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NCERT Exemplar · Class 12 Mathematics Continuity and Differentiability

106 questions · 106 still being checked

EXERCISE 5.3 41–50 (part 5 of 11)

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

    Exercise 41

    sec⁡−1(14x3−3x),0<x<12\displaystyle \sec ^{-1}\left(\frac{1}{4 x^3-3 x}\right), 0<x<\frac{1}{\sqrt{2}}

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    On this interval \(\displaystyle u=4x^3-3x\) lies in \(\displaystyle [-1,0)\), so \(\displaystyle \sec^{-1}\frac1u=\cos^{-1}u\): \[y=\cos^{-1}(4x^3-3x) \] \[\frac{dy}{dx}=\frac{-(12x^2-3)}{\sqrt{1-(4x^3-3x)^2}} \] \[1-(4x^3-3x)^2=(1-x^2)(1-4x^2)^2 \] \[\frac{dy}{dx}=\frac{-3(4x^2-1)}{\sqrt{1-x^2}\,\left|4x^2-1\right|} \] The sign of \(\displaystyle 4x^2-1\) changes at \(\displaystyle x=\tfrac12\), where \(\displaystyle y=\cos^{-1}(-1)=\pi\) has a corner: \[0<x<\tfrac12:\quad \frac{dy}{dx}=\frac{3}{\sqrt{1-x^2}} \] \[\tfrac12<x<\tfrac1{\sqrt2}:\quad \frac{dy}{dx}=-\frac{3}{\sqrt{1-x^2}} \] Answer: \(\displaystyle \dfrac{3}{\sqrt{1-x^2}}\) for \(\displaystyle 0<x<\tfrac12\); \(\displaystyle -\dfrac{3}{\sqrt{1-x^2}}\) for \(\displaystyle \tfrac12<x<\tfrac1{\sqrt2}\); not differentiable at \(\displaystyle x=\tfrac12\).

  2. Exercise 42

    tan⁡−1(3a2x−x3a3−3ax2),−13<xa<13\displaystyle \tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right), \frac{-1}{\sqrt{3}}<\frac{x}{a}<\frac{1}{\sqrt{3}}

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    NCERT’s answer
    \(\displaystyle \frac{3 a}{a^2+x^2}\)
    Put \(\displaystyle x=a\tan\theta\), so \(\displaystyle \theta=\tan^{-1}\frac xa\in\left(-\frac\pi6,\frac\pi6\right)\): \[\frac{3a^2x-x^3}{a^3-3ax^2}=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=\tan3\theta \] \[y=\tan^{-1}(\tan3\theta)=3\theta \quad\left(3\theta\in\left(-\tfrac\pi2,\tfrac\pi2\right)\right) \] \[y=3\tan^{-1}\frac xa \] \[\frac{dy}{dx}=\frac{3}{1+\dfrac{x^2}{a^2}}\cdot\frac1a=\frac{3a}{a^2+x^2} \] Answer: \(\displaystyle \dfrac{3a}{a^2+x^2}\)

  3. Exercise 43

    tan⁡−1(1+x2+1−x21+x2−1−x2),−1<x<1,x≠0\displaystyle \tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right),-1<x<1, x \neq 0

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    NCERT’s answer
    \(\displaystyle \frac{-x}{\sqrt{1-x^4}}\)
    Put \(\displaystyle x^2=\cos2\theta\), so \(\displaystyle 0<\theta<\frac\pi4\): \[\sqrt{1+x^2}=\sqrt2\cos\theta,\qquad \sqrt{1-x^2}=\sqrt2\sin\theta \] \[y=\tan^{-1}\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=\tan^{-1}\frac{1+\tan\theta}{1-\tan\theta} \] \[y=\tan^{-1}\tan\left(\frac\pi4+\theta\right)=\frac\pi4+\theta \quad\left(\tfrac\pi4+\theta\in\left(\tfrac\pi4,\tfrac\pi2\right)\right) \] \[y=\frac\pi4+\frac12\cos^{-1}x^2 \] \[\frac{dy}{dx}=\frac12\cdot\frac{-1}{\sqrt{1-x^4}}\cdot2x \] Answer: \(\displaystyle -\dfrac{x}{\sqrt{1-x^4}}\)
  4. Find \(\displaystyle \frac{dy}{dx}\) of each of the functions expressed in parametric form in Exercises from $\displaystyle 44$ to 48.

    Exercise 44

    x=t+1t,y=t−1t\displaystyle x=t+\frac{1}{t}, y=t-\frac{1}{t}

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    NCERT’s answer
    \(\displaystyle \frac{t^2+1}{t^2-1}\)
    \[\frac{dx}{dt}=1-\frac1{t^2},\qquad \frac{dy}{dt}=1+\frac1{t^2} \] \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{1+\frac1{t^2}}{1-\frac1{t^2}} \] \[\frac{dy}{dx}=\frac{t^2+1}{t^2-1} \] Answer: \(\displaystyle \dfrac{t^2+1}{t^2-1}\), \(\displaystyle t\neq0,\pm1\)

  5. Exercise 45

    x=eθ(θ+1θ),y=e−θ(θ−1θ)\displaystyle x=e^\theta\left(\theta+\frac{1}{\theta}\right), y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)

