Question

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

Answer 1

Answer: 2

Differentiate w.r.t. y keeping x constant
$xf(x,y)=\underset{1}{\overset{x}{\int }}f(t)dt+xf(y)$
$\text{ Put }y=1$
$xf(x)={\int }_1^{x}f(t)dt+3x$
$x{f}^{\prime }(x)+f(x)=f(x)+3$
$f^{\prime }(x)=\frac{3}{x}$
$f^{\prime }(x)=3\ln x+C$
$f^{\prime }(1)=3\Rightarrow c=3\Rightarrow f(x)=3\ln (ex)$
$f^{\prime }(x)=\frac{3}{x}=\frac{y}{x}\Rightarrow y=3$
$\therefore$ 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 $k=2$.