Verified Solution • 2025–26 Session moderate

QJEEM23Maaed0: JEE Main 2023 Mathematics - Class 11 JEEM23Maaed0 Solution: If the set is equal to the interval , then is equal to :

If the set {Re⁡(z−zˉ+zzˉ2−3z+5zˉ):z∈C,Re⁡(z)=3}\left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in \mathbb{C}, \operatorname{Re}(z)=3\right\} is equal to

the interval (α,β](\alpha, \beta] , then 24(β−α)24(\beta-\alpha) is equal to :

Multiple Choice Options

  • A 36
  • B 27
  • C 42
  • D 30

Step-by-Step Verified Solution

Let z1=(z−zˉ+zzˉ2−3z+5zˉ)z_1=\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right)

Let z=3+iy\mathrm{z}=3+\mathrm{iy}

zˉ=3−iy\bar{z}=3-i y

amp;z1=2iy+(9+y2)2−3(3+iy)+5(3−iy)amp;=9+y2+i(2y)8−8iyamp;=(9+y2)+i(2y)8(1−iy)amp;Re⁡(z1)=(9+y2)−2y28(1+y2)\begin{aligned}& z_1=\frac{2 i y+\left(9+y^2\right)}{2-3(3+i y)+5(3-i y)} \\\\& =\frac{9+y^2+i(2 y)}{8-8 i y} \\\\& =\frac{\left(9+y^2\right)+i(2 y)}{8(1-i y)} \\\\& \operatorname{Re}\left(z_1\right)=\frac{\left(9+y^2\right)-2 y^2}{8\left(1+y^2\right)}\end{aligned}

amp;=9−y28(1+y2)amp;=18[10−(1+y2)(1+y2)]amp;=18[101+y2−1]amp;1+y2∈[1,∞]amp;11+y2∈(0,1]amp;101+y2∈(0,10]amp;101+y2−1∈(−1,9]amp;Re⁡(z1)∈(−18,98]amp;α=−18,β=98amp;24(β−α)=24(98+18)=30\begin{aligned}& =\frac{9-y^2}{8\left(1+y^2\right)} \\\\& =\frac{1}{8}\left[\frac{10-\left(1+y^2\right)}{\left(1+y^2\right)}\right] \\\\& =\frac{1}{8}\left[\frac{10}{1+y^2}-1\right] \\\\& 1+y^2 \in[1, \infty] \\\\& \frac{1}{1+y^2} \in(0,1] \\\\& \frac{10}{1+y^2} \in(0,10] \\\\& \frac{10}{1+y^2}-1 \in(-1,9] \\\\& \operatorname{Re}\left(\mathrm{z}_1\right) \in\left(\frac{-1}{8}, \frac{9}{8}\right] \\\\& \alpha=\frac{-1}{8}, \beta=\frac{9}{8} \\\\& 24(\beta-\alpha)=24\left(\frac{9}{8}+\frac{1}{8}\right)=30\end{aligned}

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