decreasing.

Application:

Utis activity may be useful in explaining the concepts of decreasing and increasing functions.

Viva Voice:

Question 1. What is the condition for strictly increasing function f?
Answer:

f is strictly increasing in (a. b) if f'(x) > 0 for each x e (a. b).

Question 2. What do you mean by the critical point of a function?
Answer:

A point c in the domain of a function f is called a critical point at which either f’ (c) = 0 or f is not differentiable.

Question 3. What is the condition for strictly decreasing function f?
Answer:

f is strictly decreasing in (a. b) if f'(x) < 0 for each x \(\epsilon\) (a, b).

Question 4. If the curve has an upward trend as shown. Is it increasing or decreasing?
Answer:

It is an increasing function.

CBSE Class 12 Maths Practical Notes Activity 12 Application Of Derivatives

Objective:

To understand the concepts of local maxima. local minima and point of inflection.

Materials Required:

A piece of plywood, wires adhesive, white paper.

Methods Of Construction:

1. Take a piece of plywood of a convenient size and paste a white paper on it.
2. Take two pieces of wires each of length 40 cm and fix them on the paper on plywood in the form of the x-axis and y-axis.
3. Take another wire of a suitable length and bend it in the shape of a curve. Fix this curved wire on the white paper pasted on plywood
4. Take five more wires each of length say 2 cm and fix them at the points A. C. B. P and D as shown

Demonstration:

1. In the figure, wires at points A. B. C, and D represent tangents to the curve and are parallel to the axis.
2. The slopes of tangents at these points are zero, i.e….. the value of the first derivative at these points is zero. The tangent at P intersects the curve.
3. At points A and B the sign of the first derivative changes from negative to positive. So. they are the points of local minima.
4. At the points C and D. sign of the first derivative changes from positive to negative. So. they are the points of local maxima.
5. At this point, the P sign of the first derivative does not change. So. it is a point of inflection.

Observation:

1. The sign of the slope of the tangent (first derivative) at a point on the curve to the immediate left of A is negative.
2. The sign of the slope of the tangent (first derivative) at a point on the curve to the immediate right of A is positive.
3. The sign of the first derivative at a point on the curve to the immediate left of B is negative.
4. The sign of the first derivative at a point on the curve to the immediate right of B is positive.
5. The sign of the first derivative at a point on the curve to the immediate left of C is positive.
6. The sign of the first derivative at a point on the curve to the immediate right of C is negative.
7. The sign of the first derivative at a point on the curve to the immediate left of D is positive.
8. The sign of the first derivative at a point on the curve to the immediate right of D is negative.
9. The sign of the first derivative at a point immediate left of P is positive and the immediate right of P is positive.
10. A and B are points of local Minima.
11. C and D are points of local Maxima.
12. P is a point of inflection.

Result:

Points A and B are the points of local minima because as we cross the points from left to right the sign of f(x) [Slope of tangent] changes from -ve to +ve.

Points C and D are the points of local maxima because as we cross the points from left to right the value of f(x) i.e. the slope of tangent changes from +ve to -ve.

Point P is the point of inflection because the tangent line cuts the curve at point P.

Application:

This activity may help in explaining the concepts of points of local maxima, local minima, and inflection.

The concepts of maxima/minima are useful in problems of daily life such as making packages of maximum capacity at minimum cost.

CBSE Class 12 Maths Practical Notes Activity 13 Application Of Derivatives

Objective:

To construct an open box of maximum volume from a given rectangular sheet by cutting equal squares from each corner.

Material Required:

Chart papers, scissors, cellolape. calculator.

Method Of Construction:

1. Take a rectangular chart paper of size 20 cm x 10 cm and name it ABCD.
2. Cut four equal squares on each side x cm from each corner A. B. C and D.
3. Repeat the process by taking the same size of chart papers and different values of x.
4. Make an open box by folding its flaps using cello tape/adhesive.

Demonstration:

1. When x = 1. Volume of the box = 144 cm\(^3\)
2. When x = 1.5, Volume of the box = 178.5 cm\(^3\)
3. When x = 1.8, Volume of the box = 188.9 cm\(^3\).
4. When x = 2, Volume of the box = 192 cm\(^3\).
5. When x = 2.1. Volume of the box = 192.4 cm\(^3\).
6. When x = 2.2. Volume of the box = 192.2 cm\(^3\).
7. When x = 2.5, Volume of the box = 187.5 cm\(^3\).
8. When x = 3. Volume of the box = 168 cm\(^3\).
9. Clearly, the volume of the box is maximum when x = 2.1.

Observation:

1. \(V_1\)= Volume of the open box ( when x=1.6)=. .182 .784 cm\(^3\)
2. \(V_2\)= Volume of the open box ( when x =1.9)=190.836 cm \(^3\)
3. V = Volume of the open box ( when x =2.1)=.192 .444 cm \(^3\)
4. \(V_3\)= Volume of the open box ( when x =2.2)=192.192 cm \(^3\)}
5. V_4= Volume of the open box ( when x=2.4)=189.696 cm\(^3\)}
6. \(V_5\)= Volume of the open box ( when x-3.2)= 156.672 cm\(^3\)
7. Volume \(V_1\) is less than volume V.
8. Volume \(V_2\) is less than volume V.
9. Volume \(V_3\) is less than volume V.
10. Volume \(V_4\) is less than volume V.
11. Volume \(V_5\) is less than volume V.

Result:

The volume of the open box is maximum when x = 2.1 cm

Application:

1. This activity is useful in explaining the concepts of maxima/minima of functions.
2. It is also useful in making packages of maximum volume with minimum cost.
3. Let V denote the volume of the box.
4. Now V=(20-2 x)(10-2 x) x or V=200 x-60 \(x^3\)+4 x^3
5.  \(\frac{d V}{d x}=200-120 x+12 x^2\). For maxima or minima, we have, \(\frac{d V}{d x}-0\). i.e. 3 \(x^2-30\) x+50=0
6. i.e.. x=\(\frac{30 \pm \sqrt{900-600}}{6}=7.9\) or 2.1
7. Reject x=7.9.
8.  \(\frac{d^2 V}{d x^2}=-120+24 x\)
9. When \(\mathrm{x}=2.1, \frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{dx}}\) is negative.
10. Hence, V should be maximum at x =2.1.

Viva Voice:

Question 1. What do you mean by local maximum?
Answer:

f(x) is said to attain the local maximum at x = a if f (x) stops to increase and then begins to decrease as x increases through a.

Question 2. What do you mean by absolute maxima and absolute minima?
Answer:

Absolute maxima mean maximum value of f(x) in its interval or other words absolute maxima of f at x = c is f(c) \(\forall\) x s domain and absolute minima mean minimum value of f(x) in its interval in other word absolute minima of f at x = c is f(c), \(\forall\) x \(\varepsilon\)domain.

Question 3. What are the two rules to find local maxima and local minima?
Answer:

1. First-order derivative test.
2. Second-order derivative test.

Question 4. Can you construct an open box of maximum volume by cutting squares from each corner?

Answer: Yes

Question 5. What is the local maximum value?
Answer:

The value of f(x) at x = a. when function f (x) is a local maximum at x = a.

Question 6. What do you mean by local minimum?
Answer:

f (x) is said to attain the local minimum at x = a. if f(x) stops to decrease and then begins to increase as x increases or decreases through a.

Question 7. Can you find the volume of the open box, using the derivative?

Answer: Yes.

Question 8. What is the local minimum value?
Answer:

The value of f(x) at x = a. when f (x) is local minimum at x = a.

CBSE Class 12 Maths Practical Notes Activity 14 Application Of Derivatives

Objective:

To understand the concepts of absolute maximum and minimum values of a function in a given closed interval through its graph.

Material Required:

Drawing board, white chart paper, adhesive, geometry box. pencil and eraser, sketch pens, ruler, and calculator.

Method Of Construction:

1. Fix a white chart paper of convenient size on a drawing board using adhesive.
2. Draw two perpendicular lines on the squared paper as the two rectangular axes.
3. Graduate the two axes as shown in Fig
4. Let the given function be f (x)=(4\(\mathrm{x}^2\) – 9) (\(\mathrm{x}^2\) – 1) in the interval [-2. 2|.
5. Taking different values of x in |-2.2], find the values of f (x) and plot the ordered pairs (x, f (x).
6. Obtain the graph of the function by joining the plotted points by a free-hand curve as shown.
7. Demonstration:
8. Some ordered pairs satisfying f (x) are as follows:
9. Plotting these points on the chart paper and joining the points by a free-hand curve, the curve is obtained.

Observation:

The absolute maximum value of f(x) is 21 at x = ±2.

The absolute minimum value of the f(x) is -1.56 at x = ±1.27.

Result:

Hence, we have understood the concepts of absolute maximum and minimum values of a function in a given closed interval through its graph.

Application:

1. The activity is useful in explaining the concepts of the absolute maximum/minimum value of a function graphically.
2. Consider f(x)=\(\left(4 x^2-9\right)\left(x^2-1\right)\)
3. f(x)=0 gives the values of x as \(\pm \frac{3}{2}\) and \(\pm\) 1.
4. Both these values of x lie in the given closed interval [-2,2].
5. f(x)=\(\left(4 x^2-9\right) 2 x+8 x\left(x^2-1\right)=16 x^2-26 x=2 x\left(8 x^2-13\right)\)
6. f(x)=0 gives x=0, x= \(\pm \sqrt{\frac{13}{8}}- \pm 1.27\). These two values of x lie in [-2,2].
7. The function has local maxima/minima at x = 0 and x = ± 1.27. respectively.

Viva Voice:

Question 1. What is the point of absolute maximum?
Answer:

It is the point where f(x) has an absolute maximum value.

Question 2. What is the procedure to find the absolute minimum in [a, b]?
Answer:

We find the values of f(x) at all those points where fix) vanishes, we also find f(a) and f(b). Out of all these values, the least value will be the absolute minimum.

Question 3. What is the point of absolute minimum?
Answer:

It is the value of x where fix) has absolute minimum value.

Question 4. Find the absolute maximum-minimum value for the function fix) = \(x_3\) in [—2, 2].
Answer:

f(x) = 0 ⇒2x = 0 ⇒ x = 0

f(0) = 0, f(2) – 4, f(-2) = 4

Absolute maximum value of f(x) = 4 and absolute minimum value of f(x)

CBSE Class 12 Maths Practical Notes Activity 15 Application Of Derivatives

Objective:

To find the time when the area of a rectangle of given dimensions becomes maximum if the length is decreasing and the breadth is increasing at given rates.

Material Required:

Chart paper, paper cutter, scale, pencil, eraser, cardboard.

Method Of Construction:

1. Take a rectangle R of dimensions 16 cm \(\times\) 8 cm.
2. Let the length of the rectangle is decreasing at the rate of 1 cm second and the breadth is increasing at the rate of 2 cm / second.
3. Cut other rectangles \(R_2, R_2, R_1, R_4, R_4, R_4, R_4, R_4\) etc, of dimensions 15 cm \(\times\) 10 cm, 14 cm \(\times\) 12 cm, 13 cm \(\times\) 14 cm .12 cm \(\times\) 16 cm .11 cm \(\times\) 18 cm}, 10 cm \(\times\) 20 cm}, 9 cm \(\times\) 22 cm .8 cm \(\times\) 24 cm .
4. Paste these rectangles on cardboard.

Demonstration:

1. The length of the rectangle is decreasing at the rate of 1 cm/s and the breadth is increasing at the rate of 2 cm/s.
2. Area of the given rectangle \(R_4\)=16 \(\times\) 8=128 \(\mathrm{~cm}^2\).
3. Area of rectangle \(\mathrm{R}_{\mathrm{t}}=15 \times 10=150 \mathrm{~cm}^2\)( after 1 \(\mathrm{sec}\)).
4. Area of rectangle \(R_2\)=168 \(\mathrm{~cm}^2 \)(after 2 sec).
5. Area of reciangle \(\mathrm{R}_4\)=182 \(\mathrm{~cm}^2\) (after \(3 \mathrm{sec}\)).
6. Area of rectangle \(\mathrm{R}_4\)=192 \(\mathrm{~cm}^2 (after 4 \mathrm{sec}\) ).
7. Area of rectangle \(\mathrm{R}_{\mathrm{R}}\)=198 \(\mathrm{~cm}^2\) (after 5 sec ).
8. Area of rectangle \(\mathrm{R}_{\mathrm{r}}\)=200 \(\mathrm{~cm}^2\) (after 6 \(\mathrm{sec} )\).
9. Area of rectangle \(\mathrm{R}_{\mathrm{r}}\)=198 \(\mathrm{~cm}^2\) (after \(7 \mathrm{sec} \)) and so on.
10. Thus the area of the rectangle is maximum after 6 sec.

Observation:

1. Area of the rectangle \(R_2\)( after 1 sec)=150 \(\mathrm{~cm}^2\).
2. Area of the rectangle \(R_3\) (after \(\left.3
   \mathrm{sec}\right)=182 \mathrm{~cm}^2\).
3. Area of the rectangle \(R_6\) (afler \(\left.5 \mathrm{sec}\right)={198 \mathrm{~cm}^2}\).
4. Area of the rectangle R, (after 6 \(\mathrm{sec})=200 cm^2\).
5. Area of the rectangle \(\mathrm{R}_n[.latex]( after 7 [latex]\mathrm{sec})= 198 \mathrm{~cm}^2\).
6. Area of the rectangle \(R_v \)(after \(\left.8 \mathrm{sec}\right)=192 \mathrm{~cm}^2\).
7. Rectangle of Maximum area (after .6. seconds) =200 \(\mathrm{~cm}^2\).
8. The area of the rectangle is maximum after 6 sec.
9. The maximum area of the rectangle is 200 \(m^2\).

Result :

Hence, we have calculated the time when the area of a rectangle of given dimensions becomes maximum if the length is decreasing and the breadth is increasing at given rates.

Application:

1. Let the length and breadth of the rectangle be a and b.
2. The length of the rectangle after t seconds =\(\mathrm{a}-\mathrm{t}\).
3. The breadth of the rectangle after t seconds =b+2 t.
4. Area of the rectangle ( after t \sec )=\(\Delta(t)=\langle a-t)(b+2 t)=a b-b t+2 a t-2 t^2\)
5. \(A^{\prime}(t)\)=-b+2 a-4 t
6. For maxima or minima, \(\mathrm{A}^{\prime}(\mathrm{t})\)=0.

• • \(A^{\prime}(t)=0 t=\frac{2 a-b}{4}\)
  • \(A^{\prime \prime}(t)\)=-4
  • \(A^{\prime \prime}\left(\frac{2 a-b}{4}\right)\)=-4, which is negative

1. Thus, A(t) is maximum at t=\(\frac{2 a-b}{4}\) seconds.
2. Here, \(\mathrm{a}-16 \mathrm{~cm}, \mathrm{~b}=8 \mathrm{~cm}\).
3. Thus, \(\mathrm{t}=\frac{32-8}{4}=\frac{24}{4}\)=6 seconds
4. Hence, after 6 seconds, the area will become maximum.

Viva Voice

Question 1. What is the length of the rectangle after t seconds, if it is decreasing by 2 cm every second?
Answer:

The length of the rectangle after t seconds is (a – 2t) cm. where a is its initial length.

Question 2. What is the area of the rectangle after t seconds if the length is decreasing at 5 cm/sec and the breadth is increasing at 3 cm/sec?
Answer:

Area of rectangle after t seconds = (a – 5t) (b+ 3t) cm².

Where a and b are initial values of length and breadth.

Question 3. What is the value of A'(t) at the maximum or minimum area where A(t) is the area of a rectangle?
Answer:

A'(t) = 0 when the area is maximum or minimum.

Question 4. The length x of a rectangle is decreasing at the rate of cm/minute and width y is increasing at the rate of ’b’ cm/minute. When x = 3 and y = 9, find the rate of change of the area of the rectangle.
Answer:

1. Given : \(\frac{\mathrm{dx}}{\mathrm{dt}}\)= a cm / min
2. and \(\frac{\mathrm{dy}}{\mathrm{dt}}\)= b cm / min .
3. A=x y
4. \(\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{dx}}{\mathrm{dt}} \mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dt}}\)
5. =-a y+b x
6. When x-3 & y=9, then \(\frac{d A}{d t}\)=3 b-9 a.

CBSE Class 12 Maths Practical Notes Activity 16 Application Of Derivatives

Objective:

To verify that amongst all the rectangles of the same perimeter, the square has the maximum area.

Material Required:

Chart paper, paper cutter, scale, pencil, eraser cardboard, glue.

Method Of Construction :

1. Take a cardboard of a convenient size and paste a white paper on it.
2. Make rectangles for each of the perimeters (say 48 cm) on chart paper. Rectangles of different dimensions are as follows:
3. \(\mathrm{R}_1: 16 \mathrm{~cm} \times 8 \mathrm{~cm},\mathrm{R}_2: 15 \mathrm{~cm} \times 9 \mathrm{~cm}\)
4. \(\mathrm{R}_3: 14 \mathrm{~cm} \times 10 \mathrm{~cm}, \mathrm{R}_4: 13 \mathrm{~cm} \times 11 \mathrm{~cm}\)
5. \(\mathrm{R}_5: 12 \mathrm{~cm} \times 12 \mathrm{~cm}, \mathrm{R}_6: 12.5 \mathrm{~cm} \times 11.5 \mathrm{~cm}\)
6. \(\mathrm{R}_7: 10.5 \mathrm{~cm} \times 13.5 \mathrm{~cm}\)
7. Cut out these rectangles and paste them on the white paper on the cardboard
8. Repeat step 2 for more rectangles of different dimensions, each having a perimeter of 48 cm.
9. Paste these rectangles on cardboard.

Demonstration:

1. Area of rectangle of \(\mathrm{R}_1=16 \mathrm{~cm} \times 8 \mathrm{~cm}=128 \mathrm{~cm}^2\)
2. Area of rectangle \(\mathrm{R}_2=15 \mathrm{~cm} \times 9 \mathrm{~cm}=135 \mathrm{~cm}^2\)
3. Area of \(\mathrm{R}_3=140 \mathrm{~cm}^2\)
4. Area of \(R_4=143 \mathrm{~cm}^2\)
5. Area of \(\mathrm{R}_5=144 \mathrm{~cm}^2\)
6. Area of \(\mathrm{R}_{\mathrm{r}}=143.75 \mathrm{~cm}^2\)
7. Area of R =141.75 \(\mathrm{~cm}^2\)
8. The perimeter of each rectangle is the same but their area is different. The area of rectangle R5 is the maximum. It is a square of side 12 cm.

Observation:

1. Perimeter of each rectangle \(R_1, R_2, R_3, R_4, R_4, R_6, R_2\) is Same(i.e. 48 cm )
2. Area of the rectangle \(R_1\) is less than the area of rectangle R,
3. Area of the rectangle \(R_0\) is less than the area of rectangle \(R_3\),
4. The rectangle R has the dimensions 12 cm x 12cm and hence it is a Square
5. Of all the rectangles with the same perimeter, the Square has the maximum area.

Result:

Hence, amongst all rectangles, the square has the maximum area.

Application:

1. This activity is useful in explaining the idea of Maximum of a function.
2. Let the length and breadth of the rectangle be x and y.
3. The perimeter of the rectangle is P=48 cm.
4. 2(x+y)=48 or x+y=24 or y=24-x
5. Let A(x) be the area of a rectangle. then A(x) =x y=x(24-x)
6. =24 \(x-x^2\)
7. \(N^{\prime}(x)\) =24-2 x
8. 24-2 x=0, x=12
9. \(A^{\prime \prime}\)(x)=-2
10. \(A^{\prime}(12)=-2\), which is negative
11. Therefore, the area is maximum when x=12
12. \(y=x=24-12=12\)
13. So, x=y=12
14. Hence. amongst all rectangles. the square has the maximum area.

Viva Voice:

Question 1. What is the area of the rectangle with length = x cm. breadth = y cm and perimeter = 24 cm?
Answer:

Perimeter, P = 2(x + y) = 24 cm.

Area of the rectangle. A = xy

A(x) = x (12 – x) = 12x – \(x_2\)

Question 2. What is \(A^{\prime}(x)\) for \(A(x)-12 x-x^2\) ?
Answer:

A(x)=12-2 x .

Question 3. For what value of x, A(x)=48 \(x-x^2\) is maximum?
Answer:

Consider \(A^{\prime}(x)=48-2\) x=0

x=\(\frac{48}{2}=24 and \mathrm{A}^{\prime \prime}(\mathrm{x}=24)=-2<0\)

A \ x is the maximum when x=12 cm.

Question 4. What is \(A^{\prime \prime}(x) if A(x)=12-3 x^2\) ?
Answer:

⇒ \(\mathrm{A}^{\prime \prime}(\mathrm{x})=-6<0\) .

CBSE Class 12 Maths Practical Notes Activity 17 Integrals

Objective:

To evaluate the definite integral \(\int \sqrt{\left(1-x^2\right)}\) dx as the limit of a sum and verify it by actual integration.

Material Required:

Cardboard, white paper, scale, pencil, graph paper.

Method Of Construction:

1. Take a cardboard of a convenient size and paste a white paper on it.
2. Draw two perpendicular lines to represent coordinate axes XOX’ and YOV.
3. Draw a quadrant of a circle with O as the center and a radius of 1 unit,( 10 cm) as shown
4. The curve in the 1st quadrant represents the graph of the function \(\sqrt{\left(1-x^2\right)}\) in the interval [0,1].

Demonstration :

1. Let origin O be denoted by \(P_0\) and the points where the curve meets the x-axis and y-axis be denoted by \(P_{10}\) and Q, respectively.

2. Divide \(\mathrm{P}_0 \mathrm{P}_{10}\) into 10 equal parts with points of division as: \(\mathrm{P}_1, \mathrm{P}_2, \mathrm{P}_3, \ldots, \mathrm{P}_9\).

