Mathematics · 2019
JEE Main · 11 January 2019, Shift 1 · Q72
Let f(x)= {-1, -2 ≤ x<0; x^2-1, 0 ≤ x ≤ 2} and g(x)=|f(x)|+f(|x|). Then, in the interval (-2,2), g is:
Let $\displaystyle f(x)=\left\{\begin{array}{ll}-1, & -2 \leq x<0 \\ x^2-1, & 0 \leq x \leq 2\end{array}\right.$ and
$\displaystyle g(x)=|f(x)|+f(|x|)$. Then, in the interval
$\displaystyle (-2,2), g$ is :
Official answer
(2)
not differentiable at one point
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.