f(x)={\begin{matrix}\frac{\pi }{4}+{\mathrm{tan}}^{-1}x, & |x|\leq 1 \\ \frac{1}{2}(|x|-1), & |x|>1\end{matrix}
Making graph
Clearly, function is continuous on R–1 and differentiable on R––1,1.
JEE Main 2020 — Mathematics Calculus
The function f(x)={\begin{matrix}\frac{\pi }{4}+{\mathrm{tan}}^{-1}x, & |x|\leq 1 \\ \frac{1}{2}(|x|-1), & |x|>1\end{matrix} is :
Held on 4 Sept 2020 · Verified 6 Jul 2026.
continuous on R−1 and differentiable on R−−1,1.
both continuous and differentiable on R−1
continuous on R−−1and differentiable on R−−1,1
both continuous and differentiable on R−−1
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