3. From each of the points, \(\mathrm{P}_{\mathrm{t}} \mathrm{i}=1,2, \ldots\), 9 draw perpendiculars on the x-axis 10 meet the curve at the points. \(Q_1, Q_2, Q_3, \ldots Q_0\).

Measure the lengths of \(P_0, Q_0, P_1, Q_1, \ldots, P_9 Q_0\) and call them as \(y_0, y_1, \ldots, y_4\) whereas width of each part, \(P_0 P_1, P_1 P_2, \ldots\) is 0.1 units.

4. \(y_0-P_0 Q_n\)=1 units

1. • \(y_1=P_1 Q_1\)=0.99 units
   • \(y_2=P_2 Q_2\)=0.97 units
   • \(y_3=P_3 Q_1\)=0.95 units
   • \(y_4=P_4 Q_4\)=0.92 units
   • \(y_3=P_5 Q_5\)=0.87units
   • \(y_6=P_6 Q_6/\)=0.8 units
   • \(y_7=P_7 Q_7\)=0.71 units
   • \(y_8=P_3 Q_5\)=0.6  units
   • \(y_9=P_5 Q_5\)=0.43 units
   • \(y_{10}=P_{10} Q_{10}\)=which is very small near to 0 .

5.  Area of the quadrant of the circle (area bounded by the curve and the two axes) = sum of the areas of trapeziums.

1. 1. =\(\frac{1}{2} \times 0.1\left[\begin{array}{l}
      (1+0.99)+(0.99+0.97)+(0.97+0.95)+(0.95+0.92) \\
      +(0.92+0.87)+(0.87+0.8)+(0.8+0.71)+(0.71+0.6) \\
      +(0.6+0.43)+(0.43)
      \end{array}\right]\)
   2. = 0.1 [0.5 + 0.99 + 0.97 + 0.95 + 0.92 + 0.87 + 0.80 + 0.71+ 0.60 + 0.43]
   3. = 0.1 x 7.74 = 0.774 sq. units.(approx.)

Question 6. Definite integral =\(\int_0^1 \sqrt{1-x^2} d x\)
Solution:

= \(\left[\frac{x \sqrt{1-x^2}}{2}+\frac{1}{2} \sin ^{-1} x\right]_0^1=\frac{1}{2} \times \frac{\pi}{2}=\frac{3.14}{4}\)=0.785 sq. units

Thus, the area of the quadrant as a limit of a sum is nearly the same as the area obtained by actual integration.

Observation:

1. The function representing the arc of the quadrant of the circle is y=\(\sqrt{1-x^2}\).
2. Area of the quadrant of a circle with radius 1 unit =\(\int_0^1 \sqrt{1-x^2}\) d x=0.785. sq. units
3. Area of the quadrant as a limit of a sum = 0.774 sq. units.
4. The two areas are nearly equal.

Result:

Hence; we have evaluated the definite integral \(\int_a^b \sqrt{\left(1-x^2\right)}\)dx as the limit of a sum and verified it by actual integration.

Application:

This Activity can be used to demonstrate the concept of area bounded by a curve. This activity can also be applied to find the approximate value of π

Viva Voice

Question 1. What is \(\int_i^b \sqrt{1-x^2}\) d x ?
Answer:

It is an area of the region bounded by the lines x=a, x=b, and the curve y=\(\sqrt{1-x^2}\).

Question 2. What is the integrand in \(\int_0^4 \sqrt{1-x^2}\) d x?
Answer:

It is the function being integrated into the integral. Here it is \(\sqrt{1-x^2}\).

Question 3. Which process is known as anti-derivative?
Answer:

Integration or primitive.

Question 4. What is the use of integration in Mathematics?
Answer:

Integration is a way of adding slices to find the area of curves.

Question 5. Write the formula to find \(\int_i^b f(x) d x\) as the limit of the sum.
Answer:

1. \(\int_a^b f(x) d x=\lim _{a \rightarrow 0} h\{f(a)+f(a+h)+f(a+2 h)+\ldots .+f(a+(n-1) h\}\)
2. where h \(\rightarrow 0\) when \(\mathrm{n} \rightarrow \infty\) and
3. \(\mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}\).

Question 6. What is a formula for \(\int \sqrt{a^2-x^2}\) d x?
Answer:

⇒ \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+C\)

CBSE Class 12 Maths Practical Notes Activity 18 Vector Algebra

Objective:

To verify that the angle in a semi-circle is a right angle, use the vector method.

Material Required:

Cardboard, white paper, adhesive, pens, geometry box. eraser, wires, and paper arrowheads.

Method Of Construction:

Take a thick cardboard of size 30 cm x 30 cm.

On the cardboard. paste a white paper of the same size using are adhesive.

On this paper draw a circle. with center O and radius 10 cm.

Fix nails at the points O. A, B. P, and Q. Join OP. OA. OB. AP. AQ. BQ. OQ and BP using wires.

Put arrows on OA. OB. OP. Alb BP. OQ. AQ and BQ to show them as vectors, using paper arrowheads

Demonstration :

1. Using a protractor. measure the angle between the vectors \(\overrightarrow{\mathrm{AP}}\) and \(\overline{\mathrm{BP}}\). i.e. \(\angle \mathrm{APB}-90^{\circ}\).
2. Similarly, the angle between the vectors \(\overrightarrow{\mathrm{AQ}}\) and \(\overrightarrow{\mathrm{BQ}}, i.e., \angle \mathrm{AQB}-90^{\circ}\).
3. Repeat the above process by taking some more points R. S. T. … on the semi-circles, forming vectors AR. BR: AS, BS: AT. BT;.., etc.. i.e.. angle formed between two vectors in a semi-circle is a right angle.

Observation:

By actual measurement.

⇒ \(|\overline{\mathrm{OP}}|=|\overline{\mathrm{OA}}|=|\overline{\mathrm{OB}}|\)

=\(|\overline{\mathrm{OQ}}|=\mathrm{r}=\mathrm{a}=\mathrm{p}=\underline{12.5 \mathrm{~cm}}\)

⇒ \(|\overrightarrow{\mathrm{AP}}|=20 \mathrm{~cm},|\overrightarrow{\mathrm{BP}}|=15 \mathrm{~cm},|\overrightarrow{\mathrm{AB}}|=25 \mathrm{~cm}\)

⇒ \(|\overrightarrow{\mathrm{AQ}}|=24 \mathrm{~cm},|\overrightarrow{\mathrm{BQ}}|=7 \mathrm{~cm} \cdot|\overrightarrow{\mathrm{AB}}|=25 \mathrm{~cm}\)

⇒\(|\overrightarrow{\mathrm{AP}}|^2+|\overline{\mathrm{BP}}|^2\)

=\(|\overline{\mathrm{AB}}|^2,|\overrightarrow{\mathrm{AQ}}|^2+|\overline{\mathrm{BQ}}|^2\)

=\(|\overline{\mathrm{AB}}|^2\)

Similarly, for points R, S. T.

⇒ \(\angle \mathrm{ARB}=90^{\circ}, \angle \mathrm{ASB}=90^{\circ}, \angle \mathrm{ATB}=90^{\circ}, \ldots \ldots, \ldots \ldots\)

i.e., the angle in a semi-circle is a right angle.

Result:

The angle in a semi-circle is 90°.

Application:

This activity can be used to explain the concepts of

1. opposite vectors
2. vectors of equal magnitude
3. perpendicular vectors,
4. The dot product of two vectors.

Let OA=O B=a=O P=P

⇒ \(\overline{\mathrm{OA}}=-\vec{a} \cdot \overline{\mathrm{OB}}=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{p}}\)

⇒ \(\overline{\mathrm{AP}}=-\overline{\mathrm{OA}}+\overline{\mathrm{OP}}=\overrightarrow{\mathrm{a}}+\overline{\mathrm{p}}: \overline{\mathrm{BP}}\)

=\(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}}\)

⇒ \(\overline{\mathrm{AP}} \cdot \overline{\mathrm{BP}}=(\overline{\mathrm{p}}+\overrightarrow{\mathrm{a}}) \cdot(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{a}})\)

=\(|\overrightarrow{\mathrm{p}}|^2-|\overrightarrow{\mathrm{a}}|^2\)=0

= (Since }\(|\overrightarrow{\mathrm{p}}|^2=|\overrightarrow{\mathrm{a}}|^2 \)

So, the angle APB between the vectors \(\overline{\mathrm{AP}}\) and \(\overline{\mathrm{BP}}\) is at a right angle.

Similarly, \(\overrightarrow{\mathrm{AQ}} \cdot \overrightarrow{\mathrm{BQ}}\)=0, So \(\angle \mathrm{AQB}=90^{\circ}\) and so on.

Viva Voice:

Question 1. If \(\vec {a}\) and \(\vec{b}\) are two vectors such that the angle between them is \(\theta\), then what is \(\cos \theta\)?

Answer:

⇒ \(\cos \theta=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|}\)

Question 2. If \(\vec{a}\) and \(\vec{b}\) are two vectors such that the angle between them is \(\theta\) then what is \(\sin \theta\)?
Answer:

⇒ \(\sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\).

Question 3. What is the formula for the projection of \(\vec a\) on \(\vec b\)?
Answer:

Projection of \(\vec{a}\) on \(\vec{b}=\frac{a, b}{|b|}\).

Question 4. If \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other then \(\vec{a}\) . \(\vec{b}\) is equal to
Answer:

Zero.

Question 5. Is dot product in vectors commutative?
Answer:

Yes, the dot product in the vector is commutative. For example if \(\bar{a}\) and \(\vec{b}\) are two vectors. Then \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\)

CBSE Class 12 Maths Practical Notes Activity 19  Vector Algebra

Objective:

To verify geometrically that \(\vec{c} \times(\vec{a}+\vec{b})=\vec{c} \times \vec{a}+\vec{c} \times \vec{b}\)

Material Required:

Geometry box. cardboard, white paper, cutter, sketch pen. cellotape.

Method Of Construction:

1. Fix a white paper on the cardboard.
2. Draw a line segment OA (= 6 cm, say) and let it represent \(\vec{c}\)
3. Draw another line segment OB (= 4 cm, say) at an angle
   (say 60°) with OA. Let \(\overline{O B}=\vec{a}\)
4. Draw BC (= 3 cm, say) making an angle (say 30°) with \(\overline{OA}\).Let \(\overline{BC}=\vec{b}\)
5. Draw perpendiculars BM, CL, and BN.
6. Complete parallelograms OAPC, OAQB and BQPC.

Demonstration:

⇒\(\overline{\mathrm{OC}}=\overline{\mathrm{OB}}+\overline{\mathrm{BC}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\) and let \(\angle \mathrm{COA}=\alpha\).

⇒ \(|\vec{c} \times(\vec{a}+\vec{b})|=|\vec{c}||\vec{a}+\vec{b}| \sin \alpha\)= area of parallelogram OAPC.

⇒ \(|\vec{c} \times \vec{a}|\)= area of parallelogram OAQB.

⇒ \(|\vec{c} \times \vec{b}|\)= area of parallelogram BQPC.

Area of parallelogram \(\mathrm{OAPC}=(\mathrm{OA})(\mathrm{CL})\)

=\((\mathrm{OA})(\mathrm{LN}+\mathrm{NC})-(\mathrm{OA})(\mathrm{BM}+\mathrm{NC})\)

=\((\mathrm{OA})(\mathrm{BM})+(\mathrm{OA})(\mathrm{NC})\)

= Area of parallelogram OAQB + Area of parallelogram BQPC

=\(|\vec{c} \times \vec{a}|+|\vec{c} \times \vec{b}|\)

So, \(|\vec{c} \times(\vec{a}+\vec{b})|=|\vec{c} \times \vec{a}|+|\vec{c} \times \vec{b}|\)

Direction of each of these vectors \(\overrightarrow{\mathrm{c}} \times(\overrightarrow{\mathrm{a}}+\overline{\mathrm{b}}), \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}\)

and \(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}\) is perpendicular to the same plane.

So, \(\vec{c} \times(\vec{a}+\vec{b})=\vec{c} \times \vec{a}+\vec{c} \times \vec{b}\)

Observation:

⇒ \(|\mathrm{c}|=|\overrightarrow{\mathrm{OA}}|=\mathrm{OA}=\underline{21.5 {units}(\mathrm{cm})}\)

⇒ \(|\vec{a}+\vec{b}|=|\overrightarrow{O C}|\)=O C= 26 units

⇒ \(\mathrm{CL}\)=22 units BM =17.5 units, CN =4.5 units

⇒ \(|\vec{c} \times(\vec{a}+\vec{b})|\)= Area of parallelogram OAPC

=\((\mathrm{OA})(\mathrm{Cl})\)= 473 sq. units  → Equation 1

⇒ \(|\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}|\)= Area of parallelogram OAQB

⇒ \(-(\mathrm{OA})(\mathrm{BM})-\underline{21.5} \times\) 17.5-376.25 sq units → Equation 2

⇒ \(|\vec{c} \times \vec{b}|\) – Area of parallelogram BQPC

=\((\mathrm{OA})(\mathrm{CN})=\underline{21.5} \times \underline{4.5}\)=96.75 sq units  → Equation 3

From (1). (2) and (3).

Area of parallelegram ONPC = Area of parallelgram OAQB + Area of prallelgram BQPC

Thus. \(|\vec{c} \times(\vec{a}+\vec{b})|=|\vec{c} \times \vec{a}|+|\vec{c} \times \vec{b}|\)

⇒ \(\vec{c} \times \vec{a}.\vec{c} \times \vec{b}\) and \(\overrightarrow{\mathrm{c}} \times(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\) are all in the direction of \(\frac{\perp}{\text { (Perpendicular) }}\)  to the plane of paper.

Therefore: \(\vec{c} \times(\vec{a}+\vec{b})=\vec{c} \times \vec{a}+\underline{\vec{c}} \times \vec{b}\).

Result :

Hence \(\vec{c} \times(\vec{a}+\vec{b})=\vec{c} \times \vec{a}+\vec{c} \times \vec{b}\)

Application :

Through the activity, the distributive property of vector multiplication over addition can be explained.

Viva Voice

Question 1. Define \(\vec{a} \times \vec{b}\)
Answer:

1. \(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}(Where 0<\theta<\pi )\)
2. Here. \(|\vec{a}|\) is magnitude of \(\overrightarrow{\mathrm{a}}\).
3. \(|\overrightarrow{\mathrm{b}}|\) is magnitude of \(\overrightarrow{\mathrm{b}}, \hat{\mathrm{n}}\) is a unit vector perpendicular to both \(\overrightarrow{\mathrm{a}} and \vec{b}, \vec{a} \times \vec{b}\) is perpendicular to the plane of \(\vec{a}\) and \(\vec{b}\).

Question 2. If a and \bar{b} represent the adjacent sides of a parallelogram, then find its area.
Answer:

⇒ \(|\vec{a} \times \hat{b}|\)

Question 3. If \(\vec{a} \times \vec{b}=\overrightarrow{0}\), what does it mean?
Answer:

1. \(\overline{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0}\)
2. \(|\overrightarrow{\mathrm{a}}| \overrightarrow{\mathrm{b}} \mid \sin \theta \hat{\mathrm{n}}=\overrightarrow{0}\)
3. \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\) or \(\vec{a}\) and \(\vec{b}\) are collinear vectors.
4. \(\vec {a}\) is zero vector or \(\vec{b}\) is a zero vector or \(\vec{a}\) and \(\vec {b}\) are collinear.

Question 4. Write the value of \((\hat{i} \times \hat{j}) \times(\hat{j} \times \hat{k})\).
Answer:

1. \(\hat{i} \times \hat{j}=\hat{k}\) and \(\hat{j} \times \hat{k}=\hat{i}\)
2. \((\hat{i} \times \hat{j}) \times(\hat{j} \times \hat{k})=\hat{k} \times \hat{i}=\hat{j}\)

Question 5. If \(|\vec{a}|=5,|\vec{b}|=13\) and \(|\vec{a} \times \vec{b}|=25\); then find \(\vec{a} \cdot \vec{b}\)
Answer:

1. \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta=25=5 \times 13 \times \sin \theta\)
2. \(\sin \theta=\frac{5}{13} \Rightarrow \cos \theta=\frac{12}{13}\)
3. Now \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta=5 \times 13 \times \frac{12}{13} \Rightarrow \vec{a} \cdot \vec{b}=60\)

CBSE Class 12 Maths Practical Notes Activity 20 Probability

Objective:

To explain the computation of conditional probability of a given event A when event B has already occurred, through an example of throwing a pair of dice.

Material Required:

1. A piece of plywood, white paper pen/pencil, scale, a pair of dice.
2. Method Of Construction:
3. Paste a white paper on a piece of plywood of a convenient size,
4. Make a square and divide it into 36 unit squares of size lent each.
5. Write a pair of numbers as shown.

Demonstration:

1. Given above gives all possible outcomes of the given experiment. Hence, it represents the sample space of the experiment.
2. Suppose we have to find the conditional probability of an event A if an event B has already occurred, where A is the event “a number 4 appears on both the dice'” and B is the event “4 has appeared on at least one of the dice” i.e, we have to find P(A | B).
3. From the above number of outcomes favorable to A = 1
4. Number of outcomes favorable to B = 11
5. and Number of outcomes favorable to AnB = 1.

Observation:

1. Outcome(s) favourable to A: 1 , n(A)=1
2. Outcomes favorable to B: 11, n(B)=11.
3. Outcomes favouble to A \cap B: 1, n\((A \cap B)\)=1.
4. \(P(A \cap B)=1 / 36\).
5. \(\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{1 / 11}{}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)

Result:

The result of the experiment is \(\frac{1}{11}\)

Application:

This activity helps understand the concept of conditional probability, which is further used in Bayes’ Theorem.

Viva Voice

Question 1. What is conditional probability?
Answer:

If F and F are two events associated with the sample space of the random experiment, then the conditional probability of event E when F has already occurred is denoted by P(EF) and it is equal to P(E / F)=\(\frac{P(E \cap F)}{P(F)}\), where \(P(F) \neq 0\).

Question 2. What is the condition of an independent event?
Answer:

⇒ \(\mathrm{P}(\mathrm{A}) . \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})\), where A, B are two events.

Question 3. What are the properties of conditional probability?
Answer:

1. Let E and F be events associated with sample space S of an experiment then
2. \(\mathrm{P}(\mathrm{S} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{S} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\mathrm{P}(\mathrm{F})}{\mathrm{P}(\mathrm{F})}=1\)
3. \(\mathrm{P}(\mathrm{A} \cup \mathrm{B} / \mathrm{F})=\mathrm{P}(\mathrm{A} / \mathrm{F})+(\mathrm{B} / \mathrm{F})-\mathrm{P}(\mathrm{A} \cap \mathrm{B} / \mathrm{F})\)
4. Where A and B are any two events associated with S.
5. \(\mathrm{P}(\mathrm{E} / \mathrm{F})=1-\mathrm{P}(\mathrm{E} / \mathrm{F})\)

Question 4. State the theorem of total probability.
Answer:

1. If \(\left(E_1, E_2 \ldots \ldots . . E_n\right)\) is a partition of sample space, S and suppose that each of \(E_1, E_2 \ldots \ldots. E_n\) has non-zeto probability and A is the event associated with S, then
2. \(P(A)=P\left(E_1\right), P\left(A / E_1\right)+P\left(E_2\right), P\left(A / E_2\right)+\ldots .+P\left(E_{11}\right), P\left(A / E_n\right)\)

Question 5. Find P(A / B) if \(P(A)=\frac{4}{13}, P(B)=\frac{7}{13}\), and \(P(A \cap B)=\frac{4}{13}\).
Answer:

P(A / B)=\(\frac{P(A \cap B)}{P(B)}=\frac{4 / 13}{7 / 13}=\frac{4}{7}\).

CBSE Class 12 Maths Practical Notes Chapter Wise Viva-Voice Questions Chapter 1 Relations And Functions

Question 1. What do you mean by relation?
Answer:

If A and B are two non-empty sets, then any subset R of A x B is called a relation from set A to set B.

Question 2. Explain the domain and range of relation R.
Answer:

If R is the relation from set A to set B then the set of all first coordinates of R is called the domain of R and at set of all second coordinates of R is called the range of R.

Question 3. What do you mean by equivalence relation?
Answer:

Any relation R on a set A is said to be an equivalence relation if it is (1) reflexive (2) symmetric (3) transitive.

Question 4. What is an equivalence class?
Answer:

Let R be an equivalence relation on a non-empty set A, for all a ∈ A. Then the equivalence class of ‘a’ is defined as the set of all such elements of A which are related to ‘a’ under R and denoted by [a].

Question 5. What is a function?
Answer:

Let A and B be two non-empty sets. Then a rule f which associates with each element x ∈A. A unique element, denoted by f(x) of B is called a function from A to B and denoted by f: A → B.

Question 6. What is an injective (one-on-one) function?
Answer:

A function f: X → Y is said to be one-one or injective, if the image of a distinct element of X is also a distinct element of Y under rule f. i.e., if \(x_1 \cdot x_2 \in\) X such that \(f\left(x_1\right)=f\left(x_2\right) \Rightarrow x_1=x_2\) then f is one-one (injective).

Question 7. What do you mean by onto function?
Answer:

A function f: X → Y is said to be onto function if each element of Y is the image of at least one element of X i.e. \(\forall y \in Y\) there exists of x \(\in\) X.

Question 8. Let A and B be two finite sets having m and n elements respectively, then what is the possible number of functions from A to B?
Answer:

Possible number of functions from A to B be \(\mathrm{n}^{\mathrm{m}}\).

Question 9. How many onto functions are possible from the set {1, 2, 3,…, n} to itself?
Answer:

There will be n onto functions from set{1,2, 3,… n} to itself.

Question 10. A function which is one-one and onto then it is known as.
Answer:

Invertible function or Obijective function.

