Question

If f :$N\to R^{+},f\left(x\right)=\sqrt{x}$, then the value of $\frac{f\left(25\right)}{f\left(9\right)+f\left(16\right)}$ is

• A. $0$
• B. $1$
• C. $\frac{5}{7}$
• D. $9$

Answer 1

The correct option is C $\frac{5}{7}$

Given $f\left(x\right)=\sqrt{x}$

$\frac{f\left(25\right)}{f\left(9\right)+f\left(16\right)}$

$=\frac{\sqrt{25}}{\sqrt{9}+\sqrt{16}}$

$=\frac{5}{3+4}=\frac{5}{7}$

$\therefore \frac{f\left(25\right)}{f\left(9\right)+f\left(16\right)}=\frac{5}{7}$
Hence, option C.