Question

Sol. [Key Concept: $f(xy) = f(x)f(y)$ with $f(0) \neq 0$ forces $f \equiv 1$; differentiating the integral equation gives the linear ODE $g' + \frac{3}{x}g = 1$ (I.F. $x^{3}$); apply $g(1) = 0$ and find $g(2)$.]

$$\ \because f(xy) = f(x)f(y) $$$\therefore f(x) = x^{n}$ or $f(x) = k$

But $f(0) \neq 0 \Rightarrow f(x) \neq x^{n}$

For $f(x) = k \Rightarrow k = k \cdot k \Rightarrow k = 1(\because f(0) \neq 0)$

$${\Rightarrow f(x) = 1 }{x^{2}\text{ }g(x) = \int_{1}^{x}\mspace{2mu}\left( t^{2}(1) - tg(t) \right) \cdot dt }$$diff $w/r$ to $x$

$${\Rightarrow x^{2}{\text{ }g}^{'}(x) + g(x) \cdot 2x = x^{2} - xg(x) }{\Rightarrow {xg}^{'}(x) + 3\text{ }g(x) = x }{\Rightarrow g^{'}(x) + \frac{3}{x}g(x) = 1 }{IF = e^{\int\frac{3}{x}dx} = e^{3ln(x)} = x^{3} }$$Solution of DE :

$${\Rightarrow g(x) \cdot x^{3} = \int x^{3}dx }{\Rightarrow x^{3} \cdot g(x) = \frac{x^{4}}{4} + C }{\Rightarrow C = - \frac{1}{4}\{\because\text{ }g(1) = 0\} }{g(x) \cdot x^{3} = \frac{x^{4}}{4} - \frac{1}{4} }$$Put $x = 2$

(8) $g(2) = 4 - \frac{1}{4}$

$g(2) = \frac{15}{32}$.