JEE Main 2022 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 26 July 2022, Shift 1, Maths Q8

Question 8 of 90 in this shift, Maths question 8 of 30, Section A.

f(x) = \begin{cases} x + \mathrm{a}, & x \leq 0 \\ |x - 4|, & x > 0 \end{cases}

and

g(x) = \begin{cases} x + 1, & x < 0 \\ (x - 4)^2 + \mathrm{b}, & x \geq 0 \end{cases}

are continuous on

\mathbf{R}

, then

(g o f)\,(2) + (f o g)\,(-2)

is equal to :

(1) $-10$
(2) $10$
(3) $8$
(4) $-8$ Official answer

Official answer

Option 4

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.