Let f be a differentiable function satisfying f(x)= 2 √3 ∫ _0^√3f( λ ^2x 3)dλ ,x>0 and f(1)=√3. If y=f(x) passes through the point (α ,6), then α is…

Given,

$$f(x)=\frac{2}{\sqrt{3}}{\int }_{0}^{\sqrt{3}}f(\frac{{\lambda }^{2}x}{3})d\lambda$$f(x)=3​2​∫03​​f(3λ2x​)dλ

Now let $\frac{{\lambda }^{2}x}{3}=t$3λ2x​=t

$$\Rightarrow \frac{2\lambda x}{3}d\lambda =dt$$⇒32λx​dλ=dt

$$\Rightarrow d\lambda =\frac{3}{2}\cdot \frac{1\sqrt{x}}{x\cdot \sqrt{3}\sqrt{t}}dt$$⇒dλ=23​⋅x⋅3​t​1x​​dt

$$\Rightarrow d\lambda =\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{x}}\cdot \frac{dt}{\sqrt{t}}$$⇒dλ=23​​⋅x​1​⋅t​dt​

So, $\displaystyle f(x)=\frac{1}{\sqrt{x}}{\int }_{0}^{x}\frac{f(t)}{\sqrt{t}}dt$f(x)=x​1​∫0x​t​f(t)​dt

$$\sqrt{x}f(x)={\int }_{0}^{x}\frac{f(t)}{\sqrt{t}}dt$$x​f(x)=∫0x​t​f(t)​dt

Now differentiating both side we get,

$$\Rightarrow \sqrt{x}\cdot {f}^{'}(x)+\frac{f(x)}{2\sqrt{x}}=\frac{f(x)}{\sqrt{x}}$$⇒x​⋅f′(x)+2x​f(x)​=x​f(x)​

$$\Rightarrow \sqrt{x}\cdot {f}^{'}(x)=\frac{f(x)}{2\sqrt{x}}$$⇒x​⋅f′(x)=2x​f(x)​

$$\Rightarrow \frac{dy}{y}=\frac{dx}{2x}$$⇒ydy​=2xdx​

Now integrating both side we get,

$$\Rightarrow \int \frac{dy}{y}=\int \frac{dx}{2x}$$⇒∫ydy​=∫2xdx​

$$\Rightarrow \mathrm{ln}y=\frac{1}{2}\mathrm{ln}x+\mathrm{ln}c$$⇒lny=21​lnx+lnc

Given $f(1)=\sqrt{3}$f(1)=3​, so $\Rightarrow \mathrm{ln}\sqrt{3}=0+\mathrm{ln}c$⇒ln3​=0+lnc

$$\Rightarrow c=\sqrt{3}$$⇒c=3​

So, $f(x)=\sqrt{3x}$f(x)=3x​

Now given $f(x)$f(x) is passing through $(\alpha ,6)$(α,6),

So $f(\alpha )=6\Rightarrow 36=3\alpha$f(α)=6⇒36=3α

$$\Rightarrow \alpha =12$$⇒α=12