CBSE Class 12 Maths Practical Notes Chapter 2 (Inverse Trigonometric Functions)

Question 1. Why do we restrict the inverse trigonometric functions to a particular domain and codomain?
Answer:

Since trigonometric functions are periodic are not objective in general. To make the inverse trigonometric function objective (one-on-one), we restrict the domain and codomain. The inverse can be found out of bijective functions.

Question 2. Does \(\sin ^{-1} x=\frac{1}{\sin x}\)

Answer: No

Question 3. The principal value branch of see\(^{-1}\) is:
Answer:

⇒ \([0, \pi]-\left\{\frac{\pi}{2}\right\}\)

Question 4. The principal value branch of \(\sin ^{-1}\) is…
Answer:

⇒ \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

Question 5. If we interchange x and y-axis, the graph of y = sin x changes to:
Answer:

y=\(\sin ^{-1} x\)

Question 6. The domain of the function y = \(\sin ^{-1}\left(x^2\right)\) is:

Answer: [-1, 1].

CBSE Class 12 Maths Practical Notes Chapter 3 (Matrices)

Question 1. What do you mean by matrix?
Answer:

A matrix is an ordered rectangular (arrangement) array of numbers or functions subject to certain rules of operations.

Question 2. What do you mean by the order of a matrix?
Answer:

A matrix having m rows and n columns is called a matrix of order m x n.

Question 3. What is a square matrix?
Answer:

A matrix having an equal number of rows and columns is called a square matrix.

Question 4. What are symmetric matrices?
Answer:

Square matrices are said to be symmetric if A₁ = A where A₁ is the transpose of A.

Question 5. What are skew-symmetric matrices?
Answer:

Square matrices are said to be skew-symmetric matrices if \(A^{T}\), where \(A^{T}\) is the transpose of A.

Question 6. Does a rectangular matrix possess its inverse?
Answer:

No. rectangular matrices do not possess their inverse,

Question 7. If a matrix is both symmetric and skew-symmetric then it is

Answer: A zero matrix.

CBSE Class 12 Maths Practical Notes Chapter 4 (Determinants)

Question 1. What do you mean by determinant?
Answer:

Every square matrix can be associated with an expression or a number which is known as its determinant.

Question 2. What do you mean by a singular matrix?
Answer:

A square matrix is said to be singular if its corresponding determinant is zero i.e., |A| = 0.

Question 3. What is a non-singular matrix?
Answer:

A square matrix is called to be non-singular if its corresponding determinant is not equal to zero i.e. |A|=0.

Question 4. If A is a non-singular matrix of order n, then the value of |adj A| in terms of |A| is:

Answer: \(|A|^{n-1}\)

Question 5. If A is an invertible matrix, then write \(\mathrm{A}^{-1}\) in terms of ad joint A and |A|.
Answer:

⇒ \(A^{-1}=\frac{1}{|A|} \cdot adj(A)\)

Question 6. If A is a square matrix of order 3 such that |A| = 7, then what will be the value of |—A|?

Answer: -7

Question 7. What are Mu and \(C^21\) of the determinant A ? |A|=\(\left|\begin{array}{ccc}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\)

Answer: \(\mathrm{M}_{13}-30. \mathrm{C}_{21}-4\)

CBSE Class 12 Maths Practical Notes Chapter 5 Continuity And Differentiability

Question 1. Explain the continuity of a function at a point.
Answer:

Let f be a real function on a sub-set of real numbers and let c be a point in the domain of f Then f is continuous at c if \(\lim _{x \rightarrow c} \Gamma(x)=f(c)\)

or in other words, we can say if the left-hand limit, right-hand limit, and value of the function at x=c exist and are equal to each other i.e., \(\lim _{x \rightarrow-} f(x)=\lim _{x \rightarrow c} f(x)=f(c)\).

Question 2. ln Which interval of the following functions is continuous?

1. Constant function
2. Identity function
3. Polynomial functions
4. Rotational function
5. sine and cosine function
6. tangent and secant functions,

Answer: (1), (2), and (3) for all real numbers ( R )

⇒ \(R-\{x: q(x)=0\}\)

All real values

⇒ \(R-\left\{(2 n+1) \frac{\pi}{2}, n \in Z\right\}\)

Question 3. Is the function f(x)=|x|, a continuous function?
Answer:

Yes, as f(x)=\(\{\begin{array}{ccc}-x & \text { if } & x<0 \\ x & \text { if } & x \geq 0\end{array}\).

⇒ \(\lim _{x \rightarrow 5} f(x)=f(c)\)

Hence, f is a continuous function at all points.

Question 4. What do you mean by the domain of continuity?
Answer:

The Set of all real values of x for which the function f(x) is continuous is called the domain of continuity.

Question 5. What is the discontinuity of a function at a point?
Answer:

If a function is not continuous at a point then it is called discontinuous at that point, in other words. if \(\lim _{x \rightarrow a^{-}}\) f(x) is not equal to \(\lim _{x \rightarrow a^{-}}\) f(x) or f(a) then f is called discontinuous at point a.

Question 6. Explain the differentiability of a function in an

1. Open interval (a, b)
2. Closed interval [a, b}

Answer:

The function y = f(x) is said to be differentiable in an open interval (a. b) if it is
differentiable at every, point of(a, b) i.e.. we can draw unique tangents at each point of (a,b)

The function y = f(x) is said to be differentiable in the closed interval [a,b] if Rf(a) and Lf(b) exist and f(x) exists for every point of (a,b).

Question 6. Is every differentiable function continuous, what is about its converse?
Answer:

Yes. every differentiable function is continuous but the converse is not true in all cases.

Question 7. Write the derivative of \(\sqrt{\tan } \sqrt{x}\)w.r. to x.
Answer:

⇒ \(\frac{\sec ^2 \sqrt{x}}{4 \sqrt{x} \sqrt{\tan \sqrt{x}}}\)

CBSE Class 12 Maths Practical Notes Chapter 6 Application Of Derivatives

Question 1. How can you find the rate of change of y w.r. to x at x = \(x_0\) if y = f (x)?
Answer:

We find \(\frac{d y}{d x}\) at  x=\(x_0\)  i.e., f(x) at x=\(x_0\) .

Question 2. What do you understand by monotonic function?
Answer:

A function is said to be monotonic in an interval if it is either increasing or decreasing in the given interval.

Question 3. Explain the increasing function and decreasing function.
Answer:

A function f(x) is said to be decreasing in ( a, b) if

⇒ \(\mathrm{x}_1<\mathrm{x}_2 \Rightarrow \mathrm{f}\left(\mathrm{x}_1\right) \geq \mathrm{f}\left(\mathrm{x}_2\right), \forall \mathrm{x}_1, \mathrm{x}_2 \in(\mathrm{a}, \mathrm{b})\)

or \(\mathrm{f}(\mathrm{x}) is decreasing in (\mathrm{a}, \mathrm{b})\)

if \(\forall \mathrm{x} \in(\mathrm{a}, \mathrm{b}) ; \mathrm{f}(\mathrm{x})<0\)

Question 4. In the given which points are local maxima and which are of local minima?
Answer:

Points A. C and E are of local maxima and B. L) arc of local minima.

Question 5. What do you mean by a stationary point turning point or critical point to the curve y = f(x)?
Answer:

The values of x for which f'(x) = 0 are called stationary points or turning points or critical points.

CBSE Class 12 Maths Practical Notes Chapter 7 Integrals

Question 1. The value of \(\int_{-\pi / 2}^{\pi / 2} \sin ^7\) is:
Answer:

Zero, Since f(x) is an odd function.

Question 2. Why, do we write arbitrary constants in the case of Indefinite integral?
Answer:

It is called indefinite because it is not unique. There exist infinitely many integrals which can be obtained by putting C from the set of real numbers.

Question 3. If u and v are two functions of x, then what is the formula for \(\int u, v, d x\)
Answer:

Integration by parts: i.e \(u \int v d x-\int\left\{\frac{d u}{d x} \int v d x\right\}\) dx+C

Question 4. \(\int\left\{e^x f(x)+f^{\prime}(x)\right\} d x\)=?
Answer:

⇒ \(\int\left\{e^x f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C\)

Question 5. \(\int a^x d x\)=?
Answer:

⇒ \(\frac{a^x}{\log _e a}+C\)

CBSE Class 12 Maths Practical Notes Chapter 8  Application Of Integrals

Question 1. Draw the figure of curves y = sin x and y = cos x and shade the area between two curves when x \(\in\left[0, \frac{\pi}{2}\right]\)
Answer:

Shaded area between two curves y =\(\sin \mathrm{x}\) and y =\(\cos x\) when x \(\in\left[0, \frac{\pi}{2}\right]\).

Question 2. Shade the area of the region represented by \(\left\{(x, y): x^2 \leq y \leq|x|\right\}\)
Answer:

Question 3. Find the area shown shaded in the given below.

Answer:

Area of shaded portion =\(\int_0^4 \sqrt{x}\) dx i.e., \(\frac{16}{3}\) sq. units.

CBSE Class 12 Maths Practical Notes Chapter 9 Differential Equations

Question 1. What is the order and degree of differential equation
\(\log \left(\frac{d y}{d x}\right)+\frac{d^3 y}{d x^3}\)=y?

Answer:

Order: 3. degree, not defined.

Question 2. The sum of order and degree of differential equation \(\frac{d}{d x}\left\{\left(\frac{d y}{d x}\right)^2\right\}\)=0 is

Answer: 3

Question 3. How many arbitrary constants are there in the general solution of the differential equation of order three?

Answer: 3.

Question 4. Is (x-y) \(\frac{d y}{d x}\)=x+2 y a linear differential equation?
Answer:

No, it is a homogeneous differential equation,

Question 5. Is \(\frac{d y}{d x}+P y\)=Q a linear differential equation, if yes what is its Integrating factor?
Answer:

Yes it is a linear differential equation and its I.F. = \(e^{\int P d x}\)

CBSE Class 12 Maths Practical Notes Chapter 10 Vectors

Question 1. If \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\mathrm{k}\), then  \(\overrightarrow{\mathbf{a}}\) is
Answer:

⇒ \(\hat{\mathrm{a}}=\frac{2 \hat{\mathrm{i}}}{\sqrt{30}}+\frac{5 \hat{\mathrm{j}}}{\sqrt{30}}+\frac{\hat{\mathrm{k}}}{\sqrt{30}}\).

Question 2. Direction cosine of vector \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{k} are:
Answer:

⇒ \(\left.<\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right\rangle\)

Question 3. What is the projection of \(\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { on } \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) ?

Answer:  \(\frac{5}{3} \sqrt{6}\)

Question 4. If the dot product of a and h is zero, what is the relation between a and b? If \overrightarrow{\mathrm{a}} \neq 0, \overrightarrow{\mathrm{b}} \neq 0.

Answer: \(\vec{a} \perp \vec{b}\)

Question 5. What is the cosine of angle between vectors \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) and \(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)?

Answer: \(\frac{5}{7}\)

CBSE Class 12 Maths Practical Notes Chapter 11 Three-Dimensional Geometry

Question 1. What will be the vector equation of a line if it passes through two points with position vectors a and b?
Answer:

⇒ \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})\)

Question 2. What is the formula for the shortest distance between two skew lines: \(\overrightarrow{\mathbf{r}}=\vec{a}_1+\lambda \vec{b}_1\) and \(\vec{r}=\vec{a}_2+\mu \vec{b}_2\)
Answer:

⇒ \(\mathrm{S}.\mathrm{D} .=\left|\frac{\left(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1\right) \cdot\left(\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right)}{\left|\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right|}\right|\)

Question 3. If \(\theta\) is the acute angle between the lines \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}_1+\lambda \overrightarrow{\mathbf{b}}_1 ; \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}_2+\mu \overrightarrow{\mathbf{b}}_2\) then cos θ is equal to
Answer:

⇒ \(\cos \theta=\left|\frac{\vec{b}_1 \cdot b_2}{\left|\vec{b}_1\right| \cdot \mid \vec{b}_2}\right|\)

Question 4. What are skew lines?
Answer:

Those lines that are not parallel none intersecting are called skew lines. In other words, we can say those lines which are tied in different planes.

Question 5. Write the condition of perpendicular lines.
Answer:

⇒ \(a_1 a_2+b_1 b_2+c_1 c_2\)=0 in cartesian form, where \(a_1, b_1, c_1\) and \(a_2, b_2, c_2\) are direction ratios of two lines.

CBSE Class 12 Maths Practical Notes Chapter 12 Linear Programming

Question 1. What do you mean by objective function in L.P.P.?
Answer:

The linear function \(Z=c_1 x_1+c_2 x_2+\ldots+\mathfrak{c}_n x_n\) which is to be maximized or minimized is called the objective function.

Question 2. What do you mean by the feasible solution of L.P.P.?
Answer:

The solution that satisfies the constraints is called a feasible solution.

Question 3. What do you mean by the optimum solution?
Answer:

Any feasible solution which minimizes or maximizes the objective function is called optimum.

Question 4. What are the applications of linear programming?
Answer:

It is useful in solving various problems of daily life, like diet problems, transportation problems, manufacturing problems, etc.

Question 5. How many types of feasible regions?
Answer:

There are two types of feasible regions.

1. Bounded feasible region
2.  Unbounded feasible region,

CBSE Class 12 Maths Practical Notes Chapter 13 Probability

Question 1. Let E and F be two events associated with the same random experiment then condition E and F are to be independent is:
Answer:

⇒ \(P(E \cap F)=P(E)+P(F)\)

Question 2. Let X be a discrete random variable assuming values \(\mathrm{x}_1, \mathrm{x}_2 \ldots \mathrm{x}_{\mathrm{n}}\) with probability \(p_1, p_2 \ldots p_n\) respectively what is formula for the variance of X?
Answer:

⇒ \(\sigma^2=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2\)

Question 3. What is the formula of the mean in probability?
Answer:

⇒\(E(x)=\bar{x}=\mu=\sum_{i=1}^n x_i p\left(x_i\right)\)

Question 4. \(\mathrm{P}(\mathrm{B} / \mathrm{A})\) is equal to
Answer:

⇒ \(P(B / A)=\frac{P(B \cap A)}{P(A)}\), where \(P(A) \neq 0\)

Question 5. Let A and B be two given mutually exclusive events then \(\mathrm{P}(A \cap B)\) is?
Answer:

In mutually exclusive events A \(\cap \mathrm{B}=\phi\)

⇒ \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})\)=0 [Mutually exclusive event]

CBSE Class 12 Maths Chapter 10 Vector Algebra Important Questions

Question 1. Find the angle between the vectors \((\hat{\mathrm{i}}-\hat{\mathrm{j}})\) and \((\hat{\mathrm{j}}-\hat{\mathrm{k}}) \text {. }\)

Or,

Write the projection of the vector \({\vec{r}}=3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\) on

1. x-axis and
2. y-axis

Solution:

Let θ be the angle between \(\vec{a}=\hat{i}-\hat{j}\) and \(\vec{b}=\hat{j}-\hat{k}\)

Now, \(\vec{a} \cdot \vec{b}=(\hat{i}-\hat{j}) \cdot(\hat{j}-\hat{k})=-1,|\vec{a}|=\sqrt{2},|\vec{b}|=\sqrt{2}\)

∴ \(\cos \theta=-\frac{1}{2}=\cos \frac{2 \pi}{3} \Rightarrow \theta=\frac{2 \pi}{3}\)

Or,

Given, \({\vec{r}}=3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\)

Projection of \(\vec{a}\) on \(\vec{b}\) is given as \(\vec{a}\).\(\vec{b}\)

Read and Learn More CBSE Class 12 Maths Important Question and Answers

⇒ Projection of \(\vec{r}\) on x-axis = \(\frac{\vec{r} \cdot \hat{i}}{|\hat{i}|}=\frac{(3 \hat{i}-4 \hat{j}+12 \hat{k}) \hat{i}}{1}=3\)

and projection of \(\vec{r}\) on y-axis = \(\frac{{\mathrm{r}} \cdot \hat{\mathrm{j}}}{|\hat{\mathrm{j}}|}=\frac{(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}) \cdot \hat{\mathrm{j}}}{1}=-4\)

Question 2. If \(\vec{a}=\alpha \hat{i}+3 \hat{j}-6 \hat{k} \text { and } \vec{b}=2 \hat{i}-\hat{j}-\beta \hat{k} \text {. }\), find the value of α and β so that \(\vec{a}\) and \(\vec{b}\) may be collinear.
Solution:

⇒ \(\vec{a}=\alpha \hat{i}+3 \hat{j}-6 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}-\beta \hat{k} \) (given)

Since \(\vec{a}\) and \(\vec{b}\) be collinear, we must have: \(\alpha \hat{i}+3 \hat{j}-6 \hat{k}=t(2 \hat{i}-\hat{j}-\beta \hat{k})\)

On comparing both sides

α = 2t,  t = -3, -βt = -6

⇒ β = -2, α = 2 x (-3) = -6

Hence α= -6, β = -2

Question 3. Find the magnitude of vector \(\vec{a}\) given by \(\vec{a}\) = \((\hat{i}+3 \hat{j}-2 \hat{k}) \times(-\hat{i}+3 \hat{k})\)
Solution:

⇒ \({\vec{a}}=(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \times(-\hat{\mathrm{i}}+3 \hat{\mathrm{k}})\)

=\(\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & 3 & -2 \\
-1 & 0 & 3
\end{array}\right|\)

= \(9 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)

= \(|{\vec{a}}|=\sqrt{(9)^2+(-1)^2+(3)^2}=\sqrt{91} \text { units }\)

Question 4. If \(|\vec{a}|=4 \text { and }-3 \leq \lambda \leq 2\), then \(|\lambda \vec{a}|\) lies in:

1. [0,12]
2. [2,3]
3. [8,12]
4. [-12,8]

Solution: 1. [0,12]

⇒ \(|\vec{a}|=4\) (given)

Now, \(|\lambda \vec{a}|=|\lambda| \vec{a}|\Rightarrow| \lambda \vec{a}|=4| \lambda\) (because \(|\vec{a}|=4\))

Also, \(-3 \leq \lambda \leq 2\) (given)

⇒ \(0 \leq|\lambda| \leq 3 \Rightarrow 0 \leq 4|\lambda| \leq 12 \Rightarrow 0 \leq|\lambda \vec{a}| \leq 12\)

∴ \(|\lambda \vec{a}| \in[0,12]\)

Question 5. The area of a triangle formed by verticles O, A and B where, \(\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) and \(\overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k}\), is

1. 3√5 sq units
2. 5√5 sq units
3. 6√5 sq units
4. 4 sq units

Solution: 1. 3√5 sq units

⇒ \(\overrightarrow{O B}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k}\) (given)

Area of \(\triangle \mathrm{OAB}=\frac{1}{2}|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}|\)…..(1)

Now, \(\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}\)

= \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1\end{array}\right|=8 \hat{i}-10 \hat{j}+4 \hat{k}\)

⇒ \(|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}|=\sqrt{64+100+16}=\sqrt{180}\)

So, area of \(\triangle \mathrm{OAB}=\frac{1}{2} \sqrt{180}=3 \sqrt{5}\) sq. units [from (1)]

Question 6. Find the magnitude of each of the two vectors \(\vec{a}\) and \(\vec{b}\), having the same magnitude such that the angle between them is 60 and their scalar product is \(\frac{9}{2}\)
Solution:

Given, \(|\vec{a}|=|\vec{b}|\)

Also, angle between \(\vec{a}\) and \(\vec{b}\) is \(60^{\circ}\) and \(\vec{a} \cdot \vec{b}=\frac{9}{2}\)

⇒ \(|\vec{a}||\vec{b}| \cos 60^{\circ}=\frac{9}{2}\)

or \(|\vec{a}|^2 \times \frac{1}{2}=\frac{9}{2}\)

(because \(|\vec{a}|=|\vec{b}|\))

⇒ \(|\vec{a}|^2=9 \text { or }|\vec{a}|=3\)

Hence, magnitudes of \(\vec{a}\) and \(\vec{b}\) are equal to 3 .

Question 7. Write the projection of the vector \((\vec{b}+\vec{c})\) on the vector \(\vec{a}\), where \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\) and \({\vec{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \text {. }\)
Solution:

Given, \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\) and \({\vec{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)

⇒ \(\vec{\mathrm{b}}+\vec{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)

∴ Projection of \(\vec{\mathrm{b}}+\vec{\mathrm{c}}\) on \(\vec{\mathrm{a}}\) = \(\vec{\mathrm{b}}+\vec{\mathrm{c}}\)  \(\hat{\mathrm{a}}\)

= \((3 \hat{i}+\hat{j}+2 \hat{k}) \cdot \frac{(2 \hat{i}-2 \hat{j}+\hat{k})}{3}=\frac{6}{3}=2 \text { units }\)

Question 8. If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are three mutually perpendicular unit vectors, find the value of |\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\)|.
Solution:

Since \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors, we have:

⇒ |\(\vec{a}\)| = |\(\vec{b}\)|=|\(\vec{c}\)|= 1….(1)

Since \(\vec{a}\).\(\vec{b}\).\(\vec{a}\) are mutually perpendicular vectors, we have:

⇒ \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{c}\) = \(\vec{c}\).\(\vec{a}\) = 0…..(2)

Now, |\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\)|2 = (\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{a}\)).(\(\vec{a}\) + 2\(\vec{b}\) +3\(\vec{a}\)) (|\(\vec{a}\)|’=\(\vec{a}\).\(\vec{a}\))

= \(\vec{a}\).\(\vec{a}\) + 4(\(\vec{b}\).\(\vec{b}\))+9(\(\vec{c}\).\(\vec{c}\))+4\(\vec{a}\).\(\vec{b}\)+6\(\vec{a}\).\(\vec{c}\)+12\(\vec{a}\).\(\vec{c}\) (\(\vec{a}\).\(\vec{a}\) = \(\vec{b}\),\(\vec{a}\))

= |\(\vec{a}\)|2 + 4|\(\vec{b}\)|2 + 9|\(\vec{c}\)|2 + 0 [using(2)]

= 1 + 4 + 9 = 14 [using (1)]

⇒ |\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\)|2= 14

⇒ |\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\)|= V14

Question 9. If the side AB and BC of a parallelogram ABCD are represented as vectors \(\overrightarrow{\mathrm{AB}}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\overrightarrow{\mathrm{BC}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\). then find the unit vector along diagonal AC.
Solution:

⇒ \(\overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{k}, \overrightarrow{\mathrm{BC}}=\hat{i}+2 \hat{j}+3 \hat{k}\) (given)

⇒ \(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\mathrm{BC}=(2 \hat{i}+4 \hat{j}-5 \hat{k})+(\hat{i}+2 \hat{j}+3 \hat{k})=3 \hat{i}+6 \hat{j}-2 \hat{k}\)

⇒ \(|\overrightarrow{\mathrm{AC}}|=\sqrt{9+36+4}=\sqrt{49}=7\)

∴ Required unit vector along diagonal \(\mathrm{AC}=\frac{\overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AC}}|}\)

= \(\frac{3 \hat{i}+6 \hat{j}-2 \hat{k}}{7}=\frac{1}{7}(3 \hat{i}+6 \hat{j}-2 \hat{k})\)

Question 10. Find a vector 7 equally inclined to the three axes and whose magnitude is 3√3 units.

