JEE Main 2021 · MathsMultiple choiceSingle correctHardCalculation

JEE Main 22 July 2021, Shift 2, Maths Q65

Question 65 of 90 in this shift, Maths question 5 of 30, Section A.

Let

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

be defined as

f(x)=\begin{cases}\frac{x^3}{(1-\cos2x)^2}\log_e\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right), & x\neq0\\ \alpha, & x=0\end{cases}

. If

f

is continuous at

x=0

, then

\alpha

is equal to :

(1) $0$
(2) $1$ Official answer
(3) $2$
(4) $3$

Official answer

Option 2

NTA final key.

Chapter: Continuity and Differentiability
Topic: Continuity and Differentiability
Idea tested: Limits via Series Expansion for Continuity

Same idea in other shifts

Asked 4× in all

7 Apr 2025, Shift 2 · Q25If the function $f(x)=\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$ is continuous at $x=0$ , then $f(0)$ is equal to ________.HardNumerical value
5 Apr 2024, Shift 1 · Q7If the function $f(x)=\frac{\sin3x+\alpha\sin x-\beta\cos3x}{x^3}$ , $x\in\mathbb{R}$ , is continuous at $x=0$ , then $f(0)$ is equal to :MediumSingle correct
17 Mar 2021, Shift 1 · Q84If the function $f(x)=\frac{\cos(\sin x)-\cos x}{x^{4}}$ is continuous at each point in its domain and $f(0)=\frac{1}{k}$ , then k is…MediumNumerical value

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.