If Q denotes the set of all rational numbers and $f (p / q) = \sqrt{p^{2} - q^{2}}$ for any $\frac{p}{q} \in Q$ then observe the following Statements
Statement - I : $f (\frac{p}{q})$ is real for each $\frac{p}{q} \in Q$
Statement - II : $f (\frac{p}{q})$ is a complex number for each $\frac{p}{q} \in Q$

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I is true, II is false

Both I and II are true

I is false, II is true

Both I and II are false

answer is C.

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Detailed Solution

$f (\frac{p}{q}) = \sqrt{p^{2} - q^{2}} \Rightarrow f (\frac{p}{q}) = |q| \sqrt{{(\frac{p}{q})}^{2} - 1} f (\frac{p}{q}) = \{\begin{cases} r e a l i f p \geq q \\ I m a g i n a r y i f p < q \end{cases} ∴ f (\frac{p}{q}) i s a c o m p l e x n u m b e r \forall \frac{p}{q} \in Q$