JEE Mathematics 2023 Question with Solution

Given: $\Re(z)=3$

Find: The value of $24(\beta-\alpha)$ where the set of real parts equals $\left(\alpha,\beta\right]$.

Let

$$z=3+iy,\quad y\in\mathbb{R}$$

Then

$$\bar z=3-iy$$

Now simplify the numerator:

$$z-\bar z+z\bar z=(3+iy)-(3-iy)+(3+iy)(3-iy)$$ $$=2iy+9+y^2$$ $$=(9+y^2)+2iy$$

Simplify the denominator:

$$2-3z+5\bar z=2-3(3+iy)+5(3-iy)$$ $$=2-9-3iy+15-5iy$$ $$=8-8iy$$ $$=8(1-iy)$$

Form the expression and rationalize:

$$\frac{(9+y^2)+2iy}{8(1-iy)}\times\frac{1+iy}{1+iy}$$ $$=\frac{[(9+y^2)+2iy](1+iy)}{8(1+y^2)}$$

On expanding, the real part of the numerator is

$$(9+y^2)-2y^2=9-y^2$$

Hence,

$$\Re\left(\frac{z-\bar z+z\bar z}{2-3z+5\bar z}\right)=\frac{9-y^2}{8(1+y^2)}$$

Let

$$f(y)=\frac{9-y^2}{8(1+y^2)}$$

Then

$$f'(y)=\frac{-20y}{8(1+y^2)^2}$$

So the maximum occurs at $y=0$, giving

$$f(0)=\frac{9}{8}$$

As $y\to\pm\infty$,

$$f(y)\to-\frac{1}{4}$$

Therefore, the range is

$$\left(-\frac{1}{4},\frac{9}{8}\right]$$

So,

$$\alpha=-\frac{1}{4},\quad \beta=\frac{9}{8}$$

Now compute:

$$\beta-\alpha=\frac{9}{8}+\frac{1}{4}=\frac{11}{8}$$ $$24(\beta-\alpha)=24\times\frac{11}{8}=30$$

Therefore, the correct option is B.

Given: $\Re(z)=3$

Find: The interval endpoints and then $24(\beta-\alpha)$.

Use the standard substitution

$$z=3+iy$$

so the complex expression becomes a real function of $y$ after taking the real part. This works because fixing the real part leaves only one real parameter.

From simplification,

$$\Re=\frac{9-y^2}{8(1+y^2)}$$

Now observe that this function attains its maximum at $y=0$ and approaches $-\frac{1}{4}$ as $|y|\to\infty$. Hence the interval is

$$\left(-\frac{1}{4},\frac{9}{8}\right]$$

Therefore,

$$24(\beta-\alpha)=24\left(\frac{9}{8}+\frac{1}{4}\right)=30$$

So the correct option is B.