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NCERT Exemplar · Class 11 Mathematics Limits and Derivatives

80 questions · 80 still being checked

EXERCISE 13.3 41–50 (part 5 of 8)

  1. Differentiate each of the functions w. r. to \(\displaystyle x\) in Exercises $\displaystyle 29$ to 42.

    Exercise 41

    sin⁡3xcos⁡3x\displaystyle \sin ^3 x \cos ^3 x

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    NCERT’s answer
    \(\displaystyle \frac{3}{4} \sin ^2 2 x \cos 2 x\)
    Combine the powers first. \[y=(\sin x\cos x)^3=\left(\tfrac12\sin 2x\right)^3=\tfrac18\sin^3 2x \] Chain rule: \[\frac{dy}{dx}=\tfrac18\cdot 3\sin^2 2x\cdot\frac{d}{dx}(\sin 2x) \] \[=\tfrac18\cdot 3\sin^2 2x\cdot 2\cos 2x \] \[=\tfrac34\sin^2 2x\cos 2x \] \[\sin^2 2x=4\sin^2x\cos^2x\ \Rightarrow\ \frac{dy}{dx}=3\sin^2x\cos^2x\cos 2x \] Answer: \(\displaystyle \dfrac{dy}{dx}=\dfrac34\sin^2 2x\cos 2x=3\sin^2x\cos^2x\cos 2x\)

  2. Exercise 42

    1ax2+bx+c\displaystyle \frac{1}{a x^2+b x+c}

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    NCERT’s answer
    \(\displaystyle \frac{-(2 a x+b)}{\left(a x^2+b x+c\right)^2}\)
    \[y=(ax^2+bx+c)^{-1} \] Chain (power) rule: \[\frac{dy}{dx}=-(ax^2+bx+c)^{-2}\cdot\frac{d}{dx}(ax^2+bx+c) \] \[=-\frac{2ax+b}{(ax^2+bx+c)^2} \] Answer: \(\displaystyle \dfrac{dy}{dx}=-\dfrac{2ax+b}{(ax^2+bx+c)^2}\)
  3. Differentiate each of the functions with respect to ‘\(\displaystyle x\)’ in Exercises $\displaystyle 43$ to $\displaystyle 46$ using first principle.

    Exercise 43

    cos⁡(x2+1)\displaystyle \cos \left(x^2+1\right)

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    NCERT’s answer
    \(\displaystyle -2 x \sin \left(x^2+1\right)\)
    \[f(x)=\cos(x^2+1),\qquad f'(x)=\lim_{h\to0}\frac{\cos\left((x+h)^2+1\right)-\cos(x^2+1)}{h} \] Use \(\displaystyle \cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\): \[\frac{A+B}{2}=x^2+1+xh+\frac{h^2}{2},\qquad \frac{A-B}{2}=\frac{h(2x+h)}{2} \] \[f'(x)=\lim_{h\to0}\left[-2\sin\!\left(x^2+1+xh+\tfrac{h^2}{2}\right)\cdot\frac{\sin\dfrac{h(2x+h)}{2}}{\dfrac{h(2x+h)}{2}}\cdot\frac{2x+h}{2}\right] \] \[\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1,\qquad \theta=\frac{h(2x+h)}{2}\to0 \] \[f'(x)=-2\sin(x^2+1)\cdot 1\cdot x \] Answer: \(\displaystyle f'(x)=-2x\sin(x^2+1)\)

  4. Exercise 44

    ax+bcx+d\displaystyle \frac{a x+b}{c x+d}

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    NCERT’s answer
    \(\displaystyle \frac{a d-b c}{(c x+d)^2}\)
    \[f(x)=\frac{ax+b}{cx+d},\qquad f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \] \[f(x+h)-f(x)=\frac{(ax+b+ah)(cx+d)-(ax+b)(cx+d+ch)}{(cx+ch+d)(cx+d)} \] Numerator: \[ah(cx+d)-ch(ax+b)=h(ad-bc) \] \[f'(x)=\lim_{h\to0}\frac{ad-bc}{(cx+ch+d)(cx+d)} \] \[=\frac{ad-bc}{(cx+d)^2} \] Answer: \(\displaystyle f'(x)=\dfrac{ad-bc}{(cx+d)^2}\), \(\displaystyle cx+d\neq0\)

