JEE Main 2026 Jan 28 Shift 1, Mathematics Q1: Invertible And Composite Functions

JEE Main2026Jan 28, Shift 1Mathematics

If $g (x) = 3 x^{2} + 2 x - 3$ , $f (0) = - 3$ and $4 g (f (x)) = 3 x^{2} - 32 x + 72$ , then $f (g (2))$ is equal to:

Solution

$g (2) = 3 (4) + 2 (2) - 3 = 13$ , so the target is $f (13)$ .

Expand $4 g (f (x)) = 12 f (x)^{2} + 8 f (x) - 12 = 3 x^{2} - 32 x + 72$ , hence

$12 f (x)^{2} + 8 f (x) - (3 x^{2} - 32 x + 84) = 0$ .

Treat as a quadratic in $f (x)$ :

$f (x) = \frac{- 8 \pm 64 + 48 ( 3 x ^{2} - 32 x + 84 )}{24} = \frac{- 8 \pm 4 ( 3 x - 16 )}{24}$

(the discriminant simplifies because $64 + 48 (3 x^{2} - 32 x + 84) = 16 (3 x - 16)^{2}$ ).

Use the boundary condition $f (0) = - 3$ to pick the sign: $\frac{- 8 + 4 ( - 16 )}{24} = - 3$ , so the '+' branch is correct.

$∴ f (x) = \frac{- 8 + 4 ( 3 x - 16 )}{24}$ , giving $f (13) = \frac{- 8 + 4 ( 23 )}{24} = \frac{84}{24} = \frac{7}{2}$ .

Option (3).

Concept behind this question

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