JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 4)
For $$a \in \mathbb{C}$$, let $$\mathrm{A}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z}) > \operatorname{Im}(\bar{a}+z)\}$$ and $$\mathrm{B}=\{z \in \mathbb{C}: \operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)\}$$. Then among the two statements :
(S1): If $$\operatorname{Re}(a), \operatorname{Im}(a) > 0$$, then the set A contains all the real numbers
(S2) : If $$\operatorname{Re}(a), \operatorname{Im}(a) < 0$$, then the set B contains all the real numbers,
both are false
only (S1) is true
only (S2) is true
both are true
Explanation
We are given that $a \in \mathbb{C}$ and $z \in \mathbb{C}$.
Let $a = x_1 + iy_1$ and $z = x_2 + iy_2$ where $x_1, y_1, x_2, y_2 \in \mathbb{R}$
We are also given two sets A and B defined as follows :
- A is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is greater than the imaginary part of $(\overline{a} + z)$.
- B is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is less than the imaginary part of $(\overline{a} + z)$.
Statement (S1) says : If the real part and imaginary part of $a$ are both positive, then the set A contains all the real numbers.
Statement (S2) says : If the real part and imaginary part of $a$ are both negative, then the set B contains all the real numbers.
We need to determine which of these statements are true.
Let's evaluate each statement.
1. Statement (S1) : For $z \in A$,
$Re(a + \overline{z}) > Im(\overline{a} + z)$
This can be re-written as $x_1 + x_2 > y_2 - y_1$
If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 > 0$, then the condition simplifies to $x_2 > -(x_1 + y_1)$.
This indicates that A covers a part of the negative real axis, but not the entire real axis. Therefore, Statement (S1) is false.
2. Statement (S2) : For $z \in B$,
$Re(a + \overline{z}) < Im(\overline{a} + z)$
This can be re-written as $x_1 + x_2 < y_2 - y_1$
If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 < 0$, then the condition simplifies to $x_2 < -(x_1 + y_1)$.
This indicates that B covers a part of the positive real axis, but not the entire real axis. Therefore, Statement (S2) is false.
Therefore, both (S1) and (S2) are false, so the answer is Option A : both are false.
Let $a = x_1 + iy_1$ and $z = x_2 + iy_2$ where $x_1, y_1, x_2, y_2 \in \mathbb{R}$
We are also given two sets A and B defined as follows :
- A is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is greater than the imaginary part of $(\overline{a} + z)$.
- B is the set of all complex numbers $z$ for which the real part of $(a + \overline{z})$ is less than the imaginary part of $(\overline{a} + z)$.
Statement (S1) says : If the real part and imaginary part of $a$ are both positive, then the set A contains all the real numbers.
Statement (S2) says : If the real part and imaginary part of $a$ are both negative, then the set B contains all the real numbers.
We need to determine which of these statements are true.
Let's evaluate each statement.
1. Statement (S1) : For $z \in A$,
$Re(a + \overline{z}) > Im(\overline{a} + z)$
This can be re-written as $x_1 + x_2 > y_2 - y_1$
If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 > 0$, then the condition simplifies to $x_2 > -(x_1 + y_1)$.
This indicates that A covers a part of the negative real axis, but not the entire real axis. Therefore, Statement (S1) is false.
2. Statement (S2) : For $z \in B$,
$Re(a + \overline{z}) < Im(\overline{a} + z)$
This can be re-written as $x_1 + x_2 < y_2 - y_1$
If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 < 0$, then the condition simplifies to $x_2 < -(x_1 + y_1)$.
This indicates that B covers a part of the positive real axis, but not the entire real axis. Therefore, Statement (S2) is false.
Therefore, both (S1) and (S2) are false, so the answer is Option A : both are false.
Comments (0)
