If the function $f(x)=\begin{cases}\frac{1}{x}\log_e\left(\frac{1+\frac{x}{\mathrm{a}}}{1-\frac{x}{\mathrm{b}}}\right) & , \quad x 0\end{cases}$ is continuous at $x=0$, then $\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{4}{\mathrm{k}}$ is equal to :

JEE Main 2021 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 31 August 2021, Shift 1, Maths Q68

Question 68 of 90 in this shift, Maths question 8 of 30, Section A.

If the function

f(x)=\begin{cases}\frac{1}{x}\log_e\left(\frac{1+\frac{x}{\mathrm{a}}}{1-\frac{x}{\mathrm{b}}}\right) & , \quad x<0\\ \mathrm{k} & , \quad x=0\\ \frac{\cos^2x-\sin^2x-1}{\sqrt{x^2+1}-1} & , \quad x>0\end{cases}

is continuous at

x=0

, then

\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{4}{\mathrm{k}}

is equal to :

(1) $-5$ Official answer
(2) $5$
(3) $-4$
(4) $4$

Official answer

Option 1

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.