Let f(x)=\ array cl -1, & -2 ≤ x < 0 x^ 2 -1, & 0 ≤ x ≤ 2 array . and g(x)=| (x)|+f(x ) . Then, in the interval (-2,2), g is:

Correct answer: D. not differentiable at one point f(x)=\ array cc -1, & -2 ≤ x < 0 x^ 2 -1, & 0 ≤ x ≤ 2 array . Then, f(x )=\ array cc -1, & -2 ≤|x| < 0 |x|^ 2 -1, & 0 ≤|x| ≤ 2 array . f(|x|)=x^ 2 -1,-2 ≤ x ≤ 2 g(x)=\ aligned 1+x^ 2 -1, &-2 ≤ x < 0 (x^ 2 -1)+|x^ 2 -1|, & 0 ≤ x ≤ 2 aligned . =\ array cc x^ 2 , & -2 ≤ x < 0 0, & 0 ≤ x < 1 2(x^ 2 -1), & 1 ≤ x ≤ 2 array . g^ (0^ - )=0, g^ (0^ * )=0, g^ (1^ - )=0, g^ (1^ + )=4 g(x) is non-differentiable at x=1 g(x) is not differentiable at one point.

JEE Main 201911 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Continuity and Differentiability

Let f(x)=\ array cl -1, & -2 ≤ x < 0 x^ 2 -1, & 0 ≤ x ≤ 2 array ....

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:

Options

Adifferentiable at all points
Bnot continuous
Cnot differentiable at two points
Dnot differentiable at one point

Correct answer

D. not differentiable at one point

Step-by-step solution

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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