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

not differentiable at one point

not differentiable at two points

Let f(x) = begincases -1, & -2 ≤ x < 0 \ x^2 - 1, & 0 ≤ x ≤… — JEE Main 2019 Mathematics PYQ

JEE Main 2019 · 11 Jan · Shift I

Let

f(x) = \begin{cases} -1, & -2 \leq x < 0 \\ x^2 - 1, & 0 \leq x \leq 2 \end{cases}

and

g(x) = |f(x)| + f(|x|)

. Then, in the interval

(-2, 2)

g

Show answer

Correct option: A

[JEE Main 2019 · 11 Jan · Shift I]

Practise Limits Continuity and Differentiability the way it is asked

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