Question

Let $f$ be a differentiable function satisfying $f(xy)=xf(y)+yf(x)-2x-2y+2\forall x,y>0$. If $f^{\prime }(1)=1$ then
[Note: $\{k\}$ denotes fractional part function of $k$.]

• A. $\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{f\left(e^x\right)-\{x\}-2}{x^2}=-1$
• B. $\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{f\left(e^x\right)-\{x\}-2}{x^2}=1$
• C. minimum value of $f(x)$ is $2-\frac{1}{e}$
• D. minimum value of $f(x)$ occurs at $x=e$.

Answer 1

Answer: (B), (C)

$f(xy)=xf(y)+yf(x)-2(x+y)+2$
$\text{ partially differentiating w.r.t. }x$
$f^{\prime }(xy)y=f(y)+y{f}^{\prime }(x)-2$
$\text{ putting }x=1$
$f^{\prime }(y)y=f(y)+y-2$
$f^{\prime }(x)x=f(x)=x-2$
$\frac{dy}{dx}-\frac{y}{x}=1-\frac{2}{x}\text{ (linear differential equation) }$
$\text{ I.F. }=e^{\int \frac{-1}{x}dx}=e^{-\ln x}=\frac{1}{x}$
$y\cdot \frac{1}{x}=\int \frac{1}{x}\left(1-\frac{2}{x}\right)dx+C$
$\frac{y}{x}=\ln x+\frac{2}{x}+C;f(1)=2\Rightarrow C=0$
$y=x\ln x+2=f(x)$
$\text{ Now, }$
(A) &(B)$\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{f\left(e^x\right)-\{x\}-2}{x^2}=\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{x{e}^x+2-\{x\}-2}{x^2}=\underset{x\to 0^{+}}{\mathrm{Lim}}\frac{x\left(e^x-1\right)}{x^2}=1$
(C) $f(x)=x\ln x+2$
$f^{\prime }(x)=x\cdot \frac{1}{x}+\ln x\cdot 1=1+\ln x$
for $x>\frac{1}{e},f(x)$ is increasing
for $0<x<\frac{1}{e},f(x)$ is decreasing
${\therefore f(x)|}_{\min .}=f\left(\frac{1}{e}\right)=\frac{-1}{e}+2\Rightarrow$
(C) is correct
(D) $f(x)\to \infty$ as $x\to \infty$