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Let f(x)= … . and g(x)=|η(x)|+f(x mid) . Then, in the interval (-2,2), g is:

JEE Main 2019 — Mathematics Calculus

2019mcqmedium

Let f(x)={−1,−2≤x<0x2−1,0≤x≤2f(x)=\left\{\begin{array}{cl}-1, & -2 \leq x < 0 \\ x^{2}-1, & 0 \leq x \leq 2\end{array}\right.f(x)={−1,x2−1,​−2≤x<00≤x≤2​ and g(x)=∣η(x)∣+f(x∣).g(x)=|\eta(x)|+f(x \mid) .g(x)=∣η(x)∣+f(x∣). Then, in the interval (−2,2),g(-2,2), g(−2,2),g is:

Official previous-year question

Held on 11 Jan 2019 · Verified 6 Jul 2026.

Options

  1. A

    differentiable at all points

  2. B

    not continuous

  3. C

    not differentiable at two points

  4. D

    not differentiable at one point

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