JEE Main 2026 — Jan 23, Shift 1 — Mathematics solution

JEE Main2026Jan 23, Shift 1Mathematics

Let $f (x) = ⎩ ⎨ ⎧ \frac{a x ^{2} + 2 a x + 3}{4 x ^{2} + 4 x - 3}, x \neq = - \frac{3}{2}, \frac{1}{2} b, x = - \frac{3}{2}, \frac{1}{2}$ be continuous at $x = - \frac{3}{2}$ . If $f \circ f (x) = \frac{7}{5}$ , then $x$ is equal to :

A
$2$
B
$1$
C
$0$
D
$1.4$

Solution

Continuity at $x = - 3/2$ . The denominator $4 x^{2} + 4 x - 3 = (2 x - 1) (2 x + 3)$ vanishes there, so the numerator must too: $a (- 3/2)^{2} + 2 a (- 3/2) + 3 = \frac{9 a}{4} - 3 a + 3 = 0$ , giving $a = 4$ .

Simplify $f$ . Numerator $4 x^{2} + 8 x + 3 = (2 x + 1) (2 x + 3)$ , so for valid $x$ :

$f (x) = \frac{( 2 x + 1 ) ( 2 x + 3 )}{( 2 x - 1 ) ( 2 x + 3 )} = \frac{2 x + 1}{2 x - 1}$

Compute $f \circ f$ . Let $u = \frac{2 x + 1}{2 x - 1}$ . Then $f (u) = \frac{2 u + 1}{2 u - 1} = \frac{2 \cdot \frac{2 x + 1}{2 x - 1} + 1}{2 \cdot \frac{2 x + 1}{2 x - 1} - 1} = \frac{( 4 x + 2 ) + ( 2 x - 1 )}{( 4 x + 2 ) - ( 2 x - 1 )} = \frac{6 x + 1}{2 x + 3}$ .

Set $\frac{6 x + 1}{2 x + 3} = \frac{7}{5}$ : $5 (6 x + 1) = 7 (2 x + 3) \Rightarrow 30 x + 5 = 14 x + 21 \Rightarrow x = 1$ .

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