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NCERT Exemplar · Class 12 Mathematics Continuity and Differentiability

106 questions · 106 still being checked

EXERCISE 5.3 61–70 (part 7 of 11)

  1. Exercise 61

    If y=(cos⁡x)(cos⁡x)(cos⁡x)…∞\displaystyle y=(\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}, show that dydx=y2tan⁡xylog⁡cos⁡x−1\displaystyle \frac{d y}{d x}=\frac{y^2 \tan x}{y \log \cos x-1}.

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    The exponent tower is \(\displaystyle y\) itself, so\[y=(\cos x)^{y} \]\[\log y=y\log\cos x \]Differentiate with respect to \(\displaystyle x\).\[\frac1y\frac{dy}{dx}=\frac{dy}{dx}\log\cos x+y\cdot\frac{-\sin x}{\cos x} \]\[\frac{dy}{dx}\left(\frac1y-\log\cos x\right)=-y\tan x \]\[\frac{dy}{dx}\cdot\frac{1-y\log\cos x}{y}=-y\tan x \]\[\frac{dy}{dx}=\frac{-y^2\tan x}{1-y\log\cos x}=\frac{y^2\tan x}{y\log\cos x-1} \]

  2. Exercise 62

    If xsin⁡(a+y)+sin⁡acos⁡(a+y)=0\displaystyle x \sin (a+y)+\sin a \cos (a+y)=0, prove that dydx=sin⁡2(a+y)sin⁡a\displaystyle \frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}.

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    Solve for \(\displaystyle x\).\[x=-\sin a\cdot\frac{\cos(a+y)}{\sin(a+y)}=-\sin a\cot(a+y) \]Differentiate with respect to \(\displaystyle y\).\[\frac{dx}{dy}=\sin a\,\operatorname{cosec}^2(a+y)=\frac{\sin a}{\sin^2(a+y)} \]\[\frac{dy}{dx}=\frac{1}{dx/dy}=\frac{\sin^2(a+y)}{\sin a} \]

  3. Exercise 63

    If 1−x2+1−y2=a(x−y)\displaystyle \sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y), prove that dydx=1−y21−x2\displaystyle \frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}}.

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    Put \(\displaystyle x=\sin A,\ y=\sin B\) with \(\displaystyle A,B\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\).\[\cos A+\cos B=a(\sin A-\sin B) \]\[2\cos\frac{A+B}{2}\cos\frac{A-B}{2}=2a\cos\frac{A+B}{2}\sin\frac{A-B}{2} \]Cancel \(\displaystyle \cos\frac{A+B}{2}\).\[\cot\frac{A-B}{2}=a \quad\Rightarrow\quad A-B=2\cot^{-1}a \]\[\sin^{-1}x-\sin^{-1}y=2\cot^{-1}a \]\(\displaystyle a\) is constant, so differentiate with respect to \(\displaystyle x\).\[\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0 \]\[\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}} \]

  4. Exercise 64

    If y=tan⁡−1x\displaystyle y=\tan ^{-1} x, find d2ydx2\displaystyle \frac{d^2 y}{d x^2} in terms of y\displaystyle y alone.

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    NCERT’s answer
    \(\displaystyle -2 \sin y \cos ^3 y\)
    \[\frac{dy}{dx}=\frac{1}{1+x^2} \]\[\frac{d^2y}{dx^2}=\frac{-2x}{(1+x^2)^2} \]Use \(\displaystyle x=\tan y\) and \(\displaystyle 1+x^2=\sec^2y\).\[\frac{d^2y}{dx^2}=\frac{-2\tan y}{\sec^4y}=-2\sin y\cos^3y \]Answer: \(\displaystyle -2\sin y\cos^3y\) (equivalently \(\displaystyle -\sin 2y\cos^2y\))
  5. Verify the Rolle’s theorem for each of the functions in Exercises $\displaystyle 65$ to 69.

    Exercise 65

    f(x)=x(x−1)2\displaystyle f(x)=x(x-1)^2 in [0,1]\displaystyle [0,1].

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    A polynomial: continuous on \(\displaystyle [0,1]\), differentiable on \(\displaystyle (0,1)\). \[f(0)=0,\qquad f(1)=1\cdot 0^2=0 \;\Rightarrow\; f(0)=f(1) \] \[f'(x)=(x-1)^2+2x(x-1)=(x-1)(3x-1) \] \[f'(c)=0 \;\Rightarrow\; c=\tfrac13\in(0,1) \] Answer: Rolle's theorem holds, with \(\displaystyle c=\tfrac13\).

  6. Exercise 66

    f(x)=sin⁡4x+cos⁡4x\displaystyle f(x)=\sin ^4 x+\cos ^4 x in [0,π2]\displaystyle \left[0, \frac{\pi}{2}\right].

