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Question: If \(Q\) denotes the set of all rational numbers and \[f\left( \dfrac{p}{q} \right)=\sqrt{{{p}^{2}}-...

If QQ denotes the set of all rational numbers and f(pq)=p2−q2f\left( \dfrac{p}{q} \right)=\sqrt{{{p}^{2}}-{{q}^{2}}} for any pq∈Q\dfrac{p}{q}\in Q, then observe the following statements.
Statements
I. f(pq)f\left( \dfrac{p}{q} \right)is real for each pq∈Q\dfrac{p}{q}\in Q .
II. f(pq)f\left( \dfrac{p}{q} \right) Is a complex number for each pq∈Q\dfrac{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

Explanation

Solution

Here we have been given a function and we have to find whether the two statements given are correct for it or not. Firstly as we can see that the values of the function should be a rational number so we will take any rational number. Then we will find one rational number that doesn't satisfy the statements if we get that one rational number that means the statements are wrong and we will get our desired answer.

Complete answer: We have been given the function as,
f(pq)=p2−q2f\left( \dfrac{p}{q} \right)=\sqrt{{{p}^{2}}-{{q}^{2}}} ……(1)\left( 1 \right)
For any (pq∈Q)\left( \dfrac{p}{q}\in Q \right)
Now for Statement I we take,
Let p=3,q=4p=3,q=4so we have,
pq=34∈Q\dfrac{p}{q}=\dfrac{3}{4}\in Q
Now put the value in equation (1) as follows,
f(pq)=f(34)f\left( \dfrac{p}{q} \right)=f\left( \dfrac{3}{4} \right)
⇒f(34)=32−42\Rightarrow f\left( \dfrac{3}{4} \right)=\sqrt{{{3}^{2}}-{{4}^{2}}}
On simplifying we get,
⇒f(34)=9−25\Rightarrow f\left( \dfrac{3}{4} \right)=\sqrt{9-25}
⇒f(34)=−7\Rightarrow f\left( \dfrac{3}{4} \right)=\sqrt{-7}
As we have negative sign inside square root and we know i2=−1{{i}^{2}}=\sqrt{-1} we get,
⇒f(34)=7i\Rightarrow f\left( \dfrac{3}{4} \right)=\sqrt{7}i
As 7i∈C\sqrt{7}i\in C where C=C= Complex number
So for each, pq∈Q\dfrac{p}{q}\in Q f(pq)f\left( \dfrac{p}{q} \right) is not real.
So statement I is incorrect.
Next for statement II let us assume the value of the two variables as,
Let p=4,q=3p=4,q=3so we have,
pq=43∈Q\dfrac{p}{q}=\dfrac{4}{3}\in Q
Now put the value in equation (1) as follows,
f(pq)=f(43)f\left( \dfrac{p}{q} \right)=f\left( \dfrac{4}{3} \right)
⇒f(43)=42−32\Rightarrow f\left( \dfrac{4}{3} \right)=\sqrt{{{4}^{2}}-{{3}^{2}}}
On simplifying we get,
⇒f(43)=16−9\Rightarrow f\left( \dfrac{4}{3} \right)=\sqrt{16-9}
⇒f(43)=7\Rightarrow f\left( \dfrac{4}{3} \right)=\sqrt{7}
As 7∈R\sqrt{7}\in R where R=R= real number
So for each, pq∈Q\dfrac{p}{q}\in Q f(pq)f\left( \dfrac{p}{q} \right) is not complex.
So statement II is incorrect.
So both the statements I and II are incorrect.
Hence the correct option is (D).

Note:
In such type of question, always try to find one such value that will make the statement incorrect. For checking the statement I we took the numerator value smaller than the denominator value of the rational number as the function will always give negative value for it and for the statement II we took the numerator value greater than denominator value for rational number as the function will always give positive value. So always let the value according to the function given.