Let f ( x )= begin cases frac ax 2 +2 ax +3 4 x 2 +4 x -3 , & x neq-3/2, 1/2\\ b & x =-3/2, 1/2end cases be continuous at x=-3/2 . If fof (x)=7/5 , then x is equal to :

Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 23 January, Morning Shift — Question 9

Let $\mathrm{f}(\mathrm{x})= \begin{cases}\frac{\mathrm{ax}^{2}+2 \mathrm{ax}+3}{4 \mathrm{x}^{2}+4 \mathrm{x}-3}, & \mathrm{x} \neq-\frac{3}{2}, \frac{1}{2}\\ \mathrm{~b} & \mathrm{x}=-\frac{3}{2}, \frac{1}{2}\end{cases}$ be continuous at $x=-\frac{3}{2}$ . If $\operatorname{fof}(x)=\frac{7}{5}$ , then $x$ is equal to :

Option A:
$2$
Option B:
$1$
Correct
Option C:
$0$
Option D:
$1.4$

Answer: B

Step-by-step solution

f(x)= \begin{cases} \dfrac{ax^{2}+2ax+3}{(2x-1)(2x+3)}, & x \ne -\dfrac{3}{2}, \dfrac{1}{2} \\[8pt] b, & x = -\dfrac{3}{2}, \dfrac{1}{2} \end{cases}

For continuity at $x = -\dfrac{3}{2}$ ,

\lim_{x \to -\frac{3}{2}} \dfrac{ax^{2}+2ax+3}{(2x-1)(2x+3)}

Numerator must be zero:

a\left(-\frac{3}{2}\right)^2 +2a\left(-\frac{3}{2}\right)+3=0

\frac{9a}{4}-3a+3=0

\frac{3a}{4}=3 \Rightarrow a=4

\therefore f(x)= \begin{cases} \dfrac{4x^{2}+8x+3}{(2x-1)(2x+3)}, & x \ne -\dfrac{3}{2}, \dfrac{1}{2} \\[8pt] b, & x = -\dfrac{3}{2}, \dfrac{1}{2} \end{cases}

Factorising:

4x^{2}+8x+3=(2x+1)(2x+3)

\Rightarrow f(x)= \begin{cases} \dfrac{(2x+1)(2x+3)}{(2x-1)(2x+3)}, & x \ne -\dfrac{3}{2}, \dfrac{1}{2} \\[8pt] b, & x = -\dfrac{3}{2}, \dfrac{1}{2} \end{cases}

f(x)=\dfrac{2x+1}{2x-1} \quad (x \ne -\tfrac{3}{2}, \tfrac{1}{2})

f(f(x)) = f\!\left(\frac{2x+1}{2x-1}\right) = \frac{2\left(\frac{2x+1}{2x-1}\right)+1} {2\left(\frac{2x+1}{2x-1}\right)-1}

= \frac{6x+1}{2x+3}

Given

\frac{6x+1}{2x+3}=\frac{7}{5}

30x+5=14x+21

16x=16 \Rightarrow x=1

\boxed{x=1}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Continuity