JEE Mathematics 2026 Question with Solution

Given: $g(x) = 3x^2 + 2x - 3$, $f(0) = -3$, and

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

Find: $f(g(2))$

From the given relation,

$$g(f(x)) = \frac{3}{4}x^2 - 8x + 18$$

Assume

$$f(x) = ax + b$$

Since $f(0) = -3$, we get $b = -3$. Therefore,

$$f(x) = ax - 3$$

Now substitute into $g$:

$$g(f(x)) = 3(ax-3)^2 + 2(ax-3) - 3$$ $$= 3a^2x^2 - 18ax + 27 + 2ax - 6 - 3$$ $$= 3a^2x^2 - 16ax + 18$$

Comparing coefficients with

$$\frac{3}{4}x^2 - 8x + 18$$

we get

$$3a^2 = \frac{3}{4}$$

so

$$a^2 = \frac{1}{4}$$

and

$$-16a = -8$$

so

$$a = \frac{1}{2}$$

Hence,

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

Now,

$$g(2) = 3(2)^2 + 2(2) - 3 = 13$$

Therefore,

$$f(g(2)) = f(13) = \frac{13}{2} - 3 = \frac{7}{2}$$

So the correct option is D.

Given: The composite function value is provided and $f(0) = -3$.

Find: The value of $f(g(2))$.

The key idea is to first express $g(f(x))$ in simplified form and then compare coefficients.

Since

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

dividing by $4$ gives

$$g(f(x)) = \frac{3}{4}x^2 - 8x + 18$$

Let

$$f(x) = ax + b$$

Using $f(0) = -3$,

$$b = -3$$

Thus,

$$f(x) = ax - 3$$

Now evaluate $g(f(x))$ using $g(t) = 3t^2 + 2t - 3$:

$$g(f(x)) = 3(ax-3)^2 + 2(ax-3) - 3$$

Expand the square:

$$(ax-3)^2 = a^2x^2 - 6ax + 9$$

So,

$$g(f(x)) = 3(a^2x^2 - 6ax + 9) + 2ax - 6 - 3$$ $$= 3a^2x^2 - 18ax + 27 + 2ax - 6 - 3$$ $$= 3a^2x^2 - 16ax + 18$$

Now compare with

$$\frac{3}{4}x^2 - 8x + 18$$

Matching the coefficients of $x^2$ and $x$,

$$3a^2 = \frac{3}{4}, \qquad -16a = -8$$

From the linear coefficient,

$$a = \frac{1}{2}$$

Hence,

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

Next,

$$g(2) = 3\cdot 4 + 2\cdot 2 - 3 = 12 + 4 - 3 = 13$$

Finally,

$$f(g(2)) = f(13) = \frac{13}{2} - 3 = \frac{7}{2}$$

Therefore, the value of $f(g(2))$ is $\frac{7}{2}$, so the correct option is D.