Question

If $\int f\left(x\right)dx=2\left\{f\right(x){\}}^3+c$, and $f\left(x\right)\ne 0$ then $f\left(x\right)$ is

• A. $\frac{x}{2}$
• B. $x^3$
• C. $\frac{1}{\sqrt{x}}$
• D. $\sqrt{\frac{x}{3}}$

Answer 1

The correct option is D $\sqrt{\frac{x}{3}}$

$\int f\left(x\right)dx=2\left\{f\right(x){\}}^3+c$
$\Rightarrow f\left(x\right)=\frac{d\left[2\right[f\left(x\right){]}^3]}{dx}$
$\Rightarrow f\left(x\right)=2\times 3\left[f\right(x){]}^2\frac{df\left(x\right)}{dx}$
$\Rightarrow \frac{1}{6}dx=f\left(x\right)df\left(x\right)$
Applying Integration on both sides
$\int \frac{1}{6}dx=\int f\left(x\right)df\left(x\right)$
$\frac{x}{6}=\frac{\left[f\right(x){]}^2}{2}$
$f\left(x\right)=\sqrt{\frac{x}{3}}$