Mathematics · 2019
JEE Main · 10 January 2019, Shift 1 · Q71
Let f(x)= {max {|x|, x^2}, |x| ≤ 2; 8-2|x|, 2<|x| ≤ 4} Let S be the set of points in the interval ( -4,4 ) at which f is not differentiable. Then S:
Let $\displaystyle f(x)=\left\{\begin{array}{cc}\max \left\{|x|, x^2\right\}, & |x| \leq 2 \\ 8-2|x|, & 2<|x| \leq 4\end{array}\right.$
Let $\displaystyle S$ be the set of points in the interval ( $\displaystyle -4,4$ ) at which $\displaystyle f$ is not differentiable.
Then S :
Official answer
(3)
equals $\displaystyle \{-2,-1,0,1,2\}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.