JEE Main 20204 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 4 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2020 · 4 September · Evening Shift · Mathematics · Continuity and Differentiability

The function f(x) = \ cases (π)/(4) + ^ - 1 x, & |x| ≤ 1...

The function

f\left(x\right) = \left\{

\begin{cases}\dfrac{\pi}{4} + \tan^{- 1}x, & \left|x\right| \le 1 \\ \dfrac{1}{2}\left(\left|x\right| - 1\right), & \left|x\right| > 1\end{cases}$\right) is : $The function f(x) = \ cases (π)/(4) + ^ - 1 x, & |x| ≤ 1 (1)/(2)(|x| - 1), & |x| > 1 cases ) is :$

Options

Acontinuous on $R - \left\{1\right\}$ and differentiable on $R - \left\{- 1, 1\right\}$ .
Bboth continuous and differentiable on $R - \left\{1\right\}$
Ccontinuous on $R - \left\{- 1\right\}$ and differentiable on $R - \left\{- 1, 1\right\}$
Dboth continuous and differentiable on $R - \left\{- 1\right\}$

Correct answer

A. continuous on $R - \left\{1\right\}$ and differentiable on $R - \left\{- 1, 1\right\}$ .

Step-by-step solution

f\left(x\right) = \left\{ $\begin{cases}\dfrac{\pi}{4} + \tan^{- 1}x, & \left|x\right| \le 1 \\ \dfrac{1}{2}\left(\left|x\right| - 1\right), & \left|x\right| > 1\end{cases}$ \right)

Making graph

$The function f(x) = \ cases (π)/(4) + ^ - 1 x, & |x| ≤ 1 (1)/(2)(|x| - 1), & |x| > 1 cases ) is :$

Clearly, function is continuous on $R – \left\{1\right\}$ and differentiable on $R – \left\{– 1, 1\right\}$ .

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