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    NCERT’s answer
    \(\displaystyle e^{-2 \theta}\left(\frac{-\theta^3+\theta^2+\theta+1}{\theta^3+\theta^2+\theta-1}\right)\)
    \[\frac{dx}{d\theta}=e^\theta\left(\theta+\frac1\theta\right)+e^\theta\left(1-\frac1{\theta^2}\right)=\frac{e^\theta}{\theta^2}\left(\theta^3+\theta^2+\theta-1\right) \] \[\frac{dy}{d\theta}=-e^{-\theta}\left(\theta-\frac1\theta\right)+e^{-\theta}\left(1+\frac1{\theta^2}\right)=\frac{e^{-\theta}}{\theta^2}\left(-\theta^3+\theta^2+\theta+1\right) \] \[\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=e^{-2\theta}\,\frac{1+\theta+\theta^2-\theta^3}{\theta^3+\theta^2+\theta-1} \] Answer: \(\displaystyle e^{-2\theta}\,\dfrac{1+\theta+\theta^2-\theta^3}{\theta^3+\theta^2+\theta-1}\)

  6. Exercise 46

    x=3cos⁡θ−2cos⁡3θ,y=3sin⁡θ−2sin⁡3θ\displaystyle x=3 \cos \theta-2 \cos ^3 \theta, y=3 \sin \theta-2 \sin ^3 \theta.

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    NCERT’s answer
    \(\displaystyle \cot \theta\)
    \[\frac{dx}{d\theta}=-3\sin\theta+6\cos^2\theta\sin\theta=3\sin\theta\,(2\cos^2\theta-1)=3\sin\theta\cos2\theta \] \[\frac{dy}{d\theta}=3\cos\theta-6\sin^2\theta\cos\theta=3\cos\theta\,(1-2\sin^2\theta)=3\cos\theta\cos2\theta \] \[\frac{dy}{dx}=\frac{3\cos\theta\cos2\theta}{3\sin\theta\cos2\theta}=\cot\theta \] Answer: \(\displaystyle \cot\theta\), where \(\displaystyle \sin\theta\ne0,\ \cos2\theta\ne0\)

  7. Exercise 47

    sin⁡x=2t1+t2,tan⁡y=2t1−t2\displaystyle \sin x=\frac{2 t}{1+t^2}, \quad \tan y=\frac{2 t}{1-t^2}.

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    NCERT’s answer
    $\displaystyle 1$
    Put \(\displaystyle t=\tan\varphi\) and use the double-angle formulae. \[\sin x=\frac{2\tan\varphi}{1+\tan^2\varphi}=\sin2\varphi,\qquad \tan y=\frac{2\tan\varphi}{1-\tan^2\varphi}=\tan2\varphi \] \[x=y=2\varphi=2\tan^{-1}t \] \[\frac{dx}{dt}=\frac{dy}{dt}=\frac{2}{1+t^2} \] \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2/(1+t^2)}{2/(1+t^2)}=1 \] Answer: \(\displaystyle 1\)

  8. Exercise 48

    x=1+log⁡tt2,y=3+2log⁡tt\displaystyle x=\frac{1+\log t}{t^2}, \quad y=\frac{3+2 \log t}{t}.

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    NCERT’s answer
    \(\displaystyle t\)
    \[\frac{dx}{dt}=\frac{\frac1t\cdot t^2-(1+\log t)\cdot2t}{t^4}=\frac{-(1+2\log t)}{t^3} \] \[\frac{dy}{dt}=\frac{\frac2t\cdot t-(3+2\log t)\cdot1}{t^2}=\frac{-(1+2\log t)}{t^2} \] \[\frac{dy}{dx}=\frac{-(1+2\log t)/t^2}{-(1+2\log t)/t^3}=t \] Answer: \(\displaystyle t\), where \(\displaystyle 1+2\log t\ne0\)

  9. Exercise 49

    If x=ecos⁡2t\displaystyle x=e^{\cos 2 t} and y=esin⁡2t\displaystyle y=e^{\sin 2 t}, prove that dydx=−ylog⁡xxlog⁡y\displaystyle \frac{d y}{d x}=\frac{-y \log x}{x \log y}.

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    Take logarithms of the given relations.\[\log x = \cos 2t, \qquad \log y = \sin 2t \]\[\frac{dx}{dt} = e^{\cos 2t}\,(-2\sin 2t) = -2x\log y \]\[\frac{dy}{dt} = e^{\sin 2t}\,(2\cos 2t) = 2y\log x \]\[\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2y\log x}{-2x\log y} \]Answer: \(\displaystyle \dfrac{dy}{dx} = \dfrac{-y\log x}{x\log y}\)

  10. Exercise 50

    If x=asin⁡2t(1+cos⁡2t)\displaystyle x=a \sin 2 t(1+\cos 2 t) and y=bcos⁡2t(1−cos⁡2t)\displaystyle y=b \cos 2 t(1-\cos 2 t), show that (dydx)at t=π4=ba\displaystyle \left(\frac{d y}{d x}\right)_{\text {at } t=\frac{\pi}{4}}=\frac{b}{a}.

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    Differentiate both with the product rule.\[\frac{dx}{dt} = a\left[2\cos 2t\,(1+\cos 2t) - 2\sin^2 2t\right] = 2a\,(\cos 2t + \cos 4t) \]\[\frac{dy}{dt} = b\left[-2\sin 2t\,(1-\cos 2t) + 2\sin 2t\cos 2t\right] = 2b\,(\sin 4t - \sin 2t) \]At \(\displaystyle t=\dfrac{\pi}{4}\): \(\displaystyle 2t=\dfrac{\pi}{2}\), \(\displaystyle 4t=\pi\).\[\frac{dx}{dt} = 2a\left(\cos\frac{\pi}{2} + \cos\pi\right) = -2a \]\[\frac{dy}{dt} = 2b\left(\sin\pi - \sin\frac{\pi}{2}\right) = -2b \]\[\left(\frac{dy}{dx}\right)_{t=\pi/4} = \frac{-2b}{-2a} \]Answer: \(\displaystyle \dfrac{b}{a}\)