Let $f: \mathbf{R} \to \mathbf{R}$ be a function given by $f(x)=\begin{cases}\frac{1-\cos 2x}{x^2}, & x 0\end{cases}$ where $\alpha, \beta \in \mathbf{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to :

JEE Main 2024 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 4 April 2024, Shift 1, Maths Q6

Question 6 of 90 in this shift, Maths question 6 of 30, Section A.

Let

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

be a function given by

f(x)=\begin{cases}\frac{1-\cos 2x}{x^2}, & x<0\\ \alpha, & x=0,\\ \frac{\beta\sqrt{1-\cos x}}{x}, & x>0\end{cases}

where

\alpha, \beta \in \mathbf{R}

. If

f

is continuous at

x=0

, then

\alpha^2+\beta^2

is equal to :

(1) $3$
(2) $6$
(3) $12$ Official answer
(4) $48$

Official answer

Option 3

NTA final key.

Chapter: Continuity and Differentiability
Topic: Continuity and Differentiability
Idea tested: Continuity at a Point

Same idea in other shifts

Asked 8× in all

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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.