Question

Let $f: R^{+} \rightarrow R$ be a derivable function satisfying $\int \limits_1^{x y} f(t) d t=y \int \limits_1^x f(t) d t+x \int \limits_1^y f(t) d t \forall x, y>0$, if $f (1)=3$, then number of integral values of $k$ for which $f ( x )= kx$ has exactly 2 solutions.

Answer 1

Differentiate w.r.t. y keeping x constant
$x f(x, y)=\int\limits_1^x f(t) d t+x f(y) $
$\text { Put } y=1$
$x f(x)=\int_1^x f(t) d t+3 x $
$x f^{\prime}(x)+f(x)=f(x)+3 $
$f^{\prime}(x)=\frac{3}{x} $
$f^{\prime}(x)=3 \ln x+C $
$f^{\prime}(1)=3 \Rightarrow c=3 \Rightarrow f(x)=3 \ln (e x) $
$f^{\prime}(x)=\frac{3}{x}=\frac{y}{x} \Rightarrow y=3$
$\therefore$ Slope of tangent to $y =\ln ( ex )$ from origin $=\frac{3}{1}$
$\therefore$ If $y = kx$ is to intersect $y =3 \ln$ ex in two distinct points $k \in(0,3)$
$\therefore$ Number of integral values of $k=2$.