Question

Let $f$ be a differentiable function satisfying the functional rule $f ( xy )= f ( x )+ f ( y )+ xy - x - y \forall x , y >0$ and $f^{\prime}(1)=4$ then $\int \frac{f(x)}{x} d x$ is equal to
[Note: ' $C$ ' is constant of integration.]

• A. $3 \ln ^2 x+x+C$
• B. $3 \ln x + x + C$
• C. $\frac{3}{2} \ln x+x+C$
• D. $\frac{3}{2} \ln ^2 x+x+C$

Answer 1

Answer: (D)

Differentiable w.r.t. $X$
$\text { y } f^{\prime}(x y)=f^{\prime}(x)+y-1 $
$\text { Put } x=1 $
$y f^{\prime}(y)=f^{\prime}(1)+y-1 $
$y f^{\prime}(y)=y+3 $
$\therefore f^{\prime}(y)=1+\frac{3}{y}$
Integrating
$g ( y )= y +3 \ln y + C $
$\text { Also, } f (1)=1 $
$C =0$
$\therefore f ( x )=3 \ln x + x $
$\int \frac{ f ( x )}{ x } dx =\int \frac{3 \ln x }{ x } dx +\int dx =\frac{3}{2} \ln ^2 x + x + C $