Q38 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
Let $f(x)= \begin{cases}\frac{\mathrm{a} x^{2}+2 \mathrm{a} x+3}{4 x^{2}+4 x-3} &, x \neq-\frac{3}{2}, \frac{1}{2} \\ \mathrm{~b} &, x=-\frac{3}{2}, \frac{1}{2}\end{cases}$
be continuous at $x=-\frac{3}{2}$. If $f \circ f(x)=\frac{7}{5}$, then $x$ is equal to:
be continuous at $x=-\frac{3}{2}$. If $f \circ f(x)=\frac{7}{5}$, then $x$ is equal to:
- A4
- B0
- C2
- Correct answerD1
Solution
$4x^2+4x-3 = (2x+3)(2x-1)$. For continuity at $x = -3/2$, numerator must vanish there:
$a(9/4) - 3a + 3 = 0 \Rightarrow a = 4$.
$f(x) = \frac{(2x+1)(2x+3)}{(2x+3)(2x-1)} = \frac{2x+1}{2x-1}$ for $x \neq -3/2$.
$f \circ f(x) = f\!\left(\frac{2x+1}{2x-1}\right) = \frac{6x+1}{2x+3}$.
$\frac{6x+1}{2x+3} = \frac{7}{5} \Rightarrow 30x+5 = 14x+21 \Rightarrow x = 1$.
$a(9/4) - 3a + 3 = 0 \Rightarrow a = 4$.
$f(x) = \frac{(2x+1)(2x+3)}{(2x+3)(2x-1)} = \frac{2x+1}{2x-1}$ for $x \neq -3/2$.
$f \circ f(x) = f\!\left(\frac{2x+1}{2x-1}\right) = \frac{6x+1}{2x+3}$.
$\frac{6x+1}{2x+3} = \frac{7}{5} \Rightarrow 30x+5 = 14x+21 \Rightarrow x = 1$.
Hence, the correct answer is option: 'D'.