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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2026 · 23 January · Morning Shift · Mathematics · Continuity and Differentiability

Let f(x)= cases a x^ 2 +2 a x+3 4 x^ 2 +4 x-3 &, x ≠-(3)/(2),...

Let f(x)=f(x)=\begin{cases}\frac{\mathrm{a} x^{2}+2 \mathrm{a} x+3}{4 x^{2}+4 x-3} &, x \neq-\frac{3}{2}, \frac{1}{2} \\ \mathrm{~b} &, x=-\frac{3}{2}, \frac{1}{2}\end{cases}$<br>becontinuousat<br> be continuous atx=-\frac{3}{2}.If. Iff \circ f(x)=\frac{7}{5},then, thenx$ is equal to:

Options

  1. A4
  2. B0
  3. C2
  4. D1

Correct answer

D. 1

Step-by-step solution

4x2+4x−3=(2x+3)(2x−1)4x^2+4x-3 = (2x+3)(2x-1). For continuity at x=−3/2x = -3/2, numerator must vanish there:
a(9/4)−3a+3=0⇒a=4a(9/4) - 3a + 3 = 0 \Rightarrow a = 4.
f(x)=(2x+1)(2x+3)(2x+3)(2x−1)=2x+12x−1f(x) = \frac{(2x+1)(2x+3)}{(2x+3)(2x-1)} = \frac{2x+1}{2x-1} for x≠−3/2x \neq -3/2.
f∘f(x)=f ⁣(2x+12x−1)=6x+12x+3f \circ f(x) = f\!\left(\frac{2x+1}{2x-1}\right) = \frac{6x+1}{2x+3}.
6x+12x+3=75⇒30x+5=14x+21⇒x=1\frac{6x+1}{2x+3} = \frac{7}{5} \Rightarrow 30x+5 = 14x+21 \Rightarrow x = 1.
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