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NCERT Exemplar · Class 12 Mathematics Continuity and Differentiability

106 questions · 106 still being checked

EXERCISE 5.3 31–40 (part 4 of 11)

  1. Differentiate each of the following w.r.t. \(\displaystyle x\) (Exercises $\displaystyle 25$ to $\displaystyle 43$) :

    Exercise 31

    cos⁡(tan⁡x+1)\displaystyle \cos (\tan \sqrt{x+1})

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    NCERT’s answer
    \(\displaystyle \frac{-1}{2 \sqrt{x+1}} \sin (\tan \sqrt{x+1}) \sec ^2(\sqrt{x+1})\)
    \[y = \cos\left(\tan\sqrt{x+1}\right) \] \[\frac{dy}{dx} = -\sin\left(\tan\sqrt{x+1}\right)\cdot\frac{d}{dx}\tan\sqrt{x+1} \] \[\frac{d}{dx}\tan\sqrt{x+1} = \sec^2\sqrt{x+1}\cdot\frac{1}{2\sqrt{x+1}} \] \[\frac{dy}{dx} = -\frac{\sec^2\sqrt{x+1}\,\sin\left(\tan\sqrt{x+1}\right)}{2\sqrt{x+1}} \] Answer: \(\displaystyle -\dfrac{\sec^2\sqrt{x+1}\,\sin\left(\tan\sqrt{x+1}\right)}{2\sqrt{x+1}} \)

  2. Exercise 32

    sin⁡x2+sin⁡2x+sin⁡2(x2)\displaystyle \sin x^2+\sin ^2 x+\sin ^2\left(x^2\right)

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    NCERT’s answer
    \(\displaystyle 2 x \cos (x)^2+2 x \sin \left(2 x^2\right)+\sin 2 x\)
    \[y = \sin x^2 + \sin^2 x + \sin^2\left(x^2\right) \] \[\frac{d}{dx}\sin x^2 = 2x\cos x^2 \] \[\frac{d}{dx}\sin^2 x = 2\sin x\cos x = \sin 2x \] \[\frac{d}{dx}\sin^2\left(x^2\right) = 2\sin x^2\cos x^2\cdot 2x = 2x\sin 2x^2 \] \[\frac{dy}{dx} = 2x\cos x^2 + \sin 2x + 2x\sin 2x^2 \] Answer: \(\displaystyle 2x\cos x^2 + \sin 2x + 2x\sin 2x^2 \)

  3. Exercise 33

    sin⁡−1(1x+1)\displaystyle \sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)

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    NCERT’s answer
    \(\displaystyle \frac{-1}{2 \sqrt{x}(x+1)}\)
    Chain rule with \(\displaystyle u=(x+1)^{-1/2}\); \(\displaystyle \sin^{-1}u\) needs \(\displaystyle |u|<1\), so \(\displaystyle x>0\). \[\frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx} \] \[1-u^2=1-\frac{1}{x+1}=\frac{x}{x+1} \] \[\frac{du}{dx}=-\frac12\,(x+1)^{-3/2} \] \[\frac{dy}{dx}=\sqrt{\frac{x+1}{x}}\cdot\left(-\frac{1}{2(x+1)^{3/2}}\right)=-\frac{1}{2(x+1)\sqrt{x}} \] Answer: \(\displaystyle -\dfrac{1}{2(x+1)\sqrt{x}}\), for \(\displaystyle x>0\).

  4. Exercise 34

    (sin⁡x)cos⁡x\displaystyle (\sin x)^{\cos x}

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    NCERT’s answer
    \(\displaystyle (\sin x)^{\cos x}\left[\frac{\cos ^2 x}{\sin x}-\sin x . \log \sin x\right]\)
    Take logarithms (the power needs \(\displaystyle \sin x>0\)). \[\ln y=\cos x\,\ln\sin x \] \[\frac1y\frac{dy}{dx}=-\sin x\,\ln\sin x+\cos x\cdot\frac{\cos x}{\sin x} \] \[\frac{dy}{dx}=(\sin x)^{\cos x}\left[\cos x\cot x-\sin x\,\ln\sin x\right] \] Answer: \(\displaystyle (\sin x)^{\cos x}\left[\cos x\cot x-\sin x\,\ln\sin x\right]\).

  5. Exercise 35

    sin⁡mx⋅cos⁡nx\displaystyle \sin ^m x \cdot \cos ^n x

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    NCERT’s answer
    \(\displaystyle \sin ^{m x} x \cos ^n x(-n \tan x+m \cot x)\)
    Product rule, then chain rule on each factor. \[\frac{dy}{dx}=m\sin^{m-1}x\cos x\cdot\cos^n x+\sin^m x\cdot n\cos^{n-1}x\,(-\sin x) \] \[\frac{dy}{dx}=\sin^{m-1}x\,\cos^{n-1}x\left(m\cos^2x-n\sin^2x\right) \] Answer: \(\displaystyle \sin^{m-1}x\cos^{n-1}x\left(m\cos^2x-n\sin^2x\right)\).

  6. Exercise 36

    (x+1)2(x+2)3(x+3)4\displaystyle (x+1)^2(x+2)^3(x+3)^4

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    NCERT’s answer
    \(\displaystyle (x+1)(x+2)^2(x+3)^3\left[9 x^2+34 x+29\right]\)
    Logarithmic differentiation. \[\ln|y|=2\ln|x+1|+3\ln|x+2|+4\ln|x+3| \] \[\frac1y\frac{dy}{dx}=\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3} \] \[\frac1y\frac{dy}{dx}=\frac{2(x+2)(x+3)+3(x+1)(x+3)+4(x+1)(x+2)}{(x+1)(x+2)(x+3)} \] \[2(x^2+5x+6)+3(x^2+4x+3)+4(x^2+3x+2)=9x^2+34x+29 \] \[\frac{dy}{dx}=(x+1)(x+2)^2(x+3)^3\,(9x^2+34x+29) \] Answer: \(\displaystyle (x+1)(x+2)^2(x+3)^3\,(9x^2+34x+29)\).

