For $a, b > 0$, let $f(x) = \begin{cases} \frac{\tan((a+1)x) + b\tan x}{x}, & x 0 \end{cases}$ be a continous function at $x = 0$. Then $\frac{b}{a}$ is equal to :

JEE Main 2024 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 8 April 2024, Shift 2, Maths Q9

Question 9 of 90 in this shift, Maths question 9 of 30, Section A.

For

a, b > 0

, let

f(x) = \begin{cases} \frac{\tan((a+1)x) + b\tan x}{x}, & x < 0 \\ 3, & x = 0 \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b\sqrt{a}\,x\sqrt{x}}, & x > 0 \end{cases}

be a continous function at

x = 0

. Then

\frac{b}{a}

is equal to :

(1) $6$ Official answer
(2) $5$
(3) $4$
(4) $8$

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.