If the function $f:N \rightarrow N$ is defined by $f(x) = \sqrt{x}$, then $\frac{f(25)}{f(16) + f(1)}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
Given, $f(x) = \sqrt{x}$
$	herefore\ \ \ \frac{f(25)}{f(16) + f(1)} = \frac{\sqrt{25}}{\sqrt{16} + \sqrt{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = \frac{5}{4 + 1} = 1$

Q. If the function $f:N \rightarrow N$ is defined by $f(x) = \sqrt{x}$, then $\frac{f(25)}{f(16) + f(1)}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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