  7. Exercise 37

    cos⁡−1(sin⁡x+cos⁡x2),−π4<x<π4\displaystyle \cos ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right), \frac{-\pi}{4}<x<\frac{\pi}{4}

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    NCERT’s answer
    \(\displaystyle -1\)
    Reduce the argument to a single cosine. \[\frac{\sin x+\cos x}{\sqrt2}=\cos x\cos\frac{\pi}{4}+\sin x\sin\frac{\pi}{4}=\cos\left(x-\frac{\pi}{4}\right) \] \[-\frac{\pi}{4}<x<\frac{\pi}{4}\ \Rightarrow\ -\frac{\pi}{2}<x-\frac{\pi}{4}<0 \] \[\cos^{-1}(\cos\theta)=-\theta \quad (-\pi\le\theta\le0) \] \[y=-\left(x-\frac{\pi}{4}\right)=\frac{\pi}{4}-x \] \[\frac{dy}{dx}=-1 \] Answer: \(\displaystyle -1\).

  8. Exercise 38

    tan⁡−1(1−cos⁡x1+cos⁡x),−π4<x<π4\displaystyle \tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\frac{\pi}{4}<x<\frac{\pi}{4}

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Half-angle form of the radicand. \[\frac{1-\cos x}{1+\cos x}=\frac{2\sin^2\frac x2}{2\cos^2\frac x2}=\tan^2\frac x2 \] \[y=\tan^{-1}\sqrt{\tan^2\frac x2}=\tan^{-1}\left|\tan\frac x2\right|,\qquad \left|\frac x2\right|<\frac{\pi}{8} \] \[0<x<\frac{\pi}{4}:\quad y=\tan^{-1}\!\left(\tan\frac x2\right)=\frac x2,\quad \frac{dy}{dx}=\frac12 \] \[-\frac{\pi}{4}<x<0:\quad y=\tan^{-1}\!\left(\tan\left(-\frac x2\right)\right)=-\frac x2,\quad \frac{dy}{dx}=-\frac12 \] Since \(\displaystyle y=\dfrac{|x|}{2}\), there is no derivative at \(\displaystyle x=0\). Answer: \(\displaystyle \dfrac12\) for \(\displaystyle 0<x<\dfrac{\pi}{4}\); \(\displaystyle -\dfrac12\) for \(\displaystyle -\dfrac{\pi}{4}<x<0\).

  9. Exercise 39

    tan⁡−1(sec⁡x+tan⁡x),−π2<x<π2\displaystyle \tan ^{-1}(\sec x+\tan x),-\frac{\pi}{2}<x<\frac{\pi}{2}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}\)
    Write \(\displaystyle \sec x+\tan x\) as one tangent. \[\sec x+\tan x=\frac{1+\sin x}{\cos x}=\frac{1+\cos\left(\dfrac{\pi}{2}-x\right)}{\sin\left(\dfrac{\pi}{2}-x\right)} \] \[=\frac{2\cos^2\left(\dfrac{\pi}{4}-\frac x2\right)}{2\sin\left(\dfrac{\pi}{4}-\frac x2\right)\cos\left(\dfrac{\pi}{4}-\frac x2\right)}=\cot\left(\frac{\pi}{4}-\frac x2\right)=\tan\left(\frac{\pi}{4}+\frac x2\right) \] \[-\frac{\pi}{2}<x<\frac{\pi}{2}\ \Rightarrow\ 0<\frac{\pi}{4}+\frac x2<\frac{\pi}{2} \] \[y=\tan^{-1}\tan\left(\frac{\pi}{4}+\frac x2\right)=\frac{\pi}{4}+\frac x2 \] \[\frac{dy}{dx}=\frac12 \] Answer: \(\displaystyle \dfrac12\).

  10. Exercise 40

    tan⁡−1(acos⁡x−bsin⁡xbcos⁡x+asin⁡x),−π2<x<π2\displaystyle \tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right),-\frac{\pi}{2}<x<\frac{\pi}{2} and abtan⁡x>−1\displaystyle \frac{a}{b} \tan x>-1

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    NCERT’s answer
    \(\displaystyle -1\)
    Divide numerator and denominator by \(\displaystyle b\cos x\) (\(\displaystyle b\ne0\), \(\displaystyle \cos x\ne0\)). \[\frac{a\cos x-b\sin x}{b\cos x+a\sin x}=\frac{\frac ab-\tan x}{1+\frac ab\tan x} \] \[\tan^{-1}A-\tan^{-1}B=\tan^{-1}\frac{A-B}{1+AB}\quad(AB>-1) \] With \(\displaystyle A=\frac ab\), \(\displaystyle B=\tan x\), the condition \(\displaystyle \frac ab\tan x>-1\) is exactly \(\displaystyle AB>-1\). \[y=\tan^{-1}\frac ab-\tan^{-1}(\tan x)=\tan^{-1}\frac ab-x \quad\left(-\frac{\pi}{2}<x<\frac{\pi}{2}\right) \] \[\frac{dy}{dx}=-1 \] Answer: \(\displaystyle -1\).