Question

The minimum area bounded by the function $y=f\left(x\right)$ and $y=\alpha x+9\left(\alpha ϵR\right)$ where f satisfies the relation $f(x+y)=f\left(x\right)+f\left(y\right)+y\sqrt{f\left(x\right)}\forall x,yϵR$ and $f^{\prime }\left(0\right)=0\&f\left(0\right)=0$ is 9A, value of A is ___

Answer 1

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x+0)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)+h\sqrt{f\left(x\right)}-f\left(x\right)-f\left(0\right)-0\sqrt{f\left(x\right)}}{h}$

$=\underset{h\to 0}{lim}\left(\frac{f\left(h\right)-f\left(0\right)}{h-0}\right)+\sqrt{f\left(x\right)}$

$=\underset{h\to 0}{lim}\left(\frac{f^{\prime }\left(h\right)}{1}\right)+\sqrt{f\left(x\right)}$

$\Rightarrow f^{\prime }\left(x\right)=\sqrt{f\left(x\right)}$

$\int \frac{f^{\prime }\left(x\right)}{\sqrt{f\left(x\right)}}dx=\int dx$

$2\sqrt{f\left(x\right)}=x+c$

$f\left(x\right)=\frac{x^2}{4}$

when $\alpha =0$ area is minimum

required minimum area = $2{\int }_0^{9}2\sqrt{y}dy$

$\Rightarrow 4{\left(\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right)}_0^9$ = 72 sq.unit