Or,

Find the angle between unit vectors \(\vec{a}\) and \(\vec{b}\) so that v3\(\vec{a}\)– \(\vec{b}\) is also a unit vector.

Solution:

Let α be the angle made by a vector with coordinate axes.

Then, \({\vec{r}}=3 \sqrt{3}(\ell \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}})\) (because \(|{\vec{r}}|=3 \sqrt{3}\))

where l = cos α, m = cos α, n = cos α

∴ \(l^2+m^2+n^2=1 \Rightarrow 3 \cdot \cos ^2 \alpha=1 \Rightarrow \cos \alpha= \pm \frac{1}{\sqrt{3}}\)

∴ \({\vec{r}}=3 \sqrt{3}\left( \pm \frac{\hat{\mathrm{i}}}{\sqrt{3}} \pm \frac{\hat{\mathrm{j}}}{\sqrt{3}} \pm \frac{\hat{\mathrm{k}}}{\sqrt{3}}\right) \text { or } \overrightarrow{\mathrm{r}}= \pm 3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)

Or,

Given, \(\vec{a}\) and \(\vec{b}\) are unit vectors i.e. |\(\vec{a}\)| — 1 — |\(\vec{b}\)|

and \(|\sqrt{3} \vec{a}-\vec{b}|=1 \Rightarrow|\sqrt{3} \vec{a}-\vec{b}|^2=1\)

⇒ \((\sqrt{3} \vec{a}-\vec{b}) \cdot(\sqrt{3} \vec{a}-\vec{b})=1\)

or \(3|\vec{a}|^2-\sqrt{3} \vec{a} \cdot \vec{b}-\sqrt{3} \cdot \vec{a}+|\vec{b}|^2=1\)

⇒ \(3-2 \sqrt{3} \vec{a} \cdot \vec{b}+1=1\)

or \(\vec{a} \cdot \vec{b}=\frac{\sqrt{3}}{2} \Rightarrow|\vec{a}||\vec{b}| \cos \theta=\frac{\sqrt{3}}{2}\)

⇒ \(\cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6}\)

Question 11. Show that the points \(\mathrm{A}(-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})\), \(\mathrm{B}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\) and \(\mathrm{C}(7 \hat{\mathrm{i}}-\hat{\mathrm{k}})\) are collinear.

Or,

Find \(|\vec{a} \times \vec{b}| \text {, if } \vec{a}=2 \hat{i}+\hat{j}+3 \hat{k} \text { and } \vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k} \text {. }\)

Solution:

⇒ \(\vec{\mathrm{a}}=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{k}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=7 \hat{\mathrm{i}}-\hat{\mathrm{k}}\) (given)

⇒ \(\overrightarrow{\mathrm{AB}}=\vec{\mathrm{b}}-\vec{\mathrm{a}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\),

⇒ \(\overrightarrow{\mathrm{AC}}=\vec{\mathrm{c}}-\vec{\mathrm{a}}=9 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}=33 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)

⇒ \(\overrightarrow{\mathrm{AC}}=3 \cdot \overrightarrow{\mathrm{AB}}\)

⇒ \(\overrightarrow{\mathrm{AC}} \| \overrightarrow{\mathrm{AB}}\) but A is common in both \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AB}}\)

⇒ \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AB}}\) are collinear

⇒ \(\mathrm{A} \cdot \mathrm{B}\) and \(\mathrm{C}\) are collinear

Given, \(\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}\)

⇒ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 3 \\
3 & 5 & -2
\end{array}\right|-\hat{i}(-2-15)-\hat{j}(-4-9)+\hat{k}(10-3) \Rightarrow \vec{a} \times \vec{b}=-17 \hat{i}+13 \hat{j}+7 \hat{k}\)

or \(|\vec{a} \times \vec{b}|=\sqrt{(-17)^2+(13)^2+(7)^2}=\sqrt{289+169+49}=\sqrt{507}=13 \sqrt{3}\)

Question 12. If θ is the angle between two vectors \(\hat{i}-2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-2 \hat{j}+\hat{k}\). find \(\sin \theta\).
Solution:

Let \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}\)

∴ \(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{\hat{i}-2 \hat{j}+3 \hat{k} \cdot 3 \hat{i}-2 \hat{j}+\hat{k}}{\sqrt{(1)^2+(-2)^2+(3)^2} \times \sqrt{(3)^2+(-2)^2+(1)^2}}=\frac{3+4+3}{\sqrt{14} \cdot \sqrt{14}}=\frac{10}{14}=\frac{5}{7}\)

Hence, \(\cos \theta=\frac{5}{7} \Rightarrow \sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{25}{49}}\)

Hence; \(\sin \theta=\sqrt{\frac{24}{49}}=\frac{2}{7} \sqrt{6}\)

Question 13. The two adjacent sides of a parallelogram are represented by vectors \(2 \hat{i}-4 \hat{j}+5 \hat{k}\) and \(\hat{i}-2 \hat{j}-3 \hat{k}\). Find the unit vector parallel to one of its diagonals. Also, find the area of the parallelogram.

Or,

If \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k} \cdot \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}\) are such that the vector (\(\vec{a}\) + X\(\vec{b}\)). is perpendicular to vector c . then find the value of X.

Solution:

Adjacent sides of a parallelogram are given as \(\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}-3 \hat{k} \text {. }\)

Then, the diagonal of a parallelogram is given by \(\vec{a}\) + \(\vec{b}\)

⇒ \(\vec{a}+\vec{b}=(2 \hat{i}-4 \hat{j}+5 \hat{k})+(\hat{i}-2 \hat{j}-3 \hat{k})=3 \hat{i}-6 \hat{j}+2 \hat{k}\)

Thus, the unit vector parallel to the diagonal is \(\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{3^2+(-6)^2+2^2}}=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}\)

Also, Area of parallelogram = \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|\)

⇒ \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & -4 & 5 \\
1 & -2 & -3
\end{array}\right|=\hat{\mathrm{i}}(12+10)-\hat{\mathrm{j}}(-6-5)+\hat{\mathrm{k}}(-4+4)\)

⇒ \(\vec{\mathrm{a}} \times \vec{\mathrm{b}}=22 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}\)

∴ \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{22^2+11^2}=11 \sqrt{5}\)

Hence, the area of the parallelogram is 11√5 square units.

Given, \(\vec{\mathrm{a}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \vec{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\vec{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}\).

Given, \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c}\).

⇒ \((\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0 \Rightarrow \vec{a} \cdot \vec{c}+\lambda(\vec{b} \cdot \vec{c})=0 \Rightarrow \lambda=-\frac{\vec{a} \cdot \vec{c}}{b \cdot c}\)

⇒ \(\lambda=-\frac{(2 \hat{i}+2 \hat{j}+3 \hat{k}) \cdot(3 \hat{i}+\hat{j})}{(-\hat{i}+2 \hat{j}+\hat{k}) \cdot(3 \hat{i}+\hat{j})}\)

⇒ \(\lambda=-\frac{6+2+0}{-3+2+0}=8\)

Question 14. If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three vectors such that \(\vec{a}\).\(\vec{b}\) = \(\vec{a}\).\(\vec{c}\) and \(\vec{a}\)x\(\vec{b}\) = \(\vec{a}\)x\(\vec{c}\), \(\vec{a}\)*\(\vec{0}\), then show that \(\vec{b}\) = \(\vec{c}\).

Or,

If |\(\vec{a}\)| = 3. |\(\vec{a}\)| = 5. |\(\vec{c}\)| = 4 and \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = 0. then find the value of (\(\vec{a}\).\(\vec{b}\) + \(\vec{b}\).\(\vec{c}\) + \(\vec{c}\).\(\vec{a}\)).

Solution:

Given; \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}\) and \(\vec{a} \neq \vec{0}\)

⇒ \(\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}=0\) and \(\vec{a} \neq \vec{0}\)

⇒ \(\vec{a} \cdot(\vec{b}-\vec{c})=0\) and \(\vec{a} \neq \vec{0}\)

or \(\vec{b}-\vec{c}=\vec{0}\) or \(\vec{a} \perp(\vec{b}-\vec{c})\)

⇒ \(\vec{b}=\vec{c}\) or \(\vec{a} \perp(\vec{b}-\vec{c})\)

Again ; \(\vec{\mathrm{a}} \times \vec{\mathrm{b}}=\vec{\mathrm{a}} \times \vec{\mathrm{c}}\) and \(\vec{\mathrm{a}} \neq \vec{0}\)

⇒ \((\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})=\vec{0}\) and \(\vec{a} \neq \vec{0}\)

or \(\vec{\mathrm{a}} \times(\vec{\mathrm{b}}-\vec{\mathrm{c}})=\vec{0}\) and \(\vec{\mathrm{a}} \neq \vec{0}\)

⇒ \((\vec{b}-\vec{c})=\vec{0}\) or \(\vec{a} \|(\vec{b}-\vec{c})\)

⇒ \(\vec{b}=\vec{c}\) or \(\vec{a} \|(\vec{b}-\vec{c})\)

From (1) and (2); we get \(\vec{b}\) = \(\vec{c}\)

(\(\vec{a}\) cannot be both and to (\(\vec{b}\)–\(\vec{c}\) simultaneously)

Given that \(|\vec{a}|=3,|\vec{b}|=5,|\vec{c}|=4 and \vec{a}+\vec{b}+\vec{c}=\vec{0}\)

⇒ \(|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0\) (because \(|\vec{a}|^2=\vec{a} \cdot \vec{a}\))

⇒ \(|\vec{a}|^2+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{a}+|\vec{b}|^2+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+|\vec{c}|^2=0\)

⇒ \(9+25+16+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\)

(because \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\))

⇒ \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-25\)

Question 15. The scalar product of the vector \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) with a unit vector along the sum of the vectors \(\vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\vec{\mathrm{c}}=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) is equal to 1. Find the value of λ and hence find the unit vector along \(\vec{b}\) + \(\vec{c}\).
Solution:

⇒ \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\vec{c}=\lambda \hat{i}+2 \hat{j}+3 \hat{k}\)

⇒ \({\vec{b}}+\vec{\mathrm{c}}=\hat{\mathrm{i}}(2+\lambda)+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}=\vec{\mathrm{d}}\)(let)

Unit vector along \(\vec{d}= \pm \frac{\hat{i}(2+\lambda)+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}\)

⇒ \((\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot\left(\frac{\hat{\mathrm{i}}(2+\lambda)+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}\right)=1\)

(because \(\vec{\mathrm{a}} \cdot \vec{\mathrm{d}}=\hat{1})\)

⇒ \((2+\lambda)+6-2= \pm \sqrt{(2+\lambda)^2+36+4}\)

⇒ \(6+\lambda= \pm \sqrt{4+4 \lambda+\lambda^2+40}\)

⇒ \(36+\lambda^2+12 \lambda=44+4 \lambda+\hat{\lambda}^2 \text { or } 8 \lambda=8 \Rightarrow \lambda=1\)

Unit vector along \(\vec{b}+\vec{c}\)

= \(\frac{\hat{\mathrm{i}}(2+\lambda)+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}\)

= \(\frac{3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{\sqrt{9+36+4}}=\frac{3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{7}\) (because \lambda=1)

Question 16. Let \(\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}-\hat{k}\). Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{c}\) and \(\vec{b}\) and \(\vec{d}\) \(\vec{a}\) = 21.
Solution:

Given: \(\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}\),

⇒ \(\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}\)

⇒ \(\vec{c}=3 \hat{i}+\hat{j}-k \text { and } \vec{d} \cdot \vec{a}=21\)

⇒ \(\vec{c} \times \vec{b}\)

– \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & -1 \\
1 & -4 & 5
\end{array}\right|\)

– \(\hat{i}(5-4)-\hat{j}(15+1)+\hat{k}(-12-1)\)

or \(\vec{c} \times \vec{b}=\hat{i}-16 \hat{j}-13 \hat{k}\)

Now, \(\vec{d}\) is \(\perp\) to both \(\vec{c}\) and \(\vec{b}\) (given)

i.e. \(\vec{\mathrm{d}} \|(\vec{\mathrm{c}} \times \vec{\mathrm{b}})\) or \(\vec{\mathrm{d}}=\lambda \vec{\mathrm{c}} \times \vec{\mathrm{b}}\)

⇒ \(\vec{d}=\lambda(\hat{i}-16 \hat{j}-13 \hat{k})=(\lambda \hat{i}-16 \lambda \hat{j}-13 \lambda k)\)….(1)

Now, \(\vec{\mathrm{d}} \cdot \vec{\mathrm{a}}=21\)

⇒ \((\lambda \hat{\mathrm{i}}-16 \lambda \hat{\mathrm{j}}-13 \lambda \mathrm{k}) \cdot(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\mathrm{k})=21\) (from (1))

or \(4 \lambda-80 \lambda+13 \lambda=21\)

⇒ \(-63 \lambda=21 \text { or } \lambda=\frac{21}{-63}=-\frac{1}{3}\)

Put this value of \(\lambda\) in equation (1), we get :

⇒ \(\vec{\mathrm{d}}=-\frac{1}{3}(\hat{\mathrm{i}}-16 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})\)

⇒ \(\vec{\mathrm{d}}=-\frac{1}{3} \hat{\mathrm{i}}+\frac{16}{3} \hat{\mathrm{j}}+\frac{13}{3} \hat{\mathrm{k}}\).

CBSE Class 12 Maths Chapter 9 Differential Equations Important Questions

Question 1.

1. Write the order and degree of the differential equation \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\)
2. Find the general solution of the differential equation \(\frac{dy}{dx}\) = a. where a is an arbitrary constant.

Solution:

1. Given differential equation is: \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\)

⇒ Order = 2 and degree is not defined since the given equation is not a polynomial in
derivatives

2. Given differential equation is \(\frac{dy}{dx}\) = a

⇒ dy = a dx

Integrating both sides; we get: ∫l dy =∫a dx ⇒  y = ax + C

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Question 2. Find the integrating factor of x \(\frac{dy}{dx}\) + (1 + x cot x) y = x
Solution:

x\(\frac{d y}{d x}+(1+x \cot x) y=x \Rightarrow \frac{d y}{d x}+y\left(\frac{1}{x}+\cot x\right)=1\)

Comparing it with \(\frac{dy}{dx}\) + Py = Q; we get:

P = \(\left(\frac{1}{x}+\cot x\right)\) and Q=1

⇒ \(I \cdot F .=e^{\int P d x}=e^{\int\left(\frac{1}{x}+\cot x\right) d x}=e^{(\mathrm{log} x+\log \sin x)}\)

or I.F. = \(e^{\log (x \sin x)}=x \sin x\)

Question 3. Find the general solution of the differential equation: \(\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^x}\)
Solution:

Given differential equation is \(\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^{-x}}\)

⇒ \(\frac{d y}{d x}=\frac{3 e^{2 x}\left(1+e^{2 x}\right)}{e^{-x}\left(1+e^{2 x}\right)} \Rightarrow \frac{d y}{d x}=3 e^{3 x}\)

⇒ \(\int 1 \cdot d y=\int 3 e^{3 x} d x \Rightarrow y=3\left(\frac{e^{3 x}}{3}\right)+C \Rightarrow y=e^{3 x}+C\)

Question 4. Find the general solution of the differential equation: log(\(\frac{dy}{dx}\)) = ax + by.
Solution:

Given that, log(\(\frac{dy}{dx}\)) = ax + by.

⇒ \(\frac{d y}{d x}=e^{a x+b y}=e^{a x} \cdot e^{b y} \Rightarrow \frac{d y}{e^{b y}}=e^{a x} d x\)

On integrating both sides; we get:

⇒ \(\int e^{-b y} d y=\int e^{a x} d x \Rightarrow \frac{e^{-b y}}{-b}=\frac{e^{a x}}{a}+c_1 \)

⇒ \(e^{-b y}=-\frac{b}{a} e^{a x}+C \text { where } C=-b c_1\)

Question 5. Find the general solution of the differential equation: e^dy/dx = x²
Solution:

Given that e^dy/dx = x²

Taking log on both sides; we get:

⇒ log e^dy/dx = log x²

⇒ \(\frac{dy}{dx}\) = 2 logx

⇒ dy = 2logxdx

On integrating both sides; we have:

⇒ \(\int 1 \cdot d y=2 \int 1 \cdot \log x d x \Rightarrow y=2\left[\log x \int 1 \cdot d x-\int\left(\frac{1}{x} \cdot \int 1 d x\right) d x\right]\)

⇒ y = 2x logx -2x+c

Question 6. If the solution of the differential equation \(\frac{d y}{d x}=\frac{2 x y-y^2}{2 x^2} \text { is } \frac{a x}{y}=b \log |x|+C \text {; }\) then find the value of ‘a’ and ‘b’.
Solution:

Given, \(\frac{a x}{y}=b \log |x|+C\)

Differentiating with respect to x, \(a\left[\frac{y-x \frac{d y}{d x}}{y^2}\right]=\frac{b}{x}\)

⇒ \(y-x \frac{d y}{d x}=\frac{b y^2}{a x} \Rightarrow x \frac{d y}{d x}=y-\frac{b y^2}{a x} \)

⇒ \(\frac{d y}{d x}=\frac{a x y-b y^2}{a x^2}\) will be the required differential equation

Now comparing with \(\frac{d y}{d x}=\frac{2 x y-y^2}{2 x^2}\), we get a = 2 and b = 1

Question 7. Find the particular solution of the differential equation x \(\frac{dy}{dx}\) – y = x².e^x. given y(1) = 0.

Or,

Find the general solution of the differential equation x \(\frac{dy}{dx}\) = y(log y – log x +1).

Solution:

Given differential equation is x \(\frac{dy}{dx}\) – y = x² · e^x

⇒ \(\frac{dy}{dx}\) — \(\frac{y}{x}\)= x · e^x

This is a linear differential equation of the form \(\frac{dy}{dx}\) + Py = Q

where P = –\(\frac{1}{x}\) and Q = x · e^x

∴ I.F. = \(\mathrm{e}^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\)

Hence, the general solution is given by \(y\left(\frac{1}{x}\right)=\int x \cdot e^x\left(\frac{1}{x}\right) d x\)

⇒ \(\frac{y}{x}=\int e^x d x \Rightarrow \frac{y}{x}=e^x+c\)

Substituting x = 1 and y = 0, we get; 0 = e¹ + c

⇒ c = -e

∴ Required particular solution: \(\frac{y}{x}\)= e^x – e

Or,

Given differential equation is \(x \frac{d y}{d x}=y(\log y-\log x+1) \Rightarrow \frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)\)

Put y = \(v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \Rightarrow v+x \frac{d v}{d x}=v(\log v+1)\)

⇒ \(\frac{1}{v \log v} d v=\frac{d x}{x}\)

Integrating both sides: \(\int \frac{1}{v \log v} d v=\int \frac{1}{x} d x\)

⇒ log(log v) = log x + log c = log cx ⇒ log v = cx

⇒ log \(\frac{y}{x}\) = cx, which is the required general solution.

Question 8. Solve the Differential equation \(\left(1+e^{y / x}\right) d y+e^{y / x}\left(1-\frac{y}{x}\right) d x=0(x \neq 0)\)
Solution:

Given differential equation is \(\left(1+e^{y / x}\right) d y+e^{y / x}\left(1-\frac{y}{x}\right) d x=0(x \neq 0)\)

⇒ \(\left(1+e^{y / x}\right) d y=\left(\frac{y}{x}-1\right) e^{y x} d x \Rightarrow \frac{d y}{d x}=\frac{e^{y x}\left(\frac{y}{x}-1\right)}{1+e^{y x}}\)

Put \(\frac{y}{x}=v \text { i.e. } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

∴ \(v+x \frac{d v}{d x}=\frac{e^v(v-1)}{1+e^v} \Rightarrow x \frac{d v}{d x}=\frac{e^v(v-1)}{1+e^v}-v\)

⇒ \(x \frac{d v}{d x}=\frac{v \cdot e^v-e^v-v-v e^v}{1+e^v} \Rightarrow x \frac{d v}{d x}=-\frac{\left(e^v+v\right)}{1+e^v}\)

⇒ \(\int \frac{1+e^v}{v+e^v} d v=-\int \frac{d x}{x} \Rightarrow \log \left|v+e^v\right|=-\log |x|+\log C\)

⇒ \(\log \left|v+e^v\right|=\log \left|\frac{C}{x}\right| \Rightarrow v+e^v=\frac{C}{x}\)

⇒ \(x\left(\frac{y}{x}+e^{y / x}\right)=C \Rightarrow y+x e^{y / x}=C\)

Question 9. Solve the differential equation \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\)

Or,

Solve the differential equation \(\frac{d y}{d x}=-\left[\frac{x+y \cos x}{1+\sin x}\right]\)

Solution:

Given differential equation is: \(x \frac{d y}{d x}=y-x \tan (y / x) \Rightarrow \frac{d y}{d x}=\left(\frac{y}{x}\right)-\tan \left(\frac{y}{x}\right)\)

The given equation is a homogenous differential equation.