  5. Exercise 45

    x23\displaystyle x^{\frac{2}{3}}

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    NCERT’s answer
    \(\displaystyle \frac{2}{3} x^{-\frac{1}{3}}\)
    \[f(x)=x^{2/3},\qquad f'(x)=\lim_{h\to0}\frac{(x+h)^{2/3}-x^{2/3}}{h} \] Put \(\displaystyle u=(x+h)^{1/3},\ v=x^{1/3}\); then \(\displaystyle u\to v\) as \(\displaystyle h\to0\), and \[h=u^3-v^3=(u-v)(u^2+uv+v^2) \] \[f'(x)=\lim_{u\to v}\frac{u^2-v^2}{u^3-v^3}=\lim_{u\to v}\frac{(u-v)(u+v)}{(u-v)(u^2+uv+v^2)} \] \[=\frac{2v}{3v^2}=\frac{2}{3v} \] Answer: \(\displaystyle f'(x)=\dfrac{2}{3}x^{-1/3}\), \(\displaystyle x\neq0\)

  6. Exercise 46

    xcos⁡x\displaystyle x \cos x

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    NCERT’s answer
    \(\displaystyle \cos x-x \sin x\)
    \[f(x)=x\cos x,\qquad f'(x)=\lim_{h\to0}\frac{(x+h)\cos(x+h)-x\cos x}{h} \] Split the numerator: \[(x+h)\cos(x+h)-x\cos x=x\left[\cos(x+h)-\cos x\right]+h\cos(x+h) \] \[\cos(x+h)-\cos x=-2\sin\!\left(x+\tfrac h2\right)\sin\tfrac h2 \] \[f'(x)=\lim_{h\to0}\left[-x\sin\!\left(x+\tfrac h2\right)\cdot\frac{\sin(h/2)}{h/2}+\cos(x+h)\right] \] \[=-x\sin x\cdot 1+\cos x \] Answer: \(\displaystyle f'(x)=\cos x-x\sin x\)
  7. Evaluate each of the following limits in Exercises $\displaystyle 47$ to 53.

    Exercise 47

    lim⁡y→0(x+y)sec⁡(x+y)−xsec⁡xy\displaystyle \lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}

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    NCERT’s answer
    \(\displaystyle \sec x(x \tan x+1)\)
    With \(\displaystyle f(x)=x\sec x\), the limit is the definition of \(\displaystyle f'(x)\): \[\lim_{y\to0}\frac{f(x+y)-f(x)}{y}=\frac{d}{dx}(x\sec x) \] Product rule, \(\displaystyle \frac{d}{dx}\sec x=\sec x\tan x\): \[\frac{d}{dx}(x\sec x)=1\cdot\sec x+x\sec x\tan x \] Answer: \(\displaystyle \sec x+x\sec x\tan x=\sec x\,(1+x\tan x)\)

  8. Exercise 48

    lim⁡x→0(sin⁡(α+β)x+sin⁡(α−β)x+sin⁡2αx)cos⁡2βx−cos⁡2αx⋅x\displaystyle \lim _{x \rightarrow 0} \frac{(\sin (\alpha+\beta) x+\sin (\alpha-\beta) x+\sin 2 \alpha x)}{\cos 2 \beta x-\cos 2 \alpha x} \cdot x

