Question

Let $f(x)$ be a differentiable function satisfying the condition $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)},y\ne 0,f(y)\ne 0$ for all $x,y\in R$ and $f^{\prime }(1)=2$, then

• A. Area bounded by curve $y=f(x),y=2^x$ and $y$-axis in $1^{\text{st }}$ quadrant is $\frac{9-\ln 256}{\ln 8}$.
• B. Area bounded by curve $y=f(x),y=2^x$ and $y$-axis in $1^{\text{sf }}$ quadrant is $\frac{9+\ln 256}{\ln 8}$.
• C. $\underset{x\to 0}{\mathrm{Lim}}\left[\frac{f(x)}{x}\right]$ does not exist, where [ ] denotes greatest integer function.
• D. Absolute maximum value of $f(x)$ over the interval $[-3,2]$ is 9 .

Answer 1

Answer: (A), (C), (D)

$\Theta f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$
Differentiating w.r.t. $y$ keeping $x$ constant
$f^{\prime }\left(\frac{x}{y}\right)\cdot \left(\frac{-x}{y^2}\right)=\frac{-f(x)}{f^2(y)}\cdot f^{\prime }(y)$
Putting $y=1$
$x{f}^{\prime }(x)=\frac{f(x)\cdot f^{\prime }(1)}{(f(1){)}^2}$
Putting $x=y=1$ in given relation, we get $f(1)=1$
$\therefore \frac{f^{\prime }(x)}{f(x)}=\frac{2}{x}\Rightarrow \ln f(x)=2\ln x+c$
$\text{ Putting }x=1,c=0$
$\therefore f(x)=x^2$
Putting $x=1,c=0$
$\therefore f(x)=x^2$
$\text{ Area bounded }=\underset{0}{\overset{2}{\int }}\left(2^x-x^2\right)dx={\left(\frac{2^x}{\ln 2}-\frac{x^3}{3}\right)}_0^2$
$=\frac{4}{\ln 2}-\frac{8}{3}-\frac{1}{\ln 2}=\frac{9-8\ln 2}{3\ln 2}$

Clearly, L.H.L. $=\underset{x\to 0^{-}}{\mathrm{Lim}}\left[\frac{f(x)}{x}\right]=-1$
and R.H.L. $=\underset{x\to 0^{-}}{\mathrm{Lim}}\left[\frac{f(x)}{x}\right]=0$
$\therefore$ Limit does not exist.