Functions – Functional Equations – JEE Main 28 Jan 2026 Shift 1

Solution:

First, we find the value of $g(2)$:
$$g(2) = 3(2)^2 + 2(2) – 3$$
$$g(2) = 12 + 4 – 3 = 13$$

We need to find $f(g(2))$, which means we need to find $f(13)$.

Given the functional equation:
$$4g(f(x)) = 3x^2 – 32x + 72$$

Substitute $g(t) = 3t^2 + 2t – 3$ where $t = f(x)$:
$$4[3(f(x))^2 + 2f(x) – 3] = 3x^2 – 32x + 72$$
$$12(f(x))^2 + 8f(x) – 12 = 3x^2 – 32x + 72$$
$$12(f(x))^2 + 8f(x) – (3x^2 – 32x + 84) = 0$$

This is a quadratic equation in terms of $f(x)$. Using the quadratic formula:
$$f(x) = \frac{-8 \pm \sqrt{64 – 4(12)(-(3x^2 – 32x + 84))}}{2(12)}$$
$$f(x) = \frac{-8 \pm \sqrt{64 + 48(3x^2 – 32x + 84)}}{24}$$
$$f(x) = \frac{-8 \pm \sqrt{16[4 + 3(3x^2 – 32x + 84)]}}{24}$$
$$f(x) = \frac{-8 \pm 4\sqrt{4 + 9x^2 – 96x + 252}}{24}$$
$$f(x) = \frac{-8 \pm 4\sqrt{9x^2 – 96x + 256}}{24}$$
$$f(x) = \frac{-8 \pm 4\sqrt{(3x – 16)^2}}{24}$$

Given $f(0) = -3$. Let’s check the signs:
Case 1 (Positive sign):
$$f(0) = \frac{-8 + 4(3(0) – 16)}{24} = \frac{-8 – 64}{24} = \frac{-72}{24} = -3$$
This satisfies the condition.

So, the function is:
$$f(x) = \frac{-8 + 4(3x – 16)}{24}$$

Now calculate $f(13)$:
$$f(13) = \frac{-8 + 4(3(13) – 16)}{24}$$
$$f(13) = \frac{-8 + 4(39 – 16)}{24}$$
$$f(13) = \frac{-8 + 4(23)}{24}$$
$$f(13) = \frac{-8 + 92}{24} = \frac{84}{24} = \frac{7}{2}$$

Ans. (3)