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Differential Calculus

Differential Calculus Exercise 1 Multiple Choice Questions

Question 1. If y = 2^x,then \(\frac{d y}{d x}\)= ?

1. \(x\left(2^{x-1}\right)\)
2. \(\frac{2^x}{(\log 2)}\)
3. \(2^x(\log 2)\)
4. None of these

Answer:  3. \(2^x(\log 2)\)

⇒ \(\frac{d y}{d x}=2^x(\log 2)\)

Question 2. If y = \(\log _{10} x, \text { then } \frac{d y}{d x}=?\)

1. \(\frac{1}{x}\)
2. \(\frac{1}{x}\)(log 10)
3. \(\frac{1}{x(\log 10)}\)
4. None of these

Answer: 3. \(\frac{1}{x(\log 10)}\)

⇒ \(y=\frac{\log x}{\log 10}\)

Read and Learn More WBCHSE Solutions For Class 12 Maths

Question 3. \(\text { If } y=e^{1 / x} \text {, then } \frac{d y}{d x}=?\)

1. \(\frac{1}{x} \cdot e^{(1 / x-1)}\)
2. \(\frac{-e^{1 / x}}{x^2}\)
3. \(e^{1 / x} \log x\)
4. None of these

Answer: 2. \(\frac{-e^{1 / x}}{x^2}\)

⇒ \(\frac{d y}{d x}=\frac{e^{1 / x}}{-x^2}\)

Question 4. \(\text { If } y=x^x \text {, then } \frac{d y}{d x}=?\)

1. x^x log x*
2. x^x (1+log x)
3. x (1+log x)
4. None of these

Answer: 2. x^x (1+log x)

log y= x^x log x. Now differentiate w.r.t. x.

Question 5. \(\text { If } y=x^{\sin x}, \text { then } \frac{d y}{d x}=?\)

1. \((\sin x) \cdot x^{(\sin x-1)}\)
2. \((\sin x \cos x) \cdot x^{(\sin x-1)}\)
3. \(x^{\sin x}\left\{\frac{\sin x+x \log x \cdot \cos x}{x}\right\}\)
4. None of these

Answer: 3. \(x^{\sin x}\left\{\frac{\sin x+x \log x \cdot \cos x}{x}\right\}\)

logy= sin x(logx)

Question 6. \(\text { If } y=x^{\sqrt{x}} \text {, then } \frac{d y}{d x}=?\)

1. \(\sqrt{ } x \cdot x^{(\sqrt{ } x-1)}\)
2. \(\frac{x^{\sqrt{x}} \log x}{2 \sqrt{x}}\)
3. \(x^{\sqrt{x}}\left\{\frac{2+\log x}{2 \sqrt{ } x}\right\}\)
4. None of these

Answer: 3. \(x^{\sqrt{x}}\left\{\frac{2+\log x}{2 \sqrt{ } x}\right\}\)

log y= \(\sqrt{ } x(\log x)\)

Question 7. \(\text { If } y=e^{\sin \sqrt{x}} \text {, then } \frac{d y}{d x}=?\)

1. \(e^{\sin \sqrt{ } x} \cdot \cos \sqrt{x}\)
2. \(\frac{e^{\sin \sqrt{x}} \cos \sqrt{x}}{2 \sqrt{x}}\)
3. \(\)
4. None of these

Answer: 2. \(\frac{e^{\sin \sqrt{x}} \cos \sqrt{x}}{2 \sqrt{x}}\)

log y= sin \(\sqrt{ } x\)

Question 8. If y (tanx)^{cot x}, then \(\frac{d y}{d x}\)=?

1. cot x. (tan x)^{cot x-1}.sec²x
2. -(tan x)^{cot x} . cosec²x
3. (tan x)^{cot x}. cosec²x(l- log tan x)
4. None of these

Answer: 3. (tan x)^{cot x}. cosec²x(l- log tan x)

log y= cot x .log (tan x)

Question 9. If y= (sin x)log^x, then\(\frac{d y}{d x}\) =?

1. (logx). (sinx)^(logx-1). cos x
2. (sin x)^logx . \(\left\{\frac{x \log x+\log \sin x}{x}\right\}\)
3. (sin x)^logx . \(\left\{\frac{(x \log x) \cot x+\log \sin x}{x}\right\}\)
4. None of these

Answer: 3. (sin x)^logx . \(\left\{\frac{(x \log x) \cot x+\log \sin x}{x}\right\}\)

log y= (log x). (log sin x)

Question 10. If y = sin(x^x),then \(\frac{d y}{d x}\) = ?

1. x^xcos (x^x)
2. x^x cos x^x (1+logx)
3. x^xcos x^xlog ^x
4. None of these

Answer: 2. x^x cos x^x (1+logx)

Let x^x = z. Then, y= sin z. Then, \(\frac{d y}{d x}=\left(\frac{d y}{d z} \times \frac{d z}{d x}\right)\)

Question 11. If y= \(\sqrt{x \sin x}\) then \(\frac{d y}{d x}\) = ?

1. \(\frac{(x \cos x+\sin x)}{2 \sqrt{x \sin x}}\)
2. \(\frac{1}{2}(x \cos x+\sin x) \cdot \sqrt{x \sin x}\)
3. \(\frac{1}{2 \sqrt{x \sin x}}\)
4. None of these

Answer: 1. \(\frac{(x \cos x+\sin x)}{2 \sqrt{x \sin x}}\)

y² = x sin x

= 2y . \(\frac{d y}{d x}\)

= (xcos + sinx).

Question 12. If e^x+y= xy, then \(\frac{d y}{d x}\) = ?

1. \(\frac{x(1-y)}{y(x-1)}\)
2. \(\frac{y(1-x)}{x(y-1)}\)
3. \(\frac{(x-x y)}{(x y-y)}\)
4. None of these

Answer: 2. \(\frac{y(1-x)}{x(y-1)}\)

(x+ y)= log x + logy

= \(1+\frac{d y}{d x}=\frac{1}{x}+\frac{1}{y} \cdot \frac{d y}{d x}\)

Question 13. If (x + y) = sin (x+ y), then \(\frac{d y}{d x}\) = ?

1. -1
2. 1
3. \(\frac{1-\cos (x+y)}{\cos ^2(x+y)}\)
4. None of these

Answer: 1. -1

(x + y) = sin(x + y) ⇒ \(1+\frac{d y}{d x}=\cos (x+y) \cdot\left[1+\frac{d y}{d x}\right]\)

Question 14. If \(\sqrt{ } x+\sqrt{ } y=\sqrt{ } a\),then \(\frac{d y}{d x}\) = ?

1. \(\frac{-\sqrt{x}}{\sqrt{y}}\)
2. \(-\frac{1}{2} \cdot \frac{\sqrt{y}}{\sqrt{x}}\)
3. \(\frac{-\sqrt{y}}{\sqrt{x}}\)
4. None of these

Answer: 3. \(\frac{-\sqrt{y}}{\sqrt{x}}\)

⇒ \(\sqrt{ } x+\sqrt{ } y=\sqrt{ } a \Rightarrow \frac{1}{2 \sqrt{ } x}+\frac{1}{2 \sqrt{ } y} \cdot \frac{d y}{d x}\)= 0

Question 15. x^y= y^x,then \(\frac{d y}{d x}\) =?

1. \(\frac{(y-x \log y)}{(x-y \log x)}\)
2. \(\frac{y(y-x \log y)}{x(x-y \log x)}\)
3. \(\frac{y(y+x \log y)}{x(x+y \log x)}\)
4. None of these

Answer: 2. \(\frac{y(y-x \log y)}{x(x-y \log x)}\)

x^y = y^x ⇒ y log x= x logy. Now, differentiate both sides w.r.t. x.

Question 16. If x^py^q= (x + y)^y(p+q) ,then \(\frac{d y}{d x}\) = ?

1. \(\frac{x}{y}\)
2. \(\frac{y}{x}\)
3. \(\frac{x^{p-1}}{y^{q-1}}\)
4. None of these

Answer:  2. \(\frac{y}{x}\)

plog x+ qlogy= (p + q) log(x+ y).

Question 17. If y = \(x^2 \sin \frac{1}{x}\), then\(\frac{d y}{d x}\) = ?

1. \(x \sin \frac{1}{x}-\cos \frac{1}{x}\)
2. \(-\cos \frac{1}{x}+2 x \sin \frac{1}{x}\)
3. \(-x \sin \frac{1}{x}+\cos \frac{1}{x}\)
4. None of these

Answer: 2. \(-\cos \frac{1}{x}+2 x \sin \frac{1}{x}\)

⇒ \(\frac{d y}{d x}=x^2 \cdot\left(\cos \frac{1}{x}\right)\left(\frac{-1}{x^2}\right)+2 x \sin \frac{1}{x}\)

Question 18. If y = cos² x³, then \(\frac{d y}{d x}\) = ?

1. – 3x²sin(2x³)
2. -3x² sin² x³
3. -3x² cos²(2x³)
4. None of these

Answer: 1. – 3x²sin(2x³)

y = \(\left(\cos x^3\right)^2 \Rightarrow \frac{d y}{d x}\)

=2\(\left(\cos x^3\right)\left(-\sin x^3\right)\left(3 x^2\right)\)

=-3x² \(\sin \left(2 x^3\right)\)

Question 19. If y = \(\log \left(x+\sqrt{x^2+a^2}\right)\), then \(\frac{d y}{d x}\) =?

1. \(\frac{1}{2\left(x+\sqrt{x^2+a^2}\right)}\)
2. \(\frac{-1}{\sqrt{x^2+a^2}}\)
3. \(\frac{1}{\sqrt{x^2+a^2}}\)
4. None of these

Answer: 3. \(\frac{1}{\sqrt{x^2+a^2}}\)

⇒ \(\frac{d y}{d x}=\frac{1}{\left(x+\sqrt{x^2+a^2}\right.} \cdot\left\{1+\frac{1}{2 \sqrt{x^2+a^2}} \times 2 x\right\}\)

= \(\frac{1}{\sqrt{x^2+a^2}}\)

Question 20. If y \(=\log \left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right)\) then\(\frac{d y}{d x}\)= ?

1. \(\frac{1}{\sqrt{x}(1-x)}\)
2. \(\frac{-1}{x(1-\sqrt{x})^2}\)
3. \(\frac{-\sqrt{ } x}{2(1-\sqrt{ } x)}\)
4. None of these

Answer: 1. \(\frac{1}{\sqrt{x}(1-x)}\)

y= \(\log (1+\sqrt{ } x)-\log (1-\sqrt{ } x)\)

= \(\frac{d y}{d x}=\left\{\frac{1}{2 \sqrt{ } x(1+\sqrt{ } x)}+\frac{1}{2 \sqrt{ } x(1-\sqrt{ } x)}\right\}\)

= \(\frac{1}{2 \sqrt{x}} \cdot \frac{2}{(1-x)}=\frac{1}{\sqrt{x(1-x)}}\)

Question 21. If y \(=\log \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}-x}\right)\) , then \(\frac{d y}{d x}\)=?

1. \(\frac{2}{\sqrt{1+x^2}}\)
2. \(\frac{2 \sqrt{1+x^2}}{x^2}\)
3. \(\frac{-2}{\sqrt{1+x^2}}\)
4. None of these

Answer: 1. \(\frac{2}{\sqrt{1+x^2}}\)

y = \(\log \left(\sqrt{1+x^2}+x\right)-\log \left(\sqrt{1+x^2}-x\right)\), Now differentiatiate.

Question 22. If y= \(\sqrt{\frac{1+\sin x}{1-\sin x}}\), then \(\frac{d y}{d x}\)=?

1. \(\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\)
2. \(\frac{1}{2}{cosec}^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\)
3. \(\frac{1}{2}{cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right) \cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\)
4. None of these

Answer: 2. \(\frac{1}{2}{cosec}^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\)

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

= \(\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\)

Question 23. If y= \(\sqrt{\frac{\sec x-1}{\sec x+1}}\) then \(\frac{d y}{d x}\) = ?

1. sec²x
2. \(\frac{1}{2} \sec ^2 \frac{x}{2}\)
3. \(\frac{-1}{2}{cosec}^2 \frac{x}{2}\)
4. None of these

Answer: 2. \(\frac{1}{2} \sec ^2 \frac{x}{2}\)

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

= \(\tan \frac{x}{2}\)

Question 24. If y = \(=\sqrt{\frac{1+\tan x}{1-\tan x}}\), then \(\frac{d y}{d x}\) = ?

1. \(\frac{1}{2} \sec ^2 x \cdot \tan \left(x+\frac{\pi}{4}\right)\)
2. \(\frac{\sec ^2\left(x+\frac{\pi}{4}\right)}{2 \sqrt{\tan \left(x+\frac{\pi}{4}\right)}}\)
3. \(\frac{\sec ^2\left(\frac{x}{4}\right)}{\sqrt{\tan \left(x+\frac{\pi}{4}\right)}}\)
4. None of these

Answer: 2. \(\frac{\sec ^2\left(x+\frac{\pi}{4}\right)}{2 \sqrt{\tan \left(x+\frac{\pi}{4}\right)}}\)

y= \(\left\{\tan \left(x+\frac{\pi}{4}\right)\right\}^{\frac{1}{2}}\)

= \(\frac{d y}{d x}=\frac{1}{2}\left\{\tan \left(x+\frac{\pi}{4}\right)\right\}^{-\frac{1}{2}} \cdot \sec ^2\left(x+\frac{\pi}{4}\right)\)

Question 25. If y = \(\tan ^{-1}\left(\frac{1-\cos x}{\sin x}\right)\), then \(\frac{d y}{d x}\) = ?

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

Answer: 3. \(\frac{1}{2}\)

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

= \(\tan ^{-1}\left\{\tan \frac{x}{2}\right\}\)

=\(\frac{x}{2}\)

Question 26. If y= \(\tan ^{-1}\left\{\frac{\cos x+\sin x}{\cos x-\sin x}\right\}\) , then \(\frac{d y}{d x}\) = ?

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

Answer: 1. 1

y = \(\tan ^{-1}\left\{\frac{\cos x+\sin x}{\cos x-\sin x}\right\}\)

= \(\tan ^{-1}\left\{\frac{1+\tan x}{1-\tan x}\right\}\)

= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+x\right)\right\}\)

= \(\left(\frac{\pi}{4}+x\right)\)

Question 27. If y= \(\tan ^{-1}\left\{\frac{\cos x+\sin x}{\cos x-\sin x}\right\}\) ,then \(\frac{d y}{d x}\) = ?

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

Answer: 2. \(\frac{-1}{2}\)

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

= \(\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}\)

= \(\left(\frac{\pi}{4}-\frac{x}{2}\right)\)

Question 28. If y = \(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\) ,then \(\frac{d y}{d x}\) = ?

1. \(\frac{1}{2}\)
2. \(\frac{-1}{2}\)
3. \(\frac{1}{\left(1+x^2\right)}\)
4. None of these

Answer:  2. \(\frac{-1}{2}\)

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

⇒ \(\tan ^{-1}\left\{\tan \frac{x}{2}\right\}\)

= \(\frac{x}{2}\)

Question 29. If y= \(\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)\) , then \(\frac{d y}{d x}\) = ?

1. \(\frac{a}{b}\)
2. \(\frac{-b}{a}\)
3. 1
4. -1

Answer: 4. -1

\(=\tan ^{-1}\left(\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}\right)\)

= \(\tan ^{-1}\left(\frac{\tan \theta-\tan x}{1+\tan \theta \tan x}\right)\)

= \(\tan ^{-1} \tan (\theta-x)=\theta-x=\left(\tan ^{-1} \frac{a}{b}-x\right)\)

= \(\frac{d y}{d x}\)= 1

Question 30. If y= sin^-1(3x- 4X³), then \(\frac{d y}{d x}\) = ?

1. \(\frac{3}{\sqrt{1-x^2}}\)
2. \(\frac{-4}{\sqrt{1-x^2}}\)
3. \(\frac{3}{\sqrt{1+x^2}}\)
4. None of these

Answer: 1. \(\frac{3}{\sqrt{1-x^2}}\)

Putting x= sin θ, we get y= sin 1(sin 3θ) = 3θ = 3sin 1x.

Question 31. If y= cos^-1(4X³– 3x), then \(\frac{d y}{d x}\) = ?

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

Answer: 2. \(\frac{3}{\sqrt{1-x^2}}\)

Putting x= cos θ, we get y= cos^-1(cos 3θ) = 3θ= 3cos^-1x.

Question 32. If y = \(\tan ^{-1}\left(\frac{\sqrt{ } a+\sqrt{ } x}{1-\sqrt{a x}}\right)\), then \(\frac{d y}{d x}\) = ?

1. \(\frac{1}{(1+x)}\)
2. \(\frac{1}{\sqrt{x}(1+x)}\)
3. \(\frac{2}{\sqrt{x}(1+x)}\)
4. \(\frac{1}{2 \sqrt{x}(1+x)}\)

Answer: 4. \(\frac{1}{2 \sqrt{x}(1+x)}\)

Put \(\sqrt{ } a\)= tan θ and \(\sqrt{ } x\)= tan Φ.

Then

y= tan¹ {tan (θ+ Φ)>)} = θ+ Φ = tan^-1 \(\sqrt{ } a\) + tan^-1 \(\sqrt{ } x\).

Question 33. If y = tan-1\(\frac{2 x}{\left(1+x^4\right)}\), then \(\frac{d y}{d x}\) = ?

1. \(\frac{2 x}{\left(1+x^4\right)}\)
2. \(\frac{-2 x}{\left(1+x^4\right)}\)
3. \(\frac{x}{\left(1+x^4\right)}\)
4. None of these

Answer: 1. \(\frac{2 x}{\left(1+x^4\right)}\)

Putting x²= tan θ, we get:

y= \(y=\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\theta\right)\right\}=\left(\frac{\pi}{4}+\theta\right)\)

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

Question 34. If y = cos^-1 x³, then\(\frac{d y}{d x}\) = ?

1. \(\frac{-1}{(1+x)}\)
2. \(\frac{2}{\sqrt{(1+x)}}\)
3. \(\frac{-1}{2 \sqrt{ } x(1+x)}\)
4. None of these

Answer: 3. \(\frac{-1}{2 \sqrt{ } x(1+x)}\)

y = \(\cot ^{-1} \sqrt{x}\)

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

Question 35. If y = cos^-1x³, then\(\frac{d y}{d x}\) = ?

1. \(\frac{-1}{\sqrt{1-x^6}}\)
2. \(\frac{-3 x^2}{\sqrt{1-x^6}}\)
3. \(\frac{-3}{x^2 \sqrt{1-x^6}}\)
4. None of these

Answer: 2. \(\frac{-3 x^2}{\sqrt{1-x^6}}\)

y = \(\cos ^{-1} x^3\)

⇒ \(\frac{d y}{d x}=\frac{-3 x^2}{\sqrt{1-x^6}}\)

Question 36. If y= tan^-1(sec x + tan x), then \(\frac{d y}{d x}\) = ?

1. \(\frac{1}{2}\)
2. \(\frac{1}{2}\)
3. 1
4. None of these

Answer: 1. \(\frac{1}{2}\)

y= \(\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)=\tan ^{-1}\left[\frac{\{\cos (x / 2)+\sin (x / 2)\}^2}{\cos ^2(x / 2)-\sin ^2(x / 2)}\right]\)

= \(\tan ^{-1}\left\{\frac{\cos (x / 2)+\sin (x / 2)}{\cos (x / 2)-\sin (x / 2)}\right\}\)

= \(\tan ^{-1}\left\{\frac{1+\tan (x / 2)}{1-\tan (x / 2)}\right\}\)

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

= \(\left(\frac{\pi}{4}+\frac{x}{2}\right)\)

Question 37. If y \(\cot ^{-1}\left(\frac{1-x}{1+x}\right)\), then \(\frac{d y}{d x}\)

1. \(\frac{-1}{\left(1+x^2\right)}\)
2. \(\frac{1}{\left(1+x^2\right)}\)
3. \(\frac{1}{\left(1+x^2\right)^{3 / 2}}\)
4. None of these

Answer: 2. \(\frac{1}{\left(1+x^2\right)}\)

Put x= tan θ. Then, y= cot^-1.tan\(\left(\frac{\pi}{4}-\theta\right)\)

= \(\cot ^{-1}\left[\cot \left\{\frac{\pi}{2}-\left(\frac{\pi}{4}-\theta\right)\right\}\right]\)

= \(\left(\frac{\pi}{4}+\theta\right)\)

y = \(\frac{\pi}{4}+\tan ^{-1} x\)

Question 38. If y= \(y\sqrt{\frac{1+x}{1-x}}\), then \(\frac{d y}{d x}\)

1. \(\frac{2}{(1-x)^2}\)
2. \(\frac{x}{(1-x)^{3 / 2}}\)
3. \(\frac{1}{(1-x)^{3 / 2} \cdot(1+x)^{1 / 2}}\)
4. None of these

Answer: 3. \(\frac{1}{(1-x)^{3 / 2} \cdot(1+x)^{1 / 2}}\)

log y = \(\frac{1}{2}\{\log (1+x)-\log (1-x)\}\)

⇒ \(\frac{1}{y} \cdot \frac{d y}{d x}\)

= \(\frac{1}{2}\left\{\frac{1}{(1+x)}+\frac{1}{(1-x)}\right\}\)

Question 39. If y = \(\sec ^{-1}\left(\frac{x^2+1}{x^2-1}\right)\) , then \(\frac{d y}{d x}\)= ?

1. \(\frac{-2}{\left(1+x^2\right)}\)
2. \(\frac{2}{\left(1+x^2\right)}\)
3. \(\frac{-1}{\left(1-x^2\right)}\)
4. None of these

Answer: 1. \(\frac{-2}{\left(1+x^2\right)}\)

Put x= cot θ. Then

y = \(\sec ^{-1}\left(\frac{1+\tan ^2 \theta}{1-\tan ^2 \theta}\right)\)

= \(\sec ^{-1}\left(\frac{1}{\cos 2 \theta}\right)=\sec ^{-1}(\sec 2 \theta)\)

= 2θ = 2 cot x^-1

Question 40. If y = \(\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)\),then \(\frac{d y}{d x}\)= ?

1. \(\frac{-2}{\left(1+x^2\right)}\)
2. \(\frac{-2}{\left(1-x^2\right)}\)
3. \(\frac{-2}{\sqrt{1-x^2}}\)
4. None of these

Answer: 3. \(\frac{-2}{\sqrt{1-x^2}}\)

Put x= cos θ.

Then , y= \(\sec ^{-1}\left(\frac{1}{2 \cos ^2 \theta-1}\right)\)

= sec(sec 2θ) = 2 θ = 2 cos^-1 x

Question 41. If y = \(\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-1}{x}\right\}\), then \(\frac{d y}{d x}\)= ?

1. \(\frac{1}{\left(1+x^2\right)}\)
2. \(\frac{2}{\left(1+x^2\right)}\)
3. \(\frac{1}{2\left(1+x^2\right)}\)
4. None of these

Answer: 3. \(\frac{1}{2\left(1+x^2\right)}\)

Put x = tan θ.

Then, y =\(\frac{1}{2}\) tan^-1 x

Question 42. If \(y=\sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}\) , then \(\frac{d y}{d x}\) = ?

1. \(\frac{-1}{2 \sqrt{1-x^2}}\)
2. \(\frac{1}{2 \sqrt{1-x^2}}\)
3. \(\frac{1}{2\left(1+x^2\right)}\)
4. None of these

Answer: 1. \(\frac{-1}{2 \sqrt{1-x^2}}\)

Put x = cos θ. Then,

y= \(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\)

Question 43. If x= at², y= 2at, then \(\frac{d y}{d x}\) = ?

1. \(\frac{1}{t}\)
2. \(\frac{-1}{t^2}\)
3. \(\frac{-2}{t}\)
4. None of these

Answer: 1. \(\frac{1}{t}\)

⇒ \(\frac{d x}{d t}=2 a t\)

⇒  \(\frac{d y}{d t}=2 a\)

So, \(\frac{d y}{d x}=\frac{(d y / d t)}{(d x / d t)}\)

= \(\frac{1}{t}\)

Question 44. If x= a sec θ, y= b tan θ, then \(\frac{d y}{d x}\) =?

1. \(\frac{b}{a} \sec \theta\)
2. \(\frac{b}{a}{cosec} \theta\)
3. \(\frac{b}{a} \cot \theta\)
4. None of these

Answer: 2. \(\frac{b}{a}{cosec} \theta\)

⇒ \(\frac{d x}{d \theta}=a \sec \theta \tan \theta\)

= \(\frac{d y}{d \theta}=b \sec ^2 \theta\). \(\frac{d y}{d x}=\frac{(d y / d \theta)}{(d x / d \theta)}\)

= \(\frac{b}{a}{cosec} \theta\)

Question 45. If x = a cos 2θ, y= b sin 2θ, then \(\frac{d y}{d x}\) = ?

1. \(\frac{-a}{b}\)
2. \(\frac{a}{b}\) cot θ
3. \(\frac{-b}{a}\)
4. None of these

Answer: 3. \(\frac{-b}{a}\)

⇒ \(\frac{d x}{d \theta}=-2 a \cos \theta \sin \theta\),

=  \(\frac{d y}{d \theta}=2 b \sin \theta \cos \theta\)

= \(\frac{d y}{d x}=\frac{(d y / d \theta)}{(d x / d \theta)}\)

= \(\frac{-b}{a}\)

Question 46. If x= a(cos θ+ θ sin θ) and y = a(sin θ — θ cos θ), then\(\frac{d y}{d x}\) =?

1. cot θ
2. tan θ
3. cot θ
4. a tan θ

Answer:  2. tan θ

⇒ \(\frac{d y}{d x}=\frac{(d y / d \theta)}{(d x / d \theta)}\)

Question 47. If y = \(x^{x^{x \ldots \infty}}\), then \(\frac{d y}{d x}\)= ?

