Question

If $f(x+y+1)={(\sqrt{f\left(x\right)}+\sqrt{f\left(y\right)})}^2\forall x,y\in R$ and $f\left(0\right)=1,$ then the value of $f\left(\frac{1}{2}\right)+f\left(1\right)+f\left(2\right)$ is

• A. $\frac{61}{4}$
• B. $\frac{29}{2}$
• C. $\frac{31}{4}$
• D. $\frac{21}{2}$

Answer 1

The correct option is A $\frac{61}{4}$

Given $f(x+y+1)={(\sqrt{f\left(x\right)}+\sqrt{f\left(y\right)})}^2$ and $f\left(0\right)=1$
Put $x=y=0$
$f\left(1\right)={(\sqrt{f\left(0\right)}+\sqrt{f\left(0\right)})}^2=4$

Put $x=1,y=0$
$f\left(2\right)={(\sqrt{f\left(1\right)}+\sqrt{f\left(0\right)})}^2\Rightarrow f\left(2\right)=(1+2{)}^2=9$

Put $x=y=\frac{1}{2}$
$f\left(2\right)={(\sqrt{f\left(\frac{1}{2}\right)}+\sqrt{f\left(\frac{1}{2}\right)})}^2\Rightarrow f\left(2\right)=4f\left(\frac{1}{2}\right)\Rightarrow f\left(\frac{1}{2}\right)=\frac{9}{4}\therefore f\left(\frac{1}{2}\right)+f\left(1\right)+f\left(2\right)=\frac{9}{4}+9+4=\frac{61}{4}$