JEE Main 2026 — Mathematics Calculus

Given $f(xy) = f(x)f(y)$f(xy)=f(x)f(y) for all $x, y \in \mathbb{R}$x,y∈R and $f(0) \neq 0$f(0)=0.

Substituting $y = 0$y=0, we get $f(0) = f(x)f(0)$f(0)=f(x)f(0).

Since $f(0) \neq 0$f(0)=0, dividing by $f(0)$f(0) gives $f(x) = 1$f(x)=1 for all $x \in \mathbb{R}$x∈R.

The given integral equation is:

$$x^2 g(x) = \int\limits_1^x (t^2 f(t) - tg(t))\,dt$$x2g(x)=1∫x​(t2f(t)−tg(t))dt

Substituting $f(t) = 1$f(t)=1:

$$x^2 g(x) = \int\limits_1^x (t^2 - tg(t))\,dt$$x2g(x)=1∫x​(t2−tg(t))dt

Differentiating both sides with respect to $x$x using Leibniz's rule:

$$2x g(x) + x^2 g'(x) = x^2 - xg(x)$$2xg(x)+x2g′(x)=x2−xg(x)

$$x^2 g'(x) + 3xg(x) = x^2$$x2g′(x)+3xg(x)=x2

Dividing by $x$x (since $x \ge 1$x≥1):

$$x g'(x) + 3g(x) = x$$xg′(x)+3g(x)=x

$$g'(x) + \dfrac{3}{x}g(x) = 1$$g′(x)+x3​g(x)=1

This is a linear differential equation. The integrating factor is:

$$\text{IF} = e^{\int \dfrac{3}{x}\,dx} = e^{3\ln x} = x^3$$IF=e∫x3​dx=e3lnx=x3

Multiplying the differential equation by $x^3$x3:

$$x^3 g'(x) + 3x^2 g(x) = x^3$$x3g′(x)+3x2g(x)=x3

$$\dfrac{d}{dx}(x^3 g(x)) = x^3$$dxd​(x3g(x))=x3

Integrating both sides with respect to $x$x:

$$x^3 g(x) = \dfrac{x^4}{4} + C$$x3g(x)=4x4​+C

From the integral equation, substituting $x = 1$x=1 gives $\displaystyle 1^2 g(1) = \int\limits_1^1 (t^2 - tg(t))\,dt = 0$12g(1)=1∫1​(t2−tg(t))dt=0, so $g(1) = 0$g(1)=0.

Substituting $x = 1$x=1 and $g(1) = 0$g(1)=0 into the integrated equation:

$$1^3(0) = \dfrac{1^4}{4} + C \Rightarrow C = -\dfrac{1}{4}$$13(0)=414​+C⇒C=−41​

Thus, the function $g(x)$g(x) is given by:

$$x^3 g(x) = \dfrac{x^4 - 1}{4}$$x3g(x)=4x4−1​

$$g(x) = \dfrac{x^4 - 1}{4x^3}$$g(x)=4x3x4−1​

Substituting $x = 2$x=2:

$$g(2) = \dfrac{2^4 - 1}{4(2^3)} = \dfrac{16 - 1}{32} = \dfrac{15}{32}$$g(2)=4(23)24−1​=3216−1​=3215​

Answer: $\dfrac{15}{32}$3215​