1. \(\frac{y}{x(1-\log x)}\)
2. \(\frac{y^2}{x(1-\log x)}\)
3. \(\frac{y^2}{x(1-y \log x)}\)
4. None of these

Answer: 3. \(\frac{y^2}{x(1-y \log x)}\)

y = \(x^y\)

log y =  y log x

⇒ \(\frac{1}{y} \frac{d y}{d x}\)

⇒ \(\frac{y}{x}+(\log x) \frac{d y}{d x}\)

Question 48. If y = \(\sqrt{x+\sqrt{x+\sqrt{x+}}} \ldots \infty\) , then \(\frac{d y}{d x}\) = ?

1. \(\frac{1}{(2 y-1)}\)
2. \(\frac{1}{\left(y^2-1\right)}\)
3. \(\frac{2 y}{\left(y^2-1\right)}\)
4. None of these

Answer:  1. \(\frac{1}{(2 y-1)}\)

y= \(\sqrt{x+y}\)

y²= \(x+y\)

⇒ \(2 y \frac{d y}{d x}=\cos x+\frac{d y}{d x}\)

Question 49. If y = \(\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+}}} \ldots\), then \(\frac{d y}{d x}\) = ?

1. \(\frac{\sin x}{(2 y-1)}\)
2. \(\frac{\cos x}{(y-1)}\)
3. \(\frac{\cos x}{(2 y-1)}\)
4. None of these

Answer: 3. \(\frac{\cos x}{(2 y-1)}\)

y = \(e^{x+y} \Rightarrow x+y=\log y\)

y = sin x+y

= \(2 y \frac{d y}{d x}\)

= \(\cos x+\frac{d y}{d x}\)

Question 50. If y = \(e^x+e^{x+\ldots}\), then\(\frac{d y}{d x}\)

1. \(\frac{1}{(1-y)}\)
2. \(\frac{y}{(1-y)}\)
3. \(\frac{y}{(y-1)}\)
4. None of these

Answer:  2. \(\frac{y}{(1-y)}\)

y= e^(x+y)

x+y =logy

1+ \(+\frac{d y}{d x}=\frac{1}{y} \cdot \frac{d y}{d x}\)

Question 51.  The value k for which \(f(x)=\left\{\begin{array}{c}
\frac{\sin 5 x}{3 x}, \text { if } x \neq 0 \\
k, \text { if } x=0
\end{array} \text { is continuous at } x=0\right.\) is 

1. \(\frac{1}{3}\)
2. 0
3. \(\frac{3}{5}\)
4. \(\frac{5}{3}\)

Answer: 4. \(\frac{5}{3}\)

For continuity at x= 0, we must have \(\lim _{x \rightarrow 0} f(x)\) = f(0).

⇒ \(\lim _{x \rightarrow 0} f(x)\)

= \(\lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x} \times \frac{5}{3}=\frac{5}{3} \lim _{5 x \rightarrow 0} \frac{\sin 5 x}{5 x}\)

= \(\left(\frac{5}{3} \times 1\right)=\frac{5}{3}\)

∴ We must have,f(0) = \(\frac{5}{3}\)

k = \(\frac{5}{3}\)

Question 52. \(\left\{\begin{array}{l}
x \sin \frac{1}{x}, \text { if } x \neq 0 \\
0, \text { when } x=0 .
\end{array}\right.\) Then, which of the following is the true statement?

1. f(x) is not defined at x= 0
2. \(\lim _{x \rightarrow 0} f(x)\)does not exist
3. f(x) is continuous at x= 0
4. f(x) is discontinuous at x= 0

Answer: 3. f(x) is continuous at x= 0

f(0) = 0.

⇒ \(\lim _{x \rightarrow 0} f(x)\) = \(\lim _{x \rightarrow 0} x \sin \frac{1}{x}=0 \times(\text { a finite quantity })\)

= 0

∴ f(x)is continuous at x= 0

Question 53. The value of k for which f(x) = x is continuous at x= 0, is

1. 7
2. 4
3. 3
4. None of these

Answer: 1.7

f(0)=k.

⇒ \(\lim _{x \rightarrow 0} f(x)\)

= \(=\lim _{x \rightarrow 0} \frac{3 x+4 \tan x}{x}=\)

= \(\lim _{x \rightarrow 0}\left\{3+\frac{4 \tan x}{x}\right\}\)

(3+4) = 7

∴ f(x) is continuous at x= 0 ⇔ f(0) = 7 ⇔ k= 7.

Question 54. Let f(x) =x^3/2 Then, f'(0)= ?

1. \(\frac{3}{2}\)
2. \(\frac{1}{2}\)
3. Does not exist
4. None of these

Answer: 3. Does not exist

\(L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}\)

= \(\lim _{h \rightarrow 0} \frac{\left(h^{3 / 2}-0\right)}{-h}=\lim _{h \rightarrow 0}(-h)^{1 / 2}\), which does not exist, since

(-h)^1/2 is imaginary

Question 55. The function f(x) = lx I ∀ x ∈ R is

1. Continuous but not differentiable at x = 0
2. Differentiable but not continuous at x = 0
3. Neither continuous nor differentiable at x = 0
4. None of these

Answer: 1. Continuous but not differentiable at x = 0

f(0+0) = \(\lim _{h \rightarrow 0}|0+h|\)

= \(\lim _{h \rightarrow 0}|h|=0\)

f(0-0)= \(\lim _{h \rightarrow 0}|0-h|\)

= \(\lim _{h \rightarrow 0}|-h|=\lim _{h \rightarrow 0}|h|=0\) and f(0)= 0

∴ f(x) is continuous at x= 0

Rf'(0)= \(\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{f(h)-0}{h}=\lim _{h \rightarrow 0} \frac{|h|}{h}\)

= \(\lim _{h \rightarrow 0} \frac{h}{h}=1\)

lf'(0)= \(\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{f(-h)-0}{-h}\)

= \(\lim _{h \rightarrow 0} \frac{|-h|}{-h}\)

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

∴ Rf'(0) ≠ L f'(0), which shows that f(x) is not differentiable at x= 0.

Question 56. \(
\text { The function } f(x)=\left\{\begin{array}{l}
1+x, \text { when } x \leq 2 \\
5-x, \text { when } x>2
\end{array}\right. \text { is }\)

1. Continuous as well as differentiable at x = 2
2. Continuous but not differentiable at x = 2
3. Differentiablebutnot continuous at x = 2
4. None of these

Answer: 2. Continuous but not differentiable at x=2

f(2+ 0) = \(\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}\{5-(2+h)\}=3\)

f(2 – 0)= \(\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}\{1+(2-h)\}=3\) and f(2) = 3

∴ f(x) is continuous at x= 2

⇒ Rf'(2) = \(\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0} \frac{(3-h)-3}{h}\)

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

⇒ Lf'(2) = \(\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}=\lim _{h \rightarrow 0} \frac{(3-h)-3}{-h}\)

= \(\lim _{h \rightarrow 0} \frac{h}{h}=1\)

∴ f (x) is not differentiable at x= 2

Question 57. \(
\text { If the function } f(x)=\left\{\begin{array}{r}
k x+5, \text { when } x \leq 2 \\
x-1, \text { when } x>2
\end{array}\right. \text { is }\) continuous at x= 2, then k= ?

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

Answer: 2.- 2

f(2) = \(\lim _{x \rightarrow 2} f(x)\)

= \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)\)

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

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

= \(\lim _{h \rightarrow 0}\{k(2-h)+5\}\)

∴ 2k+5 = 1

k= -2

Also,f(2) = 2k+5 = 1 .Hence, k= -2

Question 58. If the function \(f(x)=\left\{\begin{array}{c}
\frac{1-\cos 4 x}{8 x^2}, x \neq 0 \\
k, x=0
\end{array}\right.\) is continuous at x= 0 , then k= ?

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

Answer: 3. \(\frac{1}{2}\)

⇒ \(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)\)

= \(\lim _{h \rightarrow 0} \frac{1-\cos 4 h}{8 h^2}\)

= \(\lim _{h \rightarrow 0} \frac{2 \sin ^2 2 h}{8 h^2}\)

= \(\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{\sin 2 h}{2 h}\right)^2\)

= \(\left(\frac{1}{2} \times 1^2\right)=\frac{1}{2}\)

⇒ \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin ^2 a x}{a^2 x^2} \times a^2\)

= \(a^2 \cdot \lim _{a x \rightarrow 0}\left(\frac{\sin a x}{a x}\right)^2\)

= \(a^2 \times 1^2=a^2\)

For continuity, we must have f(0) = \(\frac{1}{2}\)

For continuity, we must have f(0) = a²

Question 59. If the function f(x) \(\left\{\begin{array}{r}
\frac{\sin ^2 a x}{x^2}, \text { when } x \neq 0 \\
k, \text { when } x=0
\end{array}\right.\) is continuous at x= 0, then k= ?

1. a
2. a²
3. -2
4. -4

Answer: 2. a²

Question 60. If the function f(x) = \(\left\{\begin{aligned}
\frac{k \cos x}{(\pi-2 x)}, \text { when } x & \neq \frac{\pi}{2} \\
3, \text { when } x & =\frac{\pi}{2}
\end{aligned}\right.\)

1. 3
2. -3
3. -5
4. 6

Answer: 4. 6

⇒ \(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin ^2 a x}{a^2 x^2} \times a^2\)

= \(a^2 \cdot \lim _{a x \rightarrow 0}\left(\frac{\sin a x}{a x}\right)^2\)

= \(a^2 \times 1^2=a^2\)

For continuity, we must have f(x) = a²

Question 61. At x= 2,f(x) = [x] is

1. Continuous but not differentiable
2. Differentiable but not continuous
3. Continuous as well as differentiable
4. None of these

Answer: 4. None of these

f(2) = [2] = 2

f(2 + 0)= \(\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}[2+h]\)= 2

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

∴ f(x) is not continuous at x= 2

Rf'(2) = \(\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0} \frac{[2+h]-[2]}{h}\)

\(\lim _{h \rightarrow 0} \frac{(2-2)}{h}\) = 0

Lf'(2)= \(\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}=\lim _{h \rightarrow 0} \frac{[2-h]-[2]}{-h}\)

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

= \(\lim _{h \rightarrow 0} \frac{1}{h}=\infty\)

∴ f(x) is not differentiable at x = 2

Question 62. \(
\text { Let } f(x)=\left\{\begin{array}{r}
\frac{x^2-2 x-3}{x+1^2}, \text { when } x \neq-1 \\
k, \text { when } x=-1
\end{array}\right.\)

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

Answer: 2. -4

⇒ \(\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1} \frac{(x-3)(x+1)}{(x+1)}\)

= \(\lim _{x \rightarrow-1}(x-3)\) = -4

For continuity, we must have, f(-1) =-4

Question 63. The function f(x) = x³– 6x² + I5x- 12 is

1. Strictly decreasing on R
2. Strictly increasing on R
3. Increasingin (-∞, 2] and decreasing in (2, ∞)
4. None of these

Answer: 2. Strictly increasing on R

f'(x) = 3x² -12x+15 = 3(x² -4x+ 5) = 3[(x-2)² + 1] > 0.

⇒  f'(x) > 0 for all x ∈ R ⇒  f(x) is strictly increasing on R.

Question 64. The function f (x) = 4- 3x + 3X² -x³ is

1. Decreasing on R
2. Increasing on R
3. Strictly decreasing on R
4. Strictly increasing on R

Answer: 1. Decreasing on R

f'(x) =-3 + 6x- 3x²= 3(x²– 2x+ 1) = -3(x- 1)² < 0.

⇒   f'(x) < 0 for all x ∈ R ⇒  f(x) is decreasing on R.

Question 65. The function f(x) = 3x + cos 3x is

1. Increasing on R
2. Strictly increasing on R
3. Decreasing on R
4. Strictly decreasing on R

Answer: 1. Increasing on R

f'(x) = 3- 3sin 3x = 3(1- sin 3x) > 0 since -1 < sin 3x < 1.

f( x) > 0 for all x ∈ R ⇒  f(x) is increasing on R.

Question 66. The function/ f(x) = x³– 6X²+ 9x+ 3 is decreasing for

1. 1 < x < 3
2. x>1
3. x <1
4. x <1 or x > 3

Answer: 1. 1 < x < 3

f'(x) = 3x²– 12 + 9 = 3Cx²– 4x + 3) = 3(x- 1)(x- 3)

f'(x) = 0 ⇒ x=1 or x= 3.

There are two factors in f'(x), so we start with + ve sign.

∴ f(x) is decreasing for 1 < x < 3.

 

Question 67. The function f(x) -x³– 27x + 8 is increasing when

1. 1 x 1 <3
2. 1 x 1 >3
3. -3 < x < 3
4. None of these

Answer: 2. 1x 1 >3

f'( x) = 3x² -27= 3(x²– 9) = 3(x+ 3)(x- 3)

f'(x)= 0 ⇒ x=-3 or x= 3.

There are two factors in f(x), so we start with +ve sign.

f(x) is increasing when x < -3 or x > 3, i.e., when IxI >3.

 

Question 68. f(x) = sin x is increasing in

1. \(\left(\frac{\pi}{2}, \pi\right)\)
2. \(\left(\pi, \frac{3 \pi}{2}\right)\)
3. [0,π]
4. \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Answer: 4. \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

∴ f'(x)= cos x > 0 in \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

∴  f(x) is increasing in \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Question 69. f(x) = \(\frac{2 x}{\log x}\) is increasing in

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

Answer: 3. (e, ∞)

⇒ \(f^{\prime}(x)=\frac{(\log x) \cdot 2-2 x \cdot \frac{1}{x}}{(\log x)^2}\)

= \(\frac{2(\log x-1)}{(\log x)^2}\)

∴ f'(x)> 0 ⇔ log x-1>0 ⇔ log x>1 ⇔ log x>log e ⇔ x>e.

∴ f(x) is increasing (e, ∞)

Question 70. f(x) = (sin x- cos x) is decreasing

1. \(\left(0, \frac{3 \pi}{4}\right)\)
2. \(\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)\)
3. \(\left(\frac{7 \pi}{4}, 2 \pi\right)\)
4. None of these

Answer: 2. \(\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)\)

f'(x)= (cos x+ sinx) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right)\)

= \(\sqrt{2} \sin \left(\frac{\pi}{4}+x\right)\)

∴ f(x)’ \(\sin \left(\frac{\pi}{4}+x\right)<0\)

= \(\pi<\frac{\pi}{4}+x<2 \pi\)

= \(\frac{3 \pi}{4}<x<\frac{7 \pi}{4}\)

∴ f (x) is decreasing in \(\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)\)

Question 71. f(x) = \(\frac{x}{\sin x}\) is

1. Increasing in (0, 1)
2. Decreasing in (0, 1)
3. Increasingin \(\left(0, \frac{1}{2}\right)\) and decreasing \(\left(, \frac{1}{2}\right)\)in
4. None of these

Answer: 1. Increasing in (0, 1)

f'(x)= \(\frac{\sin x-x \cos x}{\sin ^2 x}\)

= \(\frac{\cos x(\tan x-x)}{\sin ^2 x}\)

0<x<1 ⇒  tan x >x and cos x >0 ⇒  cos x(tan x- x) > 0

⇒  f'(x)>0

∴ f(x) is increasing (0, 1).

Question 72. f(x) = x^x is decreasing in the interval

1. (0,e)
2. \(\left(0, \frac{1}{e}\right)\)
3. (0,1)
4. None of these

Answer: 2. \(\left(0, \frac{1}{e}\right)\)

f (x) = (1 +log x)

f'(x)< 0 ⇔ (1 +logx)<0 ⇒ logx<-1

= \(\log \frac{1}{e}\)

= x >0 and \(x<\frac{1}{e}\)

f(x) is decreasing in (0, \(\frac{1}{e}\))

Question 73. f(x) = x²e^x is increasing in

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

Answer: 2. (0,2)

f'(x) = 2xe^–x– x²e^–x= xe^–x(2- x)

∴ f'(x) > 0 < t ⇒ > 0 and (2- x) > 0 ⇔ 0< x <2.

∴ f(x) is increasing in (0, 2).

Question 74. f(x) = sin x- kx is decreasing for all x ∈ R, when

1. k<1
2. k≥3
3. k<3
4. k≤3

Answer: 3. k<3

f'(x)- (cos x-k) and therefore,

f(x) is decreasing ⇔ f'(x) <0⇒ cos x- k<0

⇒  cos x<k ⇔ k>cosx ⇒ k>1

Question 75. f(x) = (x + 1)³(x- 3)³ is increasing in

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

Answer: 4. (1,∞)

f'(x) = (x+1) (x- 3)³

f'(x) = 3(x+ 1)3(x- 3)²+ 3(x+ 1)²(x- 3)³

= 3(x+ 1)²(x- 3)²[(x+1) + (x- 3)] = 3(x+1)²(x- 3)²(x-1)

⇒ f'(x) > 0 when (x- 1) > 0, i.e., when x > 1.

∴ f(x) is increasing in (1, ∞).

Question 76. f(x) = [x(x-3)]² is increasing in

1. (0,∞)
2. (-,∞,0)
3. (1, 3)
4. (0,\(\frac{1}{3}\)∪(3,∞)

Answer: 4.  (0,\(\frac{1}{3}\))∪(3,∞)

f(x) = [x(x- 3)]² ⇒ f(x)= 2x(x- 3)(2x- 3).

f'(x) = 0 ⇒ x = 0 or x= \(\frac{1}{3}\) or x = \(\frac{1}{3}\) 3.

∴ f (x) is increasing when 0<x <-or x>3.

∴ f(x) is increasing in(0,\(\frac{1}{3}\)).

 

Question 77. If f(x) = kx³– 9x² + 9x + 3 is increasing for every real number x, then

1. k>3
2. k≥3
3. k<3
4. k≤3

Answer: 1. k>3

∴ f(x) = 3kx²– 18x+ 9= 3(kx²– 6x+ 3).

This is positive when k > 0 and (36- 12k) < 0 ⇒ k> 3

Question 78. f(x) = \(\frac{x}{\left(x^2+1\right)}\) is increasing in

1. (-1,1)
2. (-1, ∞)
3. (∞, -1)∪(1,∞)
4. None of these

Answer: 1. (-1,1)

f'(x)= \(\frac{\left(x^2+1\right) \cdot 1-x \cdot 2 x}{\left(x^2+1\right)^2}=\frac{\left(1-x^2\right)}{\left(1+x^2\right)^2}\)

f'(x)> 0 ⇔(1-x²)>0 < ⇒ x²< 1 < ⇒ -1< X <1.

∴ f(x) is increasing in (-1, 1)

Question 79. The least value of k for which f(x) = x² + kx+1 is increasing on (1, 2), is

1. -2
2. -1
4. 1
5. 2

Answer: 1. -2

f'(x)= (2x+ k).

1< x < 2 ⇒ 2< 2x < 4 ⇒ 2+k < 2x + k < 4+ k =  2+k< f(x) <4 + k

f (x) is increasing o (2x +k)>0 ⇔ 2+ k> 0 ⇔ k >-2.

Least value of k is -2.

Question 80. f(x) =IxI has

1. Minimum at x= 0
2. Maximum at x = 0
3. Neither a maximum nor a minimum at x = 0
4. None of these

Answer: 1. Minimum at x= 0

f(x) = IxI > 0 for all x ∈ R.

The least value of I x I is 0 at x = 0.

f(x) = IxI has minima at x= 0

Question 81. When x is positive, the minimum value of x^x is

1. e^e
2. e^1/e
3. e^–1/e
4. (1/e)

Answer: 3. e^–1/e

f(x) = x² ⇒  f'(x) = f(1 +log x) and f”(x) = \(x^x\left[\frac{1}{x}+(1+\log x)^2\right]\)

f'(x) = 0 ⇒ 1 +log x = 0 ⇒ log x =-l \(\log \left(\frac{1}{e}\right) \Rightarrow x=\left(\frac{1}{e}\right)\)

[f”(x)]x= (1/e) = \(e\left(\frac{1}{e}\right)^{1 / e}\)>0.

x= \(\frac{1}{e}\)is apoint ofminima.

Minimum value of x is \(\left(\frac{1}{e}\right)^{1 / e}=e^{-1 / e}\)

Question 82. The maximum value of \(\left(\frac{\log x}{x}\right)\) is

1. \(\left(\frac{1}{e}\right)\)
2. \(\left(\frac{2}{e}\right)\)
3. e
4. 1

Answer: 1. \(\left(\frac{1}{e}\right)\)

f(x) = \(\frac{\log x}{x}\)

f'(x)= \(\frac{x \cdot \frac{1}{x}-\log x \cdot 1}{x^2}=\frac{(1-\log x)}{x^2}\)

f'(x) = \(\frac{x^2 \cdot \frac{1}{x}-(1-\log x) \cdot 2 x}{x^4}=\frac{(-3+2 \log x)}{x^3}\)

f'(x) = 0 ⇒ 1-logx= 0 ⇒ log x=1 = log e ⇒ x= e.

f”(e)= \(\left(\frac{-3+2}{e^3}\right)=\frac{-1}{e^3}\)<0

x= e is a point of maxima

Maximum value of f(x) is \(\frac{1}{e}(\log e)=\frac{1}{e}\)

Question 83. f(x) = cosec x in (-π, 0) has a maxima at

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

Answer: 4. x =\(\frac{-\pi}{2}\)

f(x) = cosec x ⇒ f'(x) = -cosec x cot x’

= f”(x) = cosec³x+ cosec x (cot²x) = cosec x (cosec²x+ cot²x)

= cosec x (2cosec²x-1)

f(x) = 0 ⇒ cot x= 0 ⇒ X= \(\frac{-\pi}{2}\)

⇒ \(f^{\prime \prime}\left(\frac{-\pi}{2}\right)=\operatorname{cosec}\left(\frac{-\pi}{2}\right)\left[2 \operatorname{cosec}^2\left(\frac{-\pi}{2}\right)-1\right]\)

= (-1)(2-1)= -1 < 0.

x = \(\frac{-\pi}{2}\)is a point of maxima

Question 84. If x > 0 and xy=1, the minimum value of (x+ y) is

1. -2
2. 1
3. 2
4. None of these

Answer: 3. 2

xy = 1

y = \(\frac{1}{x}\)

Lest S = x+y = x+ \(\frac{1}{x}\)

Then , \(\frac{d s}{d x}=\left(1-\frac{1}{x^2}\right)=\frac{\left(x^2-1\right)}{x^2}\) and \(\frac{d^2 s}{d x^2}=\frac{2}{x^3}\)

\(\frac{d s}{d x}\) = 0

x²-1= 0 x = ± 1

⇒ \(\left.\left.\frac{d^2 s}{d x^2}\right]_{(x=-1)}=-2<0 \text { and } \frac{d^2 s}{d x^2}\right]_{(x=1)}\)

= 2>0

S is minimum at x=1 and minimum value of S = (1 + 1) = 2

Question 85. The minimum value of \(\left(x^2+\frac{250}{x}\right)\)

1. 0
2. 25
3. -39
4. None of these

Answer: 4. None of these

Let f(x) \(\left(x^2+\frac{250}{x}\right)\)

Then f'(x)= \(\left(2 x-\frac{250}{x^2}\right)and\) f”(x} = \(\left(2+\frac{500}{x^3}\right)\)

f'(x) = 0 ⇒ 2x³ -250= 0

x³ = \(\frac{250}{2}\)

x³ = 125

x³ = 5³

⇒ x= 5.

f”(5)= \(\left(2+\frac{500}{125}\right)\)

= 6 > 0

f(x) is minimum at x= 5 and minimum value = \(\left(25+\frac{250}{5}\right)\)= 75

Question 86. The minimum value of f(x) = 3x⁴– 8X³– 48x+ 25 on [0, 3] is

1. 16
2. 25
3. 50
4. 75

Answer: 3. 50

f(x) = 12x³– 24x²+ 24x- 48 = 12(x- 2)(x²+ 2)

f”(x)= 36x²– 48x + 24=12(3x²– 4x+ 2).

f'(x) = 0 ⇒ x= 2 and f”(2) = 12(3x⁴-4x² + 2) = 72>0

x= 2 is a point of minima.

Minimum value=min {f(0),f(2),f(3)}

= min {25, -39, 16} =-39.

Question 87. The maximum value of f(x) = (x- 2)(x- 3)² is

1. \(\frac{7}{3}\)
2. 3
3. \(\frac{4}{27}\)
4. 27

Answer: 3. \(\frac{4}{27}\)

f(x) = (x- 2)(x- 3)² ⇒ f(x)= (x- 3)(3x- 7) and f”(x) = (6x- 16).

f'(x) = 0 ⇒ x= 3 or x= \(\frac{7}{3}\)

f”(3)= 2>0 and f” (\(\frac{7}{3}\))= – 2>0.

x = \(\frac{7}{3}\) is a point of maxima.

Maximum value = (\(\frac{7}{3}\)– 2)(\(\frac{7}{3}\)-3) = \(\frac{4}{27}\)

Question 88. The least value off (x) = (e^x+ e^–x) is

1. -2
2. 0
3. 2
4. None of these

Answer: 3. 0

f'(x) = e^x– e^–x and f”(x) = e^x+ e^–x

f'(x) = 0  ⇒ e^x– e^–x = 0

⇒  e^x= e^–x

⇒ e^2x= e⁰

⇒ x = 0.

f”(0) = e°+ \(\frac{1}{e^0}\)= (1 + 1) = 2 > 0.

f(x) is minimum at x= 0 and minimum value of f(x) is 2

Algebra

Algebra Exercise 1A Review Of Facts And Formulae

1. Results on transposition of matrices

(A +B)’=(A’+B’)

(AB)’= (BA)’

(KB)’= (K.A’)

(A’)’= (A)

2.

