WBCHSE Solutions For Class 12 Maths Differential Calculus

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Differential Calculus

Differential Calculus Exercise 1 Multiple Choice Questions

Question 1. If y = 2x,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. xx log x*
  2. xx (1+log x)
  3. x (1+log x)
  4. None of these

Answer: 2. xx (1+log x)

log y= xx 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)

WBCHSE Solutions For Class 12 Maths Differential Calculus

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)logx, 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(xx),then \(\frac{d y}{d x}\) = ?

  1. xxcos (xx)
  2. xx cos xx (1+logx)
  3. xxcos xxlog x
  4. None of these

Answer: 2. xx cos xx (1+logx)

Let xx = 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 ex+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. xy= yx,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. \(\frac{y(y-x \log y)}{x(x-y \log x)}\)

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

Question 16. If xpyq= (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 = \(\cos ^{-1}\left(\frac{x^2-1}{x^2+1}\right)\), then \(\frac{d y}{d x}\) = ?

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

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

Putting x= cot θ we get:

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

= 2θ= 2 cot-1x

Question 34. 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 35. 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 36. 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 37. 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 38. 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 39. 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 40. 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 41. 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 42. 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 43. 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 44. 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 45. 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 46. 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 47. 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 48. 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 49. 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 50. 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 51. 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 52.  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 53. \(\left\{\begin{array}{l}
x \sin \frac{1}{x}, \text { if } x \neq 0 \\
0, \text { when } x=0 .
\end{array}\right.\) 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 54. 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 55. Let f(x) =x3/2 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 56. 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 57. \(
\text { The function } f(x)=\left\{\begin{array}{l}
1+x, \text { when } x \leq 2 \\
5-x, \text { when } x>2
\end{array}\right. \text { is }\)

  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 58. \(
\text { If the function } f(x)=\left\{\begin{array}{r}
k x+5, \text { when } x \leq 2 \\
x-1, \text { when } x>2
\end{array}\right. \text { is }\) 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 59. 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 60. 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 61. 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 62. At x= 2,f(x) = [x] is

  1. Continuous but not differentiable
  2. Differentiablebutnot 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 63. \(
\text { Let } f(x)=\left\{\begin{array}{r}
\frac{x^2-2 x-3}{x+1^2}, \text { when } x \neq-1 \\
k, \text { when } x=-1
\end{array}\right.\)

  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 64. 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 65. 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 66. 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 67. 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.

Class-12-Maths Differential Calculus Question 67 Decreasing Positive

 

Question 68. 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.

Class-12-Maths Differential Calculus Increasing Positive

 

Question 69. 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 70. 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 71. 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 72. 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 73. f(x) = xx 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 74. f(x) = x2ex 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 75. 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 76. 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 77. 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,\(\frac{1}{3}\)).

Class-12-Maths Differential Calculus Question 77 Increasing Positive

 

78. 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 79. 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 80. 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 81. 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 82. When x is positive, the minimum value of xx is

  1. ee
  2. e1/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 83. 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 84. 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 85. 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 86. 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 87. 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 88. 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 89. The least value off (x) = (ex+ e–x) is

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

Answer: 3. 0

f'(x) = ex– e–x and f”(x) = ex+ e–x

f'(x) = 0  ⇒ ex– e–x = 0

⇒  ex= e–x

⇒ e2x= 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

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