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JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Sets Relations and Functions

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:

(A) $\frac{25}{6}$
(B) $- \frac{25}{6}$
(C) $\frac{7}{2}$ ✓ Correct
(D) $- \frac{7}{2}$

Answer: (C)

Solution

Since

g (2) = 3 \cdot 4 + 4 - 3 = 13

, the value needed is

f (13)

. Substituting

g (f (x)) = 3 f (x)^{2} + 2 f (x) - 3

into

4 g (f (x)) = 3 x^{2} - 32 x + 72

gives

12 f (x)^{2} + 8 f (x) - 12 = 3 x^{2} - 32 x + 72

, i.e.

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

. Solving,

f (x) = \frac{- 8 \pm 144 x ^{2} - 1536 x + 4096}{24} = \frac{- 8 \pm ( 12 x - 64 )}{24}

, and the condition

f (0) = - 3

selects

f (x) = \frac{- 8 + ( 12 x - 64 )}{24} = \frac{x - 6}{2}

. Therefore

f (g (2)) = f (13) = \frac{13 - 6}{2} = \frac{7}{2}

.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option

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