A is symmetric ⇔  A’= A

A is skew-symmetric ⇔ A’=-A

Read and Learn More WBCHSE Solutions For Class 12 Maths

3.

A is idempotent ⇔ A²= A

A is nilpotent of order n = Aⁿ= 0

A is orthogonal ⇔ AA’ = A’A =I

A is involutory ⇔ A²=I

4.

A is singular ⇔ I A I = 0

A is non-singular ⇔ I A I # 0

A^-1 exists ⇔ I A I # 0

(AB)^-1 = B^-1 A^-1

(kA)^-11= \(\frac{1}{k}\)A^-1

(A’)^-1 = (A^-1)’

I A^-1I = \(\frac{1}{|A|}\)

5.

A(adj . A) = I AI I

I adj. A I = I AI^n-1

adj . AB = (adj . B) . (adj – A)

6. For square matrices A and B of the same order, we have

1. (A + B)²= A²+AB +BA + B²
2. (A- B)² = A²– AB- BA + B²
3. (A + B) . (A- B) = A²– AB +BA- B²
4. A and B anticommute AB= -BA

Class 12 Algebra Multiple Choice Questions

Question 1.  If A and B are 2 – 2-rowed square matrices such that   

Answer: 2. \(\left[\begin{array}{rr}
7 & -5 \\
1 & -5
\end{array}\right]\)

⇒ 2A= (A + B) + (A- B) and 2B = (A + B)- (A- B)

Question  2. \(\text { If }\left[\begin{array}{rr}
3 & -2 \\
5 & 6
\end{array}\right]+2 A=\left[\begin{array}{rr}
5 & 6 \\
-7 & 10
\end{array}\right], \text { then } A=?\)

1. \(\left[\begin{array}{rr}
1 & 3 \\
-5 & 4
\end{array}\right]\)

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

3. \(\left[\begin{array}{rr}
1 & 4 \\
-6 & 2
\end{array}\right]\)

4. None of these

Answer: 3. \(\left[\begin{array}{rr}
1 & 4 \\
-6 & 2
\end{array}\right]\)

⇒ 2A= \(\left[\begin{array}{rr}
5 & 6 \\
-7 & 10
\end{array}\right]\)– \(\left[\begin{array}{rr}
3 & -2 \\
5 & 6
\end{array}\right]\)= \(\left[\begin{array}{rr}
2 & 8 \\
-12 & 4
\end{array}\right]\)

Question 3. \(A=\left[\begin{array}{rr}
2 & 0 \\
-3 & 1
\end{array}\right] \text { and } B=\left[\begin{array}{rr}
4 & -3 \\
-6 & 2
\end{array}\right] \text { are such that } 4 A+3 X=5 B \text {, then } X=\text { ? }\)

1. \(\left[\begin{array}{rr}
4 & -5\\
-6 & 2
\end{array}\right]\)

2. \(\left[\begin{array}{rr}
4 & 5 \\
-6 & -2
\end{array}\right]\)

3. \(\left[\begin{array}{rr}
-4 & 5 \\
-6 & -2
\end{array}\right]\)

4. None of these

Answer: 1. \(\left[\begin{array}{rr}
4 & -5\\
-6 & 2
\end{array}\right]\)

⇒ 4A + 3X= 5B => 3X= (5B-4A) => X = \(\frac{1}{3}\)(5B-4A).

Question 4. \(\text { If }(A-2 B)=\left[\begin{array}{rr}
1 & -2 \\
3 & 0
\end{array}\right] \text { and }(2 A-3 B)=\left[\begin{array}{rr}
-2 & 2 \\
3 & -3
\end{array}\right] \text {, then } B=\text { ? }\)

1. \(\left[\begin{array}{rr}
6 & -4 \\
-3 & 3
\end{array}\right]\)

2. \(\left[\begin{array}{rr}
-4 & 6 \\
-3 & -3
\end{array}\right]\)

3. \(\left[\begin{array}{rr}
4 & -6 \\
-3 & -3
\end{array}\right]\)

4. None of these

Answer: 2. \(\left[\begin{array}{rr}
-4 & 6 \\
-3 & -3
\end{array}\right]\)

⇒ B = (2A-3B)-2(A-2B).

Question 5. \(\text { If }(2 A-B)=\left[\begin{array}{rrr}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right] \text { and }(2 B+A)=\left[\begin{array}{rrr}
3 & 2 & 5 \\
-2 & 1 & -7
\end{array}\right] \text {, then } A=\text { ? }\)

1. \(\left[\begin{array}{rrr}
-3 & 2 & 1 \\
2 & 1 & -1
\end{array}\right]\)

2. \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
2 & -1 & -1
\end{array}\right]\)

3. \(\left[\begin{array}{rrr}
3 & -2 & 1 \\
-2 & 1 & -1
\end{array}\right]\)

4. None of these

Answer: 3. \(\left[\begin{array}{rrr}
3 & -2 & 1 \\
-2 & 1 & -1
\end{array}\right]\)

⇒ 5A = 2(2A- B) + (2B + A). Then, A = \(\frac{1}{5}\)(5A)

Question 6. \(\text { If } 2\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
7 & 0 \\
10 & 5
\end{array}\right] \text {, then }\)

1. (x = -2,y = 8)

2. x =  2y = 8

3. x = 3,y = -6

4. x =-3,y = 6

Answer: 2. x =  2y = 8

⇒ [2x+1 = 5 ⇒ x= 2] and [8 + y= 0 => y= -8].

Question 7. \(\text { If }\left[\begin{array}{cc}
x-y & 2 x-y \\
2 x+z & 3 z+w
\end{array}\right]=\left[\begin{array}{rr}
-1 & 0 \\
5 & 13
\end{array}\right] \text {, then }\)

1. z = 3,w = 4

2. z = 4,w = 3

3. z =1,w = 2

2. z = 2,w =-1

Answer: 1. z = 3,w = 4

⇒ (x- y=-1 and 2x- y= 0) => (x=1, y= 2)

⇒  (2x+ z = 5 => z = 3) and (3z+ w= 13 => w= 4).

Question 8. \(\text { If }\left[\begin{array}{cc}
x & y \\
3 y & x
\end{array}\right]\left[\begin{array}{l}
1 \\
2
\end{array}\right]=\left[\begin{array}{l}
3 \\
5
\end{array}\right] \text {, then }\)

1. x = 1,y = 2

2. x = 2,y = 1

3. x =1,y = 1

4. None of these

Answer: 3. x =1,y = 1

⇒  Solve x+ 2y= 3 and 3y+ 2x = 5.

Question 9. \(\text { If the matrix } A=\left[\begin{array}{cc}
3-2 x & x+1 \\
2 & 4
\end{array}\right] \text { is singular, then } x=\text { ? }\)

1. 0

2. 1

3. -1

4. -2

Answer:  2. 1

⇒  A is singular ⇔ IAI = 0.

Question 10. \(\text { If } A_\alpha=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right], \text { then }\left(A_\alpha\right)^2=\text { ? }\)

1. \(\left[\begin{array}{cc}
\cos ^2 \alpha & \sin ^2 \alpha \\
-\sin ^2 \alpha & \cos ^2 \alpha
\end{array}\right]\)

2. \(\left[\begin{array}{cc}
\cos 2 \alpha & \sin 2 \alpha \\
-\sin 2 \alpha & \cos 2 \alpha
\end{array}\right]\)

3. \(\left[\begin{array}{cc}
2 \cos \alpha & 2 \sin \alpha \\
-\sin \alpha & 2 \cos \alpha
\end{array}\right]\)

4. None of these

Answer: 1. \(\left[\begin{array}{cc}
\cos ^2 \alpha & \sin ^2 \alpha \\
-\sin ^2 \alpha & \cos ^2 \alpha
\end{array}\right]\)

Question 11. \(\text { If } A=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \text { be such that } A+A^{\prime}=I \text {, then } \alpha=\text { ? }\)

1. π

2. \(\frac{\pi}{3}\)

3. 2π

4. \(\frac{2 \pi}{3}\)

Answer: 2. \(\frac{\pi}{3}\)

Question 12. \(\text { If } A=\left[\begin{array}{rrr}
1 & k & 3 \\
3 & k & -2 \\
2 & 3 & -4
\end{array}\right] \text { is singular, then } k=\text { ? }\)

1. \(\frac{16}{3}\)

2. \(\frac{34}{5}\)

3. \(\frac{33}{2}\)

4. None of these

Answer: 3. \(\frac{33}{2}\)

⇒  A is singular ⇔ IAI = 0.

Question 13. \(\text { If } A=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\), then adj. A= ?

1. \(\left[\begin{array}{ll}
d & -c \\
-b & a
\end{array}\right]\)

2. \(\left[\begin{array}{ll}
-d & b \\
c & -a
\end{array}\right]\)

3. \(\left[\begin{array}{ll}
d & -b \\
-c & a
\end{array}\right]\)

4. \(\left[\begin{array}{ll}
-d & -b \\
c & a
\end{array}\right]\)

Answer: 3. \(\left[\begin{array}{ll}
d & -b \\
-c & a
\end{array}\right]\)

Question 14. \(\text { If } A=\left[\begin{array}{cc}
2 x & 0 \\
x & x
\end{array}\right] \text { and } A^{-1}=\left[\begin{array}{rr}
1 & 0 \\
-1 & 2
\end{array}\right], \text { then } x=?\)

1. 1

2. 2

3. \(\frac{1}{2}\)

4. -2

Answer: 3. \(\frac{1}{2}\)

Use AA^-1 =I.

Question 15. If A and B are square matrices of the same order, then (A + B)(A — B) = ?

1. (A²– B²)

2. A² + AB + BA + B²

3. A²-AB + BA-B²

4. None of these

Answer: 3. A²-AB + BA-B²

⇒ Using distributive law, we have

(A +B)-(A-B) = A(A-B) +B(A-B) = (A²-AB +BA-B²).

Question 16. If A and B are square matrices of the same order, then (A + B)²=?

1. A²+ 2AB + B²

2.  A²-AB-BA + B²

3. A² + 2BA + B²

4. None of these

Answer: 2.  A²-AB-BA + B²

⇒ (A + B)²= (A + B)-(A + B)=A(A + B) + B(A + B) = (A²+ AB + BA + B²).

Question 17. If A and B are square matrices of the same order, then (A- B)² = ?

1. A²– 2AB + B²

2. A²+ AB-BA-B²

3. A²– 2BA + B²

4. None of these

Answer: 2. A²+ AB-BA-B²

⇒ (A- B)² = (A-B) . (A-B)= A(A- B)- B(A- B) = (A2- AB- BA + B2).

Question 18. If A and B are symmetric matrices of the same order, then (AB-BA) is
always

1. A symmetric matrix

2. A skew-symmetric matrix

3. A zero matrix

4. An identity matrix

Answer: 2. A skew-symmetric matrix

Given A’= A and B’ = B.

∴ (AB- BA)’= (AB)’- (BA)’= (B’A’- A’B’) = (BA- AB) = -(AB- BA)

∴ (AB- BA) is skew-symmetric.

Question 19. Matrices A and B are inverses of each other only when

1. AB = BA

2. AB = BA = O

3. AB = O,BA =I

4. AB = BA = I

Answer: 4. AB = BA = I

A and B are inverses of each other only when AB = BA =I.

Question 20. For square matrices A and B of the same order, we have adj (AB) = ?

1. (adj A)(adj B)

2.  (adj A)(adj B)

3. I ABI

4. None of these

Answer: 2.  (adj A)(adj B)

⇒ adj (AB)= (adj B) (adj A).

Question 21. If A is a 3-rowed square matrix and I A I =4, then adj (adj A) =?

1. 4A

2. 16A

3. 64A

4. None of these

Answer: 1. 4A

adj(adj A) = IAI^(n-1)=  IAI^(3-2).A= IAI -A = 4A.

Question 22. If A is a 3-rowed square matrix and I A I =5, then I adj AI =?

1. 5

2. 25

3. 125

4. None of these

Answer: 2. 25

I adj AI = IA^(n-1)I= I A 1²= 5²= 25.

Question 23. For any two matrices A and B

1. AB = BA is always true

2. AB = BA is never true

3. Sometimes AB = BA and sometimes AB BA

4. Whenever AB exists, then BA exists

Answer: 3. Sometimes AB = BA and sometimes AB BA

Question 24.  \(\text { If } A \cdot\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right] \text {, then } A=\text { ? }\)

1. \(\left[\begin{array}{rr}
1 & -1 \\
1 & 1
\end{array}\right]\)

2. \(\left[\begin{array}{rr}
1 & 1 \\
-1 & 1
\end{array}\right]\)

3. \(\left[\begin{array}{rr}
1 & 1 \\
1 & -1
\end{array}\right]\)

4. None of these

Answer: 3. \(\left[\begin{array}{rr}
1 & 1 \\
1 & -1
\end{array}\right]\)

⇒ Let \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{rr}
3 & 2 \\
1 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
2 & 3
\end{array}\right]\), find a,b,c and d

Question 25. If A is an invertible square matrix, then 1A^-1I =?

1. |A|

2. \(\frac{1}{|A|}\)

3. 1

4. 0

Answer: 2. \(\frac{1}{|A|}\)

AA^-1 = ⇒  IAA^-1 I = III =1

⇒  IAI. I A^-1 I  =1 = \(\frac{1}{|A|}\)

Question 26. If A and B are invertible matrices of the same order, then (AB)^-1 =?

1. (A^-1 x B^-1)

2. (A x B^-1)

3. (A^-1x B)

4. (B^-1x A^-1)

Answer: 4. (B^-1x A^-1)

⇒  (AB)^-1= B^-1A^-1

Question 27. If A and B are two non-zero square matrices of the same order such that AB = 0, then

1.  IAI =0 or IBI =0

2. IAI = 0 and IBI =0

3.  IAI ≠ 0and IB I ≠ 0

4. None of these

Answer: 2. IAI = 0 and IBI =0

⇒ [AB= 0 and A ≠ 0, B≠ 0] => IAI =0 and IBI =0.

Question 28. If A is a square matrix such that IAI ≠ 0 and A²– A + 2I = 0, then A^-1 =?

1. (I-A)

2. (I+A)

3. \(\frac{1}{2}\)(I-A)

4. \(\frac{1}{2}\)(I+A)

Answer:  3. \(\frac{1}{2}\)(I-A)

⇒ 21=(A- A² ) => 2A^-1 = A^-1 A-A^-1AA = I-IA = (I-A)

A^-1 = \(\frac{1}{2}\) (I-A)

Question 29.  \(
\text { If } A=\left[\begin{array}{lll}
1 & \lambda & 2 \\
1 & 2 & 5 \\
2 & 1 & 1
\end{array}\right] \text { is not invertible, then } \lambda=\text { ? }\)

1. 2

2. 1

3. -1

4. 0

Answer: 2. 1

⇒ A is not invertible ⇒ I A I = 0.

Question 30. \(\text { If } A=\left[\begin{array}{rr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right] \text {, then } A^{-1}=\text { ? }\)

1. A

2. -A

3. adj A-

4. -adj A

Answer: 3. adj A-

IAI =1 ⇒ A^-1 =\(
\frac{1}{|A|}
\) adj A= (adj A).

Question 31. \(
\text { The matrix } A=\left[\begin{array}{cc}
a b & b^2 \\
-a^2 & -a b
\end{array}\right] \text { is }\)

1. Idempotent

2. Orthogonal

3. Nilpotent

4. None of these

Answer:  3. Nilpotent

A²= 0  ⇒ A is nilpotent.

Question 32. \(
\text { The matrix } A=\left[\begin{array}{rrr}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right] \text { is }\)

1. Non-singular

2. Idempotent

3. Nilpotent

4. Orthogonal

Answer: 2. Idempotent

A²= A ⇒ A is idempotent.

Question 33. If A is singular, then A (adj A) =?

1. A unit matrix

2. A null matrix

3. A symmetric matrix

4. None of these

Answer: 2. A null matrix

Given IAI = 0. So, A(adj A) = IAI.I = 0-I = 0.

∴ A(adj A) is a null matrix.

Question 34. \(
\text { For any 2-rowed square matrix } A, \text { if } A \cdot({adj} A)=\left[\begin{array}{ll}
8 & 0 \\
0 & 8
\end{array}\right] \text {, then the value of }\)

1. 0

2. 8

3. 64

4. 4

Answer: 2. 8

A . adj  A = IAI. I= 8 \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) = 8I= |A|

Question 35. \(\text { If } A=\left[\begin{array}{rr}
-2 & 3 \\
1 & 1
\end{array}\right] \text {, then }\left|A^{-1}\right|=\text { ? }\)

1. -5

2. \(\frac{-1}{5}\)

3. \(\frac{1}{25}\)

4. 25

Answer: 2. \(\frac{-1}{5}\)

AA^-1  = I ⇒  I AA^-1 I = 1I1 => IAI .I A^-1 I =1 ⇒ IA^-1 I = \(\frac{1}{|A|}\)

IAI = \(\left|\begin{array}{rr}
-2 & 3 \\
1 & 1
\end{array}\right|\)=(-2- 3) = (-5)⇒I A^-1I= \(\frac{-1}{5}\)

Question 36. \(
\text { If } A=\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right] \text { and } A^2+x I=y A \text {, then the values of } x \text { and } y \text { are }\)

1.  x = 6, y = 6

2.  x = 8, y = 8

3. x = 5, y – 8

4. x = 6, y = 8

Answer: 2.  x = 8, y = 8

⇒ \(\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]+x\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=y\left[\begin{array}{ll}
3 & 1 \\
7 & 5
\end{array}\right]\)

⇒ \(\left[\begin{array}{rr}
16 & 8 \\
56 & 32
\end{array}\right]+\left[\begin{array}{ll}
x & 0 \\
0 & x
\end{array}\right]=\left[\begin{array}{rr}
3 y & y \\
7 y & 5 y
\end{array}\right]\)

⇒ \(\left[\begin{array}{cc}
16+x & 8 \\
56 & 32+x
\end{array}\right]=\left[\begin{array}{cc}
3 y & y \\
7 y & 5 y
\end{array}\right]\)

y= 8 and (16 + x= 3y= 3 x 8= 24) => x= 8.

Question 37. If matrices A and B anticommuting, then

1. AB = BA

2. AB = -BA

3. (AB) = (BA)^-1

4. None of these

Answer: 2. AB = -BA

A and B anticommute ⇔ AB= -BA.

Question 38. \(
\text { If } A=\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right], \text { then adj } A=\text { ? }\)

1. \(\left[\begin{array}{rr}
3 & -5 \\
-1 & 2
\end{array}\right]\)

2. \(\left[\begin{array}{rr}
3 & -1 \\
-5 & 2
\end{array}\right]\)

3. \(\left[\begin{array}{rr}
-1 & 2 \\
3 & -5
\end{array}\right]\)

4. None of these

Answer: 1. \(\left[\begin{array}{rr}
3 & -5 \\
-1 & 2
\end{array}\right]\)

Question 39. \(
\text { If } A=\left[\begin{array}{rr}
3 & -4 \\
-1 & 2
\end{array}\right] \text { and } B \text { is a square matrix of order } 2 \text { such that } A B=I \text {, then }\) B= ?

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

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

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

4. None of these

Answer: 3. \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\)

AB =7 ⇒ B = A^-1

Question 40. If A and B are invertible square matrices of the same order, then (AB)^-1 = ?

1. AB^-1

2. A^-1B

3. A^-1B^-1

4. B^-1A^-1

Answer: 4. B^-1A^-1

(AB)^-1 = B^-1 A^-1

Question 41. \(
\text { If } A=\left[\begin{array}{rr}
2 & -1 \\
1 & 3
\end{array}\right] \text {, then } A^{-1}=\text { ? }\)

1. \(\left[\begin{array}{cc}
\frac{3}{7} & \frac{-1}{7} \\
\frac{1}{7} & \frac{2}{7}
\end{array}\right]\)

2. \(\left[\begin{array}{cc}
\frac{3}{7} & \frac{-1}{7} \\
\frac{1}{7} & \frac{2}{7}
\end{array}\right]\)

3. \(\left[\begin{array}{cc}
\frac{3}{7} & \frac{-1}{7} \\
\frac{1}{7} & \frac{2}{7}
\end{array}\right]\)

4. None of these

Answer: 2. \(\left[\begin{array}{cc}
\frac{3}{7} & \frac{-1}{7} \\
\frac{1}{7} & \frac{2}{7}
\end{array}\right]\)

⇒  \(|A|=\left|\begin{array}{rr}
2 & -1 \\
1 & 3
\end{array}\right|=(6+1)=7 \neq 0\)

M₁₁ = 3, M₁₂= 1, M₂₁ = -1 and M₂₂= 2

∴ C₁₁= 3, C₁₂= -1, C₂₁ =1 and C₂₂= 2

⇒ A \(=\left[\begin{array}{rr}
3 & -1 \\
1 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right.]\)

Question 42. \(
\text { If }|A|=3 \text { and } A^{-1}=\left[\begin{array}{rr}
3 & -1 \\
\frac{-5}{3} & \frac{2}{3}
\end{array}\right], \text { then adj } A=?\)

1. \(\left[\begin{array}{rr}
9 & 3 \\
-5 & -2
\end{array}\right]\)

2. \(\left[\begin{array}{rr}
9 & -3 \\
-5 & 2
\end{array}\right]\)

3. \(\left[\begin{array}{rr}
-9 & 3 \\
5 & -2
\end{array}\right]\)

4. \(\left[\begin{array}{rr}
9 & -3 \\
5 & -2
\end{array}\right]\)

Answer: 2. \(\left[\begin{array}{rr}
9 & -3 \\
-5 & 2
\end{array}\right]\)

⇒ \(
A^{-1}=\frac{1}{|A|} \cdot {adj} A \Rightarrow {adj} A=|A| \cdot A^{-1}=3 A^{-1}=\left[\begin{array}{rr}
9 & -3 \\
-5 & 2
\end{array}\right]\)

Question 43. \(
\text { If } A \text { is an invertible matrix and } A^{-1}=\left[\begin{array}{ll}
3 & 4 \\
5 & 6
\end{array}\right] \text {, then } A=\text { ? }\)

1. \(\left[\begin{array}{rr}
6 & -4 \\
-5 & 3
\end{array}\right]\)

2. \(\left[\begin{array}{ll}
\frac{1}{3} & \frac{1}{4} \\
\frac{1}{5} & \frac{1}{6}
\end{array}\right]\)

3. \(\left[\begin{array}{rr}
-3 & 2 \\
\frac{5}{2} & \frac{-3}{2}
\end{array}\right]\)

4. None of these

Answer: 3. \(\left[\begin{array}{rr}
-3 & 2 \\
\frac{5}{2} & \frac{-3}{2}
\end{array}\right]\)

⇒ A = (A^-1)^-1. So, find the inverse of A^-1.

Question 44. \(
\text { If } A=\left[\begin{array}{rr}
1 & 2 \\
4 & -3
\end{array}\right] \text { and } f(x)=2 x^2-4 x+5, \text { then } f(A)=?\)

1. \(\left[\begin{array}{rr}
19 & 2 \\
4 & -3
\end{array}\right]\)

2. \(\left[\begin{array}{rr}
19 & -16 \\
-32 & 51
\end{array}\right]\)

3. \(\left[\begin{array}{rr}
19 & -11 \\
-27 & 51
\end{array}\right]\)

4. None of these

Answer: 2. \(\left[\begin{array}{rr}
19 & -16 \\
-32 & 51
\end{array}\right]\)

⇒ (A) = 2A²-4A+5I.

Question 45. \(
\text { If } A=\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right], \text { then } A^2-4 A=\text { ? }\)

1. I

2. 5I

3. 3I

4. 0

Answer: 2. 5I

⇒ \(\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]\left[\begin{array}{ll}
1 & 4 \\
2 & 3
\end{array}\right]-\left[\begin{array}{ll}
4 & 16 \\
8 & 12
\end{array}\right]=\left[\begin{array}{ll}
9 & 16 \\
8 & 17
\end{array}\right]-\left[\begin{array}{ll}
4 & 16 \\
8 & 12
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right]\)=5I

Question 46. If A is a 2-rowed square matrix and I A I =6, then A . adj A = ?

1. \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)

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

3. \(\left[\begin{array}{ll}
\frac{1}{6} & 0 \\
0 & \frac{1}{6}
\end{array}\right]\)

4. None of these

Answer: 1. \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)

⇒ A.(adj A) = IAI.

⇒ \(
I=6 \cdot\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)

Question 47. If A is an invertible square matrix and A:is a non-negative real number, (kA^-1) = ?

1. k. A^-1

2. \(\frac{1}{k} \cdot A^{-1}\)

3. – K. A^-1

4. None of these

Answer: 2. \(\frac{1}{k} \cdot A^{-1}\)

⇒ \((k A)^{-1}=\frac{1}{k} \cdot A^{-1}\) is true.

Question 48. \(
\text { If } A=\left[\begin{array}{rrr}
3 & 4 & 1 \\
1 & 0 & -2 \\
-2 & -1 & 2
\end{array}\right], \text { then } A^{-1}=\text { ? }\)

1. \(\left[\begin{array}{rrr}
2 & 9 & -8 \\
-2 & 8 & 7 \\
-1 & 5 & -4
\end{array}\right]\)

2. \(\left[\begin{array}{rrr}
2 & 9 & -8 \\
2 & 8 & 7 \\
-1 & -5 & -4
\end{array}\right]\)

3. \(\left[\begin{array}{rrr}
2 & -9 & -8 \\
2 & 8 & 7 \\
-1 & -5 & -4
\end{array}\right]\)

4. None of these

Answer: 3. \(\left[\begin{array}{rrr}
2 & -9 & -8 \\
2 & 8 & 7 \\
-1 & -5 & -4
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|}\).adj A.

