Let $f:\mathbb{R}\to\mathbb{R}$ be defined as : $f(x)=\begin{cases}\frac{a-b\cos 2x}{x^2} & ; x 1\end{cases}$ If $f$ is continuous everywhere in $\mathbb{R}$ and m is the number of points where $f$ is NOT differentiable then m+a+b+c equals :

JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 1 February 2024, Shift 1, Maths Q8

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

Let

f:\mathbb{R}\to\mathbb{R}

be defined as :

f(x)=\begin{cases}\frac{a-b\cos 2x}{x^2} & ; x<0\\ x^2+cx+2 & ; 0\leq x\leq 1\\ 2x+1 & ; x>1\end{cases}

f

is continuous everywhere in

\mathbb{R}

and m is the number of points where

f

is NOT differentiable then m+a+b+c equals :

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

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.