Question

Let $f:R^{+}\to R^{+}$ be a differentiable function satisfying $f\left(xy\right)=\frac{f\left(x\right)}{y}+\frac{f\left(y\right)}{x}$ for all $x,yϵR^{+}$. Also $f\left(1\right)=0,f^{\prime }\left(1\right)=1$. If M is the greatest value of $f\left(x\right)$ then $[m+e]$ is ___ (where [.] represents Greatest Integer Function).

• A. 3
• B. 2
• C. 0
• D. 1

Answer 1

The correct option is A

3

$f\left(xy\right)=\frac{f\left(x\right)}{y}+\frac{f\left(y\right)}{x}$

$f^{\prime }\left(xy\right)\cdot (y+x{y}^{\prime })=\frac{f^{\prime }\left(x\right)}{y}-\frac{f\left(x\right)y^{\prime }}{y^2}+\frac{f^{\prime }\left(y\right)y^{\prime }}{x}-\frac{f\left(y\right)}{x^2}$

Let $x=1$;$f^{\prime }\left(y\right)\cdot (y+y^{\prime })=\frac{f^{\prime }\left(1\right)}{y}-\frac{f\left(1\right)y^{\prime }}{y^2}+\frac{f^{\prime }\left(y\right)y^{\prime }}{1}-\frac{f\left(y\right)}{1}$

Substituting the value we can write $\frac{dy}{dx}+\frac{y}{x}=\frac{1}{x^2}$. Solving the linear differential equation we get $f\left(x\right)=\frac{\log x}{x}$ as $f\left(1\right)=0$.

Maximum value of $f\left(x\right)=\frac{1}{e}$

$\therefore [\frac{1}{e}+e]=3$