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    NCERT’s answer
    \(\displaystyle \frac{2 \alpha}{\alpha^2-\beta^2}\)
    Numerator, by \(\displaystyle \sin P+\sin Q=2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}\): \[\sin(\alpha+\beta)x+\sin(\alpha-\beta)x=2\sin\alpha x\cos\beta x \] \[N=2\sin\alpha x\,(\cos\beta x+\cos\alpha x) \] Denominator, by \(\displaystyle \cos 2\theta=1-2\sin^2\theta\): \[D=\cos 2\beta x-\cos 2\alpha x=2\left(\sin^2\alpha x-\sin^2\beta x\right) \] \[\frac{N}{D}\cdot x=\frac{\dfrac{\sin\alpha x}{x}\,(\cos\alpha x+\cos\beta x)}{\alpha^2\left(\dfrac{\sin\alpha x}{\alpha x}\right)^2-\beta^2\left(\dfrac{\sin\beta x}{\beta x}\right)^2} \] \[\lim_{x\to0}=\frac{\alpha\,(1+1)}{\alpha^2-\beta^2} \] Answer: \(\displaystyle \dfrac{2\alpha}{\alpha^2-\beta^2}\), \(\displaystyle \alpha^2\neq\beta^2\)

  9. Exercise 49

    lim⁡x→π4tan⁡3x−tan⁡xcos⁡(x+π4)\displaystyle \lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan ^3 x-\tan x}{\cos \left(x+\dfrac{\pi}{4}\right)}

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    NCERT’s answer
    -$\displaystyle 4$
    Factor, then cancel the factor that vanishes at \(\displaystyle \frac{\pi}{4}\). \[\tan^3 x-\tan x=\tan x(\tan x-1)(\tan x+1) \] \[\tan x-1=\frac{\sin x-\cos x}{\cos x} \] \[\cos\left(x+\frac{\pi}{4}\right)=\frac{\cos x-\sin x}{\sqrt2} \] \[\frac{\tan x-1}{\cos\left(x+\dfrac{\pi}{4}\right)}=\frac{-\sqrt2}{\cos x} \] \[\lim_{x\to\frac{\pi}{4}}\tan x(\tan x+1)\cdot\frac{-\sqrt2}{\cos x}=1\cdot2\cdot\frac{-\sqrt2}{1/\sqrt2} \] \[=-4 \] Answer: \(\displaystyle -4\)

  10. Exercise 50

    lim⁡x→π1−sin⁡x2cos⁡x2(cos⁡x4−sin⁡x4)\displaystyle \lim _{x \rightarrow \pi} \frac{1-\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}\left(\cos \dfrac{x}{4}-\sin \dfrac{x}{4}\right)}

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    NCERT’s answer
    \(\displaystyle \frac{1}{\sqrt{2}}\)
    Write numerator and \(\displaystyle \cos\frac{x}{2}\) through \(\displaystyle \cos\frac{x}{4}\pm\sin\frac{x}{4}\). \[1-\sin\frac{x}{2}=\left(\cos\frac{x}{4}-\sin\frac{x}{4}\right)^2 \] \[\cos\frac{x}{2}=\left(\cos\frac{x}{4}-\sin\frac{x}{4}\right)\left(\cos\frac{x}{4}+\sin\frac{x}{4}\right) \] \[\frac{1-\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}\left(\cos\dfrac{x}{4}-\sin\dfrac{x}{4}\right)}=\frac{\left(\cos\dfrac{x}{4}-\sin\dfrac{x}{4}\right)^2}{\left(\cos\dfrac{x}{4}-\sin\dfrac{x}{4}\right)^2\left(\cos\dfrac{x}{4}+\sin\dfrac{x}{4}\right)}=\frac{1}{\cos\dfrac{x}{4}+\sin\dfrac{x}{4}} \] \[\lim_{x\to\pi}=\frac{1}{\cos\dfrac{\pi}{4}+\sin\dfrac{\pi}{4}}=\frac{1}{\sqrt2} \] Answer: \(\displaystyle \dfrac{1}{\sqrt2}\)