Question

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

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

Answer 1

Answer: (A), (B), (C)

$\underset{x\to 0}{lim}\left[\frac{f\left(x\right)}{x}\right]$ does not exist, where $[.]$ denotes greatest integer function.

Given, $f\left(\frac{x}{y}\right)=\frac{f\left(x\right)}{f\left(y\right)}$
Differentiating w.r.t. $y$ keeping $x$ constant, we get
$f^{\prime }\left(\frac{x}{y}\right)\cdot (-\frac{x}{y^2})=-\frac{f\left(x\right)}{f^2\left(y\right)}\cdot f^{\prime }\left(y\right)$
Putting $y=1,$ we get
$x{f}^{\prime }\left(x\right)=\frac{f\left(x\right)\cdot f^{\prime }\left(1\right)}{\left(f\right(1){)}^2}$
Putting $x=y$ in given relation, we get $f\left(1\right)=1$
$\therefore \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{2}{x}$
$\Rightarrow \ln |f\left(x\right)|=2\ln |x|+c$
Putting $x=1$, we get $c=0$
$\therefore f\left(x\right)=x^2$ ($f\left(x\right)\ne -x^2$ because given relation is not satisfied.)

Figure:


$\therefore$ Required area is
$=\underset{0}{\overset{2}{\int }}(2^x-x^2)dx$
$={[\frac{2^x}{\ln 2}-\frac{x^3}{3}]}_0^2$
$=\frac{4}{\ln 2}-\frac{8}{3}-\frac{1}{\ln 2}$
$=\frac{9-\ln 256}{\ln 8}$ sq. units


$L.H.L.=\underset{x\to 0^{-}}{lim}\left[\frac{f\left(x\right)}{x}\right]$
$\Rightarrow L.H.L.=\underset{x\to 0^{-}}{lim}\left[\frac{x^2}{x}\right]$
$\Rightarrow L.H.L.=\underset{x\to 0^{-}}{lim}\left[x\right]=-1$ and
$R.H.L.=\underset{x\to 0^{+}}{lim}\left[x\right]=0$
Since $R.H.L.\ne L.H.L.$
$\therefore$ Limit does not exist.


Absolute maximum value of $f\left(x\right)$ over the interval $[-3,2]$ is $(-3{)}^2=9.$