JEE Mathematics 2023 Question with Solution

Given: $a \neq b$ are two non-zero real numbers and $X = \{z \in \mathbb{C}: \Re(a z^2 + b z) = a \text{ and } \Re(b z^2 + a z) = b\}$.

Find: The number of elements of $X$.

Let $z = x + iy$, where $x, y \in \mathbb{R}$. Then

Using the real-part conditions,

and

So we get the system

Subtracting the two equations,

Since $a \neq b$,

From the provided working, solving further shows that this system has no solution satisfying both conditions. Therefore, the set $X$ has no elements.

Hence, the number of elements in $X$ is $0$. Therefore, the correct option is A.

Given: Two real-part equations obtained by interchanging $a$ and $b$.

Find: The number of complex numbers $z$ satisfying both.

A quick way is to write $z = x+iy$ and compare the two real-part equations. Because the expressions are symmetric in $a$ and $b$, subtracting them immediately uses the fact that $a \neq b$ and forces a restrictive condition on $x$ and $y$.

The extracted solution states that after solving the resulting system, no valid pair $(x,y)$ exists. So there are no such complex numbers $z$.

Therefore, the number of elements is $0$, so the correct option is A.