JEE Main 2023 — Mathematics Algebra

Given that the set${Re(\frac{z-\bar{z}+z\bar{z}}{2-3z+5\bar{z}}):z\in \mathbb{C},Rez=3}$Re(2−3z+5zˉz−zˉ+zzˉ​):z∈C,Rez=3 is equal to the interval $(\alpha ,\beta ]$(α,β].

Let $z=x+iy$z=x+iy

$$=Re(\frac{x+iy-(x-iy)+{x}^{2}+{y}^{2}}{2-3(x+iy)+5(x-iy)})$$=Re(2−3(x+iy)+5(x−iy)x+iy−(x−iy)+x2+y2​)

$$=Re(\frac{{x}^{2}+{y}^{2}+i(2y)}{2+2x-8iy})$$=Re(2+2x−8iyx2+y2+i(2y)​)

$$=Re(\frac{({x}^{2}+{y}^{2}+2yi)(2(1+x)+8iy)}{{(2(1+x))}^{2}+{(8y)}^{2}})$$=Re((2(1+x))2+(8y)2(x2+y2+2yi)(2(1+x)+8iy)​)

$$=\frac{2({x}^{2}+{y}^{2})(1+x)-16{y}^{2}}{4{(1+x)}^{2}+{(8y)}^{2}}$$=4(1+x)2+(8y)22(x2+y2)(1+x)−16y2​

Now using, $Re(z)=3$Re(z)=3$\Rightarrow x=3$⇒x=3 we get,

$$=\frac{8(9+{y}^{2})-16{y}^{2}}{64+64{y}^{2}}$$=64+64y28(9+y2)−16y2​

$$\Rightarrow f(y)=\frac{1}{8}\frac{(9-{y}^{2})}{1+{y}^{2}}$$⇒f(y)=81​1+y2(9−y2)​

Let $t=\frac{1}{8}\frac{(9-{y}^{2})}{1+{y}^{2}}$t=81​1+y2(9−y2)​

$$\Rightarrow 8t+8t{y}^{2}=9-{y}^{2}$$⇒8t+8ty2=9−y2

$$\Rightarrow {y}^{2}(8t+1)+8t-9=0$$⇒y2(8t+1)+8t−9=0

Now $D\geq 0$D≥0

$$\Rightarrow {0}^{2}-4(8t+1)(8t-9)\geq 0$$⇒02−4(8t+1)(8t−9)≥0

$$\Rightarrow (t-(\frac{-1}{8}))(t-\frac{9}{8})\leq 0$$⇒(t−(8−1​))(t−89​)≤0

$\Rightarrow t\in (\frac{-1}{8},\frac{9}{8}]$⇒t∈(8−1​,89​] as $t\neq \frac{-1}{8}$t=8−1​

Range of $f(y)=(-0.125,1.125]$f(y)=(−0.125,1.125]

$$\Rightarrow \alpha =-0.125$$⇒α=−0.125

$$\Rightarrow \beta =1.125$$⇒β=1.125

$$\Rightarrow \beta -\alpha =1.25$$⇒β−α=1.25

$$\Rightarrow 24(\beta -\alpha )=30$$⇒24(β−α)=30

Hence this is the required solution.