Question

If $f\left(x\right)={\int }_0^x\frac{1}{f\left(x\right)}dx$ then the function $f\left(x\right)$ equals

• A. $2x$
• B. $\sqrt{2x}$
• C. $x\sqrt{2}$
• D. $\frac{x}{\sqrt{2}}$

Answer 1

The correct option is B $\sqrt{2x}$

$f\left(x\right)={\int }_a^x\frac{1}{f\left(x\right)}dx$
$\therefore f^{\prime }\left(x\right)=\frac{1}{f\left(x\right)}\Rightarrow 2f\left(x\right)f^{\prime }\left(x\right)=2\Rightarrow {\left\{f\right(x\left)\right\}}^2=2x+k$
$\Rightarrow f\left(x\right)=\sqrt{2x+k}\Rightarrow f\left(1\right)=\sqrt{2+k}\Rightarrow \sqrt{2}=\sqrt{2+k}\Rightarrow k=0$
By using $f\left(x\right)={\int }_a^x\frac{1}{f\left(x\right)}dx\Rightarrow f\left(1\right)={\int }_a^x\frac{1}{f\left(x\right)}dx=\sqrt{2}$ (Given)
$\therefore f\left(x\right)=\sqrt{2x}$