Let $f: R \rightarrow R$ be such that $f(x y)=f(x) f(y)$, for all $x, y \in R$ and $f(0) \neq 0$. Let $g:[1, \infty) \rightarrow R$ be a differentiable function such that $x^{2} g(x)=\int_{1}^{x}\left(t^{2} f(t)-\operatorname{tg}(t)\right) d t$ Then $\mathrm{g}(2)$ is equal to :

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Differential Equations

Let

f : R \to R

be such that

f (x y) = f (x) f (y)

, for all

x, y \in R

and

f (0) \neq = 0

. Let

g : [1, \infty) \to R

be a differentiable function such that

x^{2} g (x) = \int_{1}^{x} (t^{2} f (t) - tg (t)) d t

Then

g (2)

is equal to :

Answer: (C)

From

f (0) = f (0 \cdot 0) = f (0)^{2}

with

f (0) \neq = 0

f (0) = 1

; then

f (0) = f (x \cdot 0) = f (x) f (0)

forces

f (x) = 1

for all

x

. The relation becomes

x^{2} g (x) = \int_{1}^{x} (t^{2} - t g (t)) d t

. Differentiating,

x^{2} g^{'} + 2 xg = x^{2} - xg

, i.e.

g^{'} + \frac{3}{x} g = 1

; with integrating factor

x^{3}

\frac{d}{d x} (x^{3} g) = x^{3}

, so

x^{3} g = \frac{x ^{4}}{4} + C

. Setting

x = 1

in the integral equation gives

g (1) = 0

, hence

C = - \frac{1}{4}

, and

g (x) = \frac{x}{4} - \frac{1}{4 x ^{3}}

. Therefore

g (2) = \frac{1}{2} - \frac{1}{32} = \frac{15}{32}

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option