Mathematics - Functions Question with Solution | TestHub

Given $f(x+y+1)=(\sqrt{f(x)}+\sqrt{f(x)}{)}^2$

Putting $y=0$, we get $f(x+1)=(\sqrt{f(x)}+1{)}^2$;

Putting $x=0$, we have $f(1)=(1+1{)}^2=2^2$;

Putting $x=1,f(2)=(2+1{)}^2=3^2$;

Putting $x=2,f(3)=(\sqrt{f(2)}+1{)}^2=(3+1{)}^2=4^2$ and so on.

Proceeding in this way, we get $f(x)=(x+1{)}^2$.