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    Built from \(\displaystyle \sin x,\cos x\): continuous on \(\displaystyle [0,\tfrac{\pi}{2}]\), differentiable on \(\displaystyle (0,\tfrac{\pi}{2})\). \[f(0)=0+1=1,\qquad f\!\left(\tfrac{\pi}{2}\right)=1+0=1 \] \[f(x)=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\tfrac12\sin^2 2x \] \[f'(x)=-\tfrac12\cdot 2\sin 2x\cos 2x\cdot 2=-\sin 4x \] \[f'(c)=0 \;\Rightarrow\; \sin 4c=0 \;\Rightarrow\; 4c=\pi \;\Rightarrow\; c=\tfrac{\pi}{4}\in\left(0,\tfrac{\pi}{2}\right) \] Answer: Rolle's theorem holds, with \(\displaystyle c=\tfrac{\pi}{4}\).

  7. Exercise 67

    f(x)=log⁡(x2+2)−log⁡3\displaystyle f(x)=\log \left(x^2+2\right)-\log 3 in [−1,1]\displaystyle [-1,1].

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    Since \(\displaystyle x^2+2>0\), \(\displaystyle f\) is continuous on \(\displaystyle [-1,1]\) and differentiable on \(\displaystyle (-1,1)\). \[f(-1)=\log 3-\log 3=0,\qquad f(1)=\log 3-\log 3=0 \] \[f'(x)=\frac{2x}{x^2+2} \] \[f'(c)=0 \;\Rightarrow\; c=0\in(-1,1) \] Answer: Rolle's theorem holds, with \(\displaystyle c=0\).

  8. Exercise 68

    f(x)=x(x+3)e−x/2\displaystyle f(x)=x(x+3) e^{-x / 2} in [−3,0]\displaystyle [-3,0].

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    Polynomial times \(\displaystyle e^{-x/2}\): continuous on \(\displaystyle [-3,0]\), differentiable on \(\displaystyle (-3,0)\). \[f(-3)=(-3)(0)e^{3/2}=0,\qquad f(0)=0 \] \[f'(x)=(2x+3)e^{-x/2}-\tfrac12\,x(x+3)e^{-x/2} \] \[f'(x)=-\tfrac12 e^{-x/2}\,(x^2-x-6)=-\tfrac12 e^{-x/2}(x-3)(x+2) \] \[f'(c)=0 \;\Rightarrow\; c=-2\in(-3,0)\qquad(x=3\notin(-3,0)) \] Answer: Rolle's theorem holds, with \(\displaystyle c=-2\).

  9. Exercise 69

    f(x)=4−x2\displaystyle f(x)=\sqrt{4-x^2} in [−2,2]\displaystyle [-2,2].

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    \(\displaystyle 4-x^2\ge 0\) on \(\displaystyle [-2,2]\), so \(\displaystyle f\) is continuous there. \[f'(x)=\frac{-x}{\sqrt{4-x^2}}\ \text{ exists for every } x\in(-2,2) \] \[f(-2)=0=f(2) \] \[f'(c)=0 \;\Rightarrow\; c=0\in(-2,2) \] NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q69 Answer: Rolle's theorem holds, with \(\displaystyle c=0\).

  10. Exercise 70

    Discuss the applicability of Rolle's theorem on the function given by f(x)={x2+1, if 0≤x≤13−x, if 1≤x≤2.f(x)=\left\{\begin{array}{l} x^2+1, \text { if } 0 \leq x \leq 1 \\ 3-x, \text { if } 1 \leq x \leq 2 \end{array} .\right.

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    NCERT’s answer
    Not applicable since \(\displaystyle f\) is not differentiable at \(\displaystyle x = 1\)
    Each piece is a polynomial; check the join at \(\displaystyle x=1\). \[\lim_{x\to1^-}f(x)=1^2+1=2,\qquad \lim_{x\to1^+}f(x)=3-1=2=f(1) \] \[f(0)=1,\qquad f(2)=3-2=1 \;\Rightarrow\; f(0)=f(2) \] \[f'(1^-)=2x\Big|_{x=1}=2,\qquad f'(1^+)=-1 \] \[2\neq-1 \;\Rightarrow\; f \text{ is not differentiable at } x=1\in(0,2) \] NCERT_Solution_Class12_Maths_Exemplar_Ch5_Ex5-3_Q70 Answer: \(\displaystyle f\) is continuous on \(\displaystyle [0,2]\) with \(\displaystyle f(0)=f(2)\), but not differentiable at \(\displaystyle x=1\); Rolle's theorem is not applicable.