Question

Let $f\left(x\right)=\sqrt{x+\sqrt{0+\sqrt{x+\sqrt{0+\sqrt{x+......}}}}}$.If $f\left(a\right)=4$ and$f\left(a\right)=\frac{p}{q}$ where p and q are relatively prime natural numbers then $(a>0)(p+q)$ is equal to

Answer 1

We have

$f\left(x\right)=\sqrt{x+\sqrt{0+\sqrt{x+\sqrt{0+\sqrt{x+.......}}}}}$

$f\left(a\right)=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+.....}}}}$

$\Rightarrow 4^2=a+\sqrt{a+\sqrt{a+\sqrt{a+.......}}}$

$\Rightarrow 16=a+4$

$\Rightarrow a=16-4$

$\therefore a=12$

Now,

$f\left(a\right)=\frac{p}{q}$

$\Rightarrow 4=\frac{p}{q}$

$\Rightarrow p=4q$

$\therefore p+q=12$

Hence , the answer of given question is $12$.