Question

Let $f$ be a differentiable function satisfying $f(x)=\frac{2}{\sqrt{3}}\underset{0}{\overset{\sqrt{3}}{\int }}f\left(\frac{{\lambda }^{2}x}{3}\right)d\lambda ,x>0$ and $f(1)=\sqrt{3}$. If $y=f(x)$ passes through the point $(\alpha ,6)$, then $\alpha$ is equal to_______

Answer 1

Answer: 12

Let, $\frac{{\lambda }^{2}x}{3}=t$
$\Rightarrow \frac{2\lambda x}{3}d\lambda =dt$
$\Rightarrow d\lambda =\frac{3}{2}\cdot \frac{1\sqrt{x}}{x\cdot \sqrt{3}\sqrt{t}}dt$
$\Rightarrow d\lambda =\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{x}}\cdot \frac{dt}{\sqrt{t}}$
So, $f(x)=\frac{1}{\sqrt{x}}{\int }_0^x\frac{f(t)}{\sqrt{t}}dt$
$\Rightarrow \sqrt{x}\cdot f^{\prime }(x)+\frac{f(x)}{2\sqrt{x}}=\frac{f(x)}{\sqrt{x}}$
$\Rightarrow \sqrt{x}\cdot f^{\prime }(x)=\frac{f(x)}{2\sqrt{x}}$
$\Rightarrow \frac{dy}{y}=\frac{dx}{2x}$
$\Rightarrow \ln y=\frac{1}{2}\ln x+c\Rightarrow f(x)=\sqrt{x}$
$\Rightarrow y=\sqrt{3x}\{\text{ as }f(1)=\sqrt{3}\}$
So, $f(x)=\sqrt{3x}$
Now, $f(\alpha )=6\Rightarrow 36=3\alpha$
$\Rightarrow \alpha =12$