Question

If a continuous function $f$ satisfies the relation
$\stackrel{t}{\underset{0}{\int }}\left(f\right(x)-\sqrt{f^{\prime }\left(x\right)})dx=0$ and $f\left(0\right)=\frac{-1}{2}$
Then $f\left(x\right)$ is equal to

• A. $\frac{-1}{x+2}$
• B. $\frac{-x+2}{4}$
• C. $\frac{-1}{x^2+2}$
• D. $\frac{x^2-2}{4}$

Answer 1

The correct option is A $\frac{-1}{x+2}$

Given,

$\stackrel{t}{\underset{0}{\int }}\left(f\right(x)-\sqrt{f^{\prime }\left(x\right)})dx=0$

Now using Leinbnitz's rule of differentiation under integral sign we get,

$f\left(t\right)-\sqrt{f^{\prime }\left(t\right)}=0$

or, $\frac{f^{\prime }\left(t\right)}{\left[f\right(t){]}^2}=1$

or, $\frac{d\left[f\right(t\left)\right]}{\left[f\right(t){]}^2}=dt$

Now integrating both sides we get,

$-\frac{1}{f\left(t\right)}=t+c$ [ Where $c$ being integrating constant]......(1)

Given,
$f\left(0\right)=-\frac{1}{2}$.

Using this condition in (1) we get,

$c=2$.

So the solution takes the form,

$-\frac{1}{f\left(t\right)}=t+2$

or, $f\left(t\right)=-\frac{1}{t+2}$.

So, $f\left(x\right)=-\frac{1}{x+2}$.