JEE Main 2020 · MathsMultiple choiceSingle correctMediumApplication

JEE Main 4 September 2020, Shift 2, Maths Q59

Question 59 of 75 in this shift, Maths question 9 of 25.

The function

f(x) = \begin{cases} \frac{\pi}{4} + \tan^{-1} x, & |x| \leq 1 \\ \frac{1}{2}(|x| - 1), & |x| > 1 \end{cases}

is :

(1)both continuous and differentiable on $\mathbf{R} - \{1\}$ .
(2)continuous on $\mathbf{R} - \{-1\}$ and differentiable on $\mathbf{R} - \{-1, 1\}$ .
(3)continuous on $\mathbf{R} - \{1\}$ and differentiable on $\mathbf{R} - \{-1, 1\}$ .Official answer
(4)both continuous and differentiable on $\mathbf{R} - \{-1\}$ .

Official answer

Option 3

NTA final key (Sep 2020).

Chapter: Continuity and Differentiability
Topic: Continuity and Differentiability
Idea tested: Piecewise Differentiability

Same idea in other shifts

Asked 7× in all

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.