Question 49. If A is a square matrix, then (A + A’) is

1. Anull matrix

2. An identity matrix

3. A symmetric matrix

4. A skew-symmetric matrix

Answer: 3. A symmetric matrix

⇒ A is a square matrix ⇒ (A + A’) is symmetric.

Question 50. If A is a square matrix, then (A- A’) is

1. Anull matrix

2. An identity matrix

3. A symmetric matrix

4. A skew-symmetric matrix

Answer: 4. A skew-symmetric matrix

A is a square matrix ⇒(A -A’) is skew-symmetric.

Question 51. If A is a 3-rowed square matrix and I 3 AI = k I  A I, then K=?

1. 3

2. 9

3. 27

4. 1

Answer: 3. 27

⇒ I3AI =( 3 x 3 x 3) IAI =27. I A I.

Question 52. Which one of the following is a scalar matrix?

1. \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)

2. \(\left[\begin{array}{ll}
6 & 0 \\
0 & 3
\end{array}\right]\)

3. \(\left[\begin{array}{ll}
8 & 0 \\
0 & -8
\end{array}\right]\)

4. None of these

Answer: 3. \(\left[\begin{array}{ll}
8 & 0 \\
0 & -8
\end{array}\right]\)

⇒ A scalar matrix is a square matrix each of whose non-diagonal elements is 0 and all diagonal elements are equal.

Question 53. \(
\text { If } A=\left[\begin{array}{ll}
1 & -1 \\
2 & -1
\end{array}\right] \text { and } B=\left[\begin{array}{rr}
a & 1 \\
b & -1
\end{array}\right] \text { and }(A+B)^2=\left(A^2+B^2\right) \text {, then }\)

1. a= 2, b= -3

2. a= -2, b= 3

3. a= 1, b= 4

4. None of these

Answer:  3. a= 1, b= 4

A+ B)² = (A²+ B²) ⇒ A²+ B²+AB+ BA = (A²+ B²) ⇒ AB =-BA

⇒ \(\left[\begin{array}{cc}
a-b & 2 \\
2 a-b & 3
\end{array}\right]=\left[\begin{array}{ll}
-a-2 & a+1 \\
-b+2 & b-1
\end{array}\right]\)

Now, (a +1 = 2 and b-1 = 3)⇒ (a =1 and 6 = 4).

Algebra Exercise 1B Review Of Facts And Formulae

1.

1.\(\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right|=\left(a_{11} a_{22}-a_{12} a_{21}\right)\)

2. \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\)

= \(a_{11} \cdot\left|\begin{array}{ll}
a_{22} & a_{23} \\
a_{32} & a_{33}
\end{array}\right|-a_{12} \cdot\left|\begin{array}{ll}
a_{21} & a_{23} \\
a_{31} & a_{33}
\end{array}\right|+a_{13} \cdot\left|\begin{array}{ll}
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}\right|\)

\(=a_{11}\left(a_{22} a_{33}-a_{32} a_{23}\right)-a_{12}\left(a_{21} a_{33}-a_{31} a_{23}\right)+a_{13}\left(a_{21} a_{32}-a_{31} a_{22}\right)\)

2. 

1. Minor of a_ij is given by

M_ij= det. obtained after deleting ith row and jth column.

2. Co-factor of a_ij is given by C_ij= (-1)^i+j.M_ij

Δ= sum of the products of the elements of any row or column with their
corresponding co-factors.

Δ= a₁₁C₁₁+ a₁₂C₁₂ + a₁₃C₁₃

= a₁₁C₁₁+ a₂₁C₂₁ + a₃₁C₃₁

3. Properties of determinants

If the rows and columns of a determinant are interchange the value of the determinant remains unchanged.

⇒ R₁ ↔ R₃ shows the interchange of first and 3rd rows

If any two adjacent rows (or columns) of a determinant are interchanged, the value of the new determinant is the negative of the value of original determinant.

If all the elements of one row (or column) of a determinant are multiplied by k, the value of the original determinant is multiplied by k.

⇒ R₂—> R₂ shows the multiplication of each element of the second rowby k.

If the elements of a row (or a column) of a determinant are added k times the corresponding elements of another row (or column), the value of the determinant remains unchanged.

⇒ R_i —>R_j+ kR_j shows that k times the elements of jth row are added to the
corresponding elements of ith row.

If two rows (or columns) of a determinant are identical, the value of the
the determinant is zero.

If each element of a row (or a column) of a determinant is 0, the value of
the determinant is 0.

4.  Area of Δ A ABC with vertices A(x₁,y₁) B(x₂, y₂) and C(x₃, y₃) is given by

⇒ \(\Delta=\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\)

Points A(x₁,y₁) B(x₂, y₂) and C(x₃, y₃) are collinear ⇔ ar(Δ ABC) = 0 ⇔ Δ = 0.

Algebra Exercise 1B Multiple Choice Questions

Question 1. \(\left|\begin{array}{ll}
\cos 70^{\circ} & \sin 20^{\circ} \\
\sin 70^{\circ} & \cos 20^{\circ}
\end{array}\right|\)

1.  1
2.  0
3.  cos 50°
4. sin 50°

Answer: 2. 0

Question 2. \(\left|\begin{array}{ll}
\cos 15^{\circ} & \sin 15^{\circ} \\
\sin 15^{\circ} & \cos 15^{\circ}
\end{array}\right|\)

1. 1
2. \(\frac{1}{2}\)
3. \(\frac{\sqrt{3}}{2}\)
4. None of these

Answer:  3. \(\frac{\sqrt{3}}{2}\)

Δ = \(\left(\cos ^2 15^{\circ}-\sin ^2 15^{\circ}\right)\)

= cos (2×15° )= cos 30°

= \(\frac{\sqrt{3}}{2}\)

Question 3. \(\left|\begin{array}{rr}
\sin 23^{\circ} & -\sin 7^{\circ} \\
\cos 23^{\circ} & \cos 7^{\circ}
\end{array}\right|=?\)

1. \(\frac{\sqrt{3}}{2}\)
2. \(\frac{1}{2}\)
3. sin 16
4. cos 16

Answer: 2. \(\frac{1}{2}\)

Question 4. \(\left|\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right|\) = ?

1.  (a²+ b²– c²– d2²
2. (a²-b²+ c²-d²)
3. (a²+ b²+ c²+ d²)
4.  None of these

Answer: 3.  (a²+ b²+ c²+ d²)

Δ = \(\left|\begin{array}{cc}
a+i b & c+i d \\
-(c-i d) & a-i b
\end{array}\right|=(a+i b)(a-i b)+(c-i d)(c+i d)\)

= (a²+b²+c²+d²)

Question 5. \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)=?

1.  (a + b + c)
2.  3[a+ b+ c)
3.  3abc
4.  0

Answer: 4. 0

R₁ -+ (R₁ + R₂+ R₃) gives all zeros in R₁ and this gives Δ= 0

Question 6. \(\left|\begin{array}{ccc}
1 & 1+p & 1+p+q \\
2 & 3+2 p & 1+3 p+2 q \\
3 & 6+3 p & 1+6 p+3 q
\end{array}\right|\) =?

1. 0
2. 1
3. -1
4. None of these

Answer: 2. 1

Applying R₂→ (R₂– 2R₁) and R₃ -+ (R₃– 3R₁) and expandingby C₁ we get Δ=1.

Question 7.  \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|\) =?

1. (a-b)(b-c)(c-a)
2.  -(a-b)(b-c)(c-a)
3. (a- b)(b- c)(c- a)(a + b + c)
4. abc (a- b)(b- c){c- a)

Answer: 3. (a- b)(b- c)(c- a)(a + b + c)

Question 8. \(\left|\begin{array}{ccc}
1+a^2-b^2 & 2 a b & -2 b \\
2 a b & 1-a^2+b^2 & 2 a \\
2 b & -2 a & 1-a^2-b^2
\end{array}\right|=?\)

1. (1 + a²+ b²)
2. (1 + a2+ b²)²
3. (1 + a² + b²)³
4. None of these

Answer:  3. (1 + a² + b²)³

Apply R₁ → R₁ + bR₃ and R₂ → R₂ — aR₃

Question 9. \(\left|\begin{array}{lll}
\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\
\sin \beta & \cos \beta & \sin (\beta+\delta) \\
\sin \gamma & \cos \gamma & \sin (\gamma+\delta)
\end{array}\right|\)= ?

1. 0
2. 1
3. sin (a+ 8) + sin (P + 8) + sin (Y+ 8)
4.  None of these

Answer: 1. 0

Apply C₃→ C₃– (cos δ)C₁– (sinδ )C₂

Question 10. \(\left|\begin{array}{ccc}
x+y & x & x \\
5 x+4 y & 4 x & 2 x \\
10 x+8 y & 8 x & 3 x
\end{array}\right|\)

1. 0
2.  x³
3. y³
4. None of these

Answer: 2. x³

Take x common from C₂ and x common from C₃.

Apply R₃→R₃– 2R₂.

Question 11. \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|=?\)

1. (a + b + c)
2. (a + b + c)²
3.  0
4. None of these

Answer: 4. None of these

Apply C₁ → (C₁ + C₁₂+ C₃).

Question 12. \(\left|\begin{array}{ccc}
a & b & c \\
a-b & b-c & c-a \\
b+c & c+a & a+b
\end{array}\right|=?\)

1. (a³ + b³ + c³)
2.  (a+b+c)³
3. 3abc(a + b + c)
4. (a³ + b³ + c³– 3abc

Answer: 4. (a³ + b³ + c³– 3abc

Apply C₁ →(C₁ + C₂+ C₃)

Question 13. \(\left|\begin{array}{ccc}
0 & a-b & a-c \\
b-a & 0 & b-c \\
c-a & c-b & 0
\end{array}\right|=?\)

1. (a+b+c)
2. (a+b-c)
3. 1
4. 0

Answer:  4. 0

Apply R₁ → (R₁ − R₁₂) and → (R₃−R₂)

Question 14. \(\left|\begin{array}{ccc}
a+b & a & b \\
a & a+c & c \\
b & c & b+c
\end{array}\right|=?\)

1. a+b+c
2. abc
3. 4abc
4. a²b²c²

Answer: 3. 4abc

Apply R₁ →  R₁– (R₂+ R₃).

Question 15.  \(\left|\begin{array}{lll}
a & 1 & b+c \\
b & 1 & c+a \\
c & 1 & a+b
\end{array}\right|=?\)

1. (a+b+c)
2. 2(a+b+c)
3. 0
4. None of these

Answer: 3. 0

Apply C₁ →  (C₁ + C₃) and take (a+ b+ c) common from C₁

Question 16.  \(\left|\begin{array}{ccc}
a-b-c & 2 b & 2 c \\
2 a & b-c-a & 2 c \\
2 a & 2 b & c-a-b
\end{array}\right|=?\)

1. (a+b+c)
2. (a+b+c)²
3. (a+b+c)³
4. None of these

Answer: 3. (a+b+c)³

Apply Cx₁→  C₁+ C₂+ C₃

Question 17. \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
c a & c b & c^2+1
\end{array}\right|=?\)

1.  (a²+b²+c²)
2. ( 1+a²+b²+c²)
3. ( 3+a²+b²+c²)
4. None of these

Answer: 2. ( 1+a²+b²+c²)

⇒ \(\Delta=\left|\begin{array}{ccc}
a\left(a+\frac{1}{a}\right) & a b & a c \\
a b & \left(b+\frac{1}{b}\right) b & b c \\
a c & b c & \left(c+\frac{1}{c}\right) c
\end{array}\right|=(a b c) \cdot\left|\begin{array}{ccc}
a+\frac{a}{a} & a & a \\
b & b+\frac{1}{b} & b \\
c & c & c+\frac{1}{c}
\end{array}\right|\)

⇒ \(=(a b c)\left|\begin{array}{ccc}
\frac{\left(a^2+1\right)}{a} & a & a \\
b & \frac{b^2+1}{b} & b \\
c & c & \frac{c^2+1}{c}
\end{array}\right|=\left(\frac{a b c}{a b c}\right) \cdot\left|\begin{array}{ccc}
a^2+1 & a^2 & a^2 \\
b^2 & b^2+1 & b^2 \\
c^2 & c^2 & c^2+1
\end{array}\right|\)

Now apply, R₁→ R₁+R₂+R₃

Question 18.  \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right|=?\)

1.  0
2.  78
3. -39
4. 108

Answer: 1. 0

Applying C₁→(C₁-C₃) and C₂→(C₂-C₃) we get:

⇒ \(\Delta=\left|\begin{array}{lll}
46 & 21 & 219 \\
42 & 27 & 198 \\
38 & 17 & 181
\end{array}\right|=\left|\begin{array}{rrr}
4 & 21 & 9 \\
-12 & 27 & -72 \\
4 & 17 & 11
\end{array}\right|\)

C₁→(C₁-2C₂)

C₃→(C₃-10C₂)

Now , apply R₁→(R₁-R₃) and R₂→(R₂-3R₃)

Question 19. \(\left|\begin{array}{lll}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|=?\)

1. 8
2. -8
3. 16
4. 142

Answer: 2. -8

Question 20.\(\left|\begin{array}{lll}
1 ! & 2 ! & 3 ! \\
2 ! & 3 ! & 4 ! \\
3 ! & 4 ! & 5 !
\end{array}\right|=?\)

1. 2
2. 6
3. 24
4. 120

Answer: 3. 24

Question 21.\(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+y
\end{array}\right|\)

1.  (x+y)
2. (x-y)
3. xy
4.  None of these

Answer: 3. xy

Question 22. \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|=?\)

1. abc (a+b+c)
2. (a³ -b³-c³ + 3abc)
3. (-a³ -b³-c³ + 3abc)
4. None of these

Answer: 3. (-a³ -b³-c³ + 3abc)

Question 23. If a, b, cbe distinct positive real numbers, then the value of \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\) is

1. Positive
2. Negative
3. A perfect square
4. 0

Answer: 2. Negative

Δ= -(a+ b+ c)(a²+ b²+ c²-ab-bc-ca)

=- \(\frac{1}{2}\) (a+ b + c)[(a- b)²+ (b- c)²+ (c- a)²], which is negative.

Question 24. \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|=?\)

1. 0
2. abc
3. 4 a²b²c²
4. None of these

Answer: 3. 4 a²b²c²

Question 25. \(\left|\begin{array}{lll}
b c & b+c & 1 \\
c a & c+a & 1 \\
a b & a+b & 1
\end{array}\right|=?\)

1. (a-b)(b-c)(c-a)
2. – (a-b)(b-c)(c-a)
3. (a+b)(b+c)(c+a)
4.  None of these

Answer: 1. (a-b)(b-c)(c-a)

Question 26. \(\left|\begin{array}{ccc}
a^2+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|=?\)

1. (a-1)²
2. (a-1)³
3. (a-1)
4. None of these

Answer: 2. (a-1)³

Apply R₁→ (R₁ — R₂) and R₃ → (R₃ — R₂)

Question 27. \(\left|\begin{array}{lll}
\frac{1}{a} & a^2 & b c \\
\frac{1}{b} & b^2 & a c \\
\frac{1}{c} & c^2 & a b
\end{array}\right|=?\)

1. 0
2. 1
3. -1
4. None of these

Answer: 1.0

⇒ \(\Delta=\frac{1}{a b c}\left|\begin{array}{ccc}
1 & a^3 & a b c \\
1 & b^3 & a b c \\
1 & c^3 & a b c
\end{array}\right|=\left(\frac{a b c}{a b c}\right) \cdot\left|\begin{array}{ccc}
1 & a^3 & 1 \\
1 & b^3 & 1 \\
1 & c^3 & 1
\end{array}\right|=0\)

Question 28.\(\left|\begin{array}{ccc}
x+1 & x+2 & x+4 \\
x+3 & x+5 & x+8 \\
x+7 & x+10 & x+14
\end{array}\right|=?\)

1.  -2
2. 2
3. x²– 2
4. x + 2

Answer: 1. -2

Apply C₂→(C₂– C₁) and C₃ → (C₃ — C₁).

Question 29.\(\left|\begin{array}{ccc}
b+c & a & a \\
b & c+a & b \\
c & c & a+b
\end{array}\right|=?\)

1.  4abc
2.  2(a + b + c)
3.  (ab + bc + ca)
4.  None of these

Answer: 1. 4abc

Apply R₁→R₁– (R₂+R₃) and then apply R₂ → (R₂-R₁) and R₃ → (R₃-R₁).

Question 30. \(\left|\begin{array}{ccc}
a & a+2 b & a+2 b+3 c \\
3 a & 4 a+6 b & 5 a+7 b+9 c \\
6 a & 9 a+12 b & 11 a+15 b+18 c
\end{array}\right|=?\)

1.  a³
2. -a³
3. 0
4. None of these

Answer: 2. -a³

Apply R₂ → R₂ − 3R₁ and R₃→ R₃ − 6R₁.

Question 31. \(\left|\begin{array}{lll}
1 & b c & b c(b+c) \\
1 & c a & c a(c+a) \\
1 & a b & a b(a+b)
\end{array}\right|=?\)

1.  abc
2.  2 abc
3. abc(a+b+c)
4. 0

Answer: 4. 0

Apply R₂ → R₂– R₁ and R₃ → R₃-R₁.

Question 32. The value of \(\left|\begin{array}{ccc}
\cos (\theta+\phi) & -\sin (\theta+\phi) & \cos 2 \phi \\
\sin \theta & \cos \theta & \sin \phi \\
-\cos \theta & \sin \theta & \cos \phi
\end{array}\right|\) is

1.  Independent of o only
2.  Independent of o only
3.  Independent of o only
4.  Dependent of o only

Answer: 1. Independent of o only

Apply R₁ → R₁+ (sin Φ)R₂ − (cos Φ)R₃.

Take (2cosΦ) common from R₁. Now, apply R₁ →4 (R₁ +R₃).

Question 33. \(\left|\begin{array}{lll}
b+c & a & b \\
c+a & c & a \\
a+b & b & c
\end{array}\right|=?\)

1. (a + b+ c)(a- c)
2.  (a + b + c)(b- c)
3.  (a + b + c)(a-cf
4.  (a + b + c)(b- c)

Answer: 3. (a + b + c)(a-cf)

Express A as the sum of two determinants and simplify each.

Question 34. If co is a complex root of unity, then \(\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|=?\)

1. 1
2. -1
3.  0
4. None of these

Answer: 3. 0

Apply R₁→R₁+ R₂+ R₃ and use the result (1 + ω+ ω²) = 0.

Question 35. If ω is a complex cube root of unity, then the value of \(\left|\begin{array}{ccc}
1 & \omega & 1+\omega \\
1+\omega & 1 & \omega \\
\omega & 1+\omega & 1
\end{array}\right|\)

1. 2
2.  4
3.  0
4. -3

Answer: 2. 4

1 +ω+ ω² => (1 + ω) = – ω². Put (1 + ω) = -ω² and expand.

Question 36. If \(\left|\begin{array}{lll}
a+b & b+c & c+a \\
b+c & c+a & a+b \\
c+a & a+b & b+c
\end{array}\right|=k\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|, \text { then } k=?\)

1. 0
2.  1
3.  2
4. -2

Answer:  3. 2

Question 37. \(\text { The solution set of the equation }\left|\begin{array}{lll}
x & 3 & 7 \\
2 & x & 2 \\
7 & 6 & x
\end{array}\right|=0 \text { is }\)

1. {2,-3,7}
2.  {2,7,-9}
3. {-2,3,-7}
4.  None of these

Answer: 2. {2,7,-9}

Apply C₁ → C₁  −C₃ and take (x-7) common from C₁

Question 38. The solution set of the equation \(\left|\begin{array}{ccc}
x-2 & 2 x-3 & 3 x-4 \\
x-4 & 2 x-9 & 3 x-16 \\
x-8 & 2 x-27 & 3 x-64
\end{array}\right|\) is

1.  {4}
2.  {2,4}
3.  {2,8}
4.  {4,8}

Answer: 1. {4}

Apply C₂ → C₂ – 2C₁ and C₃→ C₃– 3C₁

Then, apply R₂ → R₂− R₁ and R₃→ R₃− R₁

Question 39. The solution set of the equation \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|=0 \text { is }\) = 0 is

1. {1,0}
2.  {3a,0}
3.  {a,3a}
4.  None of these

Answer: 2. {3a,0}

Apply C₁ → (C₁ + C₂+ C₃) and take (3a- x) common from C₁

Question 40. The solution set of the equation \(\left|\begin{array}{rrr}
5 & 3 & -1 \\
-7 & x & 2 \\
9 & 6 & -2
\end{array}\right|=\) =0 is

1.  {0}
2.  {6}
3.  {-6}
4.  {0,9}

Answer: 3. {-6}

Question 41. The solution set of the equation \(\left|\begin{array}{ccc}
3 x-8 & 3 & 3 \\
3 & 3 x-8 & 3 \\
3 & 3 & 3 x-8
\end{array}\right|\) = 0 is

1.  \(\left\{\frac{2}{3}, \frac{8}{3}\right\}\)
2.  \(\left\{\frac{2}{3}, \frac{11}{3}\right\}\)
3.  \(\left\{\frac{3}{2}, \frac{8}{3}\right\}\)
4. None of these

Answer: 2. \(\left\{\frac{2}{3}, \frac{11}{3}\right\}\)

Question 42. The vertices of a A ABC are A(-2, 4), B(2, -6) and C(5, 4). The area of Δ ABC is

1.  17.5 sq units
2.  35 sq units
3. 32 sq units
4. 28 squats

Answer: 2. 35 sq units

⇒ \(\Delta=\frac{1}{2}\left|\begin{array}{rrr}
-2 & 4 & 1 \\
2 & -6 & 1 \\
5 & 4 & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{rrr}
-2 & 4 & 1 \\
4 & -10 & 0 \\
7 & 0 & 0
\end{array}\right|\)

= 35 sq units

Question 43. If the points A(3, -2), B(k, 2) and C(8, 8) are collinear, then the value of k is

1. 2
2.  -3
3. 5
4. -4

Answer: 3. 5

⇒ \(\text { If } \Delta=\left|\begin{array}{rrr}
3 & -2 & 1 \\
k & 2 & 1 \\
8 & 8 & 1
\end{array}\right| \text {, then we must have } \Delta=0 \text {. }\)

Relations And Functions

Relations And Functions Exercise 1A Review Of Facts And Formulae

1. A relation R in a set A is a subset of A x A

If (a, b) ∈ R, then a R b.

If (a, b) ∉ R, then a is not related to b.

Domain (R) = {a: (a, b) ∈ R] and Range (R) = {b: (a, b) ∈ R}.

2. Let R be a relation on a non-empty set A. Then, R is said to be:

1. Reflexive, if (a, a) ∈ R for each a ∈ A

i.e., if a R a for each a ∈ R.

2. Symmetric, if (a, b) ∈ R ⇒ (b, a) ∈ R

i.e., a R b ⇒ b Ra

3. Transitive, if (a, b) e R and (b, c)e R ⇒ (a, c) ∈ R

Read and Learn More WBCHSE Solutions For Class 12 Maths

i.e., a R b, b R c ⇒  a R c.

3. A relation R on a set A is said to be an equivalence relation if it is reflexive, symmetric, and transitive.

4. Equivalence class, determined by an element a is defined as a

⇒ [a] = [b ∈ A : (a, b) ∈ R}

Relations And Functions Exercise 1A Multiple Choice Questions

Question 1. Let A = {1, 2, 3} and let R = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 2), (1, 2)}. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 2. Reflexive and transitive but not symmetric

R is reflexive and transitive but not symmetric

 

Question 2. Let A = {a, b, c} and let R= {(a, a), (a, b), (b, a)}. Then, Ris

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 3. Symmetric and transitive but not reflexive

R is symmetric and transitive but not reflexive

Question 3. Let A = {1,2, 3} and let R= {(1,1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}. Then, R is

1. Reflexive and symmetric but not transitive
2. Symmetric and transitive but not reflexive
3. Reflexive and transitive but not symmetric
4. An equivalence relation

Answer: 1. Reflexive and symmetric but not transitive

(1,2)∈ R and (2,3) ∈ R, But, (1,3) ∉ R So, R is not transitive.

Question 4. Let S be the set of all straight lines in a plane. Let R be a relation on S Defined by a R b ⇔ a ⊥ b. Then, R is

1. Reflexive but neither symmetric nor transitive
2. Symmetric but neither reflexive nor transitive
3. Transitive but neither reflexive nor symmetric
4. An equivalence relation

Answer: 2. Symmetric but neither reflexive nor transitive

a ⊥ b is not true. So R is not reflexive

a ⊥ b and b ⇒ b ⊥ a is always true.

Question 5. Let S be the set of all straight lines in a plane. Let R be a relation on S Defined by a R b ⇔ a || b. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 6. Let Z be the set of all integers and let R be a relation on Z defined by a R b o (a-b) is divisible by 3. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 7. Let R be a relation on the set N of all natural numbers, defined by a R b ⇔  a is a factor of b. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 2. Reflexive and transitive but not symmetric

Question 8. Let Z be the set of all integers and let R be a relation on Z defined by a R b ⇔  a>b. Then, R is

1. Symmetric and transitive but not reflexive
2. Reflexive and symmetric but not transitive
3. Reflexive and transitive but not symmetric
4. An equivalence relation

Answer: 3. Reflexive and transitive but not symmetric

Question 9. Let S be the set of all real numbers and let R be a relation on S defined by a R b ⇔ IaI ≤  b. Then, R is

1. Reflexive but neither symmetric nor transitive
2. Symmetric but neither reflexive nor transitive
3. Transitive but neither reflexive nor symmetric
4. None of these

Answer: 3. Transitive but neither reflexive nor symmetric

⇒  I-3I ≤ -3 is not true. So, R is not reflexive.

