Let f be a differentiable function satisfying f(x) = 2 3 ∫ _ 0 ^ 3...

Given,

$f\left(x\right) = \dfrac{2}{\sqrt{3}}{∫}_{0}^{\sqrt{3}}f\left(\dfrac{{\lambda}^{2}x}{3}\right)d\lambda$

Now let $\dfrac{{\lambda}^{2}x}{3} = t$

$⇒ \dfrac{2\lambdax}{3}d\lambda = dt$

$⇒ d\lambda = \dfrac{3}{2} \cdot \dfrac{1\sqrt{x}}{x \cdot \sqrt{3}\sqrt{t}}dt$

$⇒ d\lambda = \dfrac{\sqrt{3}}{2} \cdot \dfrac{1}{\sqrt{x}} \cdot \dfrac{dt}{\sqrt{t}}$

So, $f\left(x\right) = \dfrac{1}{\sqrt{x}}{∫}_{0}^{x}\dfrac{f\left(t\right)}{\sqrt{t}}dt$

$\sqrt{x}f\left(x\right) = {∫}_{0}^{x}\dfrac{f\left(t\right)}{\sqrt{t}}dt$

Now differentiating both side we get,

$⇒ \sqrt{x} \cdot {f}^{'}\left(x\right) + \dfrac{f\left(x\right)}{2\sqrt{x}} = \dfrac{f\left(x\right)}{\sqrt{x}}$

$⇒ \sqrt{x} \cdot {f}^{'}\left(x\right) = \dfrac{f\left(x\right)}{2\sqrt{x}}$

$⇒ \dfrac{dy}{y} = \dfrac{dx}{2x}$

Now integrating both side we get,

$⇒ ∫ \dfrac{dy}{y} = ∫ \dfrac{dx}{2x}$

$⇒ \ln y = \dfrac{1}{2}\ln x + lnc$

Given $f$\left(1\right) = \sqrt{3}$, so$⇒ ln\sqrt{3} = 0 + lnc$</p><p>$⇒ c = \sqrt{3}$</p><p>So,$f\left(x\right) = \sqrt{3x}$</p><p>Now given$f\left(x\right)$is passing through \left(\alpha, 6\right)$,

So $f\left(\alpha\right) = 6 ⇒ 36 = 3\alpha$

$⇒ \alpha = 12$