Put \(\frac{y}{x}=v \text { i.e. } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)

∴ \(v+x \frac{d v}{d x}=v-\tan v \Rightarrow x \frac{d v}{d x}=-\tan v \Rightarrow \frac{d v}{\tan v}=-\frac{d x}{x}\)

Integrating both sides; we get:

⇒ \(\int \frac{d v}{\tan v}=\int \frac{-d x}{x}\)

⇒ \(\int \cot v d v=-\int \frac{1}{x} d x \Rightarrow \log |\sin v|=-\log |x|+\log C\)

⇒ \(\log |\sin v|=\log \left|\frac{C}{x}\right|\)

⇒ \(\sin \left(\frac{y}{x}\right)=C / x \Rightarrow x \sin \left(\frac{y}{x}\right)=C\)

Or,

Given differential equation is: \(\frac{d y}{d x}=-\left[\frac{x+y \cos x}{1+\sin x}\right] \Rightarrow \frac{d y}{d x}+\left(\frac{\cos x}{1+\sin x}\right) y=\frac{-x}{1+\sin x}\)…..(1)

Equation(1) is a linear differential equation of the form \(\frac{dy}{dx}\)+ Py = Q;

where, P = \(\left(\frac{\cos x}{1+\sin x}\right)\) and \(Q=\left(\frac{-x}{1+\sin x}\right)\)

⇒ I.F. =\(\mathrm{e}^{\int^{P d x}}=\mathrm{e}^{\int \frac{\cos x}{1+\sin x} d x}=e^{\mathrm{sin}|+\sin x|}=1+\sin x\)

Hence, the solution is y(I.F) = ∫(Q.I.F)dx + C

⇒ \(y(1+\sin x)=\int\left(\frac{-x}{1+\sin x}\right) \times(1+\sin x) d x+C\)

⇒ \(y(1+\sin x)=\int-x d x+C \quad \text { or } y(1+\sin x)=\frac{-x^2}{2}+C\)

Question 10. Find the particular solution of the differential equation e^x tan y dx + (2 – e^x) sec²ydy = 0, given that y = π/4 when x = 0.

Or,

Find the particular solution of the differential equation \(\frac{dy}{dx}\) + 2ytanx = sin x, given that y = 0 when x = π/4.

Solution:

The given differential equation is e^x tan ydx + (2- e^x)sec² ydy = 0

∴ \(\left(2-e^x\right) \sec ^2 y d y=-e^x \tan y d x \Rightarrow \frac{\sec ^2 y}{\tan y} d y=\frac{-e^x}{2-e^x} d x\)

Integrating both sides, we get \(\int \frac{\sec ^2 y}{\tan y} d y=\int \frac{-e^x}{2-e^x} d x\)

⇒ log(tany) = log(2-e^x) + logC ⇒ log (tan y) = log[C(2 — e^x)]

⇒ tan y = C (2  – e^x)

y = π/4 when x = 0 ⇒ C = 1

So, the required particular solution is tan y = 2-e^x

The given differential equation is \(\frac{dy}{dx}\) + 2y tax = sin x

This is a linear equation of the form \(\frac{dy}{dx}\) + Py = Q

where P = 2 tan x and Q = sin x

Now, I.F. \(=\mathrm{e}^{\int \mathrm{Pdx} x}=\mathrm{e}^{\int 2 \mathrm{tan} x \mathrm{xdx}}=\mathrm{e}^{2 \log \sec x}=\mathrm{e}^{\log \sec ^2 x}=\sec ^2 \mathrm{x}\)

The general solution of the given differential equation is y (I.F) = ∫(Q x I.F) dx + C

⇒ y(sec²x)= ∫(sin x.sec²x)dx + C

⇒ ysec²x = ∫(sec x.tanx)dx + C

⇒ y sec² x = sec x + C…..(1)

Now, y = 0 when x = π/3

Therefore, 0 x sec² π/3 = sec π/3 + C ⇒ 0 = 2 + C ⇒ C = -2

Substituting C = -2 in equation (1). we get: ysec²x – sec x – 2 ⇒ y = cos x – 2cos²x

Hence, the required solution of the given differential equation is y = cos x – 2cos²x.

CBSE Class 12 Maths Chapter 7 Integrals Important Questions

Question 1. Find \(\int e^x\left(\log \sqrt{x}+\frac{1}{2 x}\right) d x\) Or,

Find \(\int \mathrm{e}^{2 \log x} \mathrm{dx}\)

Solution:

Let I = \(\int e^x\left(\log \sqrt{x}+\frac{1}{2 x}\right) d x\)

Let f(x) = \(\log \sqrt{x}\)

⇒ \(f^{\prime}(x)=\frac{1}{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}=\frac{1}{2 x}\)

⇒ \(\mathrm{I}=\mathrm{e}^{\mathrm{x}} \log \sqrt{\mathrm{x}}+\mathrm{c}\) (because \(\int \mathrm{e}^{\mathrm{x}}\left(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right) \mathrm{dx}=\mathrm{e}^{\mathrm{x}} \cdot \mathrm{f}(\mathrm{x})+\mathrm{c}\))

Or,

∴ I = \(\int e^{2 \log x} d x\)

I = \(\int e^{\log x^2} d x=\int x^2 d x\) (because \(e^{\log x}=x\))

= \(\frac{x^3}{3}+C\)

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Question 2. Evalute: \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos ^2 x d x\)
Solution:

Let I = \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos ^2 x d x\)

Here f(x) = x.cos²x

f(-x) = (-x) cos²(-x)

f(-x) = -x cos²x

f(-x) = -f(x)

i.e. f(x) is odd function

∴ I = 0 (\(\int_{-a}^a\) f(x)dx = 0, if f(x) is odd function)

Question 3. Find: \(\int \frac{d x}{x^2-6 x+13}\)
Solution:

I = \(\int \frac{d x}{x^2-6 x+13}=\int \frac{1}{(x-3)^2+2^2} d x\)

= \(\frac{1}{2} \tan ^{-1}\left(\frac{x-3}{2}\right)+C\)

(because \(\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\))

Question 4. Find: \(\int \frac{\tan ^3 x}{\cos ^3 x} d x\)
Solution:

Let \(I=\int \frac{\tan ^3 x}{\cos ^3 x} d x=\int \frac{\sin ^3 x}{\cos ^6 x} d x=\int \frac{\sin ^2 x \cdot \sin x}{\cos ^6 x} d x=\int \frac{\left(1-\cos ^2 x\right) \cdot \sin x}{\cos ^6 x} d x\)

Put \(\cos x=t \Rightarrow-\sin x d x=d t \Rightarrow \sin x d x=-d t\)

Now I = \(-\int \frac{\left(1-t^2\right)}{t^6} d t=-\int\left(\frac{1}{t^6}-\frac{1}{t^4}\right) d t=-\left[\frac{-1}{5 t^5}+\frac{1}{3 t^3}\right]+C=\frac{1}{5 \cos ^5 x}-\frac{1}{3 \cos ^3 x}+C\)

Question 5. Find: \(\int \frac{x-5}{(x-3)^3} e^x d x\)
Solution:

⇒ \(\int \frac{x-5}{(x-3)^3} e^x d x =\int e^x\left[\frac{(x-3)-2}{(x-3)^3}\right] d x=\int e^x\left[\frac{(x-3)}{(x-3)^3}-\frac{2}{(x-3)^3}\right] d x\)

= \(\int e^x\left[\frac{1}{(x-3)^2}+\left\{\frac{-2}{(x-3)^3}\right\}\right] d x\) [because \(\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+C)\)

= \(\frac{e^x}{(x-3)^2}+C\)

Question 6. Find: \(\int \frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x} d x\) Or,

Find: \(\int \frac{x-3}{(x-1)^3} e^x d x\)

Solution:

⇒ \(\int \frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cdot \cos ^2 x} d x=\int\left(\frac{\sin x}{\cos ^2 x}+\frac{\cos x}{\sin ^2 x}\right) d x\)

= \(\int(\tan x \sec x+\mathrm{cosec} x \cdot \cot x) d x=\sec x-\mathrm{cosec} x+C\)

Or

⇒ \(\int \frac{x-3}{(x-1)^3} e^x d x=\int e^x\left(\frac{x-1-2}{(x-1)^3}\right) d x\)

= \(\int \mathrm{e}^x\left(\frac{1}{(\mathrm{x}-1)^2}+\left(\frac{-2}{(\mathrm{x}-1)^3}\right)\right) \mathrm{dx}\)

(because \(\int \mathrm{e}^{\mathrm{x}}\left(\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right) \mathrm{dx}=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})+C\)

= \(\frac{e^x}{(x-1)^2}+C\)

Question 7. Evalute \(\int \frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x} d x\)
Solution:

⇒ \(\int \frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x} d x=\int \frac{1-2 \sin ^2 x+2 \sin ^2 x}{\cos ^2 x} d x\)

= \(\int \frac{1}{\cos ^2 x} d x=\int \sec ^2 x d x=\tan x+C\)

Question 8. Evaluate \(\int_0^{2 \pi} \frac{\mathrm{dx}}{1+\mathrm{e}^{\sin x}}\)
Solution:

Let I = \(\int_0^{2 \pi} \frac{1}{1+e^{\sin x}} d x\)

⇒ I = \(\int_0^{2 \pi} \frac{1}{1+e^{\sin (2 \pi-x)}} d x\) (\(because \int_0^a f(x) d x=\int_0^a f(a-x) d x\))

⇒ I = \(\int_0^{2 \pi} \frac{1}{1+e^{-\sin x}} d x \quad \Rightarrow I=\int_0^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x\)

From (1) and (2), we get :

⇒ 2I = \(\int_0^{2 \pi} \frac{1}{1+e^{\sin x}} d x+\int_0^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x \Rightarrow 2 I=\int_0^{2 \pi} 1 \cdot d x\)

⇒ 2I = \([x]_0^{2 \pi} \Rightarrow 2 I=2 \pi \Rightarrow I=\pi\)

Question 9. Evaluate: \(\int_{-1}^2\left|x^3-x\right| d x\)
Solution:

We know that x³ – x ≥ 0 on [0, 1] and x³ – x ≤ 0 on [0, 1] and that x³ – x ≥ 0 on [1, 2], so by property of definite integral we get

⇒ \(\int_{-1}^2\left|x^3-x\right| d x=\int_{-1}^0\left(x^3-x\right) d x+\int_0^1-\left(x^3-x\right) d x+\int_1^2\left(x^3-x\right) d x\)

= \(\int_{-1}^0\left(x^3-x\right) d x+\int_0^1\left(x-x^3\right) d x+\int_1^2\left(x^3-x\right) d x\)

= \(\left[\frac{x^4}{4}-\frac{x^2}{4}\right]_{-1}^0+\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1+\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_1^2\)

= \(-\left(\frac{1}{4}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+(4-2)-\left(\frac{1}{4}-\frac{1}{2}\right)\)

= \(-\frac{1}{4}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+2-\frac{1}{4}+\frac{1}{2}=\frac{3}{2}-\frac{3}{4}+2=\frac{11}{4}\)

Question 10. Evaluate: \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\sin |x|+\cos |x|) d x\)
Solution:

Let \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(\sin |x|+\cos |x|) d x\)

∵ f(x) == sin |x| + cos |x|

f(-x) = sin |-x| + cos |-x|

f(-x) = sin |x| + cos |x| = f(x)

∴ f(x) is even function

Therfore by prop \(\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x\)

I = \(2 \int_0^{\pi / 2}\{\sin |x|+\cos |x|\} d x\)

I = \(2 \int_0^{\pi / 2}\{\sin x+\cos x\} d x\)

I = \(2[-\cos x+\sin x]_0^{\pi / 2}\)

I = \(2[0+1-(-1)-0]=2 \times 2=4\)

Question 11. Find: \(\int \frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)} d x\) or,

Evaluate : \(\int_{-2}^1 \sqrt{5-4 x-x^2} d x\)

Solution:

I = \(\int \frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)} d x\)

Let \(x^2=y\)

Let \(\frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)}=\frac{y}{(y+1)(3 y+4)}=\frac{A}{(y+1)}+\frac{B}{(3 y+4)}\)

⇒ \(\mathrm{y}=\mathrm{A}(3 \mathrm{y}+4)+\mathrm{B}(\mathrm{y}+1)\)

from equation (1)

Put y=-1 ⇒ A=-1

Put y=-4/3 ⇒ B=4

∴ \(\frac{y}{(y+1)(3 y+4)}=\frac{-1}{y+1}+\frac{4}{3 y+4}\)

i.e. \(\frac{x^2}{\left(x^2+1\right)\left(3 x^2+4\right)}=\frac{-1}{x^2+1}+\frac{4}{3 x^2+4}\)

⇒ I = \(\int\left(\frac{-1}{x^2+1}+\frac{4}{3 x^2+4}\right) d x=-\int \frac{1}{x^2+1} d x+\frac{4}{3} \int \frac{1}{x^2+4 / 3} d x\)

= \(-\tan ^{-1} x+\frac{4}{3} \cdot \frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{x \sqrt{3}}{2}\right)+C\)

=\(\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x \sqrt{3}}{2}\right)-\tan ^{-1} x+C\)

Or,

Let I = \(\int_{-2}^1 \sqrt{5-4 x-x^2} d x\)

= \(\int_{-2}^1 \sqrt{3^2-(x+2)^2} d x\) (because \(\int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c\)

= \(\left[\frac{x+2}{2} \sqrt{3^2-(x+2)^2}+\frac{9}{2} \sin ^{-1}\left(\frac{x+2}{3}\right)\right]_{-2}^1\)

= \(\frac{9}{2} \sin ^{-1} 1=\frac{9 \pi}{4}\)

Question 12 Find \( \int\frac{x^2}{(x-1)(x+1)^2} d x\)
Solution:

let I = \( \int\frac{x^2}{(x-1)(x+1)^2} d x\)

Let \(\frac{x^2}{(x-1)(x+1)^2}=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(x+1)^2}\)

⇒ x³ = A(x + 1)² + B(x – 1 )(x + 1) + C(x – 1)

Put x = 1, ⇒ 1 = 4A ⇒ A = 1/4

Put x = -1, ⇒ 1 = C(-2) ⇒ C = -1/2

Comparing the coefficients of x² on both sides:

1 = A + B ⇒ B = 1 – A = 1-1/4 = 3/4

Hence, \(I=\frac{1}{4} \int \frac{1}{(x-1)} d x+\frac{3}{4} \int \frac{1}{(x+1)} d x-\frac{1}{2} \int \frac{1}{(x+1)^2} d x\)

I = \(\frac{1}{4} \log |x-1|+\frac{3}{4} \log |x+1|+\frac{1}{2(x+1)}+C\)

Question 13. Find 

1. \(\int \frac{x^2+2}{x^2+1} d x\)
2. \(\int_{-1}^1 \frac{|x|}{x} d x\)

Or,

Evaluate \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x\)

Solution:

1. I = \(\int \frac{x^2+2}{x^2+1} d x=\int\left(\frac{x^2+1+1}{x^2+1}\right) d x=\int \frac{x^2+1}{x^2+1} d x+\int \frac{1}{x^2+1} d x\)

= \(\int 1 \cdot d x+\int \frac{1}{x^2+1} d x=x+\tan ^{-1} x+C\)

2. Let \(I=\int_{-1}^1 \frac{|x|}{x} d x\)

Now, \(\frac{|x|}{x}= \begin{cases}\frac{-x}{x}=-1, & x<0 \\ \frac{x}{x}=1 & , x \geq 0\end{cases}\)

⇒ \(\mathrm{I} =\int_{-1}^0(-1) \cdot \mathrm{dx}+\int_0^1 1 \cdot \mathrm{dx}=[-\mathrm{x}]_{-1}^0+[\mathrm{x}]_0^1=-[0+1]+[1-0]\)

=-1+1=0

Or,

Let I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x\)….(1)

Using property \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\);

I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right) d x\)

⇒ I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x\)….(2)

Adding (1) and (2), we get

2I = \(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x+\int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x\)

2I =\(\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x} \cdot \frac{4+3 \cos x}{4+3 \sin x}\right) d x\)

2I =\(\int_0^{\pi / 2} \log 1 d x \Rightarrow 2 I=0\) (because log 1=0)

⇒ I =0

Question 14. Find \(\int \frac{x^3+1}{x^3-x} d x\)
Solution:

I = \(\int \frac{x^3+1}{x^3-x} d x\)

I = \(\int \frac{x^3+1-x+x}{x^3-x} d x \text { or } I=\int\left(1+\frac{1+x}{x^3-x}\right) d x\)

= \(\int\left(1+\frac{1+x}{x\left(x^2-1\right)}\right) d x=\int\left(1+\frac{1}{x(x-1)}\right) d x\)

= \(\int\left(1+\frac{1}{x-1}-\frac{1}{x}\right) d x=x+\log |x-1|-\log |x|+C\)

= \(x+\log \left|\frac{x-1}{x}\right|+C\)

Question 15. Evaluate: \(\int \frac{2 \cos x}{(1-\sin x)\left(2-\cos ^2 x\right)} d x\)
Solution:

I = \(\int \frac{2 \cos x}{(1-\sin x)\left(2-\cos ^2 x\right)} d x\)=\(\int \frac{2 \cos x}{(1-\sin x)\left(1+1-\cos ^2 x\right)} d x\)=\(\int \frac{2 \cos x}{(1-\sin x)\left(1+\sin ^2 x\right)} d x\)

put \(\sin \mathrm{x}=\mathrm{t} \Rightarrow \cos \mathrm{dx}=\mathrm{dt}\)

= \(\int \frac{2 d t}{(1-t)\left(1+t^2\right)}\)

⇒ \(\mathrm{I}=-2 \mathrm{I_1}\)….(1)

Now; \(I_1=\int \frac{d t}{(t-1)\left(t^2+1\right)}\)

Let \(\frac{1}{(t-1)\left(t^2+1\right)}=\frac{A}{(t-1)}+\frac{B t+C}{\left(t^2+1\right)}\)

I =A(t²+1)+B t(t-1)+C(t-1)

Putting t = 1

⇒ 1 = A x 2

⇒ A = 1/2

Comparing the coefficient of ‘t²’ on both sides we get 0 = A + B

B = -1/2

Putting t = 0, 1 = A x 1 + C x (-1)

⇒ 1 = A – C

C = A – 1 = -1/2

⇒ \(I_1=\int \frac{d t}{(t-1)\left(t^2+1\right)}=\frac{1}{2} \int \frac{d t}{t-1}-\frac{1}{2} \int \frac{t+1}{t^2+1} d t\)

= \(\frac{1}{2} \log |t-1|-\frac{1}{4} \int \frac{2 t}{t^2+1} d t-\frac{1}{2} \int \frac{d t}{t^2+1}+C_1\)

⇒ \(I_1=\frac{1}{2} \log |t-1|-\frac{1}{4} \log \left|t^2+1\right|-\frac{1}{2} \tan ^{-1} t+C_1\)

I = \(-2 I_1=-2\left[\frac{1}{2} \log |t-1|-\frac{1}{4} \log \left|t^2+1\right|-\frac{1}{2} \tan ^{-1} t+C_1\right]\)

= \(-\log |t-1|+\frac{1}{2} \log \left|t^2+1\right|+\tan ^{-1} t-2 C_1\)

⇒ I =\(-\log |\sin x-1|+\frac{1}{2} \log \left|1+\sin ^2 x\right|+\tan ^{-1}(\sin x)+C, \text { where } C=-2 C_1\)

Question 16. Prove that \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a-x) dx and hence evaluate \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x} d x\)
Solution:

Let I = \(\int_0^a f(a-x) d x\)

Put \(\mathrm{a}-\mathrm{x}=\mathrm{t} \Rightarrow \mathrm{dx}=-\mathrm{dt}\)

When \(\mathrm{x}=0, \mathrm{t}=\mathrm{a} \& \mathrm{x}=\mathrm{a}, \mathrm{t}=0\)

⇒ I = \(-\int_a^0 f(t) d t\)

or I = \(\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{t}) \mathrm{dt}\)

(because \(\int_a^b f(x) dx=-\int_b^a f(x) d x\))

⇒ \(\mathrm{I}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\)

(because \(\int_a^b f(x) d x=\int_a^b f(t) d t\))

Now; I = \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x} d x\)….(1)

I = \(\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x\)

= \(\int_0^{\pi / 2} \frac{\frac{\pi}{2}-x}{\cos x+\sin x} d x\)….(2)

⇒ (1) + (2) gives:

2I = \(\int_0^{\pi / 2} \frac{\frac{\pi}{2}}{\cos x+\sin x} d x\)

∴ 2I = \(\frac{\pi}{2} \int_0^{\pi / 2} \frac{1}{\sin x+\cos x} d x\)

⇒ I = \(\frac{\pi}{4 \sqrt{2}} \int_0^{\pi / 2} \frac{1}{\sin \left(x+\frac{\pi}{4}\right)} d x\)

Or, I = \(\frac{\pi}{4 \sqrt{2}} \int_0^{\pi / 2} \mathrm{cosec}\left(x+\frac{\pi}{4}\right) d x\)

⇒ I = \(\frac{\pi}{4 \sqrt{2}}\left[\log \left(\mathrm{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)\right]_0^{\pi / 2}\)

or I = \(\frac{\pi}{4 \sqrt{2}}\left[\log \left(\mathrm{cosec}\left(\frac{\pi}{2}+\frac{\pi}{4}\right)-\cot \left(\frac{\pi}{2}+\frac{\pi}{4}\right)\right)-\log \left(\mathrm{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right)\right]\)

∴ I = \(\frac{\pi}{4 \sqrt{2}}\left[\log \left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)-\log \left(\mathrm{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right)\right]\)

or I = \(\frac{\pi}{4 \sqrt{2}}[\log (\sqrt{2}+1)-\log (\sqrt{2}-1)] \Rightarrow I=\frac{\pi}{4 \sqrt{2}} \log \frac{\sqrt{2}+1}{\sqrt{2}-1}\)

Question 17. Evaluate: \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x} d x\)
Solution:

Let \(I=\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x} d x\)=\(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9\left\{1-(\sin x-\cos x)^2\right\}} d x\)

Put sin x – cos x = t ⇒ (cos x + sin x)dx = dl

when x = 0, t = -1 and x = π/4, t = 0

∴ I = \(\int_{-1}^0 \frac{\mathrm{dt}}{16+9\left(1-\mathrm{t}^2\right)}=\int_{-1}^0 \frac{\mathrm{dt}}{16+9-9 \mathrm{t}^2}\)

= \(\int_{-1}^0 \frac{\mathrm{dt}}{25-9 \mathrm{t}^2}=\frac{1}{9} \int_{-1}^0 \frac{\mathrm{dt}}{\frac{25}{9}-\mathrm{t}^2}=\frac{1}{9} \int_{-1}^0 \frac{\mathrm{dt}}{\left(\frac{5}{3}\right)^2-(\mathrm{t})^2}\)

(because \(\int \frac{1}{\mathrm{a}^2-\mathrm{x}^2} \mathrm{dx}=\frac{1}{2 \mathrm{a}} \log \left|\frac{\mathrm{a}+\mathrm{x}}{\mathrm{a}-\mathrm{x}}\right|+\mathrm{c})\)

= \(\frac{1}{9} \cdot \frac{1}{2} \times \frac{3}{5}\left[\log \left(\frac{\frac{5}{3}+\mathrm{t}}{\frac{5}{3}-\mathrm{t}}\right)\right]_{-1}^0\)

= \(\frac{1}{30}\left[\log (1)-\log \left(\frac{1}{4}\right)\right]=\frac{1}{30} \log 4\)

CBSE Class 12 Maths Chapter 6 Applications Of Derivatives Important Questions

Question 1. A function f: R → R is defined as f(x) = x³ + 1. Then the function has

1. No minimum value
2. No maximum value
3. Both maximum and minimum values
4. Neither maximum value nor minimum value

Solution: 4. Neither maximum value nor minimum value

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Given ; f(x) = x³ + 1, f: R →  R

f(x) = 3x² = 0

⇒ x = 0

In (0, ∞); f (x) > 0 and In (-∞, 0); f'(x) > 0

Hence; the function has neither maximum nor minimum value.