⇒ I-1I ≤ 1 ⇒ (-1) R1. But I-1I ≤ 1 is not true.

∴ R is not symmetric

3. a R b, b R c ⇒ IaI ≤  b and IbI ≤  c ⇒ IaI ≤  c.

Question 10. Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ Ia-bI ≤ 1. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 1. Reflexive and symmetric but not transitive

⇒  Ia- aI = 0 ≤ 1 is always true.

⇒  a R b ⇒ Ia- b I≤1 ⇒ I-(a-b)I<1 ⇒ Ib-aI <1 ⇒  b R a.

⇒  2 R 1 and 1 R \(\frac{1}{2}\).

But, 2 is not related to \(\frac{1}{2}\).  So, R is not transitive.

Question 11. Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔(1 + ab) > 0. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. None of these

Answer: 1. Reflexive and symmetric but not transitive

a R a, since (1 + a²) > 0.

a R b ⇒ (1 + ab) > 0 ⇒ (1 + ba) > 0 ⇒b R a.

Let a = \(\frac{-1}{2}\), b = \(\frac{-1}{2}\) and c = R.

Then, a R b and b R c. But, a is not related to c.

Question 12. Let S be the set of all triangles in a plane and let R be a relation on S defined by Δ₁S Δ₂ ⇔ Δ₁ ≡ Δ₂. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 13. Let S be the set of all real numbers and let R be a relation on S defined by a R b ⇔ a² + b² = 1. Then, R is

1. Symmetric but neither reflexive nor transitive
2. Reflexive but neither symmetric nor transitive
3. Transitive but neither reflexive nor symmetric
4. None of these

Answer: 1. Symmetric but neither reflexive nor transitive

(l² + 1²) ≠ 1. So,1 R 1 is not true.

a R b ⇒ a²+ b² =1 ⇒ b² + a² =1 ⇒  b R a.

1 R 0 and 0 R 1. But,1 is not related to 1.

Question 14. Let JR be a relation on N x N, defined by (a, b) R (c, d)⇔ a+d= b+ c. Then, R is 

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 15. Let A be the set of all points in a plane and let O be the origin. Let R = {(P, Q): OP= OQ}. Then, R is

1. Reflexive and symmetric but not transitive
2. Reflexive and transitive but not symmetric
3. Symmetric and transitive but not reflexive
4. An equivalence relation

Answer: 4. An equivalence relation

Question 16. Let Q be the set of all rational numbers and * be the binary operation, defined by a * b = a + 2b, then

1. * is commutative but not associative
2. * is associative but not commutative
3. * is neither commutative nor associative
4. * is both commutative and associative

Answer: 3. * is neither commutative nor associative

Question 17. Let a * b = a + ab for all a, b ∈ Q. Then,

1. * Is not a binary composition
2. * Is not commutative
3. * Is commutative but not associative
4. * Is both commutative and associative

Answer: 2. * Is not commutative

Question 18. Let Q⁺ be the set of all positive rationals. Then, the operation * on Q⁺ defined by a * b \(=\frac{a b}{2}\) for all a, b ∈ Q⁺ is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 4. Both commutative and associative

Question 19. Let Z be the set of all integers and let a*b = a – b + ab. Then, * is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 3. Neither commutative nor associative

Question 20. Let Z be the set of all integers. Then, the operation * on Z defined by a*b= a + b – ab is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 4. Both commutative and associative

Question 21. Let Z+be the set of all positive integers. Then, the operation * on Z+ defined by a * b = a^b is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 3. Neither commutative nor associative

Question 22. Define * on Q- {-1} by a * a+ b+ ab. Then, * on Q- {-1} is

1. Commutative but not associative
2. Associative but not commutative
3. Neither commutative nor associative
4. Both commutative and associative

Answer: 4. Both commutative and associative

Relations And Functions Exercise 1B Review Of Facts And Formulae

1. Function: Let A and B be two non-empty sets. Then, a rule / which associates to each x e A, a unique element fix) of B, is called a function from A to B, and we write, \(f\): A → B.

Dom (f) = {x ∈ A :f(x) ∈ B}, range  (f)= {f(x): x ⇒ A}.

2. Let f: A→ B. Then

⇒ f is one-one ⇔ {f(x₁) =f(x₂) ⇒ x₁ = x₂}

⇒  f is onto ⇔ range (f) = B.

3. Two functions/ and g are said to be equal if they have the same domain and f(x)=g(x) ∀ x.

4. Invertible functions: A function is said to be invertible if/is one-one and

onto.f{x) = y ⇒  f^-1(y) = x.

5. Composite of two functions:

⇒ Let/: A → B and g: B→ C, then g o f: A →  C is called the composite of f and g.

⇒ Dom (g o f) = dom (f)

g o f is defined only when range (f) ⊆ dom (g)

⇒ Dom (f o g) = dom(g).

f o g is defined only when range (g) ⊆ dom (f)

Relations And Functions Exercise 1B Multiple Choice Questions

Question 1. f : N → N : f(x) = 2x is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 2. One-one and into

⇒ 2x =  3

⇒ x =  \(\frac{3}{2}\)

∴ f is into.

Question 2. f : N → N : f(x) = x²+ x+1 is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 2. One-one and into

⇒  f(x₁) = f(x₂)

= x²₁+x₁ +1

= x²₂+x₂ +1

= (x²₁– x²₂) + (x₁-x₂) = 0

⇒ (x₁-x₂)(x₁+x₂+1) = 0

= x₁– x₂= 0

=  x₁ = x₂.

∴ f is one-one.

f(x) =1 ⇒  x²+ x+1 =1 ⇒  x(x+ 1) = 0 ⇒ x= 0 or x=-1.

And, none of 0 and -1 is in N. So, f is into.

Question 3. f: R→ R : f(x) = x² is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 4. Many-one and into

Question 4. f: R→ R : f(x) = x³ is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 1. Many-one and into

Question 5. f : R⁺→+R⁺: f(x) = e^x is

1. Many-one and into
2. Many-one and onto
3. One-one and into
4. One-one and onto

Answer: 4. One-one and onto

⇒ f(x₁) = f(x₂) => e^x₁ = e^x₂

⇒ x₁ = x₂

∴ f is one-one.

For each x ∈ R⁺ ∃ log x ∈ R⁺ s.t. f(log x)= x.

So,/is onto.

Question 6. \(f:\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \rightarrow[-1,1]: \) f(x) = sin x is

1. One-one and into
2. One-one and onto
3. Many-one and into
4. Many-one and onto

Answer: 2. One-one and onto

Question 7. f : R → R :f(x) = cos x is

1. One-one and into
2. One-one and onto
3. Many-one and into
4. Many-one and onto

Answer: 3. Many-one and into

Cos (2π- θ) = cos θ ⇒ f is Many-one.

Range (f) = [-1, 1] ⊂ R =>  f is into.

Question 8. f : C→ R : f(z) = IzI is

1. One-one and into
2. One-one and onto
3. Many-one and into
4. Many-one and onto

Answer: 3. Many-one and into

i ≠-i. But f(i) = f(-i)=1. So, fis many-one.

-1 ∈ R having no pre-image in C. So, fis into.

Question 9. Let A = R- {3} and B =R- {1}. Then f: A → B : f(x) = \(\frac{(x-2)}{(x-3)}\) is 

1. One-one and into
2. One-one and onto
3. Many-one and into
4. Many-one and onto

Answer: 2. One-one and onto

⇒ f(x₁)= f(x₂)

⇒ \(\frac{\left(x_1-2\right)}{\left(x_1-3\right)}=\frac{\left(x_2-2\right)}{\left(x_3-3\right)}\)

⇒ x₁= x₂, So f is one one.

Let \(\frac{x-2}{x-3}\). Then, x\(=\frac{3 y-2}{y-1}\). Clearly y ≠1 and x≠3

Question 10.  Let f: N → N: f(n)=

\(\frac{1}{2}\)(n+1) when n is odd

\(\frac{n}{2}\),when n is even

1. One-one and into
2. One-one and onto
3. Many-one and onto
4. Many-one and into

Answer: 4. Many-one and into

f (1) = f(2) shows that/is many-one.

If n is odd, then (2n- 1) is odd, and f(2n- 1) =n.

If n is even, then 2n is even, and f (2n) = n.

∴ f is onto.

Question 11. Let A and B be two non-empty sets and let f: (A x B) → (B x A): f(a, b) = (b, a). Then, f is

1. One-one and onto
2. One-one and into
3. Many-one and onto
4. Many-one and into

Answer: 1. One-one and onto

Question 12. Let f: Q→ Q: f(x) = (2x + 3). Then,f^-1(y) = ?

1. (2y- 3)
2. \(\\)
3. \(\frac{1}{2}\)(y-3)
4. None of these

Answer: 3. \(\frac{1}{2}\)(y-3)

y = 2x+3

x = \(\frac{1}{2}(y-3)\)

⇒ \(f^{-1}(y)=\frac{4 y}{(4-3 y)}\)

Question 13.\(\text { Let } f: R-\left\{\frac{-4}{3}\right\} \rightarrow R-\left\{\frac{4}{3}\right\}: f(x)=\frac{4 x}{(3 x+4)} \text {. Then, } f^{-1}(y)=\text { ? }\)

1. \(\frac{4 y}{(4-3 y)}\)
2. \(\frac{4 y}{(4+3 y)}\)
3. \(\frac{4 y}{(3 y-4)}\)
4. None of these

Answer: 1. \(\frac{4 y}{(4-3 y)}\)

⇒ \(y\frac{4 x}{3 x+4}\)

⇒ \(x\frac{4 y}{(4-3 y)}\)

⇒ \(f^{-1}(y)=\frac{4 y}{(4-3 y)}\)

Question 14. Let f : N→X : f(x) = 4x² + 12x+ 15. Then f^-1(y) =?

1. \(\frac{1}{2}(\sqrt{y-4}+3)\)
2. \(\frac{1}{2}(\sqrt{y-6}-3)\)
3. \(\frac{1}{2}(\sqrt{y-4}+5)\)
4. None of these

Answer: 2. \(\frac{1}{2}(\sqrt{y-6}-3)\)

y = 4x² +12x+15

= (2x+3)² +6

⇒ \(x\frac{1}{2}(\sqrt{y-6}-3)\)

⇒ \(f^{-1}(y)=\frac{1}{2}(\sqrt{y-6}-3)\)

Question 15. \(\text { If } f(x)=\frac{(4 x+3)}{(6 x-4)}, x \neq \frac{2}{3} ; \text { then }(f \circ f)(x)=\text { ? }\)

1. x
2. (2x-3)
3. \(\frac{4 x-6}{3 x+4}\)
4. None of these

Answer: 1. x

f(x)= \(\frac{4 x+3}{6 x-4}\) = y(say)

Then f(y)= \(\frac{4 y+3}{6 y-4}\)

= \(=\frac{4\left(\frac{4 x+3}{6 x-4}\right)+3}{6\left(\frac{4 x+3}{6 x-4}\right)-4}\)

= \(=\frac{34 x}{34}\)

= x

⇒ f[f(x)] = x ⇒ (f o f) (x) = x

Question 16. If f(x) = (x² — 1) and g(x) = (2x + 3), then (g o f)(x) = ?

1. (2X²+ 3)
2. (3X² + 2)
3. (2x² + 1)
4. None of these

Answer: 3. (2x²+ 1)

(g o f )(x) = g [f(x)] = g(x²-1)

= 2(x²-1)+3 = (2x²+1)

Question 17. \(\text { If } f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}, \text { then } f(x)=?\)

1. x²
2. x-1
3. x-2
4. None of these

Answer: 3. x-2

Let  \(x+\frac{1}{x}\)= z Then,

f(z) = \(f\left(x+\frac{1}{x}\right)=\left(x^2+\frac{1}{x^2}\right)=\left(x+\frac{1}{x}\right)^2-2\)= (z-2)

f(z)= (x²-2)

Question 18. \(\text { If } f(x)=\frac{1}{(1-x)^{\prime}} \text {, then }(f \circ f \circ f)(x)=\text { ? }\)

1. \(\frac{1}{(1-3 x)}\)
2. \(\frac{x}{(1+3 x)}\)
3. x
4. None of these

Answer: 3. x

(f o f)(x) = f{f(x)}

= \(f\left(\frac{1}{1-x}\right)=\frac{1}{\left(1-\frac{1}{1-x}\right)}\)

= \(\frac{1-x}{-x}=\frac{x-1}{x}\)

{f o(f o f)}(x) ={f o(f o f)(x)}]= \(=f\left(\frac{x-1}{x}\right)=\frac{1}{1-\frac{x-1}{x}}=x^1/3\)

Question 19. \(\text { If } f(x)=\sqrt[3]{3-x^3}, \text { then }(f \circ f)(x)=\text { ? }\)

1. x1/3
2. x
3. (1-x1/3)
4. None of these

Answer: 2. x

(fof)(x) =f{f(x)}- {(3- x3) ^1/3} = f/(y), where y= (3- x³) ^1/3

= (3- y³)^1/3= [3- {3- x³}]^1/3= (x³)^1/3= x

Question 20. If f(x) = x²– 3x + 2, then (f o f)(x) = ?

1. x
2. x⁴-6x³
3. x⁴– 6x³+10x²
4. None of these

Answer: 4. None of these

(fof)(x) =f{f(x)} =f{x²– 3x + 2) =f(y), where y= (x²– 3x+ 2)

= y²– 3y+ 2= (x²– 3x+ 2)²– 3(x²– 3x+ 2) + 2= (x⁴– 6x³ + 10x²– 3x).

Question 21. If f(x) = 8x³ and g(x) = x^1/3, then (g o f)(x) = ?

1. x
2. 2x
3. \(\frac{x}{2}\)
4. 3×2

Answer: 2. 2x

(g of)(x) = g{f(x)] = g(8x³) = (8x³)^1/3= 2x.

Question 22. If f (x) = x², g(x) = tan x and h(x) = log x, then {ho(gof}}\(\left(\sqrt{\frac{\pi}{4}}\right)\)

1. 0
2. 1
3. \(\frac{1}{x}\)
4. \(\frac{1}{2} \log \frac{\pi}{4}\)

Answer: 1. 0

{ho(g o f)}(x) = (ho g){f(x)} = (h o g)(x²)

= h{g(x²)} = h(tan x²) = log(tan x²)

∴ \(\{h \circ(g \circ f)\} \sqrt{\frac{\pi}{4}}=\log \left(\tan \frac{\pi}{4}\right)\)

= log 1

= 0

Question 23. If/= {(1, 2), (3, 5), (4, 1)} andg= {(2, 3), (5, 1), (1, 3)}, then (g o f) = ?

1. {(3,1), (1,3), (3, 4)}
2. {(1,3), (3, 1), (4, 3)}
3. {(3,1), (1,3), (3, 4)}
4. {(2, 5), (5, 2), (1,5)}

Answer: 2. {(1,3), (3, 1), (4, 3)}

⇒ Dom (gof) = Dom (f) = {1, 3, 4}

⇒ (g o f )(l) =g{f(l)} = g(2) = 3,(g of)(3) =g{f(3)} =g(5) =1

⇒ (g o f)(4) = g[f(4)) =g(1) = 3

⇒ g o f= {(1/3), (3, 1), (4, 3)}.

Question 24. Let/(x) = \(\sqrt{9-x^2}\). Then, dom (f) =?

1. [-3, 3]
2. (-∞, -3]
3. [3, ∞)
4. (-∞, -3] u (4, ∞)

Answer: 1. [-3, 3]

f(x) is defined only when 9- x² > 0 => x² <9 =» -3 < x < 3.

∴ Dom (f) =  [-3, 3].

Question 25. Let f(x) \(\sqrt{\frac{x-1}{x-4}}\). Then, dom (f) = ?

1. [1,4)
2. [1,4]
3. (-∞,4]
4. (-∞,1]∪ (4,∞)

Answer: 4. (-∞,1]∪ (4,∞)

f(x) is defined when-4≠ 0 and \(\frac{x-1}{x-4}\)

⇒ x ≠ 4 and (x ≥ 4 or x<l) => (x>4 or x<l)

⇒  Dom (f) = (-∞, 1]u (4, ∞).

Question 26. Let f(x) \(f(x)=e^{\sqrt{x^2-1}}\). Then, dom (f) =?

1. (∞-,1]
2. [-1,∞)
3. (1,∞)
4. (-∞,-1](1,∞)

Answer: 3. (1,∞)

f(x) is defined only when (x²-1) > 0 and (x- 1) > 0

⇒ (x – l)(x+ 1) ≥ 0 and (x- 1) > 0 => x+1 > 0 and x-1 > 0⇒ x >1

∴ dom (f) = (1, ∞).

Question 27. Let f(x) = \(\frac{x}{\left(x^2-1\right)}\) . Then, dom (f) =?

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

Answer: 4. R- {-1,1}

f(x) is not defined when (x- 1) = 0, i.e., when (x- l)(x+ 1) = 0,

i.e., when x=1 or x= -1.

dom (f) = R—{1, -1}.

Question 28. Let f(x)= \(\). Then ,dom(f) = ?

1. (-1,1)
2. [-1/ 1]
3. [-l, 1]- {0}
4. None of these

Answer: 3. [-l, 1]- {0}

⇒ \(\frac{\sin ^{-1} x}{x}\) is defined only when x≠ 0 and x ∈ [-1,1].

∴ dom (f)= [-1,1]- {0}.

Question 29. Let f(x) = cos^-12x. then, dom (f) =?

1. [-1,1]
2. \(\left[\frac{-1}{2}, \frac{1}{2}\right]\)
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
4. \(\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]\)

Answer: 2. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

sin^-12x is defined only when 2x ∈ [-1, 1] => x \(\left[\frac{-1}{2}, \frac{1}{2}\right]\)

Question 30. Let f(x) = Cos^-1(3x-1) . Then, dom (f) =?

1. \(\left(0, \frac{2}{3}\right)\)
2. \(\left[0, \frac{2}{3}\right]\)
3. \(\left[\frac{-2}{3}, \frac{2}{3}\right]\)
4. None of these

Answer: 2. \(\left[0, \frac{2}{3}\right]\)

cos^-1 (3x- 1) is defined only when (3x-1) ∈ [-1, 1]

⇒  3x ∈ [0, 2] => x ∈ \(\left[0, \frac{2}{3}\right]\)

⇒ dom(f)= \(\left[0, \frac{2}{3}\right]\)

Question 31. Let f(x)= \(\sqrt{\cos x}\). Then, dom (f) = ?

1. \(\left[0, \frac{\pi}{2}\right]\)
2. \(\left[\frac{3 \pi}{2}, 2 \pi\right]\)
3. \(\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]\)
4. None of These

Answer: 3. \(\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]\)

f(x) is defined only when cos x > 0.

⇒ x liesin 1st or 4th quadrant.

⇒  \(=\left[0, \frac{\pi}{2}\right] \cup\left[\frac{3 \pi}{2}, 2 \pi\right]\)

Question 32. Let f(x) = \(\sqrt{\log \left(2 x-x^2\right)}\). then, dom (f) = ?

1. (0,2)
2. [1,2]
3. (-,1)
4. None of these

Answer: 4. None of these

f (x) is defined only when log (2x- x²) > 0

⇒  (2x- x²) >1 => (1 + x²– 2x) < 0

⇒  (1- x)² < 0

⇒  (1- x) = 0

⇒   x=1.

⇒ dom (f) = {1}.

Question 33. Let f(x) = x². Then, dom (/) and range (/) are respectively

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

Answer: 3. R and R⁺

f(x) = x² is clearly defined for each x ∈ R.

So, dom (f)=R. y= x² x=± \(x= \pm \sqrt{ } y .\)

When y < 0, there is no real value of x. So, y ≥ 0.

∴Range (f) = R.

Question 34. Let f(x) = x³. Then, dom (f) and range (f) are respectively.

1. R and R
2. R⁺ and R⁺
3. R and R⁺
4. R⁺ and R

Answer: 1. R and R

f (x) = x³ is defined for each x ∈ R.

So, dom (f)=R.

For each y∈ R,y^1/3 ∈ R and so x= is real.

∴ range (f) =R.

Question 35. Let f(x)= \(\log (1-x)+\sqrt{x^2-1}\).Then ,dom(f) = ?

1. (1,∞)
2. (-∞,-1)
3. [-1,1]
4. (0,1)

Answer: 2. (-∞,-1)

Let f (x) = g{x) + h(x), where g(x) = log (1- x) and h(x) = \(\sqrt{x^2-1}\)

g(x) is defined only whenl-x>0 => x<l.So, dom (g) = (∞, 1)

h(x) is defined only when x2-1 > 0 => x >1 or x <-1

∴ dom (h) = (-∞, -1]∪[1, ∞)

∴ dom (f) = dom (g)∩ dom(h) = (-∞, -1].

Question 36. Let \(=\frac{1}{\left(1-x^2\right)}\). Then ,range(f)= ?

1. [-∞,1)
2. [1,∞)
3. [-1,1]
4. None of these

Answer: 2. [0,1)

y =  \(\frac{1}{\left(1-x^2\right)}\)

x = \(\sqrt{1-\frac{1}{y}}\)

Clearly, x is not defined when y = 0 or 1\(-\frac{1}{y}\) < 0, i.e., y = 0 or y < 1.

∴ range (f) = [1, ∞).

Question 37. Let f(x)\(=\frac{x^2}{\left(1+x^2\right)}\)

1. [1,∞]
2. [0,1)
3. [-1,1]
4. (0,1)

Answer: 2. [0,1)

y = \(\frac{x^2}{\left(1+x^2\right)} \Rightarrow x=\sqrt{\frac{y}{1-y}}\)

Clearly, x is defined only when \(\frac{y}{(1-y)}\) and(1-y)≠ 0, i.e., when 0 ≤ y ≤ 1.

So, range (f) = [0, 1).

Question 38. The range of f(x) = \(x+\frac{1}{x}\) is

1. [-2,2]
2. [2,∞)
3. (-∞,-2)
4. None of these

Answer: 4. None of these

y\(=\frac{x^2+1}{x}\)

⇒ x²– xy + 1= 0

x= \(\frac{y \pm \sqrt{y^2-4}}{2}\)

Question 39. The range of f(x) = a^x, Where a> 0 is

1. ]-∞,0]
2. ]∞-,0)
3. [0,∞)
4. (0,∞)

Answer: 4. (0,∞)

Clearly, a^x> 0 whatever may be the value of x.

∴ Range (f) = (0, ∞)

Relations And Functions Exercise 1C Results At A Glance

1. sin^-1x = θ => x = sin θ

cos^-1x = θ => x = cos θ

tan^-1x = θ =» x = tan θ

2. Domain and range of inverse functions

 

3. sin (sin^-1x) = x, if-1 ≤ x ≤ 1

cos (cos^-1x) = x, if-1 ≤ x ≤ 1

tan (tan^-1x) = x, if ∞ < x < ∞

4.

sin^-1(sin x) = x, if x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

cos ^-1(cos x) = x, if x ∈ [0, π]

tan^-1(tan x) = x, if x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

5.

sin^-1x = cosec^-1\(\left(\frac{1}{x}\right)\) , x≠0

cos^-1x = sec^-1\(\left(\frac{1}{x}\right)\) , x≠0

tan^-1x= cot^-1\(\left(\frac{1}{x}\right)\) , x≠0

cosec^-1x= sin^-1 \(\left(\frac{1}{x}\right)\),(x ≤ -1 or x ≥ 1)

sec^-1x= cos^-1\(\left(\frac{1}{x}\right)\), when (x ≤1 or x ≥ 1)

cot^-1x = tan^-1\(\left(\frac{1}{x}\right)\), when x > 0

6. sin^-1(-x) = -sin^-1x

cos^-1(-x) = 7i- cos^-1x

tan^-1(-x) = -tan^-1x

cot^-1(-x) =n- cot^-1x

sec^-1(-x) =n- sec^-1x

cosec^-1(-x) = -cosec^-1x

7. sin^-1-x + cos^-1-x =\(\frac{\pi}{2}\) -1 ≤ x ≤1

tan^-1-x + cot^-1-x = \(\frac{\pi}{2}\) , x ≠ 0

sec ^-1x + cosec ^-1x = \(\frac{\pi}{2}\),  (x ≤ -1 or x ≥ 1)

8.

1. For 0 <x< 1, we have

sin^-1-x=  \(\cos ^{-1} \sqrt{1-x^2}\)

= \(\tan ^{-1} \frac{x}{\sqrt{1-x^2}}=\cot ^{-1} \frac{\sqrt{1-x^2}}{x}\)

= \(\sec ^{-1} \frac{1}{\sqrt{1-x^2}}\)

= cosec^-1 \(\frac{1}{x}\)

2. Cos^-1-x= \(\sin ^{-1} \sqrt{1-x^2}\)

= \(\tan ^{-1} \frac{\sqrt{1-x^2}}{x}\)

= \(\cot ^{-1} \frac{x}{\sqrt{1-x^2}}\)

= \(\sec ^{-1} \frac{1}{x}\)

= cosec^-1 \(\frac{1}{\sqrt{1-x^2}}\)

3. For x>0, We Have

tan^-1x= \(\sin ^{-1} \frac{x}{\sqrt{1+x^2}}\)

= \(\cos ^{-1} \frac{1}{\sqrt{1+x^2}}=\cot ^{-1} \frac{1}{x}\)

= \(\sec ^{-1} \sqrt{1+x^2}\)

= cosec^-1 \(\frac{\sqrt{1+x^2}}{x}\)

9. 

2sin^-1x = sin^-1(\(2 x \sqrt{1-x^2}\)) = cos^-1 (1- 2x²)

2cos^-1x = cos^-1(2x²– 1) = sin ^-1\(\left(2 x \sqrt{1-x^2}\right.\))

2tan^-1x= tan^-1\(\left(\frac{2 x}{1-x^2}\right)\)  = cos^-1\(\left(\frac{1-x^2}{1+x^2}\right)\)

10.

