Q44 / 90 · JEE Main 2019 (11 Jan Shift 2) · Mathematics

Let $\mathrm{K}$ be the set of all real values of $x$ where the function $f(x)=\sin |x|-|x|+2(x-\pi) \cos |x|$ is not differentiable. Then the set $K$ is equal to :

  • Correct answer
    A
    $\phi$ (an empty set)
  • B
    $(\pi\}$
  • C
    $\{0\}$
  • D
    $\{0, \pi\}$

Solution

$f(x)=\sin |x|-|x|+2(x-\pi) \cos |x|$ There are two cases, Case $(1), x>0$ $f(x)=\sin x-x+2(x-\pi) \cos x$ $f^{\prime}(x)=\cos x-1+2(1-0) \cos x-2 \sin (x-\pi)$ $f^{\prime}(x)=3 \cos x-2(x-\pi) \sin x-1$ Then, function $f(x)$ is differentiable for all $x>0$ Case $(2) x < 0$ $f(x)=-\sin x+x+2(x-\pi) \cos x$ $f^{\prime}(x)=-\cos x+1-2(x-\pi) \sin x+2 \cos x$ $f^{\prime}(x)=\cos x+1-2(x-\pi) \sin x$ Then, function $f(x)$ is differentiable for all $x < 0$ Now check for $x=0$ $f^{\prime}\left(0^{*}\right) \mathrm{R} \cdot \mathrm{HD}=3-1=2$ $f^{\prime}(0) \mathrm{L} . \mathrm{H.D.}=1+1=2$ $\mathrm{L} . \mathrm{HD}=\mathrm{R} . \mathrm{H.D}$ Then, function $f(x)$ is differentiable for $x=0 .$ So it is differentiable everywhere Hence, $k=\phi$

Hence, the correct answer is option: 'A'.

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