Let $$f: \mathbb{R} \to \mathbb{R}$$ be such that $$f(xy) = f(x)f(y)$$, for all $$x, y \in \mathbb{R}$$ and $$f(0) \ne 0$$. Let $$g: [1, \infty) \to \mathbb{R}$$ be a differentiable function such that $$x^2 g(x) = \int_1^x (t^2 f(t) - tg(t))\,dt.$$ Then $$g(2)$$ is equal to :

$$f(xy) = f(x)f(y)$$

$$f(0) = f(x)f(0)$$

$$f(0)(1 - f(x)) = 0$$

$$1 - f(x) = 0 \implies f(x) = 1 \quad (\text{for all } x \in \mathbb{R})$$

$$x^2 g(x) = \int_{1}^{x} (t^2 - t g(t)) \, dt$$

$$\frac{d}{dx}[x^2 g(x)] = \frac{d}{dx}\left[\int_{1}^{x} (t^2 - t g(t)) \, dt\right]$$

$$2x g(x) + x^2 g'(x) = x^2 - x g(x)$$

$$x^2 g'(x) + 3x g(x) = x^2$$

$$g'(x) + \frac{3}{x}g(x) = 1$$

Integrating Factor ($$\text{I.F.}$$) $$= e^{\int \frac{3}{x} \, dx} = e^{3 \ln x} = x^3$$

$$g(x) \cdot x^3 = \int 1 \cdot x^3 \, dx$$

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

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

$$1^2 \cdot g(1) = \int_{1}^{1} (t^2 - tg(t)) \, dt \implies g(1) = 0$$

$$g(1) = \frac{1}{4} + \frac{C}{1^3} = 0 \implies C = -\frac{1}{4}$$

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

$$g(2) = \frac{1}{2} - \frac{1}{32} = \frac{16 - 1}{32} = \frac{15}{32}$$