JEE Mathematics 2023 Question with Solution

Given: $a = x_1 + i y_1$ and $z = x + i y$.

Find: Which of the statements (S1) and (S2) is true.

From the solution working,

so the inequality used there becomes

for (S1).

To test whether all real numbers belong to the set, take $y = 0$. For (S1), the working chooses

Then

which gives

This is false. Hence (S1) is false.

For (S2), the working uses

and chooses

Then

which gives

This is false. Hence (S2) is false.

Therefore, both statements are false, so the correct option is D.

Note: The listed options contain a discrepancy because the text of option B says "Both are false" while the solution marks option D as correct. Following the solution, the answer is recorded as D.

Given: The statements claim that the sets contain all real numbers.

Find: Whether a counterexample real number exists in each case.

A statement of the form "contains all real numbers" is disproved by finding even one real number not belonging to the set.

For (S1), the solution picks a valid case with $\operatorname{Re}(a) > 0$ and $\operatorname{Im}(a) > 0$, namely $a = 2 + 10i$, and the real number $z = -12$. Substitution leads to equality instead of strict inequality, so $z \notin A$. Therefore (S1) is false.

For (S2), the solution picks a valid case with $\operatorname{Re}(a) < 0$ and $\operatorname{Im}(a) < 0$, namely $a = -2 - 10i$, and the real number $z = 12$. Again, substitution gives equality instead of the strict inequality, so $z \notin B$. Therefore (S2) is false.

Hence both statements are false, and the solution concludes the correct option is D.