Mathematics · 2020
JEE Main · 4 September 2020, Shift 2 · Q59
The function f(x)= {(π)/4+ tan^-1 x, |x| ≤ 1; 1/2(|x|-1), |x|>1} is:
The function $\displaystyle f(x)= \begin{cases}\frac{\pi}{4}+\tan ^{-1} x, & |x| \leq 1 \\ \frac{1}{2}(|x|-1), & |x|>1\end{cases}$ is :
Official answer
(3)
continuous on $\displaystyle \mathbf{R}-\{1\}$ and differentiable on $\displaystyle \mathbf{R}-\{-1,1\}$.
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.