If Q denotes the set of all rational numbers | vedclass

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(D) Given,$f\left(\frac{p}{q}\right)=\sqrt{p^2-q^2}$ for $\frac{p}{q} \in Q$.If we take $\frac{p}{q} = \frac{1}{2}$,then $p=1$ and $q=2$.$f\left(\frac

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Both $I$ and $II$ are false

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(D) Given,$f\left(\frac{p}{q}\right)=\sqrt{p^2-q^2}$ for $\frac{p}{q} \in Q$.If we take $\frac{p}{q} = \frac{1}{2}$,then $p=1$ and $q=2$.$f\left(\frac{1}{2}\right) = \sqrt{1^2 - 2^2} = \sqrt{1 - 4} = \sqrt{-3} = i\sqrt{3}$.Since $\sqrt{-3}$ is not a real number,statement $I$ is false.Since $\sqrt{-3}$ is a complex number,statement $II$ is true for this specific case,but the statement says it is complex for *each* $\frac{p}{q} \in Q$. If we take $\frac{p}{q} = \frac{2}{1}$,then $f(2) = \sqrt{2^2 - 1^2} = \sqrt{3}$,which is a real number (and every real number is also a complex number).However,if the question implies $f$ must be defined for all rational numbers,the function is not well-defined as a real-valued function. Given the options,$I$ is false because it is not real for all $p < q$,and $II$ is false because it is not complex for all cases (it is real for $p \ge q$). Thus,both are false.

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