Question

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

• A. $\underset{x \rightarrow 0^{+}}{\text{Lim}} \frac{f\left(e^x\right)-\{x\}-2}{x^2}=-1$
• B. $\underset{x \rightarrow 0^{+}}{\text{Lim}} \frac{f\left(e^x\right)-\{x\}-2}{x^2}=1$
• 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(x y)=x f(y)+y f(x)-2(x+y)+2 $
$\text { partially differentiating w.r.t. } x $
$f^{\prime}(x y) 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 } $
$y \cdot \frac{1}{x}=\int \frac{1}{x}\left(1-\frac{2}{x}\right) d x+C$
$\frac{ y }{ x }=\ln x +\frac{2}{ x }+ C ; f (1)=2 \Rightarrow C =0$
$y = x \ln x +2= f ( x )$
$\text { Now, } $
(A) &(B)$ \underset{x \rightarrow 0^{+}} {\text{Lim}} \frac{f\left(e^x\right)-\{x\}-2}{x^2}=\underset{x \rightarrow 0^{+}} {\text{Lim}} \frac{x e^x+2-\{x\}-2}{x^2}=\underset{x \rightarrow 0^{+}} {\text{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 $\left.\therefore f ( x )\right|_{\min .}= f \left(\frac{1}{ e }\right)=\frac{-1}{ e }+2 \Rightarrow$
(C) is correct
(D) $f ( x ) \rightarrow \infty$ as $x \rightarrow \infty$