Question

Let $f ( x )$ be a derivable non-decreasing function such that $\int\limits_0^{ x }( f ( t ))^3 dt =\frac{1}{ x ^2}\left(\int\limits_0^{ x } f ( t ) dt \right)^3 \forall x \in R -\{0\}$ and $f (1)=1$ If $\int\limits_0^x f(t) d t=g(x)$ then $\frac{x g^{\prime}(x)}{g(x)}$ is

• A. always equal to 1
• B. always equal to -2
• C. may be 1 or -2
• D. not independent of $x$

Answer 1

Answer: (A)

$ \Theta (f(x))^3=\frac{1}{x^2} \times 3\left(\int\limits_0^x f(t) d t\right)^2 \cdot f(x)-\frac{2}{x^3}\left(\int\limits_0^x f(t) d t\right)^3 $
$\Rightarrow \left(g^{\prime}(x)\right)^3=\frac{3 g^{\prime}(x) \cdot(g(x))^2}{x^2}-\frac{2 g^3(x)}{x^3} $
$\Rightarrow \left(\frac{x g^{\prime}(x)}{g(x)}\right)^3=3\left(\frac{x g^{\prime}(x)}{g(x)}\right)-2$
Let $\frac{x g^{\prime}(x)}{g(x)}=t$
$\therefore t ^3-3 t +2=0 \Rightarrow( t -1)\left( t ^2+ t -2\right)=0 $
$\Rightarrow ( t -1)( t +2)( t -1)=0$
$\therefore \frac{ xg ( x )}{ g ( x )}=1 \text { or }-2$
(i) if $\frac{x^{\prime}(x)}{g(x)}=1 \Rightarrow x f(x)=\int\limits_0^x f(t) d t$
$ \Rightarrow f(x)+x f^{\prime}(x)=f(x) \Rightarrow f^{\prime}(x)=0 $
$\Rightarrow f(x)=C, C=\text { constant } $
$\text { (ii) If } \frac{x(1)=1}{} \frac{x g^{\prime}(x)}{g(x)}=-2 \Rightarrow x f(x)=-2 \int_0^x f(t) d t $
$ \Rightarrow f(x)+x f^{\prime}(x)=-2 f(x) $
$ \Rightarrow \frac{f^{\prime}(x)}{f(x)}=-\frac{3}{x} \Rightarrow \ln f(x)=-3 \ln x+\ln K$
$ \Rightarrow f(x)=\frac{K}{x^3}$
$\Theta f (1)=1 $
$\therefore K =1$
$\therefore f ( x )=\frac{1}{ x ^3} $
$ f ^{\prime}( x )=-\frac{3}{ x ^4}<0$
$\therefore f ( x )=\frac{1}{ x ^3} \text { is decreasing }$
$\text { but } f ( x ) \text { is non-decreasing } $
$\therefore \frac{ xg ( x )}{ g ( x )}=1$