3sin^-1x = sin^-1(3x- 4x³)

3cos^-1x = cos^-1(4x³– 3x)

3tan^-1x = tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\)

11.

When x > 0, y > 0 and xy < 1, we have

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

When x > 0, y > 0 and xy > 1, we have

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

12.

sin^–1x + sin^–1y = sin^–1(\(x \sqrt{1-y^2}+y \sqrt{1-x^2}\))

sin^–1x – sin^–1y = sin^–1(\(x \sqrt{1-y^2}-y \sqrt{1-x^2}\))

cos^–1x + cos^–1y= cos^–1(\(x y-\sqrt{1-x^2} \cdot \sqrt{1-y^2}\))

cos^–1x – cos^–1y = cos^–1(\(x y+\sqrt{1-x^2} \cdot \sqrt{1-y^2}\))

cot ^–1x+ cot ^–1x= cot^–1\(\left(\frac{x y-1}{x+y}\right)\)

cot ^–1x- cot ^–1y = cot^–1\(\left(\frac{x y+1}{x-y}\right)\)

Relations And Functions Exercise 1C Multiple Choice Questions

Question 1. The principal value of \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

1. \(\frac{\pi}{6}\frac{\pi}{6}\)
2. \(\frac{5 \pi}{6}\)
3. \(\frac{7 \pi}{6}\)
4. None of these

Answer: 1. \(\frac{\pi}{6}\frac{\pi}{6}\)

1. Let cos^-1\(\left(\frac{\sqrt{3}}{2}\right)\), where x ∈ [0, π].

Then, cos x=\(\frac{\sqrt{3}}{2}\)= cos\(\frac{\pi}{6}\)

⇒  x= \(\frac{\pi}{6}\)

Question 2. The principal value of cosec^-1(2) is

1. \(\frac{\pi}{3}\)
2. \(\frac{\pi}{6}\)
3. \(\frac{2 \pi}{3}\)
4. \(\frac{5 \pi}{6}\)

Answer: 2. \(\frac{\pi}{6}\)

Let cosec^-1(2) = x, where x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)-{0}

Then, cosec x= 2cosec \(\frac{\pi}{6}\)

⇒ x= \(\frac{\pi}{6}\)

Question 3. The principal value of \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)

1. \(\frac{-\pi}{4}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{3 \pi}{4}\)
4. \(\frac{5 \pi}{4}\)

Answer: 3. \(\frac{3 \pi}{4}\)

Let cos^-1 \(\left(\frac{-1}{\sqrt{2}}\right)\) = x, where x∈ [0,π]

Then,cos x = \(\frac{-1}{\sqrt{2}}\)

=  \(-\cos \frac{\pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right)\)

= \(\cos \frac{3 \pi}{4} \Rightarrow x=\frac{3 \pi}{4}\)

Question 4. The principal value of \(\sin ^{-1}\left(\frac{-1}{2}\right)\) is

1. \(\frac{-\pi}{6}\)
2. \(\frac{5 \pi}{6}\)
3. \(\frac{7 \pi}{6}\)
4. None of these

Answer: 1. \(\frac{-\pi}{6}\)

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

Then, sin x  = \(\frac1{-1}{2}\)

= \(-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right)\)

⇒ x= \(frac{-\pi}{6}\)

Question 5. The principal value of \(\cos ^{-1}\left(\frac{-1}{2}\right)\) is

1. \(\frac{-\pi}{3}\)
2. \(\frac{2 \pi}{3}\)
3. \(\frac{4 \pi}{3}\)
4. \(\frac{\pi}{3}\)

Answer: 2. \(\frac{2 \pi}{3}\)

Let cos^-1 \(\frac{-1}{2}\) = x where x∈[0,π]

Then, cos x  = \(\frac{-1}{2}\)

= \(-\cos \frac{\pi}{3}\)

= \(\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3}\)

x = \(\frac{2 \pi}{3}\)

Question 6. The principal value of tan \(\tan ^{-1}(-\sqrt{3})\)

1. \(\frac{2 \pi}{3}\)
2. \(\frac{4 \pi}{3}\)
3. \(\frac{-\pi}{3}\)
4. None of these

Answer: 3. \(\frac{-\pi}{3}\)

Let tan^-1\(-\sqrt{3}\)= x, where x ∈\(\left(\frac{-\pi}{2},\frac{\pi}{2}\right)\)

Then, tan x= \(-\sqrt{3}\)

= \(-\tan \frac{\pi}{3}=\tan \left(\frac{-\pi}{3}\right)\)

⇒ \(x=\frac{-\pi}{3}\)

Question 7. The principal value of cot^-1(-1) is

1. \(\frac{-\pi}{4}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{5 \pi}{4}\)
4. \(\frac{3 \pi}{4}\)

Answer: 4. \(\frac{3 \pi}{4}\)

Let co^-1(-1) = x, where x ∈ [0, π].

Then, cot x=-1

= \(-\cot \frac{\pi}{4}=\cot \left(\pi-\frac{\pi}{4}\right)\)

= \(\cot \frac{3 \pi}{4}\)

⇒ x= \(\frac{-\pi}{3}\)

Question 9. The principal value of cosec^-1 \((-\sqrt{2})\) is

1. \(\frac{-\pi}{4}\)
2. \(\frac{3 \pi}{4}\)
3. \(\frac{5 \pi}{4}\)
4. None of these

Answer: 1. \(\frac{-\pi}{4}\)

Let cosec\((-\sqrt{2})\)= x, where x ∈\(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)– {0}

Then , cosec x \(-\sqrt{2}\)

= cosec \(\frac{-\pi}{4}\)

= cosec \(\frac{-\pi}{4}\)

x = \(\frac{-\pi}{4}\)

Question 10. The principal value of \(\cot ^{-1}(-\sqrt{3})\)

1. \(\frac{-\pi}{6}\)
2. \(\frac{-\pi}{6}\)
3. \(\frac{7 \pi}{6}\)
4. \(\frac{5 \pi}{6}\)

Answer: 4. \(\frac{5 \pi}{6}\)

Let cot\((-\sqrt{3})\)= x, where x [0,π]

Then , cot x \(-\sqrt{3}\)

= -cot \(\frac{\pi}{6}\)

\(=\cot \left(\pi-\frac{\pi}{6}\right)=\cot \frac{5 \pi}{6}\)

x= \(\frac{5 \pi}{6}\)

Question 11. The value of \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\)

1. \(\frac{2 \pi}{3}\)
2. \(\frac{5 \pi}{3}\)
3. \(\frac{\pi}{3}\)
4. None of these

Answer: 3. \(\frac{\pi}{3}\)

Let sin^-1\(\left(\sin \frac{2 \pi}{3}\right)\) = x , where x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

Then,sin x = \(\sin \frac{2 \pi}{3}=\sin \left(\pi-\frac{\pi}{3}\right)=\sin \frac{\pi}{3}\)

x= \(\frac{\pi}{3}\)

Question 12. The value of \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)\)

1. \(\frac{13 \pi}{6}\)
2. \(\frac{7 \pi}{6}\)
3. \(\frac{5 \pi}{6}\)
4. \(\frac{\pi}{6}\)

Answer: 4. \(\frac{\pi}{6}\)

Let cos \(\left(\cos \frac{13 \pi}{6}\right)\)= x, where x [0,]

Then,cos x = \(\cos \frac{13 \pi}{6}=\cos \left(2 \pi+\frac{\pi}{6}\right)\)

x= \(\frac{\pi}{6}\)

Question 13. The value of \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)\)

1. \(\frac{7 \pi}{6}\)
2. \(\frac{5 \pi}{6}\)
3. \(\frac{\pi}{6}\)
4. None of these

Answer: 3. \(\frac{\pi}{6}\)

Let tan^-1\(\left(\tan \frac{7 \pi}{6}\right)\) = x,where x ∈ \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Then, tan x = tan \(\left(\tan \frac{7 \pi}{6}\right)\)

= tan \(\left(\pi+\frac{\pi}{6}\right)=\tan \frac{\pi}{6}\)

x=  \(\frac{\pi}{6}\)

Question 14. The value of \(\cot ^{-1}\left(\cot \frac{5 \pi}{4}\right)\)

1. \(\frac{\pi}{4}\)
2. \(\frac{-\pi}{4}\)
3. \(\frac{3 \pi}{4}\)
4. None of these

Answer: 1. \(\frac{\pi}{4}\)

Let cos^-1\(\left(\cot \frac{5 \pi}{4}\right)\) = x, where x [0,π]

Then ,cot x = cot \(\frac{5 \pi}{4}\)

= \(\cot \frac{5 \pi}{4}=\cot \left(\pi+\frac{\pi}{4}\right)\)

x\(=\frac{\pi}{4}\)

Question 15. The value of \(\sec ^{-1}\left(\sec \frac{8 \pi}{5}\right)\)

1. \(\frac{2 \pi}{5}\)
2. \(\frac{3 \pi}{5}\)
3. \(\frac{8 \pi}{5}\)
4. None of these

Answer: 1. \(\frac{2 \pi}{5}\)

Let sec^-1\(\left(\sec \frac{8 \pi}{5}\right)\)= x where x∈[0,π]- \(\left\{\frac{\pi}{2}\right\}\)

Then, sec x \(\frac{8 \pi}{5}\)

= sec \(\sec \left(2 \pi-\frac{2 \pi}{5}\right)=\sec \frac{2 \pi}{5}\)

x = \(\frac{2 \pi}{5}\)

Question 16. The value of cosec^-1 (cosec\(\frac{4 \pi}{3}\)) is

1. \(\frac{\pi}{3}\)
2. \(\frac{-\pi}{3}\)
3. \(\frac{2 \pi}{3}\)
4. None of these

Answer: 2. \(\frac{-\pi}{3}\)

Let cosec^-1 \(\frac{4 \pi}{3}\)= x where x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)

Then,cosec x= cosec \(\frac{4 \pi}{3}\)

= cosec\(\left(\pi+\frac{\pi}{3}\right)\)

=- cosec \(\frac{\pi}{3}\)

=cosec \(\frac{-\pi}{3}\)

x =\(\frac{-\pi}{3}\)

Question 17. The value of \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\)

1. \(\frac{3 \pi}{4}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{-\pi}{4}\)
4. None of these

Answer: 3. \(\frac{-\pi}{4}\)

Let tan^-1\(\left(\tan \frac{3 \pi}{4}\right)\)= x, where x ∈ \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Then,tan x= \(\tan \frac{3 \pi}{4}\)

= tan\(\left(\pi-\frac{\pi}{4}\right)=\tan \left(\frac{-\pi}{4}\right)\)

x = \(\frac{-\pi}{4}\)

Question 18. \(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\)

1. 0
2. \(\frac{2 \pi}{3}\)
3. \(\frac{\pi}{2}\)
4. π

Answer: 3. \(\frac{\pi}{2}\)

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

Then , sin x = \(-\frac{1}{2}\)

-sin = \(\frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right)\)

x= \(\frac{-\pi}{6}\)

Given example = \(\frac{\pi}{3}-\left(\frac{-\pi}{6}\right)=\left(\frac{\pi}{3}+\frac{\pi}{6}\right)\) Question

= \(\frac{3 \pi}{6}=\frac{\pi}{2}\)

Question 19. If x ≠ 0, then cos (tan^-1x+ cot^-1x)

1. -1
2. 1
3. 0
4. None of these

Answer: 3. 0

⇒ \(\cos \left(\tan ^{-1}+\cot ^{-1} x\right)\)

= \(\cos \frac{\pi}{2}\)

= 0

Question 20. The value of sin \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\)

1. \(\frac{2}{5}\)
2. \(\frac{4}{5}\)
3. \(\frac{-2}{5}\)
4. None of these

Answer: 2. \(\frac{4}{5}\)

cos^-1 x \(=\sin ^{-1} \sqrt{1-x^2}\)

= \(\cos ^{-1} \frac{3}{5}=\sin ^{-1} \sqrt{1-\frac{9}{25}}\)

= \(\sin ^{-1} \frac{4}{5}\)

∴  \(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=\frac{4}{5}\)

Question 21. \(\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\)

1. \(\frac{4 \pi}{3}\)
2. \(\frac{\pi}{2}\)
3. \(\frac{3 \pi}{4}\)
4. π

Answer: 4. \(\frac{3 \pi}{4}\)

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

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

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

= \(\left(\frac{2 \pi}{3}+\frac{\pi}{3}\right)=\pi\)

Question 22. \(\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)\)

1. \(\frac{\pi}{3}\)\(\frac{-\pi}{3}\)
2. \(\frac{5 \pi}{3}\)
3. None of these

Answer: 2. \(\frac{-\pi}{3}\)

⇒ \(\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)\)

= \(\tan ^{-1} \sqrt{3}-\left(\pi-\sec ^{-1} 2\right)\)

= \(\frac{\pi}{3}-\pi+\frac{\pi}{3}=\frac{-\pi}{3}\)

Question 23. \(\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}\)

1. \(\frac{2 \pi}{3}\)
2. \(\frac{3 \pi}{2}\)
3. 2π
4. None of these

Answer: 1. \(\frac{2 \pi}{3}\)

= \(\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}\)

= \(\frac{\pi}{3}+\left(2 \times \frac{\pi}{6}\right)=\left(\frac{\pi}{3}+\frac{\pi}{3}\right)\)

= \(\frac{2 \pi}{3}\)

Question 24. \(\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)\)

1. π
2. \(\frac{2 \pi}{3}\frac{2 \pi}{3}\)
3. \(\frac{3 \pi}{4}\)
4. \(\frac{\pi}{2}\)

Answer: 3. \(\frac{3 \pi}{4}\)

⇒ \(\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)\)

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

= \(\left(\frac{\pi}{4}+\pi-\frac{\pi}{3}-\frac{\pi}{6}\right)\)

= \(\frac{3 \pi}{4}\)

Question 25. tan[2 tan^-1\(\frac{1}{5}-\frac{\pi}{4}\)]= ?

1. \(\frac{7}{17}\)
2. \(\frac{-7}{17}\)
3. \(\frac{7}{12}\)
4. \(\frac{-7}{12}\)

Answer: 2. \(\frac{-7}{17}\)

Let 2\(\tan ^{-1} \frac{1}{5}=\theta\) .

Then \(\tan ^{-1} \frac{1}{5}=\frac{1}{2} \theta \Rightarrow \tan \frac{1}{2} \theta=\frac{1}{5}\)

∴ \(\tan \theta=\frac{2 \tan \frac{1}{2} \theta}{1-\tan ^2 \frac{1}{2} \theta}=\frac{\left(2 \times \frac{1}{5}\right)}{\left(1-\frac{1}{25}\right)}\)

= \(\left(\frac{2}{5} \times \frac{25}{24}\right)=\frac{5}{12}\)

∴ \(\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]=\tan \left(\theta-\frac{\pi}{4}\right)=\frac{\tan \theta-\tan \frac{\pi}{4}}{1+\tan \theta \cdot \tan \frac{\pi}{4}}\)

= \(\frac{\left(\frac{5}{12}-1\right)}{\left(1+\frac{5}{12} \times 1\right)}=\frac{\left(\frac{-7}{12}\right)}{\left(\frac{17}{12}\right)}=\frac{-7}{17}\)

Question 26. \(\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)\)= ?

1. \(\frac{(3-\sqrt{5})}{2}\)
2. \(\frac{(3+\sqrt{5})}{2}\)
3. \(\frac{(5-\sqrt{3})}{2}\)
4. \(\frac{(5+\sqrt{3})}{2}\)

Answer: 1. \(\frac{(3+\sqrt{5})}{2}\)

Let \(\cos ^{-1} \frac{\sqrt{5}}{3}=\theta\).Then,cos \(\cos \theta=\frac{\sqrt{5}}{3}\)

⇒ \(\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)=\tan \frac{1}{2} \theta=\frac{\sin (\theta / 2)}{\cos (\theta / 2)}\)

⇒ \(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\left\{\frac{\left(1-\frac{\sqrt{5}}{3}\right)}{\left(1+\frac{\sqrt{5}}{3}\right)}\right\}^{1 / 2}\)

= \(\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)^{1 / 2}=\left\{3-\frac{\sqrt{5}}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3+\sqrt{5}}\right\}^{1 / 2}\)

= \(\frac{(3-\sqrt{5})}{2}\)

Question 27. \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\)

1. \(\frac{3}{4}\)
2. \(\frac{4}{5}\)
3. \(\frac{3}{5}\)
4. None of these

Answer: 2. \(\frac{4}{5}\)

Let cos^-1 \(\frac{3}{5}\)= x ∈ where [0,π] .Then ,cos x= \(\frac{3}{5}\)

∴ Since x[0,π],sinx > 0

∴ Since x\(\sqrt{1-\frac{9}{25}}\)

= \(\sqrt{\frac{16}{25}}=\frac{4}{5}\)

⇒ \(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\frac{4}{5}\)

Question 28. \(\cos \left(\tan ^{-1} \frac{3}{4}\right)\)

1. \(\frac{3}{5}\)
2. \(\frac{4}{5}\)
3. \(\frac{4}{9}\)
4. None of these

Answer: 2. \(\frac{4}{5}\)

Let \(\tan ^{-1} \frac{3}{4}\) = x where x ∈ \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

∴tan x= \(\frac{3}{4}\) and since x ∈ \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\),we have cos x>0

∴cos x = \(\frac{1}{\sec x}=\frac{1}{\sqrt{1+\tan ^2 x}}\)

= \(\frac{1}{\sqrt{1+\frac{9}{16}}}=\frac{4}{5}\)

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

cos x=\(\frac{4}{5}\)

Question 29. \(\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}\)

1. 1
2. 0
3. \(\frac{-1}{2}\)
4. None of these

Answer: 1. 1

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

sin = \( \left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin \frac{\pi}{2}\)

= 1

Question 30. \(\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)\)

1. \(\frac{1}{\sqrt{5}}\)
2. \(\frac{2}{\sqrt{5}}\)
3. \(\frac{1}{\sqrt{10}}\)
4. \(\frac{2}{\sqrt{10}}\)

Answer: 3. \(\frac{1}{\sqrt{10}}\)

Let \(\cos ^{-1} \frac{4}{5}=\) = x where x∈ [0,π].

Then cos x= \(\frac{4}{5}\)

Since x∈ [0,π] ⇒ \(\frac{1}{2} x \in\left[0, \frac{\pi}{2}\right] \Rightarrow \sin \frac{1}{2} x>0\)

sin = \( \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\sin \frac{1}{2} x\)

= \(\sqrt{\frac{1-\cos x}{2}}=\sqrt{\frac{\left(1-\frac{4}{5}\right)}{2}}=\)

= \(\frac{1}{\sqrt{10}}\)

Question 31. \(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}\)

1. \(\frac{\pi}{3}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{3 \pi}{4}\)
4. \(\frac{2 \pi}{3}\)

Answer: 2. \(\frac{\pi}{4}\)

⇒ \(\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}\)

= \(\tan ^{-1}\left\{2 \cos \left(2 \times \frac{\pi}{6}\right)\right\}\)

=  \(\tan ^{-1}\left\{2 \times \frac{1}{2}\right\}\)

tan^-1 1= \(\frac{\pi}{4}\)

Question 32. \(\text { If } \cot ^{-1}\left(\frac{-1}{5}\right)=x, \text { then } \sin x=?\)

1. \(\frac{1}{\sqrt{26}}\)
2. \(\frac{5}{\sqrt{26}}\)
3. \(\frac{1}{\sqrt{24}}\)
4. None of these

Answer: 2. \(\frac{5}{\sqrt{26}}\)

cot^-1\(\left(\frac{-1}{5}\right)\)

cot x= \(\left(\frac{-1}{5}\right)\),where x∈ (0,π)

sin x>0 in (0,π)

sin x= 1/ cosecx

= \(\frac{1}{\sqrt{1+\cot ^2 x}}\)

⇒\(\frac{1}{\sqrt{1+\frac{1}{25}}}=\frac{5}{\sqrt{26}}\)

Question 33. \(\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=?\)

1. \(\frac{\pi}{2}\)
2. π
3. \(\frac{3 \pi}{2}\)
4. None of these

Answer: 3. \(\frac{3 \pi}{2}\)

Range of \(\cos ^{-1} \text { is }\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

cos^-1= \(\left(\frac{-\sqrt{3}}{2}\right)\)

x = \(\frac{-\sqrt{3}}{2}\)

– cos = \(\frac{\pi}{6}=\cos \left(\pi-\frac{\pi}{6}\right)=\cos \frac{5 \pi}{6}\)

x= \(\frac{5 \pi}{6}\)

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

= \(-\sin ^{-1} \frac{1}{2}+2 \times \frac{5 \pi}{6}\)

= \(-\frac{\pi}{6}+\frac{5 \pi}{3}=\frac{4 \pi}{6}\)

=\(\frac{3 \pi}{2}\)

Question 34. \(\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{ } 2}\right)=?\)

1. \(\frac{\pi}{2}\)
2. π
3. \(\frac{3 \pi}{2}\)
4. \(\frac{2 \pi}{2}\)

Answer: 1. \(\frac{\pi}{2}\)

Range of \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

⇒ \(\tan ^{-1}(-1)=x \Rightarrow \tan x=-1\)

= – \(\tan \frac{\pi}{4}=\tan \left(\frac{-\pi}{4}\right) \Rightarrow\)

= \(x\frac{-\pi}{4}\)

The range of cos^-1 is [0,π]

= \(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\) = y

cos y = \(\frac{-1}{\sqrt{2}}\)

= \(-\cos \frac{\pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \frac{3 \pi}{4}\)

y = \(\frac{3 \pi}{4}\)

Given example = \(-\frac{\pi}{4}+\frac{3 \pi}{4}=\frac{2 \pi}{4}=\frac{\pi}{2}\)

Question 35. \(\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)=?\)

1. 1
2. \(\frac{1}{2}\)
3. 0
4. None of these

Answer: 3. 0

Given example = \(\cot \frac{\pi}{2}\)= 0

Question 36. \(\tan ^{-1} 1+\tan ^{-1} \frac{1}{3}=?\)

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

Answer: 3. \(\tan ^{-1} 2\)

⇒ \(\tan ^{-1} 1+\tan ^{-1} \frac{1}{3}\)

\(=\tan ^{-1}\left\{\frac{1+\frac{1}{3}}{1-\frac{1}{3}}\right\}\)

= \(\tan ^{-1} 2\)

Question 37. \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=?\)

1. \(\frac{\pi}{3}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{\pi}{2}\)
4. \(\frac{2 \pi}{3}\)

Answer: 2. \(\frac{\pi}{4}\)

⇒ \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}\)

tan^-1\(\left\{\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right\}\)

tan^-1\(=\frac{\pi}{4}\)

Question 38. \(2 \tan ^{-1} \frac{1}{3}=?\)

1. \(\tan ^{-1} \frac{3}{2}\)
2. \(\tan ^{-1} \frac{3}{4}\)
3. \(\tan ^{-1} \frac{4}{3}\)
4. None of these

Answer: 2. \(\tan ^{-1} \frac{3}{4}\)

Use \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)

Question 39. \(\cos \left(2 \tan ^{-1} \frac{1}{2}\right)=?\)

1. \(\frac{3}{5}\)
2. \(\frac{4\frac{3}{5}[}{5}\)
3. \(\frac{7}{8}\)
4. [None of these

Answer: 1. \(\frac{3}{5}\)

Use \(2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\)

Question 40. \(\sin \left[2 \sin ^{-1} \frac{4}{5}\right]\)

1. \(\frac{12}{25}\)
2. \(\frac{80}{89}\)
3. \(\frac{75}{128}\)
4. None of these

Answer: 2. \(\frac{80}{89}\)

Use \(2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)

Question 41. \(\sin \left[2 \sin ^{-1} \frac{4}{5}\right]\)

1. \(\frac{2}{25}\)
2. \(\frac{16}{25}\)
3. \(\frac{24}{25}\)
4. None of these

Answer: 3. \(\frac{16}{25}\)

Use \(2 \sin ^{-1} x=\sin ^{-1}\left[2 x \sqrt{1-x^2}\right]\)

Question 42. \(\text { If } \tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3} \text {, then } x=?\)

1. \(\frac{1}{2}\)
2. \(\frac{1}{4}\)
3. \(\frac{1}{6}\)
4. None of these

Answer: 1. \(\frac{1}{2}\)

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

= \(\frac{1}{2}\)

Question 43. \(\text { If } \tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2} \text {, then } x=\text { ? }\)

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

Answer: 3. 0

We know that \(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)

= \(\tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\frac{\pi}{2}\)

= \(\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}\)

(1-x)= \(\frac{1}{(1+x)}\)(1-x) = 1

x= 0

Question 44. \(\text { If } \sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}, \text { then }\left(\cos ^{-1} x+\cos ^{-1} y\right)=?\)

1. \(\frac{\pi}{6}\)
2. \(\frac{\pi}{3}\)
3. \(\pi\)
4. \(\frac{2 \pi}{3}\)

Answer: 2. \(\frac{\pi}{3}\)

⇒ \(\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}\)

∴ \(\left(\frac{\pi}{2}-\cos ^{-1} x\right)+\left(\frac{\pi}{2}-\cos ^{-1} y\right)=\frac{2 \pi}{3}\)

⇒ \(\cos ^{-1} x+\cos ^{-1} y=\left(\pi-\frac{2 \pi}{3}\right)=\frac{\pi}{3}\)

Question 45. \(\left(\tan ^{-1} 2+\tan ^{-1} 3\right)=?\)

1. \(\frac{-\pi}{4}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{3 \pi}{4}\)
4. \(\pi\)

