Question

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

$\Theta (f(x){)}^3=\frac{1}{x^2}\times 3{\left(\underset{0}{\overset{x}{\int }}f(t)dt\right)}^2\cdot f(x)-\frac{2}{x^3}{\left(\underset{0}{\overset{x}{\int }}f(t)dt\right)}^3$
$\Rightarrow {\left(g^{\prime }(x)\right)}^3=\frac{3g^{\prime }(x)\cdot (g(x){)}^2}{x^2}-\frac{2g^3(x)}{x^3}$
$\Rightarrow {\left(\frac{x{g}^{\prime }(x)}{g(x)}\right)}^3=3\left(\frac{x{g}^{\prime }(x)}{g(x)}\right)-2$
Let $\frac{x{g}^{\prime }(x)}{g(x)}=t$
$\therefore t^3-3t+2=0\Rightarrow (t-1)\left(t^2+t-2\right)=0$
$\Rightarrow (t-1)(t+2)(t-1)=0$
$\therefore \frac{xg(x)}{g(x)}=1\text{ or }-2$
(i) if $\frac{x^{\prime }(x)}{g(x)}=1\Rightarrow xf(x)=\underset{0}{\overset{x}{\int }}f(t)dt$
$\Rightarrow f(x)+x{f}^{\prime }(x)=f(x)\Rightarrow f^{\prime }(x)=0$
$\Rightarrow f(x)=C,C=\text{ constant }$
$\text{ (ii) If }\frac{x(1)=1}{}\frac{x{g}^{\prime }(x)}{g(x)}=-2\Rightarrow xf(x)=-2{\int }_0^{x}f(t)dt$
$\Rightarrow f(x)+x{f}^{\prime }(x)=-2f(x)$
$\Rightarrow \frac{f^{\prime }(x)}{f(x)}=-\frac{3}{x}\Rightarrow \ln f(x)=-3\ln x+\ln K$
$\Rightarrow f(x)=\frac{K}{x^3}$
$\Theta f(1)=1$
$\therefore K=1$
$\therefore f(x)=\frac{1}{x^3}$
$f^{\prime }(x)=-\frac{3}{x^4}<0$
$\therefore f(x)=\frac{1}{x^3}\text{ is decreasing }$
$\text{ but }f(x)\text{ is non-decreasing }$
$\therefore \frac{xg(x)}{g(x)}=1$