Question

Let f be a real valued function satisfying $f\left(\frac{x}{y}\right)=f\left(x\right)-f\left(y\right)$ and $\underset{x\to 0}{lt}\frac{f(1+x)}{x}=3$. Then find the area bounded by the curve y=f(x), the y-axis and the line y=3 in sq units

• A. $e^3$
• B. $3e$
• C. $e+3$
• D. $3^e$

Answer 1

The correct option is B $3e$

$f\left(\frac{x}{y}\right)=f\left(x\right)-f\left(y\right)$
$x=y=1\Rightarrow f\left(1\right)=0$
$f^1\left(x\right)=\underset{h\to 0}{lt}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{x\to 0}{lt}\frac{f(1+\frac{h}{x})}{h}=\frac{3}{x}$
$\therefore f\left(x\right)=3logx$
area $={\int }_{-\infty }^{3}xdy$
$={\int }_{-\infty }^3{e}^{\frac{y}{3}}dy=3e$