Question

If $Q$ denotes the set of all rational numbers and $f\left(\frac{p}{q}\right)=\sqrt{p^2-q^2}$ for any $\frac{p}{q}\in Q$, then observe the following statements.

I. $f\left(\frac{p}{q}\right)$is real for each $\frac{p}{q}\in Q$.

II. $f\left(\frac{p}{q}\right)$ is a complex number for each $\frac{p}{q}\in Q$.

• A. Both I and II are correct
• B. I is correct, II is incorrect
• C. I is incorrect, II is correct
• D. Both I and II are incorrect

Answer 1

The correct option is D

Both I and II are incorrect

Explanation for the correct option.

Step 1: For statement I.

Given that, $f\left(\frac{p}{q}\right)=\sqrt{p^2}-\sqrt{q^2}$ and $\frac{p}{q}\in Q$.

Assume $\frac{p}{q}=\frac{4}{5}$ which clearly is rational.

$\begin{array}{rcl}f\left(\frac{4}{5}\right)& =& \sqrt{4^2-5^2}\\ & =& \sqrt{16-25}\\ & =& 3i\notin R\left(\mathrm{real}\right)\end{array}$

Hence, the statement I is false.

Step 2: For statement II.

Assume $\frac{p}{q}=\frac{5}{4}$which is rational.

$\begin{array}{rcl}f\left(\frac{5}{4}\right)& =& \sqrt{5^2-4^2}\\ & =& \sqrt{25-16}\\ & =& 3\notin \mathrm{Complex}\end{array}$

Hence, statement II is incorrect.

Therefore, Both statements are incorrect.

Hence, option D is correct.