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NCERT Exemplar · Class 12 Mathematics Continuity and Differentiability

106 questions · 106 still being checked

EXERCISE 5.3 81–90 (part 9 of 11)

  1. Exercise 81

    If x=sin⁡t\displaystyle x=\sin t and y=sin⁡pt\displaystyle y=\sin p t, prove that (1−x2)d2ydx2−xdydx+p2y=0\displaystyle \left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+p^2 y=0.

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    \[\frac{dx}{dt}=\cos t,\qquad \frac{dy}{dt}=p\cos pt \] \[\frac{dy}{dx}=\frac{p\cos pt}{\cos t} \] \[\frac{d^2y}{dx^2}=\frac{d}{dt}\!\left(\frac{p\cos pt}{\cos t}\right)\cdot\frac{1}{\cos t}=\frac{-p^2\sin pt\cos t+p\cos pt\sin t}{\cos^3 t} \] \[1-x^2=1-\sin^2 t=\cos^2 t \] \[(1-x^2)\frac{d^2y}{dx^2}=-p^2\sin pt+p\tan t\cos pt \] \[x\frac{dy}{dx}=\sin t\cdot\frac{p\cos pt}{\cos t}=p\tan t\cos pt \] \[(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+p^2y=-p^2\sin pt+p\tan t\cos pt-p\tan t\cos pt+p^2\sin pt=0 \] Answer: the left side equals \(\displaystyle 0\).

  2. Exercise 82

    Find dydx\displaystyle \frac{d y}{d x}, if y=xtan⁡x+x2+12\displaystyle y=x^{\tan x}+\sqrt{\frac{x^2+1}{2}}.

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    NCERT’s answer
    \(\displaystyle x^{\tan x}\left(\sec ^2 x \log x+\frac{\tan x}{x}\right)+\frac{x}{\sqrt{2} \sqrt{x^2+1}}\)
    Write \(\displaystyle y=u+v\) with \(\displaystyle u=x^{\tan x}\) (for \(\displaystyle x>0\)) and \(\displaystyle v=\sqrt{\dfrac{x^2+1}{2}}\). \[\ln u=\tan x\,\ln x \] \[\frac1u\frac{du}{dx}=\sec^2x\,\ln x+\frac{\tan x}{x} \] \[\frac{du}{dx}=x^{\tan x}\left(\sec^2x\,\ln x+\frac{\tan x}{x}\right) \] \[\frac{dv}{dx}=\frac12\left(\frac{x^2+1}{2}\right)^{-1/2}\cdot x=\frac{x}{\sqrt{2(x^2+1)}} \] Answer: \(\displaystyle \dfrac{dy}{dx}=x^{\tan x}\left(\sec^2x\,\ln x+\dfrac{\tan x}{x}\right)+\dfrac{x}{\sqrt{2(x^2+1)}}\)
  3. Choose the correct answers from the given four options in each of the Exercises $\displaystyle 83$ to 96.

    Exercise 83

    If f(x)=2x\displaystyle f(x)=2 x and g(x)=x22+1\displaystyle g(x)=\frac{x^2}{2}+1, then which of the following can be a discontinuous function
    (A)
    f(x)+g(x)\displaystyle f(x)+g(x)
    (B)
    f(x)−g(x)\displaystyle f(x)-g(x)
    (C)
    f(x).g(x)\displaystyle f(x) . g(x)
    (D)
    g(x)f(x)\displaystyle \frac{g(x)}{f(x)}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{g(x)}{f(x)}\) \(\displaystyle f\) and \(\displaystyle g\) are polynomials, so \(\displaystyle f+g\), \(\displaystyle f-g\), \(\displaystyle fg\) are continuous. The quotient is \[\frac{g(x)}{f(x)}=\frac{x^2+2}{4x} \] At \(\displaystyle x=0\) the denominator is \(\displaystyle 0\) while the numerator is \(\displaystyle 2\), so it is discontinuous there.

  4. Exercise 84

    The function f(x)=4−x24x−x3\displaystyle f(x)=\frac{4-x^2}{4 x-x^3} is
    (A)
    discontinuous at only one point
    (B)
    discontinuous at exactly two points
    (C)
    discontinuous at exactly three points
    (D)
    none of these

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    NCERT’s answer
    C
    (C) discontinuous at exactly three points \[f(x)=\frac{4-x^2}{x(4-x^2)} \] \[x(4-x^2)=0\;\Rightarrow\; x=0,\;2,\;-2 \] \(\displaystyle f\) is a rational function, so it is undefined, hence discontinuous, exactly at these three points.

  5. Exercise 85

    The set of points where the function f\displaystyle f given by f(x)=∣2x−1∣sin⁡x\displaystyle f(x)=|2 x-1| \sin x is differentiable is
    (A)
    R\displaystyle \mathbf{R}
    (B)
    R−{12}\displaystyle \mathbf{R}-\left\{\frac{1}{2}\right\}
    (C)
    (0,∞)\displaystyle (0, \infty)
    (D)
    none of these

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    NCERT’s answer
    B
    (B) \(\displaystyle \mathbf{R}-\left\{\tfrac12\right\}\) For \(\displaystyle x\neq\tfrac12\), \(\displaystyle |2x-1|\) and \(\displaystyle \sin x\) are differentiable, so \(\displaystyle f\) is. At \(\displaystyle x=\tfrac12\), \(\displaystyle f(\tfrac12)=0\): \[f'(\tfrac12^+)=\lim_{h\to0^+}\frac{2h\sin(\tfrac12+h)}{h}=2\sin\tfrac12 \] \[f'(\tfrac12^-)=\lim_{h\to0^-}\frac{-2h\sin(\tfrac12+h)}{h}=-2\sin\tfrac12 \] Since \(\displaystyle \sin\tfrac12\neq0\), the two values differ.

