Question

Let $f(x)$ be a differentiable function satisfying the condition $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}, y \neq 0, f(y) \neq 0$ 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. $\underset{x \rightarrow 0}{\text{Lim}} \left[\frac{f(x)}{x}\right]$ 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 }=\int\limits_0^2\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. $=\underset{x \rightarrow 0^{-}}{\operatorname{Lim}}\left[\frac{f(x)}{x}\right]=-1$
and R.H.L. $=\underset{x \rightarrow 0^{-}}{\operatorname{Lim}}\left[\frac{f(x)}{x}\right]=0$
$\therefore $ Limit does not exist.