Question

Let f(x) is a continuous function which takes positive values for x (x>0), and satisfy ${\int }_0^{x}f\left(t\right)dt=x\sqrt{f\left(x\right)}$ with $f\left(1\right)=\frac{1}{2}$. Then the value of $f(\sqrt{2}+1)$ equals

• A. 1
• B. $\sqrt{2}-1$
• C. $\frac{1}{4}$
• D. $\frac{1}{\sqrt{2}-1}$

Answer 1

The correct option is C $\frac{1}{4}$

We have ${\int }_0^{x}f\left(t\right)dt=x\sqrt{f\left(x\right)}$ ....(1)
Differentiating both the sides of equation (1) w.r.t. x, we get
$f\left(x\right)=\frac{x{f}^{\prime }x}{2\sqrt{f\left(x\right)}}+\sqrt{f\left(x\right)};Letf\left(x\right)=y^2\Rightarrow f^{\prime }\left(x\right)=2y\frac{dy}{dx}{y}^2=x.2y.\frac{dy}{dx}.\frac{1}{2y}+y\Rightarrow y^2=x.\frac{dy}{dx}+y\Rightarrow y^2-y=x.\frac{dy}{dx}\int \frac{dy}{y(y-1)}=\int \frac{dx}{x}\Rightarrow \int \frac{y-(y-1)}{y(y-1)}dy=\int \frac{dx}{x};ln\frac{(y-1)}{y}=lncx\Rightarrow \frac{(y-1)}{y}=cx$
$\Rightarrow 1-\frac{1}{y}=cx\Rightarrow \frac{1}{y}=1-cx\Rightarrow y=\frac{1}{1-cx}\therefore \sqrt{f\left(x\right)}=\frac{1}{1-cx}$ ....(1)
If $x=1,f\left(1\right)=\frac{1}{2}$ (given)
$1-c=\sqrt{2};c=1-\sqrt{2}\sqrt{f\left(x\right)}=\frac{1}{1+(\sqrt{2}-1)x}\Rightarrow f\left(x\right)=\frac{1}{[1+(\sqrt{2}-1)x{]}^2}\Rightarrow f(\sqrt{2}+1)=\frac{1}{4}$