  6. Exercise 86

    The function f(x)=cot⁡x\displaystyle f(x)=\cot x is discontinuous on the set
    (A)
    {x=nπ:n∈Z}\displaystyle \{x=n \pi: n \in \mathbf{Z}\}
    (B)
    {x=2nπ:n∈Z}\displaystyle \{x=2 n \pi: n \in \mathbf{Z}\}
    (C)
    {x=(2n+1)π2;n∈Z}\displaystyle \left\{x=(2 n+1) \frac{\pi}{2} ; n \in \mathbf{Z}\right\}
    (iv)
    {x=nπ2;n∈Z}\displaystyle \left\{x=\frac{n \pi}{2} ; n \in \mathbf{Z}\right\}

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    NCERT’s answer
    A
    (A) \(\displaystyle \{x=n\pi:\ n\in\mathbf{Z}\}\) \[\cot x=\frac{\cos x}{\sin x} \] \[\sin x=0\iff x=n\pi,\quad n\in\mathbf{Z} \] \(\displaystyle \cot x\) is undefined, hence discontinuous, exactly there.

  7. Exercise 87

    The function f(x)=e∣x∣\displaystyle f(x)=e^{|x|} is
    (A)
    continuous everywhere but not differentiable at x=0\displaystyle x=0
    (B)
    continuous and differentiable everywhere
    (C)
    not continuous at x=0\displaystyle x=0
    (D)
    none of these.

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    NCERT’s answer
    A
    (A) continuous everywhere but not differentiable at \(\displaystyle x=0\) \(\displaystyle e^{|x|}\) is a composition of continuous functions, so it is continuous everywhere. At \(\displaystyle x=0\): \[f'(0^+)=\lim_{h\to0^+}\frac{e^{h}-1}{h}=1 \] \[f'(0^-)=\lim_{h\to0^-}\frac{e^{-h}-1}{h}=-1 \] The one-sided derivatives differ.

  8. Exercise 88

    If f(x)=x2sin⁡1x\displaystyle f(x)=x^2 \sin \frac{1}{x}, where x≠0\displaystyle x \neq 0, then the value of the function f\displaystyle f at x=0\displaystyle x=0, so that the function is continuous at x=0\displaystyle x=0, is
    (A)
    0\displaystyle 0 (B) - 1\displaystyle 1 (C) 1\displaystyle 1 (D) none of these

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    NCERT’s answer
    A
    (A) \(\displaystyle 0\) \[\left|x^2\sin\frac1x\right|\le x^2 \] \[\lim_{x\to0}x^2\sin\frac1x=0 \] Continuity needs \(\displaystyle f(0)=\lim_{x\to0}f(x)=0\).

  9. Exercise 89

    If f(x)={mx+1, if x≤π2sin⁡x+n, if x>π2\displaystyle f(x)=\left\{\begin{array}{l}m x+1, \text { if } x \leq \frac{\pi}{2} \\ \sin x+n, \text { if } x>\frac{\pi}{2}\end{array}\right., is continuous at x=π2\displaystyle x=\frac{\pi}{2}, then
    (A)
    m=1,n=0\displaystyle m=1, n=0
    (B)
    m=nπ2+1\displaystyle m=\frac{n \pi}{2}+1
    (C)
    n=mπ2\displaystyle n=\frac{m \pi}{2}
    (D)
    m=n=π2\displaystyle m=n=\frac{\pi}{2}

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    NCERT’s answer
    C
    (C) \(\displaystyle n=\frac{m\pi}{2}\)Continuity at \(\displaystyle x=\frac{\pi}{2}\) needs left limit \(\displaystyle =\) right limit \(\displaystyle =f\left(\frac{\pi}{2}\right)\).\[\lim_{x\to\frac{\pi}{2}^-}(mx+1)=\frac{m\pi}{2}+1=f\!\left(\frac{\pi}{2}\right) \]\[\lim_{x\to\frac{\pi}{2}^+}(\sin x+n)=\sin\frac{\pi}{2}+n=1+n \]\[\frac{m\pi}{2}+1=1+n \]\[n=\frac{m\pi}{2} \]

  10. Exercise 90

    Let f(x)=∣sin⁡x∣\displaystyle f(x)=|\sin x|. Then
    (A)
    f\displaystyle f is everywhere differentiable
    (B)
    f\displaystyle f is everywhere continuous but not differentiable at x=nπ,n∈Z\displaystyle x=n \pi, n \in \mathbf{Z}.
    (C)
    f\displaystyle f is everywhere continuous but not differentiable at x=(2n+1)π2\displaystyle x=(2 \mathrm{n}+1) \frac{\pi}{2}, n∈Z\displaystyle n \in \mathbf{Z}.
    (D)
    none of these

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    NCERT’s answer
    B
    (B) continuous everywhere, not differentiable at \(\displaystyle x=n\pi\).\(\displaystyle \sin x\) and \(\displaystyle |x|\) are continuous, so their composition \(\displaystyle f\) is continuous. Away from \(\displaystyle n\pi\), \(\displaystyle f=\pm\sin x\) locally, which is differentiable. At \(\displaystyle x=n\pi\), put \(\displaystyle x=n\pi+h\):\[f(n\pi+h)=|(-1)^n\sin h|=|\sin h|,\quad f(n\pi)=0 \]\[f'(n\pi^+)=\lim_{h\to0^+}\frac{\sin h}{h}=1 \]\[f'(n\pi^-)=\lim_{h\to0^-}\frac{-\sin h}{h}=-1 \]\[1\neq-1 \]