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:
- \(\frac{5}{x \log \left(x^5\right) \log \left(\log x^5\right)}\)
- \(\frac{5}{x \log \left(\log x^5\right)}\)
- \(\frac{5 x^4}{\log \left(x^5\right) \log \left(\log x^5\right)}\)
- \(\frac{5 x^4}{\log \left(\log x^5\right)}\)
Solution: 1. \(\frac{5}{x \log \left(x^5\right) \log \left(\log x^5\right)}\)
Let y = \(\log \left[\log \left(\log x^5\right)\right]\)
⇒ \(\frac{d y}{d x}=\frac{1}{\log \left(\log x^5\right)} \times \frac{1}{\log \left(x^5\right)} \times \frac{1}{x^5} \times 5 x^4=\frac{5}{x \log \left(x^5\right) \log \left(\log x^5\right)}\)
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Question 2. If siny = x cos(a+y), then \(\frac{dx}{dy}\) is:
- \(\frac{\cos a}{\cos ^2(a+y)}\)
- \(\frac{-\cos a}{\cos ^2(a+y)}\)
- \(\frac{\cos a}{\sin ^2 y}\)
- \(\frac{-\cos a}{\sin ^2 y}\)
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:
- \(\frac{y+4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}\)
- \(\frac{y-4 x\left(x^2+y^2\right)}{x+4\left(x^2+y^2\right)}\)
- \(\frac{y-4 x\left(x^2+y^2\right)}{4 y\left(x^2+y^2\right)-x}\)
- \(\frac{4 y\left(x^2+y^2\right)-x}{y-4 x\left(x^2+y^2\right)}\)
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:
- 3
- 5
- 2
- 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:
- \(\frac{\cos \theta+\cos 2 \theta}{\sin \theta-\sin 2 \theta}\)
- \(\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta}\)
- \(\frac{\cos \theta-\cos 2 \theta}{\sin \theta-\sin 2 \theta}\)
- \(\frac{\cos 2 \theta-\cos \theta}{\sin 2 \theta+\sin \theta}\)
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:
- 2
- 3
- 6
- -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?
- \(\left(1-x^2\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+m^2 y=0\)
- \(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+m^2 y=0\)
- \(\left(1+x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0\)
- \(\left(1+x^2\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-m^2 x=0\)
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:
- Continuous as well as differentiable at x = 1
- Not continuous but differentiable at x = 1
- Continuous but not differentiable at x = 1
- 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 ecosx.
Solution:
Let y = sin²x and z = eco sex
⇒ \(\frac{dy}{dx}\) = 2 sin x cos x…(1)
(differentiating with respect x)
and \(\frac{dz}{dx}\) ecosx (-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 xsin 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
⇒ \(\frac{d y}{d x}=\frac{4 a \sin \theta \cos \theta}{4 a \sin ^2 \theta} \Rightarrow \frac{d y}{d x}=\frac{\cos \theta}{\sin \theta}=\cot \theta\)
∴ \(\left(\frac{d y}{d x}\right)_{\theta-\frac{\pi}{3}}=\cot \left(\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}}\).