Question 2. The function y = x²e^-x is decreasing in the interval :

1. (0,2)
2. (2,  ∞)
3. (-∞, 0)
4. (-∞, 0) ∪ (2,  ∞)

Solution: 4. (-∞, 0) ∪ (2,  ∞)

Given, y = x² e^-x

⇒ y’ = 2x² e^-x+ x² (-e^-x) = e^-x {2x – x²} = e^-x. x(2 – x) < 0

(-∞, 0) ∪ (2, ∞)

Question 3.The function f(x) = 2x³ – 15x² + 36x + 6 is increasing in the interval :

1. (-∞, 2)  ∪ (3, ∞)
2. (-∞, 2)
3. (-∞, 2] ∪ [3, ∞)
4. [3, ∞)

Solution: 3. (-∞, 2] ∪ [3, ∞)

⇒ f(x) = 2X³ – 15x² + 36x +6

⇒ f(x) = 6x² – 30x + 36 = 6(x – 2) (x – 3) > 0

Hence; f(x) is increasing in (-∞, 2] ∪ [3, ∞) [f'(x)≥0]

Question 4. The maxiumu value of \({\frac{1}{x}^x}\) is:

1. \(e^{1 / e}\)
2. \(\mathrm{e}\)
3. \(\left(\frac{1}{e}\right)^{1 / e}\)
4. \(e^e\)

Solution:

Let \(y=\left(\frac{1}{x}\right)^x \Rightarrow \log y=x \log \left(\frac{1}{x}\right)\)

⇒ \(\frac{1}{y} \frac{d y}{d x}=x \times \frac{1}{1 / x} \times\left(\frac{-1}{x^2}\right)+\log \left(\frac{1}{x}\right) \cdot 1=-1+\log \left(\frac{1}{x}\right)\)….(1)

⇒ \(\frac{d y}{d x}=\left(\frac{1}{x}\right)^x\left[-1+\log \left(\frac{1}{x}\right)\right]\)

If \(\frac{d y}{d x}=0 \Rightarrow \log \left(\frac{1}{x}\right)=1=\log e\)

(because \(\left(\frac{1}{x}\right)^x \neq 0\))

⇒ x = \(\frac{1}{e}\)

Now differentiating equation (1) with respect to x, we get \(\frac{1}{y} \frac{d^2 y}{d x^2}-\frac{1}{y^2}\left(\frac{d y}{d x}\right)^2=x \times\left(-\frac{1}{x^2}\right)\)

⇒ \(\frac{1}{y} \frac{d^2 y}{d x^2}-\frac{1}{y^2}\left(\frac{d y}{d x}\right)^2=-\frac{1}{x}\)

⇒ \(\frac{1}{y} \frac{d^2 y}{d x^2}=\left(\log \left(\frac{1}{x}\right)-1\right)^2-\frac{1}{x}\)

⇒ \(\frac{d^2 y}{d x^2}=\left(\frac{1}{x}\right)^x\left\{\left(\log \left(\frac{1}{x}\right)-1\right)^2-\frac{1}{x}\right\}\)

because \(\frac{d^2 y}{d x^2} \text { at } x=\frac{1}{e}<0\)

hence x = \(\frac{1}{e}\) is a point of maxima.

⇒ Maximum value of \(\left(\frac{1}{x}\right)^x=(e)^{1 / e}\)

Question 5. The absolute maximum value of the function f(x) = 4x – 1/2 x² in the interval (-2, 9/2) is:

1. 8
2. 9
3. 6
4. 10

Solution:

f(x) = \(4 x-\frac{1}{2} x^2 \text { in }\left[-2, \frac{9}{2}\right]\)

⇒ \(f^{\prime}(x)=4-\frac{2 x}{2}=4-x\)

Put \(f^{\prime}(x)=0 \Rightarrow x=4\)

Now; \(f(-2)=4 \times(-2)-\frac{1}{2} \times 4=-8-2=-10 \text {, }\)

f(4) = \(4 \times 4-\frac{1}{2} \times 16=16-8=8\)

and \(f\left(\frac{9}{2}\right)=4 \times \frac{9}{2}-\frac{1}{2} \times\left(\frac{9}{2}\right)^2\)

= \(\frac{36}{2}-\frac{81}{8}=\frac{144-81}{8}=\frac{63}{8}\)

Hence, absolute maximum value of f(x) = 8 at x = 4

Question 6. In a sphere of radius r, a right circular cone of height h having maximum curved surface area is inscribed. The expression for the square of the curved surface of the cone is:

1. \(2 \pi^2 r h\left(2 r h+h^2\right)\)
2. \(\pi^2 \mathrm{rh}\left(2 \mathrm{rh}+\mathrm{h}^2\right)\)
3. \(2 \pi^2 r\left(2 h^2-h^3\right)\)
4. \(2 \pi^2 \mathrm{r}^2\left(2 \mathrm{rh}-\mathrm{h}^2\right)\)

Solution: 3. \(2 \pi^2 r\left(2 h^2-h^3\right)\)

As per the given figures,

r² = R² + (h – r)²

R² = r²- (h – r)² = h(2r – h)….(1)

h² + R² = 2rh….(2)

Now; the curved surface area of a cone is S = πRl

S² = π²R²l²

Or, S² = π²[r²-(h-r)²] (R² + h²) (from (1))

= π²h(2r-h) (2rh) (from (2))

= 2π²r(2rh² – h³)

Question 7. Show that the function f(x) = 3/x + 7 is strictly decreasing for x ∈ R-{0}.
Solution:

Given, \(f(x)=\frac{3}{x}+7 \Rightarrow f^{\prime}(x)=-\frac{3}{x^2}\)

∵ \(\mathrm{f}^{\prime}(\mathrm{x})<0\)  For all x ∈ R-{0}

Hence, The given functions strictly decreasing.

Question 8. The interval in which the function f given by f(x) = x² e^-x is strictly increasing is

1. (-∞,∞)
2. (-∞,0)
3. (2,∞)
4. (0,2)

Solution: 4. (0,2)

f(x) = \(x^2 \cdot e^{-x}\) (Differentiating with respect to x)

⇒ \(f^{\prime}(x)=-x^2 e^x+2 x e^{-x}\)

⇒ \(\mathrm{f}^{\prime}(\mathrm{x})>0\) (because \(\mathrm{f}(\mathrm{x})\) is strictly increasing)

⇒ \(-\mathrm{x}^2 \cdot \mathrm{e}^{-\mathrm{x}}+2 \mathrm{x} \cdot \mathrm{e}^{-\mathrm{x}}>0\)

⇒ \(\mathrm{e}^{-\mathrm{x}}\left(2 \mathrm{x}-\mathrm{x}^2\right)>0\)

⇒ \(e^{-x} x(2-x)>0\)

(because \(\mathrm{e}^{-x} \neq 0\) as \(\mathrm{e}^{-\mathrm{x}}\) is always positive for all x in R)

Now,

when x ∈ (0, 2), f'(x) > 0

So, f(x) is strictly increasing function interval (0, 2)

Question 9. If the radius of the circle is increasing at the rate of 0.5 cm/sec, then the rate of increase of its circumference is
Solution:

At any time t let r be the radius and c be the circumference of the circle \(\frac{dr}{dt}\) = 0.5 cm/sec (given)

∵ c = 2πr

⇒ \(\frac{dc}{dt}\) = 2π(0.5) (differentiating with respect to t)

⇒ \(\frac{dc}{dt}\) = π cm /sec

The rate of increase of its circumference is cm/sec.

Question 10. In a residential society comprising 100 houses, there were 60 children between the ages of 10-15 years. Their teachers inspired them to start composting to ensure that biodegradable waste is recycled. For this purpose, instead of each child doing it for only his/her house, children convinced the Residents Welfare Association to do it as a society initiative. For this, they identified a square area in the local park. Local authorities charged an amount of Rs. 50 per square metre for space so that there is no misuse of the space and the Resident welfare association takes it seriously. The association hired a labourer for digging and takes it seriously. The association hired a labourer to dig out 250 m³ and he charged Rs.400 x (depth)². The association would like to have a minimum cost.

1. Let the side of the square plot be x m and its depth is h metres, then cost c for the pit is

1. \(\frac{50}{h}+400 h^2\)
2. \(\frac{12500}{h}+400 h^2\)
3. \(\frac{250}{h}+h^2\)
4. \(\frac{250}{h}+400 h^2\)

Solution: 2. \(\frac{12500}{h}+400 h^2\)

Since the depth of the plot is h metres and the side is x metres,

⇒ Volume of plot is x² x h ⇒ 250 = x² x h

⇒ x² = 250/h = Area of plot

∵ The charge of space is ₹ 50 per m

So, cost of plot is \(\frac{50 \times 250}{h}=₹ \frac{12500}{h}\)

and the charge of digging is 400 x h

So total cost is c = 12500/h + 400 h² ….(1)

2. Value of h(in m) for which = \(\frac{dc}{dh}\) = 0 is

1. 1.5
2. 2
3. 2.5
4. 3

Solution: 3. 2.5

From equation (1); we have

c = \(\frac{12500}{h}+400 h^2\)

Differentiating with respect to ‘h’, \(\frac{\mathrm{dc}}{\mathrm{dh}}=\frac{-12500}{\mathrm{~h}^2}+800 \mathrm{~h}=0\)

⇒ \(\frac{12500}{\mathrm{~h}^2}=800 \mathrm{~h} \Rightarrow \mathrm{h}^3=\frac{125}{8} \Rightarrow \mathrm{h}=\frac{5}{2}=2.5 \mathrm{~m}\)

3. \(\frac{\mathrm{d}^2 \mathrm{c}}{\mathrm{dh}^2}\) is given by

1. \(\frac{25000}{h^3}+800\)
2. \(\frac{500}{h^3}+800\)
3. \(\frac{100}{h^3}+800\)
4. \(\frac{500}{h^3}+2\)

Solution: 1. \(\frac{25000}{h^3}+800\)

∵ \(\frac{\mathrm{dc}}{\mathrm{dh}}=\frac{-12500}{\mathrm{~h}^2}+800 \mathrm{~h}\)

Again differentiating with respect to h; we get \(\frac{\mathrm{d}^2 \mathrm{c}}{\mathrm{dh}^2}=\frac{25000}{\mathrm{~h}^3}+800\)

4. The value of x (in m) for minimum cost is:

1. 5
2. \(10 \sqrt{\frac{5}{3}}\)
3. \(5 \sqrt{5}\)
4. 10

Solution: 4. 10

∵ h = 2.5

⇒ \(\frac{\mathrm{d}^2 \mathrm{c}}{\mathrm{dh}^2}\) at \(\mathrm{h}=2.5\) is >0

So h=2.5 is the point of minima.

because \(x^2=\frac{250}{h}\)

⇒ \(x^2=\frac{250}{2.5} \Rightarrow x^2=100 \Rightarrow x=10 \mathrm{~m}\)

5. Total minimum cost of digging the pit (in Rupees) is:

1. 4,100
2. 7,500
3. 7,850
4. 3, 220

Solution: 2. 7,500

c = \(\frac{12500}{h}+400 h^2 \text { and } h=2.5\)

c = \(\frac{12500}{2.5}+400 \times(2.5)^2\)

c = \(\frac{12500}{25}+\left(400 \times \frac{25}{4}\right)\)

c = 5000+2500 = ₹7500

Question 11. A factory makes an open cardboard box for a jewellery shop from a square sheet of side 18 cm by cutting off squares from each corner and folding up the flaps

Based on the above information answer any four of the following five questions, if x is the length of each square cut from corners.

Solution:

Let the length of the side of the cutting square be x cm, then the length and the breadth of the box will be (18 – 2x) cm each and the height of the box is x cm.

Let V be the volume of the open box formed by folding up the flaps, then V = x(18 – 2x) (18 – 2x) = 4x(9 – x)²= 4(x³ – 18x² + 81 x)

Now, differentiating with respect to x, we get \(\frac{d V}{d x}=4\left(3 x^2-36 x+81\right)=12\left(x^2-12 x+27\right) \text { and } \frac{d^2 V}{d x}=12(2 x-12)=24(x-6)\)

For maxima or minima \(\frac{dV}{dx}\) = 0

⇒ 12(x -12x+27) = 0 or (x-3)(x-9) = 0

x = 3, 9 (but x = 9 is not possible) therefore x = 3

∴ \(\left(\frac{d^2 V}{d x}\right)_{x=3}=24(3-6)=-72<0 \text { (Maxima) }\)

1. The volume of the open box is

1. 4x(x²- 18x + 81)
2. 2x(2x² + 36x + 162)
3. 2x(2x² + 36x – 162)
4. 4x(x²+ 18x + 81)

Answer: 1. 4x(x- 18x + 81)

The volume of the open box is

V = x(18 – 2x) (18 – 2x) = 4x(x² – 18x + 81) = 432 cm

2. The condition for the volume (V) to be maximum is

1. \(4 x\left(x^2-18 x+81\right)\)
2. \(2 x\left(2 x^2+36 x+162\right)\)
3. \(2 x\left(2 x^2+36 x-162\right)\)
4. \(4 x\left(x^2+18 x+81\right)\)

Answer: 1. \(4 x\left(x^2-18 x+81\right)\)

⇒ \(\frac{\mathrm{dV}}{\mathrm{dx}}=0 \text { and } \frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{dx}^2}<0\)

3. What should be the side of the square to be cut off so that the volume is maximum?

1. 6 cm
2. 9 cm
3. 3 cm
4. 4 cm

Answer: 3. 3 cm

4. Maximum volume of the open box is

1. 423 cm³
2. 432 cm³
3. 400 cm³
4. 64 cm³

Answer: 2. 432 cm³

Maximum volume of the open box

V = 4 x 3 (9 – 18 x 3 + 81)= 12 (9 – 54 + 81) = 432 cm³

5. The total area of the removed squares is:

1. 324 cm²
2. 144 cm²
3. 36 cm²
4. 64 cm²

Answer: 3. 36 cm²

Total area of removed square = 4x² = 4 x 9 = 36 cm²

Question 12. Find the intervals in which the function f(x)= \(\frac{x^4}{4}-x^3-5 x^2+24 x+12\) is

1. Strictly increasing
2. Strictly decreasing

Solution:

Given \(f(x)=\frac{x^4}{4}-x^3-5 x^2+24 x+12\)

⇒ \(f^{\prime}(x)=x^3-3 x^2-10 x+24\)

⇒ \(f^{\prime}(x)=(x-2)(x-4)(x+3)\)

therefore f'(x) = 0, gives x = – 3, 2, 4 points divides the real line into four disjoint intervals.

namely (-∞, -3), (-3, 2). (2, 4) and (4,∞).

f'(K) > 0. when x∈(-3, 2)∪(4,∞) and f'(x) < 0. when x ∈ (-∞, -3)∪(2, 4)

Hence f(x) is strictly increasing in (-3, 2) ∪ (4, ∞) and strictly decreasing in (-∞, -3) ∪ (2, 4)

Question 13. An open tank with a square base and vertical sides is to be constructed from a metal sheet to hold a given quantity of water. Show that the cost of material will be least when the depth of the tank is half of its width. If the cost is to be borne by nearby settled lower-income families, for whom water will be provided?
Solution:

Let v and A be the volume and area of the tank, x be the side of a square base and y be the height of the tank, then v = x²y….(1)

To show (minimum)A = x² + 4xy at y = x/2

⇒ A = \(x^2+4 x y \quad \Rightarrow A=x^2+4 x\left(\frac{v}{x^2}\right) \Rightarrow A=x^2+\frac{4 v}{x}\)

⇒ \(\frac{d A}{d x}=2 x-\frac{4 v}{x^2}\)

for maxima or minima \(\frac{\mathrm{dA}}{\mathrm{dx}}=0\)

⇒ \(2 \mathrm{x}-\frac{4 \mathrm{v}}{\mathrm{x}^2}=0 \Rightarrow 2 \mathrm{x}=\frac{4 \mathrm{v}}{\mathrm{x}^2} \Rightarrow 2 \mathrm{v}=\mathrm{x}^3\)

From equation (1) \(v=x^2 y \Rightarrow 2 x^2 y=x^3 \Rightarrow y=\frac{x}{2}\)

Now \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dx}^2}=2+\frac{8 \mathrm{v}}{\mathrm{x}^3}\)

⇒ \(\left(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dx}^2}\right)>0\) when \(\mathrm{x}^3=2 \mathrm{v}\)

Hence cost of material will be least at y=x/2

Question 14. Amongst all open (from the top) right circular cylindrical boxes of volume 1 25TC cm-1. find the dimensions of the box which has the least surface area.
Solution:

Let r be the radius, h be the height and V be the volume of the cylindrical box respectively

∵ V = 125 π cm³ (given)

∴ \(\pi \mathrm{r}^2 \mathrm{~h}=125 \pi \Rightarrow \mathrm{h}=\frac{125}{\mathrm{r}^2}\)….(1)

Let S be the surface area of the cylindrical box

∴ S = \(2 \pi r h+\pi r^2=\pi r(2 h+r)\)

⇒ S = \(\pi r\left(\frac{250}{r^2}+r\right)\) [from equation(1)]

S = \(\frac{250 \pi}{r}+\pi r^2\)

⇒ \(\frac{\mathrm{dS}}{\mathrm{dr}}=-\frac{250 \pi}{\mathrm{r}^2}+2 \pi r\) and \(\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{dr}^2}=\frac{500 \pi}{\mathrm{r}^3}+2 \pi\)

for maxima or minima \(\frac{\mathrm{dS}}{\mathrm{dr}}=0\)

⇒ \(2 \pi r=\frac{250 \pi}{r^2} \Rightarrow r^3=125 \Rightarrow r=5 \Rightarrow\left(\frac{d^2 S}{d^2}\right)_{r=5}>0\)

So the surface area S of the cylinder is least at r = 5

Hence, dimensions of box are r = 5 cm and h = 5 cm

Question 15. A tank with a rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If the building of the tank costs Rs. 70 per sq metre for the base and Rs 45 per square metre for the sides. What is the cost of the least expensive tank?
Solution:

Given: A tank with rectangular base and rectangular sides, open at the top and Depth of tank = 2 m

Let x m be the length and y m be the breadth of the base of the tank.

Volume of tank (= /bh) = x.y.2 = 8 m³ (given)

∴ y = \(\frac{8}{2 x}=\frac{4}{x}\)……(1)

Now cost of building the* base of the tank at the given rate of Rs 70 per square metre is Rs 70 xy……(2)

Again the cost of building the four sides (walls), of the tank at the rate of Rs 45 per square metre.

= 45(x.2 + x.2 + y.2 + y.2) = 45(4x + 4y)

= Rs (1 80x + 180y)….(3)

Let z denote the total cost of building the tank.

