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

<p>$\because$ $$f(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {f\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda ,\,x > 0} $$</p> <p>On differentiating both sides w.r.t., x, we get</p> <p>$$f'(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {{{{\lambda ^2}} \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $$</p> <p>$$f'(x) = {1 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {\lambda \,.\,{{2\lambda } \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $$</p> <p>$\therefore$ $$\sqrt 3 f'(x) = \left[ {{\lambda \over x}\,.\,f\left( {{{{\lambda ^2}x} \over 3}} \right)} \right]_0^{\sqrt 3 } - \int\limits_0^{\sqrt 3 } {{1 \over x}f\left( {{{{\lambda ^2}x} \over 3}} \right)dx} $$</p> <p>$\sqrt 3 x\,f'(x) = \sqrt 3 f(x) - {{\sqrt 3 } \over 2}f(x)$</p> <p>$xf'(x) = {{f(x)} \over 2}$</p> <p>On integrating we get : $\ln y = {1 \over 2}\ln x + \ln c$</p> <p>$\because$ $f(1) = \sqrt 3$ then $c = \sqrt 3$</p> <p>$\therefore$ ($\alpha$, 6) lies on</p> <p>$\therefore$ $y = \sqrt {3x}$</p> <p>$\therefore$ $6 = \sqrt {3\alpha } \Rightarrow \alpha = 12$.</p>