Answer: 3. \(\frac{3 \pi}{4}\)

( x = 2,y = 3) = xy>1

∴  \(\pi+\tan ^{-1}\left(\frac{2+3}{1-2 \times 3}\right)=\pi+\tan ^{-1}(-1)\)

= \(\pi-\tan (1)=\left(\pi-\frac{\pi}{4}\right)\)

= \(\frac{3 \pi}{4}\)

Question 46. \(\text { If } \tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8, \text { then } x=\text { ? }\)

1. \(\frac{1}{3}\)
2. \(\frac{1}{5}\)
3. 3
4. 5

Answer: 2. \(\frac{1}{5}\)

⇒ \(\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8\)

= \(\frac{3+x}{1-3 x}\)= 8

3+x = 8-24x

x= \(\frac{1}{5}\)

Question 47. \(\text { If } \tan ^{-1} 3 x+\tan ^{-1} 2 x=\frac{\pi}{4} \text {, then } x=\text { ? }\)

1. \(\frac{1}{2} \text { or }-2\)
2. \(\frac{1}{3} \text { or }-3\)
3. \(\frac{1}{4} \text { or }-2\)
4. \(\frac{1}{6} \text { or }-1\)

Answer: 4. \(\frac{1}{6} \text { or }-1\)

⇒ \(\tan ^{-1}\left(\frac{3 x+2 x}{1-6 x^2}\right)=\frac{\pi}{4}\)

∴ \(\frac{5 x}{1-6 x^2}=\tan \frac{\pi}{4}=1\)

= \(\frac{5 x}{1-6 x^2}=\tan \frac{\pi}{4}\)= 1

⇒ 6x²+5x-1 = 0

(x+1)(6x-1)= 0

x= -1 or x= \(\frac{1}{6}\)

Question 48. \(\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}=?\)

1. \(\frac{13}{6}\)
2. \(\frac{17}{6}\)
3. \(\frac{19}{6}\)
4. \(\frac{23}{6}\)

Answer: 2. \(\frac{17}{6}\)

cos^-1 x= \(\frac{\sqrt{1-x^2}}{x}\)

cos^-1\(\frac{4}{5}\) = tan^-1\(\frac{\sqrt{1-\frac{16}{25}}}{(4 / 5)}\)

tan^-1 = \(\frac{3}{4}\)

∴ \(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\)

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

tan^-1 = \(\frac{17}{6}\)

\(\cos ^{-1} \frac{4}{5}=\tan ^{-1} \frac{\sqrt{1-\frac{16}{25}}}{(4 / 5)}=\tan ^{-1} \frac{3}{4} \)

∴ Given example = \(\tan \left\{\tan ^{-1} \frac{17}{6}\right\}=\frac{17}{6}\)

Question 49. cot^-1 + cosec^-1\(\frac{\sqrt{41}}{4}\)

1. \(\frac{\pi}{6}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{\pi}{3}\)
4. \(\frac{3 \pi}{4}\)

Answer: 2. \(\frac{\pi}{4}\)

cosec x^-1 = \(\cot ^{-1} \sqrt{x^2-1}\)

cosec x^-1 = \(\frac{\sqrt{41}}{4}\)

cot^-1 = \(\frac{\sqrt{41}}{4}\) -1

cot^-1 = \(\frac{5}{4}\)

= \(\cot ^{-1} 9+\cot ^{-1} \frac{5}{4}\)

= \(\tan ^{-1} \frac{1}{9}+\tan ^{-1} \frac{4}{5}\)

Question 50. Range of sin^-1x is

1. \(\left[0, \frac{\pi}{2}\right]\)
2. [0, π]
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
4. None of these

Answer: 3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

Question 51. The range of cos^-1x is

1. [0, π]
2. \(\left[0, \frac{\pi}{2}\right]\)
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
4. None of these

 Answer: 1. [0, π]

Question 52. The range of tan^-1x is

1. \(\left(0, \frac{\pi}{2}\right)\)
2. \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
4. None of these

Answer: 2. \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)

Question 53. The range of sec^-1x is

1. \(\left[0, \frac{\pi}{2}\right]\)
2. [0, π]
3. \([0, \pi]-\left\{\frac{\pi}{2}\right\}\)
4. None of these

Answer: 3. \([0, \pi]-\left\{\frac{\pi}{2}\right\}\)

Question 54. Range of cosec^-1x is

1. \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
2. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)
4. None of these

Answer: 3. \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)

Question 55. Domain of cos^-1x is

1. [0,1]
2. [-1,1]
3. [-1,0]
4. None of these

Answer: 2. [-1,1]

Question 56. Domain of sec^-1 x is

1. [-1,1]
2. R- {0}
3. R- [-1,1]
4. R- {-1,1}

Answer: 3. R- [-1,1]

Binomial Distribution

Success And Failure In An Experiment: There are certain kinds of experiments that have two possible outcomes. One of these two outcomes is called a success, while the other is called a failure.

For example, in tossing a coin, we get either a head or a tail. If getting head is taken as a success then getting a tail is a failure.

Bernoulli’s Theorem Let there be n independent trials in an experiment and let the random variable X denote the number of successes in these trials.

Let the probability of getting a success in a single trial be p and that of getting a failure be q so that p + q = 1. Then,

P(X =r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}.\)

Proof: Let us denote a success by S and a failure by F.

Read and Learn More WBCHSE Solutions For Class 12 Maths

A number of ways of getting r successes in n trials = ⁿC_r.

∴ P(X=r) = \({ }^n C_r\) • {P(S) • P(S) …r times} x {P(F) • P(F)… (n – r) times}

= \({ }^n C_r\) (p • p • p …r times) x[q • q • q … (n-r) times]

= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}\)

Hence, P(X = r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}\)

Remark: We have

Conditions for the Applicability of a Binomial Distribution

1. The experiment is performed for a finite and fixed number of trials.
2. Each trial must give either a success or a failure.
3. The probability of success in each trial is the same.

Binomial Distribution Solved Examples

Example 1. A coin is tossed 4 times. If X is the number of heads observed, find the probability distribution of X.

Solution:

When a coin is tossed, we have S = {H, T}.

P(getting a head) = 1/2, and

P(not getting a head) = (1-1/2) = 1/2

Let X be the random variable denoting the number of heads.

In 4 trials, we may get 0 or 1 or 2 or 3 or 4 heads.

So, X may assume the values 0,1, 2, 3, 4.

P(X = 0) = \({ }^4 C_0 \cdot\left(\frac{1}{2}\right)^0 \cdot\left(\frac{1}{2}\right)^{(4-0)}=\frac{1}{16}\)

P(X = 1) = \({ }^4 C_1 \cdot\left(\frac{1}{2}\right)^1 \cdot\left(\frac{1}{2}\right)^{(4-1)}=\frac{1}{4}\)

P(X = 2) = \({ }^4 C_2 \cdot\left(\frac{1}{2}\right)^2 \cdot\left(\frac{1}{2}\right)^{(4-2)}=\frac{3}{8}\)

P(X = 3) = \({ }^4 C_3 \cdot\left(\frac{1}{2}\right)^3 \cdot\left(\frac{1}{2}\right)^{(4-3)}=\frac{1}{4}\)

P(X = 4) = \({ }^4 C_4 \cdot\left(\frac{1}{2}\right)^4 \cdot\left(\frac{1}{2}\right)^{(4-4)}=\frac{1}{16}\)

Hence, the required probability distribution is given by

⇒ \(\left(\begin{array}{llllll}
X: & 0 & 1 & 2 & 3 & 4 \\
P(X): & \frac{1}{16} & \frac{1}{4} & \frac{3}{8} & \frac{1}{4} & \frac{1}{16}
\end{array}\right)\)

Example 2. Find the probability distribution of the number of suffixes in three tosses of a die.

Solution:

When a die is tossed, we have S = {1, 2, 3, 4, 5, 6}.

∴ P(getting a six) = 1/6 and P(not getting a six) = (1-1/6) = 5/6

Let X be the random variable denoting the number of sixes.

In 3 trials, the number of sixes may be 0 or 1 or 2 or 3.

So, X may assume the values 0, 1, 2, 3.

P(X=0) = \({ }^3 C_0 \cdot\left(\frac{1}{6}\right)^0 \cdot\left(\frac{5}{6}\right)^{(3-0)}=\frac{125}{216}\)

P(X=1 )= \({ }^3 C_1 \cdot\left(\frac{1}{6}\right)^1 \cdot\left(\frac{5}{6}\right)^{(3-1)}=\frac{25}{72}\)

P(X=2) = \({ }^3 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^{(3-2)}=\frac{5}{72}\)

P(X=3) = \({ }^3 C_3 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^{(3-3)}=\frac{1}{216}\)

The required probability distribution of X is given below:

⇒ \(\left(\begin{array}{lcccc}
X: & 0 & 1 & 2 & 3 \\
P(X): & \frac{125}{216} & \frac{25}{72} & \frac{5}{72} & \frac{1}{216}
\end{array}\right) .\)

Example 3. Find the probability distribution of the number of doublets in four throws of a pair of dice.

Solution:

When a pair of dice is thrown, there are 36 possible outcomes.

∴ n(S)= 36.

All possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

P(getting a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\), and

P(not getting a doublet) = (1-\(\frac{1}{6}\)) = \(\frac{5}{6}\)

Let X denote the number of doublets.

In 4 throws, we can have 0 or1 2 or 3, or 4 doublets

P(X=0) = \({ }^4 C_0 \cdot\left(\frac{1}{6}\right)^0 \cdot\left(\frac{5}{6}\right)^4=\frac{625}{1296}\)

P(X=1) = \({ }^4 C_1 \cdot\left(\frac{1}{6}\right)^1 \cdot\left(\frac{5}{6}\right)^3=\frac{125}{324}\)

P(X=2) = \({ }^4 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^2=\frac{25}{216}\)

P(X=3) = \({ }^4 C_3 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^1=\frac{5}{324}\)

P(X=4) = \({ }^4 C_4 \cdot\left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^0=\frac{1}{1296}\)

The required probability distribution is given below:

⇒ \(\left(\begin{array}{lccccc}
X: & 0 & 1 & 2 & 3 & 4 \\
P(X): & \frac{625}{1296} & \frac{125}{324} & \frac{25}{216} & \frac{5}{324} & \frac{1}{1296}
\end{array}\right) .\)

Example 4. An unbiased coin is tossed 6 times. Find, using the binomial distribution, the probability of getting at least 5 heads.

Solution:

In a single throw of a coin, we have S = {H, T}.

P(getting a head) = \(\frac{1}{2}\) and P(not getting a head) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

∴ p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\)

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(6-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^6\)

∴ P(getting at least 5 heads) = P(X≥5)

= P(X = 5) + P(X = 6)

= \({ }^6 C_5 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_6 \cdot\left(\frac{1}{2}\right)^6=\left(\frac{3}{32}+\frac{1}{64}\right)=\frac{7}{64}\)

Hence, the required probability is \(\frac{7}{64}\)

Example 5. An unbiased coin is tossed 8 times. Find, by using binomial distribution, the probability of getting at least 3 heads.

Solution:

In a single throw of a coin, we have S = {H, T}.

P(getting a head) = \(\frac{1}{2}\) and P(not getting a head) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

∴ p = \(\frac{1}{2}\) and \(\frac{1}{2}\)

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^8 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(8-r)}={ }^8 C_r\left(\frac{1}{2}\right)^8\)

∴ P(getting at least 3 heads) = P(X≥3)

= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

= \(1-\left[{ }^8 C_0 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_1 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_2 \cdot\left(\frac{1}{2}\right)^8\right]\)

= \(1-\frac{1}{256} \cdot(1+8+28)=\left(1-\frac{37}{256}\right)=\frac{219}{256}\)

Hence, the required probability is \(\frac{219}{256}\).

Example 6. Six coins are tossed simultaneously. Find the probability of getting

1. 3 Heads
2. No head
3. At least one head
4. Not more than 3 heads.

Solution:

The experiment may be taken as throwing a single coin 6 times.

In a single throw of a coin, we have S = {H, T}.

P(getting a head) = \(\frac{1}{2}\) and P(not getting a head) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

Let X be the random variable showing the number of heads.

P(X = r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{6-r}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^6\)

1. P(getting 3 heads) = P(X = 3) = \(={ }^6 C_3 \cdot\left(\frac{1}{2}\right)^6=\frac{5}{16}\)

2. P(getting no head) = P(X = 0) = \({ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6=\frac{1}{64}\)

3. P(getting at least 1 head)

= \(1-P(X=0)=1-\left[{ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6\right]\)

= \(\left(1-\frac{1}{64}\right)=\frac{63}{64}\)

4. P(getting not more than 3 heads) = P(no head or 1 head or 2 heads or 3 heads)

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= \({ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_1 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_2 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_3 \cdot\left(\frac{1}{2}\right)^6\)

= \(\left(\frac{1}{2}\right)^6 \cdot(1+6+15+20)=\left(\frac{1}{64} \times 42\right)=\frac{21}{32}\)

Example 7. A die is thrown 5 times. If getting an odd number is a success, find the probability of getting at least 4 successes.

Solution:

When a die is thrown, we have S = {1,2,3,4,5,6}.

∴ P(getting an odd number) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

∴ P(a success) = \(\frac{1}{2}\), and P(not a success) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

Let X be the random variable showing the number of successes.

P(X=r) = \(={ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^5 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(5-r)}={ }^5 C_r \cdot\left(\frac{1}{2}\right)^5\)

P(at least 4 successes) = P(4 successes or 5 successes)

= P(X = 4) + P(X = 5)

= \({ }^5 C_4 \cdot\left(\frac{1}{2}\right)^5+{ }^5 C_5 \cdot\left(\frac{1}{2}\right)^5=\left(\frac{5}{32}+\frac{1}{32}\right)=\frac{6}{32}=\frac{3}{16}\)

Example 8. In 4 throws with a pair of dice, what is the probability of throwing doublets at least twice?

Solution:

In a single throw of a pair of dice, the number of all possible outcomes is 36.

All doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

P(getting a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\) and,

P(not getting a doublet) = (1-\(\frac{1}{6}\)) = \(\frac{5}{6}\)

Let X be the random variable denoting the number of doublets.

Then, P(X = r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^4 C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(4-r)} .\)

P(at least 2 doublets) = P(X = 2) + P(X = 3) + P(X = 4)

= \({ }^4 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^{(4-2)}+{ }^4 C_3 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^{(4-3)}+{ }^4 C_4 \cdot\left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^{(4-4)}\)

= \(6 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^2+4 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^1+\left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^0\)

= \(\left(\frac{25}{216}+\frac{5}{324}+\frac{1}{1296}\right)=\frac{171}{1296}\)

Example 9. The bulbs produced in a factory are supposed to contain 5% defective bulbs. What is the probability that a sample of 10 bulbs will contain no more than 2 defective bulbs?

Solution:

P(getting a defective bulb) = \(\frac{5}{100}\) = \(\frac{1}{20}\), and

P(getting a non-defective bulb) = (1-\(\frac{1}{20}\)) = \(\frac{19}{20}\), and

Then, p = \(\frac{1}{20}\) and q = \(\frac{19}{20}\)

Let X denote the number of defective bulbs.

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}\)=\({ }^{10} C_r \cdot\left(\frac{1}{20}\right)^r \cdot\left(\frac{19}{20}\right)^{(10-r)}\)

P(getting not more than 2 defective bulbs)

= P(X = 0 or X = 1 or X = 2)

= P(X = 0) + P(X = 1) + P(X = 2)

= \({ }^{10} C_0 \cdot\left(\frac{1}{20}\right)^0 \cdot\left(\frac{19}{20}\right)^{10}+{ }^{10} C_1 \cdot\left(\frac{1}{20}\right)^1 \cdot\left(\frac{19}{20}\right)^9+{ }^{10} C_2 \cdot\left(\frac{1}{20}\right)^2 \cdot\left(\frac{19}{20}\right)^8\)

= \(\left(\frac{19}{20}\right)^{10}+\frac{1}{2} \cdot\left(\frac{19}{20}\right)^9+\frac{9}{80} \cdot\left(\frac{19}{20}\right)^8=\left(\frac{19}{20}\right)^8 \cdot\left(\frac{149}{100}\right)\)

Let A = \(\left(\frac{19}{20}\right)^8 \cdot\left(\frac{149}{100}\right) \cdot\) Then,

log A = 8(log 19 – log 20) + log 149 – log 100

= 8(1.2788 – 1.3010) + 2.1732 – 2 = -0.0044 = 1.9956.

∴ A = antilog (1.9956) = 0.99.

Hence, the required probability = \(\frac{99}{100}\)

Example 10. If on average, out of 10 ships, one gets drowned then what is the probability that out of 5 ships at least 4 reach the shore safely?

Solution:

Probability of a ship to reach the shore safely = \(\frac{9}{10}\)

Probability that a ship gets drowned = (1-\(\frac{9}{10}\)) = \(\frac{1}{10}\)

Let X be the random variable showing the number of ships reaching the shore safely.

∴ P(at least 4 reaching safely)

= P(4 reaching safely or 5 reaching safely)

= P(4 reaching safely) + P(5 reaching safely)

= \({ }^5 C_4 \cdot\left(\frac{9}{10}\right)^4 \cdot\left(\frac{1}{10}\right)^{(5-4)}+{ }^5 C_5 \cdot\left(\frac{9}{10}\right)^5 \cdot\left(\frac{1}{10}\right)^0\)

= \(\frac{1}{2} \cdot\left(\frac{9}{10}\right)^4+\left(\frac{9}{10}\right)^5=\left(\frac{9}{10}\right)^4\left(\frac{1}{2}+\frac{9}{10}\right)=\frac{7}{5} \cdot\left(\frac{9}{10}\right)^4\)

A = \(\frac{7}{5} \cdot\left(\frac{9}{10}\right)^4=\frac{7 \times(9)^4}{5 \times(10)^4}\)

⇒ logA = log7 + 4xlog9-log5-4xloglO = (0.8451 + 4 x 0.9542 – 0.6990- 4) = -0.0371 = 1.9629

⇒ A = antilog (1.9629) = 0.9181.

Hence, the required probability is 0.9181.

Mean and Variance of a Binomial Distribution

Mean: If a random variable X assumes the values x₁, x₂,… ,x_n with probabilities p₁, p₂, ……., p_n respectively then the mean of X is defined by

∴ \(\mu=\sum_{i=1}^n x_i p_i .\)

To Find the Mean of a Binomial Distribution

For the binomial distribution P(X = r) = P(r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}\), where r = 0,1,2,… ,n the mean, p, is given by

⇒ \(\mu =\sum_{r=0}^n r \cdot p(r)=\sum_{r=0}^n r \cdot{ }^n C_r \cdot p^r \cdot q^{(n-r)}\)

= \({ }^n C_1 \cdot p \cdot q^{n-1}+2 \cdot{ }^n C_2 \cdot p^2 \cdot q^{(n-2)}+\ldots+n \cdot{ }^n C_n \cdot p^n q^0\)

= \(1 \cdot n p \cdot q^{n-1}+n(n-1) \cdot p^2 \cdot q^{(n-2)}+\ldots+n p^n\)

= \(n p \cdot\left[{ }^{(n-1)} C_0 \cdot p^0 \cdot q^{(n-1)}+{ }^{(n-1)} C_1 \cdot p^1 \cdot q^{(n-2)}+\ldots+{ }^{(n-1)} C_{(n-1)} \cdot p^{n-1} \cdot q^0\right]\)

= \((n p) \cdot(q+p)^{n-1}=(n p)\) [because q+p=1]

Hence, the mean is given by μ= np.

The variance =σ² is given by

⇒ \(\sigma^2=\sum_{r=0}^n r^2 \cdot p(r)-(\text { mean })^2\)

= \(\sum_{r=0}^n r^2 \cdot{ }^n C_r \cdot p^r q^{(n-r)}-(n p)^2\) [because mean=np]

= \(\sum_{r=0}^n\{r+r(r-1)\} \cdot{ }^n C_r p^r q^{(n-r)}-(n p)^2\)

= \(\sum_{r=0}^n r \cdot{ }^n C_r p^r q^{(n-r)}+\sum_{r=0}^n r(r-1) \cdot{ }^n C_r \cdot p^r \cdot q^{(n-r)}-(n p)^2\)

= \(n p+\sum_{r=2}^n r(r-1) \cdot \frac{n(n-1)}{r(r-1)} \cdot{ }^{(n-2)} C_{(r-2)} \cdot p^2 \cdot p^{(r-2)} \cdot q^{(n-r)}-(n p)^2\)

= \(n p+n(n-1) \cdot p^2 \cdot\left(\sum_{r=2}^n \sum_{r=0}^n r \cdot{ }^n C_r \cdot p^r \cdot q^{(n-r)}=\text { mean }=n p\right]\)

= \(n p+n(n-1) p^2(q+p)^{(n-2)}-n^2 p^2\)

= \(n p+n(n-1) p^2-n^2 p^2\) [because q+p=1]

= \(n p-n p^2=n p(1-p)=n p q\).

Hence, variance = npq.

∴ standard deviation = √npq.

Recurrence Relation for a Binomial Distribution

We have

P(r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)} \text { and } P(r+1)={ }^n C_{(r+1)} \cdot p^{(r+1)} \cdot q^{n-r-1} \text {. }\)

∴ \(\frac{P(r+1)}{P(r)}=\frac{{ }^n C_{(r+1)} \cdot p^{(r+1)} \cdot q^{n-r-1}}{{ }^n C_r \cdot p^r \cdot q^{(n-r)}}\)

= \(\frac{(n !)}{(r+1) ! \cdot(n-r-1) !} \cdot \frac{(r) ! \cdot(n-r) !}{(n) !} \cdot \frac{p}{q}=\frac{(n-r)}{(r+1)} \cdot \frac{p}{q}\)

∴ P(r+1) = \(\frac{(n-r)}{(r+1)} \cdot \frac{p}{q} \cdot P(r)\)

Example 11. If X follows a binomial distribution with mean 3 and variance (3/2), find

1. P(X≥1)
2. P(X≤5).

Solution:

We know that mean = np and variance = npq.

∴ np = 3 and npq = \(\frac{3}{2}\)

⇒ 3q = \(\frac{3}{2}\)

⇒ q = \(\frac{1}{2}\)

Now, np = 3 and p = \(\frac{1}{2}\)

⇒ n X \(\frac{1}{2}\) = 3

⇒ n = 6.

So, the binomial distribution is given by

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(6-r)}={ }^6 C_r\left(\frac{1}{2}\right)^6 \cdot\)

1. P(X≥1) =1-P(X=0)

= \(1-{ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6=\left(1-\frac{1}{64}\right)=\frac{63}{64}\)

2. P(X≤5) =1-P(X= 0)

= \(1-{ }^6 C_6\left(\frac{1}{2}\right)^6=\left(1-\frac{1}{64}\right)=\frac{63}{64}\)

Example 12. If X follows a binomial distribution with mean 4 and variance 2, find P(X≥5).

Solution:

We know that mean = np and variance = npq.

∴ np = 4 and npq = 2.

Now, np = 4 and npq= 2

⇒ 4q = 2 ⇒ q = \(\frac{1}{2}\)

Now,np = 4 and p = \(\frac{1}{2}\)

⇒ \(\frac{1}{2}\) n = 4 ⇒ n = 8.

So, the binomial distribution is given by

P(X= r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^8 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(8-r)}={ }^8 C_r \cdot\left(\frac{1}{2}\right)^8\)

∴ P(X > 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X= 8)

= \({ }^8 C_5 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_6 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_7 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_8 \cdot\left(\frac{1}{2}\right)^8\)

= \(\left[{ }^8 C_3+{ }^8 C_2+{ }^8 C_1+1\right]\left(\frac{1}{2}\right)^8\)

= \((56+28+8+1) \cdot \frac{1}{256}=\frac{93}{256}\)

Example 13. Find the binomial distribution for which the mean and variance are 12 and 3 respectively.

Solution:

Let X be a binomial variate for which mean = 12 and variance = 3.

Then, np =12 and npq = 3

⇔ 12xq = 3 ⇒ q = \(\frac{1}{4}\)

∴ p = (1-q) = (1-\(\frac{1}{4}\)) = \(\frac{3}{4}\)

And, np =12 ⇔ n = \(\frac{12}{p}\) = 12 x \(\frac{4}{3}\) = 16.

Thus, n =16, p = \(\frac{3}{4}\) and q = \(\frac{1}{4}\)

Hence, the binomial distribution is given by

P(X=r) = \({ }^{16} C_r \cdot\left(\frac{3}{4}\right)^r \cdot\left(\frac{1}{4}\right)^{(16-r)}\) where r = 0, 1, 2, 3, …,15.

Example 14. If the sum of the mean and variance of a binomial distribution for 5 trials is 1.8, find the distribution.

Solution:

We know that

mean = np and variance = npq.

It is being given that n = 5 and mean + variance = 1.8.

∴ np + npq = 1.8, where n = 5 +

⇔ 5p + 5pq = 1.8

⇔ p + p(1 -p) = 0.36 [q = (1-p)]

⇔ p²-2p + 0.36 = 0

⇔100p² – 200p + 36 = 0

⇔ 25p² – 50p + 9 = 0

⇔ 25p² -45p -5p + 9 = 0

⇔ 5p(5p – 9) – (5p – 9) = 0

⇔ (5p – 9)(5p -1) = 0 1

⇔ p = \(\frac{1}{5}\) = 0.2 [p cannot exceed 1].

Thus, n = 5, p = 0.2, and q = (1-p) = (1- 0.2) = 0.8.

Let X denote the binomial variate. Then, the required distribution is

P(X = r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^5 C_r \cdot(0.2)^r \cdot(0.8)^{(5-r)} \text {, }\) where r = 0,1,2,3,4,5.