Adding (2) and (3), z = 70xy + 180x + 180y

Putting y = 4/x from (1), z = 70x. 4/x + 180x + 180. 4/x

or z = 280 + 180x + 720/x…..(4)

∴ \(\frac{\mathrm{dz}}{\mathrm{dx}}=0+180-\frac{720}{\mathrm{x}^2} \text { and } \frac{\mathrm{d}^2 \mathrm{z}}{\mathrm{dx}^2}=\frac{1440}{\mathrm{x}^3}\)

Putting \(\frac{\mathrm{dz}}{\mathrm{dx}}\) = 0 to find turning points, we have

180 \(-\frac{720}{x^2}=0\)

180 = \(\frac{720}{x^2}\)

180 \( x^2=720\)

⇒ \(x^2=\frac{720}{180}=4 \Rightarrow x=2\)(because x being length can’t negative )

At x=2, \(\frac{\mathrm{d}^2 \mathrm{z}}{\mathrm{dx}^2}=\frac{1440}{\mathrm{x}^3}=\frac{1440}{8}=180(+\mathrm{ve})\)

∴ z is minimum at x=2

Putting x=2 in (4), minimum cost

z = \(280+180(2)+\frac{720}{2}=280+360+360=280+720=\text { Rs. } 1000 .\).

CBSE Class 12 Maths Chapter 5 Continuity And Differentiability Important Questions

Question 1. Differential of \(\left[\log \left(\log x^5\right)\right]\) with respect to x is:

1. \(\frac{5}{x \log \left(x^5\right) \log \left(\log x^5\right)}\)
2. \(\frac{5}{x \log \left(\log x^5\right)}\)
3. \(\frac{5 x^4}{\log \left(x^5\right) \log \left(\log x^5\right)}\)
4. \(\frac{5 x^4}{\log \left(\log x^5\right)}\)

Solution: 1. \(\frac{5}{x \log \left(x^5\right) \log \left(\log x^5\right)}\)

Let y = \(\log \left[\log \left(\log x^5\right)\right]\)

⇒ \(\frac{d y}{d x}=\frac{1}{\log \left(\log x^5\right)} \times \frac{1}{\log \left(x^5\right)} \times \frac{1}{x^5} \times 5 x^4=\frac{5}{x \log \left(x^5\right) \log \left(\log x^5\right)}\)

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Question 2. If siny = x cos(a+y), then \(\frac{dx}{dy}\) is:

1. \(\frac{\cos a}{\cos ^2(a+y)}\)
2. \(\frac{-\cos a}{\cos ^2(a+y)}\)
3. \(\frac{\cos a}{\sin ^2 y}\)
4. \(\frac{-\cos a}{\sin ^2 y}\)

Solution: 1. \(\frac{\cos a}{\cos ^2(a+y)}\)

⇒ x = \(\frac{\sin y}{\cos (a+y)}\)

⇒ \(\frac{d x}{d y}=\frac{\cos (a+y) \cdot \cos y+\sin y \cdot \sin (a+y)}{\cos ^2(a+y)}\)

⇒ \(\frac{d x}{d y}=\frac{\cos [(a+y)-y]}{\cos ^2(a+y)}=\frac{\cos a}{\cos ^2(a+y)}\)

Question 3. If (x² +y²)= xy, then \(\frac{dy}{dx}\) is:

1. \(\frac{y+4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}\)
2. \(\frac{y-4 x\left(x^2+y^2\right)}{x+4\left(x^2+y^2\right)}\)
3. \(\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}\)
4. \(\frac{4 y\left(x^2+y^2\right)-x}{y-4 x\left(x^2+y^2\right)}\)

Solution: 3. \(\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}\)

⇒ \(2\left(x^2+y^2\right)\left[2 x+2 y \frac{d y}{d x}\right]=x \frac{d y}{d x}+y \cdot 1\)

⇒ \(\frac{d y}{d x}\left[4\left(x^2 y+y^3\right)-x\right]=y-4\left(x^3+x y^2\right)\)

⇒ \(\frac{d y}{d x}=\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}\)

Question 4. The function \(f(x)=\left\{\begin{array}{cc}
\frac{e^{3 x}-e^{-5 x}}{x} & ; \text { if } x \neq 0 \\
k & ; \text { if } x=0
\end{array}\right.\) is continuous at x=0 for the value of k, as:

1. 3
2. 5
3. 2
4. 8

Solution: 4. 8

The given function is continuous at x = 0,

therefore \(\lim _{h \rightarrow 0} f(0+h)=f(0)\)

⇒ \(\lim_{h \rightarrow 0} \frac{e^{3 h}-e^{-5 h}}{h}=k \Rightarrow \lim_{h \rightarrow 0} 3\left(\frac{e^{3 h}-1}{3 h}\right)-\lim_{h \rightarrow 0}(-5)\left(\frac{e^{-5 h}-1}{-5 h}\right)=k\)

⇒ k = 3 + 5 = 8

Question 5. If x = 2 cosθ – cos2θ and y = 2 sinθ – sinθ, then \(\frac{dy}{dx}\) is:

1. \(\frac{\cos \theta+\cos 2 \theta}{\sin \theta-\sin 2 \theta}\)
2. \(\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta}\)
3. \(\frac{\cos \theta-\cos 2 \theta}{\sin \theta-\sin 2 \theta}\)
4. \(\frac{\cos 2 \theta-\cos \theta}{\sin 2 \theta+\sin \theta}\)

Solution: 2. \(\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta}\)

x = 2 cosθ – cosθ

⇒ \(\frac{dx}{d \theta}\) = -2 sinθ + 2 sin 2θ

y = 2 sinθ – sin 2θ

⇒  \(\frac{dy}{d \theta}\) = -2 cosθ + 2 cos 2θ

Hence, \(\frac{d y}{d x}=\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta}\)

Question 6. If a function f defined by: \(f(x)=\left\{\begin{array}{cc}
\frac{k \cos x}{\pi-2 x} & : \text { if } x \neq \frac{\pi}{2} \\
3 & ; \text { if } x=\frac{\pi}{2}
\end{array}\right.\) is continuous at x = \(\frac{\pi}{2}\) then the value of k is:

1. 2
2. 3
3. 6
4. -6

Solution: 3. 6

Given f(x) is continuous at x = \(\frac{\pi}{2}\)

⇒ R.H.L. = \(f\left(\frac{\pi}{2}\right) \quad \Rightarrow \lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right)=3\)

⇒  \(\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}=3\)

⇒  \(\lim _{h \rightarrow 0} \frac{-k \sinh }{-2 h}=3\)

⇒  \(\frac{k}{2}\left(\lim _{h \rightarrow 0} \frac{\sinh }{h}\right)=3\)

⇒  k= 6 (because \(\lim _{h \rightarrow 0} \frac{\sinh }{h}=1\))

Question 7. If y = sin (m sin^-1 x), which of the following equations is true?

1. \(\left(1-x^2\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+m^2 y=0\)
2. \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+m^2 y=0\)
3. \(\left(1+x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0\)
4. \(\left(1+x^2\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-m^2 x=0\)

Solution: 2. \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+m^2 y=0\)

y = \(\sin \left(m \sin ^{-1} x\right)\)

⇒ \(\frac{d y}{d x}=\cos \left(m \sin ^{-1} x\right) \times\left(\frac{m}{\sqrt{1-x^2}}\right)\)

⇒ \(\sqrt{1-x^2} \frac{d y}{d x}=m \cos \left(m \sin ^{-1} x\right)\)

∴ \(\sqrt{1-x^2} \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right) \cdot \frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x)=-m^2 \sin \left(m \sin ^{-1} x\right) \times \frac{1}{\sqrt{1-x^2}}\)

⇒ \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+m^2 y=0\)

Question 8. The function f: R → R given by f(x) = |x — 1| is:

1. Continuous as well as differentiable at x = 1
2. Not continuous but differentiable at x = 1
3. Continuous but not differentiable at x = 1
4. Neither continuous nor differentiable at x = 1

Solution: 3. Continuous but not differentiable at x = 1

Given; f: R → R, f(x) =-|x-1|

Here, f(x) is a modulus function.

We know that the modulus function is always continuous in R but not differentiable when its value becomes zero.

Hence; f(x) is continuous but not differentiable at x = 1.

Question 9. Differentiate sec² (x²) concerning x². or, If y = f(x²) and f'(x) =e^√x, then find \(\frac{dy}{dx}\)
Solution:

Let u = sec² (x²); and v = x²

Now, \(\frac{d u}{d v}=\frac{d u / d x}{d v / d x} \Rightarrow \frac{d u}{d v}=\frac{2 \sec \left(x^2\right) \cdot \sec \left(x^2\right) \cdot \tan \left(x^2\right) \cdot 2 x}{2 x}\)

⇒ \(\frac{d u}{d v}=2 \sec ^2\left(x^2\right) \tan \left(x^2\right)\)

So, \(\frac{d\left(\sec ^2\left(x^2\right)\right)}{d\left(x^2\right)}=2 \sec ^2\left(x^2\right) \cdot \tan \left(x^2\right)\)

Or,

Given, y = \(f\left(x^2\right)\) and \(f^{\prime}(x)=e^{\sqrt{x}}\)

Now; \(\frac{d y}{d x}=f^{\prime}\left(x^2\right) 2 x\) and \(f^{\prime}\left(x^2\right)=e^{\sqrt{x^2}}\)

⇒ \(\frac{d y}{d x}=e^{\sqrt{x^2}} \cdot 2 x=2 x \cdot e^x\)

Question 10. Find the value of k, so that the function \(f(x)=\left\{\begin{array}{ccc}
k x^2+5 & \text { if } & x \leq 1 \\
2 & \text { if } & x>1
\end{array}\right.\) is continous at x = 1.

Solution:

Given that f(x) = \(\left\{\begin{array}{ccc}
k x^2+5 & \text { if } & x \leq 1 \\
2 & \text { if } & x>1
\end{array}\right.\)

f(x) is continous at x = 1

So \(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1)\)

⇒ \(\lim _{x \rightarrow 1^{+}} 2=\lim _{x \rightarrow 1^{+}} k x^2+5=k+5\)

⇒ \(2=k+5 \Rightarrow k=-3\)

Question 11. If y = \(\cos (\sqrt{3 x})\), then find \(\frac{d y}{d x}\).
Solution:

y = \(\cos (\sqrt{3 x}) \Rightarrow \frac{d y}{d x}=-\sin (\sqrt{3 x}) \times \frac{1}{2 \sqrt{3 x}} \times \sqrt{3}\)

= \(\frac{-\sqrt{3}}{2 \sqrt{3 x}} \sin (\sqrt{3 x})\)

Question 12. Find the relationship between a and b so that the function f defined by \(f(x)= \begin{cases}a x+1 & \text { if } x \leq 3 \\ b x+3 & \text { if } x>3\end{cases}\)

Is continuous at x = 3 Or, Check the differentiability of f(x) = |x-3| at x = 3

Solution:

Given, \(f(x)=\left\{\begin{array}{ll}
a x+1 & \text { if } x \leq 3 \\
b x+3 & \text { if } x>3
\end{array} .\right.\) is continuois at x = 3.

⇒ L.H.L = R.H.L

⇒ \(\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0} f(3+h) \Rightarrow \lim _{h \rightarrow 0} a(3-h)+1=\lim _{h \rightarrow 0} b(3+h)+3\)

⇒  3a + 1 = 3b + 3

⇒  3a -3b = 2 or a-b = 2/3

Or,

Given, \(f(x)=|x-3|, f(x)= \begin{cases}-(x-3) ; & x<3 \\ (x-3) ; & x \geq 3\end{cases}\)

L.H.D. at x=3

= \(\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{-h}=\lim _{h \rightarrow 0} \frac{|3-h-3|-|0|}{-h}=\lim _{h \rightarrow 0} \frac{h}{(-h)}=-1\)

And R.H.D. at x=3

= \(\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}=\lim _{h \rightarrow 0} \frac{|3+h-3|-|0|}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=1\)

∴ L.H.D ≠ R.H.D

∴ Given function f(x) is not differentiable at x = 3

Question 13. If f(x) = \(\sqrt{\frac{\sec x-1}{\sec x+1}} \text {, find } f^{\prime}\left(\frac{\pi}{3}\right)\)

Or, Find f'(x) if f(x) = (tan x)^{tan x}

Solution:

f(x) = \(\sqrt{\frac{\sec x-1}{\sec x+1}} \Rightarrow f(x)=\sqrt{\frac{1-\cos x}{1+\cos x}} \Rightarrow f(x)=\sqrt{\frac{2 \sin ^2 x / 2}{2 \cos ^2 x / 2}}\)

f(x) = \(\left|\tan \frac{x}{2}\right|\)

f(x) = \(\tan x / 2\)

⇒ \(f^{\prime}(x)=\sec ^2 x / 2 \times \frac{1}{2}\) (because x lie in 1st guard)

⇒ \(f^{\prime}(\pi / 3)=\sec ^2 \pi / 6 \times \frac{1}{2} \Rightarrow f^{\prime}(\pi / 3)=\frac{4}{3} \times \frac{1}{2}=\frac{2}{3}\)

Or,

Given, f(x) = (tan x)^{tan x}

Taking log on both sides; we get:

log f(x) = log (tan x)^{tan x} ⇒ log f(x) = tan x log (tan x)

Differentiating concerning x; we get,

⇒ \(\frac{1}{f(x)} \cdot f^{\prime}(x)=\tan x \times \frac{1}{\tan x} \times \sec ^2 x+\log \tan x \times \sec ^2 x\)

⇒ \(\frac{f^{\prime}(x)}{f(x)}=\sec ^2 x+\sec ^2 x \log \tan x\)

⇒ \(f^{\prime}(x)=f(x) \cdot\left[\sec ^2 x(1+\log \tan x)\right]\)

⇒ \(f^{\prime}(x)=(\tan x)^{\tan x} \cdot\left[\sec ^2 x(1+\log \tan x)\right]\)

Question 14. Differentiate \(\tan ^{-1}\left(\frac{1+\cos x}{\sin x}\right)\) with respect to x.
Solution:

Let y = \(\tan ^{-1}\left[\frac{1+\cos x}{\sin x}\right]\)

∴ y = \(\tan ^{-1}\left[\frac{2 \cos ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right]\)

∴ y = \(\tan ^{-1}\left[\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right]=\tan ^{-1}\left[\cot \frac{x}{2}\right]\)

Hence, \(y=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right] \Rightarrow y=\frac{\pi}{2}-\frac{x}{2}\)

⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=0-\frac{1}{2}=-\frac{1}{2}\) (Differentiating with respect to x )

Question 15. If x = acosθ + bsinθ, y = asinθ – bcosθ then show that \(\frac{dy}{dx}=-frac{x}{y}\) and hence show that \(y^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+y=0\)

or,

If \(e^{y-x}=y^x \text {; Prove that } \frac{d y}{d x}=\frac{y(1+\log y)}{x \log y}\)

Solution:

Given, x = acosθ + bsinθ, and y = asinθ – bcosθ

Now, x² + y²= (acosθ + bsinθ)² + (asinθ – bcosθ)²

= {a² cos²θ + b sin²θ + 2abcosθsinθ}+ {a² sin²θ + b² cos²θ-2absin θcosθ} = a² + b²

x²+ y²= a + b

⇒ 2x + 2y\(\frac{dy}{dx}\) = 0

(differentiating both sides concerning x)

⇒ \(\frac{d y}{d x}=\frac{-x}{y} \Rightarrow \frac{d^2 y}{d x^2}=\frac{y(-1)+x \frac{d y}{d x}}{y^2}\)

(Again differentiating both sides concerning x)

⇒ \(y^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+y=0\)

or,

Given, \(e^{y-x}=y^x\)

⇒ \((\mathrm{y}-\mathrm{x}) \log \mathrm{e}=\mathrm{x} \log \mathrm{y}\)

(Taking log on both sides)

(y-x) =x log y

⇒ \(\frac{d y}{d x}-1=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y \cdot 1\)

(Differentiating both sides for x)

⇒ \(\frac{d y}{d x}\left[\frac{x}{y}-1\right]=-(1+\log y)\)

⇒ \(\frac{d y}{d x}=\frac{(1+\log y)}{\left(1-\frac{x}{y}\right)}=\frac{y(1+\log y)}{(y-x)}\)

⇒ \(\frac{d y}{d x}=\frac{y(1+\log y)}{x \log y}\)(from 1)

Question 16. Differentiate sin²x concerning e^cosx.
Solution:

Let y = sin²x and z = eco sex

⇒ \(\frac{dy}{dx}\) = 2 sin x cos x…(1)

(differentiating with respect x)

and \(\frac{dz}{dx}\) e^cosx (-sin x)…(2)

(differentiating with respect x)

⇒ \(\frac{d y}{d z}=\frac{d y / d x}{d z / d x}=\frac{2 \sin x \cdot \cos x}{e^{\cos x}(-\sin x)}\)

(from equations (1) and (2))

⇒ \(\frac{d y}{d z}=\frac{-2 \cos x}{e^{\cos x}}=-2 \cos x e^{-\cos x}\)

Question 17. If \(\tan ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2+y^2}\), prove that \(\frac{d y}{d x}=\frac{x+y}{x-y}\).

Or,

If y = \(e^{a \cos ^{-1} x},-1<x<1\), then show that \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-a^2 y=0\)

Solution:

Given: \(\tan ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2+y^2}\)

Differentiating with respect to x we get: \(\frac{1}{\left(1+\frac{y^2}{x^2}\right)} \times\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{\sqrt{x^2+y^2}} \times \frac{\left(2 x+2 y \frac{d y}{d x}\right)}{2 \sqrt{x^2+y^2}}\)

⇒ \(\frac{x^2}{\left(x^2+y^2\right)} \times\left(x \cdot \frac{d y}{d x}-y\right) \times \frac{1}{x^2}=\frac{1}{\left(x^2+y^2\right)}\left(x+y \frac{d y}{d x}\right)\)

⇒ \(x \frac{d y}{d x}-y=x+y \frac{d y}{d x}\)

⇒ \((x-y) \frac{d y}{d x}=x+y \Rightarrow \frac{d y}{d x}=\frac{x+y}{x-y}\)

Or,

Given \(y=e^{a \cos ^{-1} x}\) …(1)

differentiating with respect to x, \(\frac{d y}{d x}=-e^{a \cos ^{-1} x} \cdot a \times \frac{1}{\sqrt{1-x^2}}\)….(2)

⇒ \(\sqrt{1-x^2} \cdot \frac{d y}{d x}=-a y\) [from (1)]

⇒ \(\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=a^2 y^2\) (Squaring on both sides)

Again, differentiating concerning x, we get

⇒ \(2\left(1-x^2\right)\left(\frac{d y}{d x}\right)\left(\frac{d^2 y}{d x^2}\right)-2 x\left(\frac{d y}{d x}\right)^2=2 a^2 y \frac{d y}{d x}\)

⇒ \(2\left(\frac{d y}{d x}\right)\left[\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-a^2 y\right]=0\)

⇒ \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-a^2 y=0\)

(because \(\frac{d y}{d x} \neq 0\), as y is not constant)

Question 18. If \(x=a e^t(\sin t+\cos t) \text { and } y=a e^t(\sin t-\cos t)\), then show that \(\frac{d y}{d x}=\frac{x+y}{x-y}\).

Or,

Differentiate x^{sin x}+ (sin x)^{cos x} to x.

Solution:

x = \(a e^t(\sin t+\cos t)\)

y = \(a e^t(\sin t-\cos t)\)

(Differentiating concerning

∴ \(\frac{d x}{d t}=a\left[e^t(\cos t-\sin t)+(\sin t+\cos t) e^t\right]=a e^t(\sin t+\cos t)-a e^t(\sin t-\cos t)\)

⇒ \(\frac{d x}{d t}=x-y\)…(1)

and, \(\frac{d y}{d t}=a\left[e^t(\cos t+\sin t)+(\sin t-\cos t) e^t\right]=a e^t(\cos t+\sin t)+a e^t(\sin t-\cos t)\)

⇒ \(\frac{d y}{d t}=x+y\)

Hence; \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}-\mathrm{y}}\) [From equation (1) and (2)]

Or,

Let y = \(x^{\sin x}+(\sin x)^{\cos x}\)

y = u + v Where \(u=x^{\sin x} and v=(\sin x)^{\cos x}\)

\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)…(1) Differentiating with respect to x)

Now; \(\mathrm{u}=\mathrm{x}^{\sin \mathrm{x}}\)

⇒ log u = \(\sin x \cdot \log x\) (Taking log on both sides)

⇒ \(\frac{1}{u} \frac{d u}{d x}=\sin x \cdot \frac{1}{x}+\log x \cdot \cos x\) (Differentiating with respect to x)

⇒ \(\frac{d u}{d x}=u\left[\frac{\sin x}{x}+\log x \cdot \cos x\right]\)

⇒ \(\frac{d u}{d x}=x^{\sin x}\left[\frac{\sin x}{x}+\log x \cdot \cos x\right]\)….(2)

and \(\mathrm{v}=(\sin \mathrm{x})^{\cos \mathrm{x}}\)

⇒ \(\log \mathrm{v}=\cos \mathrm{x} \cdot \log (\sin \mathrm{x})\)

(Taking logs on both sides)

⇒ \(\frac{1}{v} \frac{d v}{d x}=\cos x \cdot \frac{1}{\sin x} \cdot \cos x+\log (\sin x) \cdot(-\sin x)\)

From equations (1), (2) and (3), we get;

⇒ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)

⇒ \(\frac{d y}{d x}=x^{\sin x}\left[\frac{\sin x}{x}+\log x \cdot \cos x\right]+(\sin x)^{\cos x}[\cos x \cdot \cot x-\sin x \cdot \log (\sin x)]\)

Question 19. If (x-a)² = (y-b)² = c², for some c > 0, prove that \(\frac{\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}}{\frac{d^2 y}{d x^2}}\) is a constant independent of a and b.
Solution:

Given, (x-a)² + (y-b)² = c²; for c > 0

Differentiating concerning x, we get,

2(x-a)+2(y-b) \(\frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{(x-a)}{(y-b)}\)….(1)

Again, differentiating with respect to x we get: \(\frac{d^2 y}{d x^2}=-\left[\frac{(y-b)(1-0)-(x-a) \frac{d y}{d x}}{(y-b)^2}\right]\)…(2)

Putting the value of dy/dx from \(\frac{d^2 y}{d x^2}=-\left[\frac{(y-b)+(x-a) \frac{(x-a)}{(y-b)}}{(y-b)^2}\right]=-\left[\frac{(x-a)^2+(y-b)^2}{(y-b)^3}\right]=-c^2 /(y-b)^3\)……(3)

Now, \(\frac{\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}}{\frac{d^2 y}{d x^2}}=\frac{\left[1+\frac{(x-a)^2}{(y-b)^2}\right]^{3 / 2}}{\frac{-c^2}{(y-b)^3}}\) (from (1) and (3)]

= \(\frac{\left[\frac{(x-a)^2+(y-b)^2}{(y-b)^2}\right]^{3 / 2}}{-c^2 /(y-b)^3}=\frac{c^3}{-c^2}=-c\); which is a constant independent of a and b.

