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Question: Answered & Verified by Expert
Let $f(x)=\left\{\begin{array}{cl}-1, & -2 \leq x < 0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.$ and

$g(x)=|\eta(x)|+f(x \mid) .$ Then, in the interval $(-2,2), g$ is:
MathematicsContinuity and DifferentiabilityJEE MainJEE Main 2019 (11 Jan Shift 1)
Options:
  • A differentiable at all points
  • B not continuous
  • C not differentiable at two points
  • D not differentiable at one point
Solution:
1199 Upvotes Verified Answer
The correct answer is: not differentiable at one point
$f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x < 0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.$

$\text { Then, } f(x \mid)=\left\{\begin{array}{cc}

-1, & -2 \leq|x| < 0 \\

|x|^{2}-1, & 0 \leq|x| \leq 2

\end{array}\right.$

$\Rightarrow f(|x|)=x^{2}-1,-2 \leq x \leq 2$

$\Rightarrow g(x)=\left\{\begin{aligned} 1+x^{2}-1, &-2 \leq x < 0 \\\left(x^{2}-1\right)+\left|x^{2}-1\right|, & 0 \leq x \leq 2 \end{aligned}\right.$

$=\left\{\begin{array}{cc}x^{2}, & -2 \leq x < 0 \\ 0, & 0 \leq x < 1 \\ 2\left(x^{2}-1\right), & 1 \leq x \leq 2\end{array}\right.$

$g^{\prime}\left(0^{-}\right)=0, g^{\prime}\left(0^{*}\right)=0, g^{\prime}\left(1^{-}\right)=0, g^{\prime}\left(1^{+}\right)=4$

$\Rightarrow g(x)$ is non-differentiable at $x=1$ $\Rightarrow g(x)$ is not differentiable at one point.

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