Example 15. The sum and the product of the mean and variance of a binomial distribution are 24 and 128 respectively. Find the distribution.

Solution:

We have np + npq – 24 and np x npq = 128

⇔ (np)( 1 + q) = 24 and n²p² x q = 128

⇔ n²p² = \(\frac{576}{(1+q)^2}\) and n²p² x q =128

⇔ \(\frac{576}{(1+q)^2}=\frac{128}{q}\)

⇔ 2(1 + q² + 2q) = 9q

⇔ 2q²-5q + 2 = 0

⇔ (2q – 1)(q – 2) = 0

⇔ q = \(\frac{1}{2}\) ⇔ q = 2

∴ p = (1-q) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

Now, np(1+q) = 24

⇔ n x \(\frac{1}{2}\)(1+\(\frac{1}{2}\)) = 24

⇔ n = 32

Hence, the requied probability distribution is given by P(X=r) = \({ }^{32} C_r \cdot\left(\frac{1}{2}\right)^{32}\)

Example 16. In a binomial distribution, prove that mean > variance.

Solution:

Let X be a binomial variate with parameters n and p.

Then, mean = np and variance = npq.

∴ (mean)- (variance) = (np- npq) = np(1 -q) = np²> 0 [(1- q) = p and np²>0 as n∈N]

⇒ [(mean)- (variance)] > 0

⇒ mean > variance.

Hence, mean > variance.

Example 17. A die is tossed thrice. Getting an even number is considered a success. What is the variance of the binomial distribution?

Solution:

Here, n = 3.

Let p = probability of getting an even number in a single throw

⇒ p = \(\frac{3}{6}\) = \(\frac{1}{2}\)

⇒ q = (1-p) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

∴ variance = npq = (3x\(\frac{1}{2}\)x\(\frac{1}{2}\)0 = \(\frac{3}{4}\)

Example 18. A die is rolled 20 times. Getting a number greater than 4 is a success. Find the mean and variance of the number of successes.

Solution:

In a single throw of a die, we have

p = probability of getting a number greater than 4 = \(\frac{2}{6}\) = \(\frac{1}{3}\)

∴ q = (1-p) = (1-\(\frac{1}{3}\)) = \(\frac{2}{3}\)

Also, n = 20 (given).

∴ mean = np =(20x\(\frac{1}{3}\))= 6.67, and

variance = npq =(20x\(\frac{1}{3}\)x\(\frac{2}{3}\))= 4.44.

Example 19. A die is tossed 180 times. Find the expected number (μ) of times the face with the number 5 will appear. Also, find the standard deviation (σ), and variance (σ²).

Solution:

In a single throw of a die, S = {1, 2, 3, 4, 5 6}.

p = (probability of getting the number 5) = \(\frac{1}{6}\)

∴ q = (1-p) = (1-\(\frac{1}{6}\)) = \(\frac{5}{6}\)

Thus, n = 180, p = \(\frac{1}{6}\) and q = \(\frac{5}{6}\)

∴ μ = np = (180x\(\frac{1}{6}\))=30.

Variance =  σ² = npq = (l80x\(\frac{1}{6}\)x\(\frac{5}{6}\)) = 25

Standard deviation = σ = √25 = 5

Binomial Distribution Exercise

Question 1. A coin is tossed 6 times. Find the probability of getting at least 3 heads.

Solution: 21/32

Hint: In a single toss, P(H) 1/2 and P(not H) = 1/2

⊂ p = 1/2, q = 1/2 and n = 6.

Let X show the number of heads. Then,

P(X= r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r\left(\frac{1}{2}\right)^6 .\)

Required probability = P(X= 3) +P(X= 4) +P(X= 5) +P(X= 6).

Question 2. A coin is tossed 5 times. What is the probability that a head appears an even number of times?

Solution: 1/2

Hint: Here,p = 1/2, q = 1/2 and n = 5

Let X show the number of heads. Then,

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^5 C_r\left(\frac{1}{2}\right)^5\).

Required probability=P(X= 0) + P(X= 2) + P(X= 4)

Question 3. 7 coins are tossed simultaneously. What is the probability that a tail appears an odd number of times?

Solution: 1/2

Hint: 7 coins being tossed simultaneously is the same as one coin being tossed 7 times.

∴ p = 1/2, q = 1/2 and n = 7.

Let X show the number of tails. Then,

P(X= r) = \({ }^n C_r \cdot p^r \cdot q^{n-r}={ }^7 C_r \cdot\left(\frac{1}{2}\right)^7\)

Required probability=P(X= 1) +P(X= 3) + P(X= 5) + P(X= 7).

Question 4. A coin is tossed 6 times. Find the probability of getting

1. Exactly 4 heads
2. At least 1 head
3. At most 4 heads.

Solution:

1. 15/64
2. 63/64
3. 57/64

Hint: P(a head) = 1/2 and P(not a head) = 1/2

∴ p = 1/2, q=2 and n = 6.

∴ P(X =r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^6 .\)

1. P(exactly 4 heads) = \({ }^6 C_4 \cdot\left(\frac{1}{2}\right)^6\)

2. P(at least1 head) =1- P(no head)

=1- P(0 head) = \(1-P(0 \text { head })=\left[1-{ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6\right]\)

3. P(at the most 4 heads) = P(4 or less than 4 heads)

= 1-P[5 heads or 6 heads] = \(1-\left[\left({ }^6 C_5+{ }^6 C_6\right)\left(\frac{1}{2}\right)^6\right] .\)

Question 5. 10 coins are tossed simultaneously. Find the probability of getting

1. Exactly 3 heads
2. Not more than 4 heads
3. At least 4 heads.

Solution:

1. 15/128
2. 193/512
3. 53/64

Hint: 10 coins being tossed simultaneously is the same as one coin being tossed 10 times.

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^{10} C_r \cdot\left(\frac{1}{2}\right)^{10}\)

1. P(exactly 3 heads) = \({ }^{10} C_3 \cdot\left(\frac{1}{2}\right)^{10} .\)

2. P(not more than 4 heads)

= P(X ≤ 4)

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= \(\left({ }^{10} C_0+{ }^{10} C_1+{ }^{10} C_2+{ }^{10} C_3+{ }^{10} C_4\right)\left(\frac{1}{2}\right)^{10}\)

3. P(at least 4 heads)

= P(4 heads or 5 heads or … or 10 heads)

= 1- P(0 head or 1 head or 2 heads or 3 heads)

= 1- [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

= \(1-\left({ }^{10} C_0+{ }^{10} C_1+{ }^{10} C_2+{ }^{10} C_3\right)\left(\frac{1}{2}\right)^{10}\)

Question 6. A die is thrown 6 times. If getting an even number is a success, find the probability of getting

1. Exactly 5 successes
2. At least 5 successes
3. At most 5 successes.

Solution:

1. 3/32
2. 7/64
3. 63/64

Hint: p = \(\frac{3}{6}=\frac{1}{2}, q=\left(1-\frac{1}{2}\right)=\frac{1}{2} \text { and } n=6 . \)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{n-r}={ }^6 C_5\left(\frac{1}{2}\right)^6\)

1. P(exactly 5 successes) = \({ }^n C_r \cdot p^r \cdot q^{n-r}={ }^6 C_5\left(\frac{1}{2}\right)^6\)

2. P(at least 5 successes) = P[(5 successes) or (6 successes)]

\(\left({ }^6 C_5+{ }^6 C_6\right)\left(\frac{1}{2}\right)^6\)

3. P(at most 5 successes) = P[0 or 1 or 2 or 3 or 5 successes]

= 1- P(6 successes) = \(\left[1-{ }^6 C_6 \cdot\left(\frac{1}{2}\right)^6\right]\)

Question 7. A die is thrown 4 times. “Getting a 1 or a 6” is considered a success. Find the probability of getting

1. Exactly 3 successes
2. At least 2 successes
3. At most 2 successes.

Solution:

1. 8/81
2. 11/27
3. 8/9

Hint: p= \(\frac{2}{6}=\frac{1}{3}, q=\left(1-\frac{1}{3}\right)=\frac{2}{3}\) and n=4.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^4 C_r \cdot\left(\frac{1}{3}\right)^r \cdot\left(\frac{2}{3}\right)^{(4-r)}\).

1. P(exactly 3 successes)= \({ }^4 C_3 \cdot\left(\frac{1}{3}\right)^3 \cdot\left(\frac{2}{3}\right)^1\)

2. P(at least 2 successes) = [P(X = 2) or P(X = 3) or P(X = 4)] =1-[P(X = 0) + P(X=1)].

3. P(at most 2 successes) = P[(X = 0) or (X = l) or (X = 2)].

Question 8. Find the probability of a 4 turning up at least once in two tosses of a fair die.

Solution: 11/36

Hint: p= \(\frac{1}{6}, q=\left(1-\frac{1}{6}\right)=\frac{5}{6}\) and n=2.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^2 C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(2-r)}\).

P[at least one 4] = P(X=1 or X=2) = P(X =1) + P(X = 2)

Question 9. A pair of dice is thrown 4 times. If “getting a doublet’ is considered a success, find the probability of getting 2 successes.

Solution: 25/216

Hint: p= \(\frac{6}{36}=\frac{1}{6}, q=\left(1-\frac{1}{6}\right)=\frac{5}{6} \text { and } n=4 . \)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^4 C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(4-r)}\)

∴ P(X=2)= \({ }^4 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^2\).

Question 10. A pair of dice is thrown 7 times. If ‘getting a total of 7’ is considered a success, find the probability of getting

1. No success
2. Exactly 6 successes
3. At least 6 successes
4. At most 6 successes.

Solution:

1. (5/6)²
2. 35/6⁷
3. 1/6⁵
4. (1-1/6⁷)

Hint: Let E={(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}.

∴ p= \(\frac{6}{36}=\frac{1}{6}, q=\left(1-\frac{1}{6}\right)=\frac{5}{6}\), n=7

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^7 C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(7-r)}\)

1. P(X=0)= \({ }^7 C_0 \cdot\left(\frac{5}{6}\right)^7\).

2. P(X=6)= \({ }^7 C_6 \cdot\left(\frac{1}{6}\right)^6 \cdot\left(\frac{5}{6}\right)\).

3. P(at least 6 successes)=P(6 successes or 7 successes)

= P(X=6)+P(X=7)

= \({ }^7 C_6 \cdot\left(\frac{1}{6}\right)^6 \cdot\left(\frac{5}{6}\right)+{ }^7 C_7 \cdot\left(\frac{1}{6}\right)^7\)

4. P(at most 6 successes)= P(X \≤ 6)

=[1-P(X=7)]= \(1-{ }^7 C_7 \cdot\left(\frac{1}{6}\right)^7\)

Question 11. There are 6% defective items in a large bulk of items. Find the probability that a sample of 8 items will include not more than one defective item.

Solution: (47/50)⁷x(71/50)

Hint: p= \(\frac{6}{100}=\frac{3}{50}, q=\left(1-\frac{3}{50}\right)=\frac{47}{50} \text { and } n=8 \text {. }\)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^8 C_r \cdot\left(\frac{3}{50}\right)^r \cdot\left(\frac{47}{50}\right)^{(8-r)}\)

Required probability = P(0 defective or 1 defective)

=P(X=0)+P(X=1) = \({ }^8 C_0 \cdot\left(\frac{47}{50}\right)^8+{ }^8 C_1 \cdot\left(\frac{3}{50}\right)\left(\frac{47}{50}\right)^7 . \)

Question 12. In a box containing 60 bulbs, 6 are defective. What is the probability that out of a sample of 5 bulbs

1. None is defective,
2. Exactly 2 are defective?

Solution:

1. (9/10)⁵
2. 729/10000

Hint: p = \(\frac{6}{60}=\frac{1}{10}, q=\left(1-\frac{1}{10}\right)=\frac{9}{10}\) and n=5

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{n-r}={ }^5 C_r \cdot\left(\frac{1}{10}\right)^r \cdot\left(\frac{9}{10}\right)^{(5-r)}\)

1. P(none is defective)= \(P(X=0)={ }^5 C_0 \cdot\left(\frac{9}{10}\right)^5\).

2. P(exactly 2 are defective)= \(P(X=2)={ }^5 C_2 \cdot\left(\frac{1}{10}\right)^2 \cdot\left(\frac{9}{10}\right)^3\).

Question 13. The probability that a bulb produced by a factory will fuse after 6 months of use is 0.05. Find the probability that out of 5 such bulbs

1. None will fuse after 6 months of use
2. At least one will fuse after 6 months of use
3. Not more than one will fuse after 6 months of use.

Solution:

1. (19/20)⁵
2. 1-(19/20)⁵
3. (19/20)⁴x(6/5)

Hint: p= \(\frac{1}{20}, q=\left(1-\frac{1}{20}\right)=\frac{19}{20}\) and n=5.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^5 C_r \cdot\left(\frac{1}{20}\right)^r \cdot\left(\frac{19}{20}\right)^{(5-r)}\)

1. P(X=0)= \({ }^5 C_0 \cdot\left(\frac{19}{20}\right)^5\).

2. \(P(X≥1)=1-P(X<1)=1-P(X=0)\)

= \(1-{ }^5 C_0 \cdot\left(\frac{19}{20}\right)^5\) \text

3. \(P(X \leq 1)=P(X=0)+P(X=1)\)

= \({ }^5 C_0 \cdot\left(\frac{19}{20}\right)^5+{ }^5 C_1 \cdot\left(\frac{1}{20}\right)\left(\frac{19}{20}\right)^4\)

Question 14. In the items produced by a factory, there are 10% defective items. A sample of 6 items is randomly chosen. Find the probability that this sample contains

1. Exactly 2 defective items,
2. Not more than 2 defective items,
3. At least 3 defective items.

Solution:

1. 3/20 x (9/10)⁴
2. 3/2 x (19/20)⁴
3. 1-[3/2x(19/20)⁴]

Hint: p= \(\frac{10}{100}\)=\(\frac{1}{10}\), q= \(\left(1-\frac{1}{10}\right)\)=\(\frac{9}{10}\) and n=6

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{10}\right)^r \cdot\left(\frac{9}{10}\right)^{(6-n)}\)

1. P(X=2)= \({ }^6 C_2 \cdot\left(\frac{1}{10}\right)^2 \cdot\left(\frac{9}{10}\right)^4\).

2. \(P(X \leq 2)={ }^6 C_0 \cdot\left(\frac{9}{10}\right)^6+{ }^6 C_1 \cdot \frac{1}{10} \times\left(\frac{9}{10}\right)^5+{ }^6 C_2 \cdot \frac{1}{100} \times\left(\frac{9}{10}\right)^4\).

3. P(X≥3)=1-P(X<3)=1-P(X≤2)

Question 15. Assume that on average one telephone number out of 15, called between 3 p.m. and 4 p.m. on weekdays, will be busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?

Solution: 1-(14/15)⁴ .(59/45)

Hint: p= \(\frac{1}{15}, q=\left(1-\frac{1}{15}\right)=\frac{14}{15} \text { and } n=6\)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n}={ }^6 C_r \cdot\left(\frac{1}{15}\right)^r \cdot\left(\frac{14}{15}\right)^{(6-n)} \text {. }\)

P(X≥3)=1-P(X<3)=1-[P(X=0)+P(X=1)+P(X=2)]

= \(1-\left[{ }^6 C_0 \cdot\left(\frac{14}{15}\right)^6+{ }^6 C_1 \cdot\left(\frac{1}{15}\right) \cdot\left(\frac{14}{15}\right)^5+{ }^6 C_2 \cdot\left(\frac{1}{15}\right)^2 \cdot\left(\frac{14}{15}\right)^4\right]\)

= \(1-\left(\frac{14}{15}\right)^4\left(\frac{59}{45}\right)\)

Question 16. Three cars participate in a race. The probability that any one of them has an accident is 0.1. Find the probability that all the cars reach the finishing line without any accident.

Solution: 729/1000

Hint: p= \(\frac{1}{10}, q=\left(1-\frac{1}{10}\right)=\frac{9}{10} \text { and } n=3 \)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^3 C_r \cdot\left(\frac{1}{10}\right)^r \cdot\left(\frac{9}{10}\right)^{\beta-n}\)

P(X=0)= \({ }^3 C_0 \cdot\left(\frac{9}{10}\right)^3\)

Question 17. Past records show that 80% of the operations performed by a certain doctor were successful. If he performs 4 operations in a day, what is the probability that at least 3 operations will be successful?

Solution: 512/625

Hint: p= \(\frac{80}{100}=\frac{4}{5}, q=\left(1-\frac{4}{5}\right)=\frac{1}{5} \text { and } n=4 \text {. }\)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^4 C_r \cdot\left(\frac{4}{5}\right)^r \cdot\left(\frac{1}{5}\right)^{(4-n)} \text {. }\)

P(X≥3)=P(X=3)+P(X=4)

= \(\left[{ }^4 C_3 \cdot\left(\frac{4}{5}\right)^3 \cdot\left(\frac{1}{5}\right)+{ }^4 C_4 \cdot\left(\frac{4}{5}\right)^4\right]\)

Question 18. The probability of a man hitting a target is (1/4). If he fires 7 times, what is the probability of his hitting the target at least twice?

Solution: 4547/8192

Hint: p=\(\frac{1}{4}, q=\left(1-\frac{1}{4}\right)=\frac{3}{4}\) and n=7

P(X=r)=\({ }^n C_r \cdot p^r \cdot q^{(n-n}={ }^7 C_r \cdot\left(\frac{1}{4}\right)^r \cdot\left(\frac{3}{4}\right)^{(n-n)}\)

P(X≥2)=1-P(X<2) =1-P[(X = 0) or (X=1)] =1-[P(X = 0) + P(X=1)].

Question 19. In a hurdles race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is (5/6). What is the probability that he will knock down fewer than 2 hurdles?

Solution: \(\frac{5^{10}}{\left(2 \times 6^9\right)}\)

Hint: p=P(knocking down 1 hurdle)= \(\left(1-\frac{5}{6}\right)=\frac{1}{6}\), q=\(\frac{5}{6}\) and n=10.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^{10} C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(10-r)}\)

P(X<2)=P(X=0)+P(X=1)

= \(\left[{ }^{10} C_0 \cdot\left(\frac{5}{6}\right)^{10}+{ }^{10} C_1 \cdot\left(\frac{1}{6}\right)^1 \cdot\left(\frac{5}{9}\right)^9\right]\)

Question 20. A man can hit a bird, once in 3 shots. On this assumption he fires 3 shots. What is the chance that at least one bird is hit?

Solution: 19/27

Hint: p = \(\frac{1}{3}\), q= \(\left(1-\frac{1}{3}\right)=\frac{2}{3}\) and n=3.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^3 C_r \cdot\left(\frac{1}{3}\right)^r \cdot\left(\frac{2}{3}\right)^{(3-r)} \cdot\)

⇒ \(P(X \geq 1) =P(X=1)+P(X=2)+P(X=3)\)

= \(\left[{ }^3 C_1 \times \frac{1}{3} \times\left(\frac{2}{3}\right)^2\right]+\left[{ }^3 C_2 \times\left(\frac{1}{3}\right)^2 \times \frac{2}{3}\right]+\left[{ }^3 C_3 \times\left(\frac{1}{3}\right)^3\right]\)

= \(\left(\frac{4}{9}+\frac{2}{9}+\frac{1}{9}\right)=\frac{7}{9}\)

Question 21. If the probability that a man aged 60 will live to be 70 is 0.65, what is the probability that out of 10 men, now 60, at least 8 will live to be 70?

Solution: 0.2615

Hint: \(P(X \geq 8)=P(X=8)+P(X=9)+P(X=10)\)

= \(\left[{ }^{10} C_8 \cdot\left(\frac{13}{20}\right)^8 \cdot\left(\frac{7}{20}\right)^2+{ }^{10} C_9 \cdot\left(\frac{13}{20}\right)^9 \cdot\left(\frac{7}{20}\right)+{ }^{10} C_{10} \cdot\left(\frac{13}{20}\right)^{10}\right]\)

= \(\left[{ }^{10} C_2 \cdot\left(\frac{13}{20}\right)^8 \cdot \frac{49}{400}+{ }^{10} C_1 \cdot\left(\frac{13}{20}\right)^9 \cdot \frac{7}{20}+\left(\frac{13}{20}\right)^{10}\right] \text {. }\)

∴ P=\(\left(\frac{13}{20}\right)^8 \cdot\left[\frac{441}{80}+\frac{91}{40}+\frac{169}{400}\right]=\left[\left(\frac{13}{20}\right)^8 \times \frac{821}{100}\right] \text {. }\)

⇒ \(\log P=[8(\log 13-\log 20)+\log 821-\log 100]\)

=[-1.4968+2.9143-2]=-0.5825 = \(\overline{1}\).4175

⇒ P= antilog \((\overline{1} .4175)\)

Question 22. A bag contains 5 white, 7 red, and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that

1. None is white,
2. All are white,
3. At least one is white.

Solution:

1. 81/256
2. 1/256
3. 175/256

Hint:

P(white)=\(\frac{5}{20}=\frac{1}{4}, P(\text { non-white })=\frac{3}{4}\)

∴ p= \(\frac{1}{4}, q=\frac{3}{4} \text { and } n=4\)

P(X=r)=\({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^4 C_r \cdot\left(\frac{1}{4}\right)^r \cdot\left(\frac{3}{4}\right)^{(4-r)}\)

1. P(X=0)= \({ }^4 C_0 \cdot\left(\frac{3}{4}\right)^4\).

2. P(X=4)=\({ }^4 C_4 \cdot\left(\frac{1}{4}\right)^4\).

3. \(P(X \geq 1)=1-P(X<1)=1-P(X=0)\).

Question 23. A policeman fires 6 bullets at a burglar. The probability that the burglar will be hit by a bullet is 0.6. What is the probability that the burglar is still unhurt?

Solution: 0.004096

Hint: p=\(\frac{6}{10}=\frac{3}{5}, q=\left(1-\frac{3}{5}\right)=\frac{2}{5}\) and n=6.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{3}{5}\right)^r \cdot\left(\frac{2}{5}\right)^{(6-r)}\)

∴ P(X=0)= \({ }^6 C_0 \cdot\left(\frac{2}{5}\right)^6\)

Question 24. A die is tossed thrice. A success is 1 or 6 on a toss. Find the mean and variance of successes.
Solution: μ = 1, σ² = 2/3

Hint: p= \(\frac{2}{6}=\frac{1}{3}, q=\left(1-\frac{1}{3}\right)=\frac{2}{3} \text { and } n=3\)

∴ μ= np and σ²= npq

Question 25. A die is thrown 100 times. Getting an even number is considered a success. Find the mean and variance of sucess.

Solution: μ = 50, σ² = 25

Question 26. Determine the binomial distribution whose mean is 9 and variance is 6.

Solution: \({ }^{27} C_r \cdot\left(\frac{1}{3}\right)^r \cdot\left(\frac{2}{3}\right)^{(27-r)}, \text { where } r=0,1,2,3, \ldots, 27\)

Question 27. Find the binomial distribution whose mean is 5 and variance is 2.5.

Solution: \({ }^{10} C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(10-r)}, 0 \leq r \leq 10\)

Question 28. The mean and variance of a binomial distribution are 4 and (4/3) respectively. Find P(X≥1).

Solution: 728/729

Hint: \(\left(n p=4 \text { and } n p q=\frac{4}{3}\right) \Rightarrow q=\frac{1}{3}\)

∴ p=(1-q)=\(\left(1-\frac{1}{3}\right)=\frac{2}{3}\)

and \(n=\frac{4}{p}=\left(4 \times \frac{3}{2}\right)=6 \)

∴ P(X=r)=\({ }^n C_r \cdot p^r \cdot q^{(n-r)}\)

= \(P(X \geq 1)=1-P(X=0) \cdot\left(\frac{2}{3}\right)^r \cdot\left(\frac{1}{3}\right)^{(6-r)}\)

= \(1-{ }^6 C_0 \cdot\left(\frac{2}{3}\right)^0 \cdot\left(\frac{1}{3}\right)^6\)

= \(\left(1-\frac{1}{3^6}\right)=\frac{728}{729}\)

Question 29. For a binomial distribution, the mean is 6 and the standard deviation is √2 ‘. Find the probability of getting 5 successes.

Solution: \({ }^9 C_5 \cdot\left(\frac{2}{3}\right)^5 \cdot\left(\frac{1}{3}\right)^4\)

Hint: np = 6 and npq = 2

⇒ q = \(\frac{1}{3}\) and p = \(\frac{2}{3}\)

⇒ \(\left(n \times \frac{2}{3}\right)=6 \Rightarrow n=\left(6 \times \frac{3}{2}\right)=9\)

∴ \(P(5 \text { successes })={ }^9 C_5 \cdot\left(\frac{2}{3}\right)^5 \cdot\left(\frac{1}{3}\right)^4\)

Question 30. In a binomial distribution, the sum and the product of the mean and the variance are (25/3) and (50/3) respectively. Find the distribution.

Solution: \({ }^{15} C_r \cdot\left(\frac{1}{3}\right)^r \cdot\left(\frac{2}{3}\right)^{(15-r)}\)

Question 31. Obtain the binomial distribution whose mean is 10 and the standard deviation is 2√2.

Solution: \({ }^{50} C_r \cdot\left(\frac{1}{5}\right)^r \cdot\left(\frac{4}{5}\right)^{(50-r)}, 0 \leq r \leq 50\)

Question 32. Bring out the fallacy, if any, in the following statement: ‘The mean of a binomial distribution is 6 and its variance is 9’.

Solution: The probability of getting a failure (i.e., q) cannot be greater than1

Hint: np = 6 and npq = 9

⇒ q = \(\frac{n p q}{n p}=\frac{9}{6}=\frac{3}{2}\)

But, q cannot be greater than 1.