Let $f : \mathbf{R} \to \mathbf{R}$ be defined as $f(x)=\begin{cases} 2\sin\left(-\frac{\pi x}{2}\right), & \text{if } x 1 \end{cases}$ If $f(x)$ is continuous on $\mathbf{R}$, then $a+b$ equals :

JEE Main 2021 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 26 February 2021, Shift 2, Maths Q77

Question 77 of 90 in this shift, Maths question 17 of 30, Section A.

Let

f : \mathbf{R} \to \mathbf{R}

be defined as

f(x)=\begin{cases} 2\sin\left(-\frac{\pi x}{2}\right), & \text{if } x<-1 \\ |ax^2+x+b|, & \text{if } -1 \le x \le 1 \\ \sin(\pi x), & \text{if } x>1 \end{cases}

f(x)

is continuous on

\mathbf{R}

, then

a+b

equals :

(1) $-3$
(2) $-1$ Official answer
(3) $1$
(4) $3$

Official answer

Option 2

NTA final key.

Chapter: Continuity and Differentiability
Topic: Continuity and Differentiability
Idea tested: Finding Unknown Constants for Continuity

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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.