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(B) Given the functional equation $f(xy) = xf(y) + yf(x) - 2xy$. Divide by $xy$: $\frac{f(xy)}{xy} = \frac{f(y)}{y} + \frac{f(x)}{x} - 2$. Let $g(x) =

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$f(x) = x \ln x + 2x$

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(B) Given the functional equation $f(xy) = xf(y) + yf(x) - 2xy$. Divide by $xy$: $\frac{f(xy)}{xy} = \frac{f(y)}{y} + \frac{f(x)}{x} - 2$. Let $g(x) = \frac{f(x)}{x}$. Then $g(xy) = g(x) + g(y) - 2$. Let $h(x) = g(x) - 2$. Then $h(xy) + 2 = h(x) + 2 + h(y) + 2 - 2$,which simplifies to $h(xy) = h(x) + h(y)$. This is Cauchy's functional equation,so $h(x) = c \ln x$. Thus,$\frac{f(x)}{x} - 2 = c \ln x$,or $f(x) = cx \ln x + 2x$. Given $f'(x) = c \ln x + c + 2$. At $x = 1$,$f'(1) = c(0) + c + 2 = 3$,so $c = 1$. Therefore,$f(x) = x \ln x + 2x$. Checking the options,$f(x) = x \ln x + 2x$ matches option $B$.

Let $f$ be a differentiable function satisfying the relation $f(xy) = xf(y) + yf(x) - 2xy$ for all $x, y > 0$ and $f'(1) = 3$. Which of the following statements is true?

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