Question 20. If y = sin(sin x), prove that \(\frac{d^2 y}{d x^2}+\tan x \frac{d y}{d x}+y \cos ^2 x=0\)
Solution:

Given y = sin(sin x)….(1)

⇒ \(\frac{d y}{d x}=\cos (\sin x) \cdot \cos x\) (Differentiating with respect to x)….(2)

⇒  \(\frac{d}{d x}\left(\frac{d y}{d x}\right)=\cos (\sin x) \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x} \cos (\sin x)\) (Again differentiating with respect to x)

⇒ \(\frac{d^2 y}{d x^2}=-\sin x \cdot \cos (\sin x)+\cos x \cdot(-\sin \{\sin x\} \cdot \cos x)\)….(3)

Put sin(sin x)=y from (1) and \(\cos (\sin x)=\frac{1}{(\cos x)} \cdot \frac{d y}{d x}\) from (2) in (3)

⇒  \(\frac{d^2 y}{d x^2}=\frac{-\sin x}{\cos x} \cdot \frac{d y}{d x}-y \cos ^2 x \Rightarrow \frac{d^2 y}{d x^2}+\tan x \frac{d y}{d x}+y \cos ^2 x=0\)

Question 21. If (x²+y²) = xy, find \(\frac{dy}{dx}\)

Or,

If x = a(2θ-sin2θ) and y = a(1-cosη2). find \(\frac{dy}{dx}\), when θ = π/3.

Solution:

Given, (x²+y²)² = xy

⇒ x⁴+ y⁴ + 2x²y² = xy

(Differentiating both sides concerning x)

⇒ \(4 x^3+4 y^3 \frac{d y}{d x}+2\left[2 x y^2+x^2 \cdot 2 y \frac{d y}{d x}\right]=x \frac{d y}{d x}+y\)

⇒ \(\frac{d y}{d x}\left(4 y^3+4 x^2 y-x\right)=\left(y-4 x^3-4 x y^2\right)\)

⇒ \(\frac{d y}{d x}=\frac{y-4 x^3-4 x y^2}{4 y^3+4 x^2 y-x}\)

Or,

Given: x = a(2θ – sin 2θ) and y = a(1-cos2θ)

Function is parametric, therefore \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{d} \theta}{\mathrm{dx} / \mathrm{d} \theta}\)….(1)

Now, x = a(2θ – sin2θ)

Differentiating concerning θ we get,

⇒ \(\frac{d x}{d \theta}=a(2-\cos 2 \theta \times 2) \Rightarrow \frac{d x}{d \theta}=2 a(1-\cos 2 \theta)\)

⇒ \(\frac{d x}{d \theta}=2 a\left(2 \sin ^2 \theta\right) \Rightarrow \frac{d x}{d \theta}=4 a \sin ^2 \theta\)….(2)

y = \(a(1-\cos 2 \theta)\)

Differentiating concerning θ we get,

⇒ \(\frac{d y}{d \theta}=2 a \sin 2 \theta \quad \Rightarrow \frac{d y}{d \theta}=4 a \sin \theta \cos \theta\)…(3)

Put the value of equations (2) and (3) in (1), and we get

⇒ \(\frac{d y}{d x}=\frac{4 a \sin \theta \cos \theta}{4 a \sin ^2 \theta} \Rightarrow \frac{d y}{d x}=\frac{\cos \theta}{\sin \theta}=\cot \theta\)

∴ \(\left(\frac{d y}{d x}\right)_{\theta-\frac{\pi}{3}}=\cot \left(\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}}\).

CBSE Class 12 Maths Chapter 1 Relations And Functions Important Questions

Question 1. Let set X = {1, 2, 3} and a relation R is defined in X as R = {(1, 3), (2, 2), (3,2)}, then minimum ordered pairs which should be added in relation R to make it reflexive and symmetric are:

1. {(1,1), (2. 3), (1,2)}
2. {(3, 3), (3, 1), (1,2)}
3. {(1, 1), (3, 3), (3. 1), (2, 3)}
4. {(1,1), (3, 3), (3, 1), (1,2)}

Solution: 3. {(1, 1), (3, 3), (3. 1), (2, 3)}

Given, R= {(1,3), (2,2), (3,2)} on set X = {1.2, 3}

Then: if R is reflexive and symmetric, then R= {(1, 1), (2, 2), (3, 3), (1,3), (3, 2), (3, 1), (2, 3)}

Hence: the ordered pairs to be added are: {(1, 1), (3, 3), (3, 1), (2, 3)}

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Question 2. If R = {(x, y); x, y ∈ Z, x² + y² ≤ 4} is a relation in set Z. then the domain of R is :

1. {0, 1,2}
2. {-2,-1, 0, 1,2}
3. {0,-1,-2}
4. {-1,0, 1}

Solution: 2. {-2,-1, 0, 1,2}

Given: R = {(x, y) : x, y ∈ Z, x² + y² ≤ 4} is a relation in set Z.

Hence: according to the given relation.

Domain of R= {-2, —1, 0. 1.2}

Question 3. Let X = {x²: x ∈ N} and the function f: N →X is defined by f(x) = x², x ∈ N. Then, this function is :

1. Injective only
2. Not objective
3. Surjective only
4. Bijective

Solution: 4. Bijective

f(x) = x²; x ∈ N; f : N → X where X = {x²: x ∈ N}

Now; f(x₁) = f(x₂)

⇒ x²₁ = x²₂

⇒ (x₁ + x₂) (x₁ – x₂) = 0

Now : x₁ + x₂ ≠ 0 (x₁,x₂ ∈ N)

⇒ x₁ – x₂ = 0 ⇒ x₁ = x₂

Hence; f is one-one.

Let y = f(x) ⇒ y = x²

⇒ x = √y

∀ y ∈ X, there exists x ∈ N

Hence, every element of the co-domain has pre-images in the domain.

⇒ Range = Co-domain

∴  f is onto.

∴ The given function is bijective.

Question 4. A function f: R → R defined by f(x) = 2 + x² is :

1. Not one-one
2. One-one
3. Not onto
4. Neither one-one nor onto

Solution: 4. Neither one-one nor onto

f(x) = 2 + x² f: R → R

One-one: Let x₁ and x₂ ∈ R such that

f(x₁) = f(x₂) => x²₁ + 2 = x²₂ = 2

⇒ x₁ = ±x₂ ⇒  f is not one-one

Onto: Let y = f(x) ⇒ y = x² + 2 ⇒ x \(=\sqrt{y-I} \in R\)

∴ y – 2 ≥ 0 ⇒ y ≥ 2 ⇒ y ∈ R [2, ∞)

Range 5≠ co-domain

Also: f is not onto as there is no pre-image for negative real numbers.

Hence; f is neither one-one nor onto.

Question 5. Write the smallest reflexive relation on set A = {a, b, c}.
Solution:

For reflexive relation, each and every element of the given set must be related to itself, at least i.e. the smallest reflexive relation on set A will be {(a, a), (b, b), (c, c)}.

Question 6. If f = {(1.2), (2, 4), (3, 1), (4, k)} is a one-one function from set A to A, where A = {1, 2, 3, 4}; then find the value of k.
Solution:

In a one-on-one function, different elements must have different images. So, according to the given function, f = {(1,2), (2. 4), (3, 1), (4, k)}, the value of k must be 3.

Question 7.

1. Check whether the relation R defined on the set {1, 2, 3, 4} as R = {(a, b); b = a + 1} is transitive. Justify your answer. Or
2. If the relation R on the set A = {x : 0 ≤ x ≤12} given by R = {(a, b): a = b} is an equivalence relation, then find the set of all elements related to 1.

Solution:

1. Given. R = ((a, b): b = a + 1} defined on the set A = {1, 2, 3, 4}

⇒ R = {(1, 2), (2, 3), (3, 4)}

By definition, for transitive relation, if a R b and b R c ⇒ a R c ∀ a, b, c ∈ A

For 1,2, 3 ∈ A; (1,2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R

Therefore, R is not transitive.

Aliter: Given; R = {(a, b): b = a + 1} defined on the set A – {1,2, 3,4}

For transitive relation: Let (a, b) ∈ R and (b, c) ∈ R ∀ a, b, c ∈ A

⇒ b = a + 1 and c = b + 1

⇒ c = a + 2

or (a, c) ∉ R

∴ R is not transitive

or

2. The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.

Aliter: Let x ∈ A which is related to 1 by the given relation R = {(a, b): a = b}

x R 1 ⇒ (x, 1) ∈ R ⇒ x = 1

∴ [1] = {1}

Question 8. The relation R in the set {1.2,3} given by R = {(1, 2), (2, 1), (1, 1)} is:

1. Symmetric and transitive, but not reflexive
2. Reflexive and symmetric, but not transitive
3. Symmetric, but neither reflexive nor transitive
4. An equivalence relation

Solution: 3. Symmetric, but neither reflexive nor transitive

Let A= {1,2, 3} and R = {(1, 2), (2, 1), (1, 1)}

R is not reflexive as (2, 2) and (3,3) ∉ R.

R is a symmetric relation as (a, b) ∈ R and (b, a) ∈ R ∀ a, b ∈ A.

R is not a transitive relation because (2, 1) ∈ R and (1,2) ∈ R but (2, 2) ∉ R.

Hence; R is symmetric, but neither reflexive nor transitive.

Question 9. Let A = {1,3, 5}, Then the number of equivalence relations in A containing (1,3) is

1. 1
2. 2
3. 3
4. 4

Solution: 2. 2

R₁ = {(1, 1), (3, 3), (5, 5), (1, 3), (3, 1)}

and R₂ = {(1, 1), (3. 3), (5, 5), (1, 3), (3, 1), (1,5), (5, 1), (3, 5), (5, 3)}

Hence, the number of equivalence relations in A containing (1, 3) is 2.

Question 10. Check whether the relation R in the set N of natural numbers given by R = {(a, b): a is divisor of b} is reflexive, symmetric or transitive. Also, determine whether R is an equivalence relation.
Solution:

The relation R in the set N is given by:

R = {(a, b): a is divisor of b}

⇒ aRb => \(\frac{b}{a}\) = K ∈ 1 ∀ a, b ∈ N,

Reflexive; aRa ∀ a∈N (By definition)

aRa ⇒ \(\frac{a}{a}\)= 1 ∈ I which is true.

So, R is reflexive relation.

Symmetric: we have

aRb ⇒ \(\frac{b}{a}\) = K ∈ I ∀ a, b ∈ N

⇒ \(\frac{a}{b}=\frac{1}{K} \notin I\)

⇒ b K a ∀ a, b∈ N

∴ If aRb ⇒ bRa ∀ a, b ∈ N, then R is not a symmetric relation.

Transitive: If aRb ⇒ \(\frac{b}{a}=K_1\) ∈ I ∀ a, b ∈ N ….(1)

bRc ⇒ \(\frac{c}{b}=K_2\) ∈ I ∀ b, c ∈ N ….(2)

from equation (1) and (2) \(\frac{b}{a} \times \frac{c}{b}=K_1 \times K_2 \Rightarrow \frac{c}{a}-\left(K_1 K_2\right) \in I \Rightarrow a R c\)

So, R is a transitive relation

∴ R is reflexive and transitive but not symmetric

R is not an equivalence relation on N.

Question 11. Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric, or transitive.
Solution:

Given, R = {(a, b): b = a + 1} is defined on the set A = {1, 2, 3, 4, 5, 6}

⇒ R= {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

Reflexive: By definition, a R a ∀a, b ∈ A

Now, 1 ∈ A but (1, 1) ∉ R.

Therefore, R is not reflexive.

Symmetric: By definition, if a R b ⇒ b R a ∀ a, b ∈ A

For 1,2 ∈ A, It can be observed that (1,2) ∈ R but (2, 1) ∉ R.

Therefore, R is not symmetric.

Transitive: By definition, if a R b and b R c ⇒ a R c ∀ a, b, c ∈ A

For 1,2, 3 ∈ A, (1,2) ∈ R and (2, 3) ∈ R but (1, 3)  ∉ R

Therefore, R is not transitive.

Hence, R is neither reflexive nor symmetric nor transitive.

Aliter: Given; R = {(a, b): b = a + 1} defined on the set A = {1,2, 3, 4, 5, 6}

Reflexive: (a, a) ∈ R ∀ a ∈ A

⇒ a = a + 1; which is not true;

∴ R is not reflexive.

Symmetric: Let (a, b) ∈ R ∀ a, b ∈ A

⇒ b = a + 1

⇒ (b, a) ∉ R [a ≠ b + 1]

∴ R is not symmetric

Transitive: Let (a, b) ∈ R and (b, c) ∈ R ∀ a, b, c ∈ A

⇒ b = a + 1 and c = b + 1

⇒ c = a + 2 or (a, c ) ∉ R

∴ R is not transitive.

Hence; R is neither reflexive nor symmetric nor transitive.

Question 12. Let A = {x ∈ Z : 0 ≤ x ≤12}. Show that R = {(a, b): a, b ∈ A, |a – b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also; write the equivalence class [2]. or, Show that the function f: R → R defined by \(f(x)=\frac{x}{x^2+1}\). ∀ x ∈R is neither one-one nor onto.
Solution:

Given, R = {(a, b): a, b ∈ A, |a – b| is divisible by 4}

Reflexive: Let a ∈ A

Now, |a — a| = 0. which is divisible by 4

So, (a, a) ∈ R ∀ a ∈ A

Hence, R is reflexive.

Symmetric: Let a, b ∈ A such that (a, b) ∈ R i.e. |a — b| is divisible by 4.

⇒ |-(b – a)| = |b — a| is also divisible by 4.

Hence; (b, a) ∈ R.

So, R is symmetric.

Transitive: Let a, b, c ∈ A such that (a, b), (b, c) ∈ R

i.e. |a – b| and |b – c| is divisible by 4.

Let \(|\mathrm{a}-\mathrm{b}|=4 \mathrm{k}_1\)

⇒ \(\mathrm{~b}-\mathrm{c}|=4 \mathrm{k}_2\)

⇒ \((\mathrm{a}-\mathrm{b})= \pm 4 \mathrm{k}_1\)….(1)

⇒ \((b-c)= \pm 4 k_2\)….(2)

Adding equations (1) and (2); \((\mathrm{a}-\mathrm{b})+(\mathrm{b}-\mathrm{c})= \pm 4 \mathrm{k}_1 \pm 4 \mathrm{k}_2= \pm 4\left(\mathrm{k}_1+\mathrm{k}_2\right)\)

⇒ a – c is divisible by 4.

⇒ |a – c| is divisible by 4.

Hence; (a, c) ∈ R

So, R is transitive.

Hence; R is an equivalence relation.

Further, let (x, 1) ∈ R ∀ x ∈ A ⇒ |x  – 1| is divisible by 4

⇒ x – 1 = 0, 4, 8, 12

⇒ x= 1, 5, 9 [x = 13 ∉ A]

Equivalence class of [1] = {1,5,9}

[The set of all elements related to 1 represents its equivalence class]

Now, we will find the equivalence class of [2]

Let (x, 2) ∈ R ∀ x ∈ A

⇒ |x — 2| = 0, 4, 8, 12

⇒ x = 2, 6, 10 [x = 14 ∉ A]

∴ Equivalence class of [2] = {2, 6, 10}. or,

f : \(R \rightarrow R ; f(x)=\frac{x}{x^2+1}\)

1. One-one function: we have \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right), \forall, \mathrm{x}_1, \mathrm{x}_2 \in \mathrm{R}\)

⇒ \(\frac{x_1}{x_1^2+1}=\frac{x_2}{x_2^2+1}\)

⇒ \(x_1 x_2^2+x_1=x_2 x_1^2+x_2 \quad \Rightarrow\left(x_1 x_2^2-x_2 x_1^2\right)+\left(x_1-x_2\right)=0\)

⇒ \(x_1 x_2\left(x_2-x_1\right)+\left(x_1-x_2\right)=0 \Rightarrow\left(x_1-x_2\right)\left(-x_1 x_2+1\right)=0\)

⇒ \(x_1=x_2 \text { and } x_1 x_2=1 \text { or } x_1=\frac{1}{x_2}\)

One-one function Example: Let \(\mathrm{x}_1=2 \in \mathrm{R} ; \mathrm{x}_2=1 / 2 \in \mathrm{R}\)

Then, \(f\left(x_1\right)=f(2)=2 / 5\) and \(f\left(x_2\right)=f(1 / 2)=2 / 5\)

⇒ latex]x_1 \neq x_2[/latex] but \(f\left(x_1\right)=f\left(x_2\right)\)

∴ f is not a one-on-one function

2. Onto function: Let y = f(x)

⇒ y = \(\frac{x}{x^2+1} \Rightarrow x^2 y-x+y=0\)

⇒ x = \(\frac{1 \pm \sqrt{1-4 y^2}}{2 y} \in R \text { if } 1-4 y^2 \geq 0 \text { and } y \neq 0\)

⇒ \(4 y^2-1 \leq 0 \text { or }(2 y+1)(2 y-1) \leq 0\)

⇒ \(x \in\left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{1}{2}\right]\)

Thus, every element of the Co-domain does not have its pre-image in the domain.

Hence, f: R → R is not onto.

∴ f is neither one-one nor onto.

CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions Important Questions

Question 1. The principal value of \(\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\) is

1. \(\frac{\pi}{12}\)
2. \(\pi\)
3. \(\frac{\pi}{3}\)
4. \(\frac{\pi}{6}\)

Solution: 1. \(\frac{\pi}{12}\)

⇒ \(\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{\pi}{3}+\left(-\frac{\pi}{4}\right)\)

= \(\frac{4 \pi-3 \pi}{12}=\frac{\pi}{12}\)

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Question 2. The principal value of \(\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)\) is:

1. \(\frac{\pi}{8}\)
2. \(\frac{3 \pi}{8}\)
3. \(-\frac{\pi}{8}\)
4. \(-\frac{3 \pi}{8}\)

Solution: 1. \(\frac{\pi}{8}\)

⇒ \(\tan ^{-1}\left[\tan \left(\frac{9 \pi}{8}\right)\right]=\tan ^{-1}\left[\tan \left(\pi+\frac{\pi}{8}\right)\right]\)

= \(\tan ^{-1}\left[\tan \left(\frac{\pi}{8}\right)\right]=\frac{\pi}{8} \in\left(-\frac{\pi}{2} \cdot \frac{\pi}{2}\right)\)

Question 3. What is the domain of the function cos^-1(2x-3)?

1. [-1, 1]
2. (1,2)
3. (-1, 1)
4. [1,2]

Solution: 4. [1,2]

For the given function: -1 ≤ 2x – 3 ≤ 1

⇒ 2 ⇒ 2x ⇒ 4

⇒ 1 ≤ x ≤ 2

Question 4. The principal value of [tan^-1 √3 – cot^-1 (-√3)] is

1. π
2. \(-\frac{\pi}{2}\)
3. 0
4. \(2 \sqrt{3}\)

Solution: 2. \(-\frac{\pi}{2}\)

⇒ \(\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})=\frac{\pi}{3}-\left(\pi-\cot ^{-1}(\sqrt{3})\right)\)

=\(\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right)=\frac{\pi}{3}-\frac{5 \pi}{6}=\frac{-3 \pi}{6}=\frac{-\pi}{2}\)

Question 5. The range or the principal valurbrance of the function y = sec^-1 x is ____

or,

The principal value of \(\cos ^{-1}\left(-\frac{1}{2}\right)\) is

Solution:

y = \(\sec ^{-1} x\) (given)

Range of \(\sec ^{-1} x\) is [0, π]-{π/2}

Or

Let \(\cos ^{-1}\left(-\frac{1}{2}\right)=y\)

⇒ \(\cos y=-\frac{1}{2} \Rightarrow \cos y=\cos \left[\pi-\frac{\pi}{3}\right]\)

⇒ \(\cos y=\cos \frac{2 \pi}{3} \Rightarrow y=\frac{2 \pi}{3} \in[0, \pi]\)

Question 6. Simplify \(\sec ^{-1}\left(\frac{1}{2 x^2-1}\right), 0<x<\frac{1}{\sqrt{2}} \text {. }\)
Solution:

Let y = \(\sec ^{-1}\left(\frac{1}{2 x^2-1}\right), 0<x<\frac{1}{\sqrt{2}}\)

Put x = \(\cos \theta\); then \(\theta=\cos ^{-1} x\)

y = \(\sec ^{-1}\left(\frac{1}{2 \cos ^2 \theta-1}\right)\)

= \(\sec ^{-1}\left(\frac{1}{\cos 2 \theta}\right)=\sec ^{-1}(\sec 2 \theta)=2 \theta=2 \cos ^{-1} x\)

Question 7. Prove that: \(3 \sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^3\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]\)
Solution:

To prove: \(3 \sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^3\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]\)

R.H.S = sin^-1(3x – 4x³)

Put x = sinθ :  θ = sin^-1x

⇒ R.H.S = sin^-1 (3sinθ – 4sin³θ) – sin^-1(sin3θ) = 3θ = 3sin^-1x = L.H.S

Hence, proved.