If the set { R e ( z - z + z z 2 - 3 z + 5 z ) : z C , R e ( z ) = 3 } is equal to the interval , then 24 is equal to [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If the set ${R e (\frac{z - \bar{z} + z \bar{z}}{2 - 3 z + 5 \bar{z}}) : z \in C, R e (z) = 3}$ is equal to the interval $(α, β]$ , then 24 $(β - α)$ is equal to [2023]

1 27

2 36

3 42

4 30

Ans.
(4)

Let $z_{1} = (\frac{z - \bar{z} + z \bar{z}}{2 - 3 z + 5 \bar{z}})$

Let $z = 3 + i y$ $[∵ Re (z) = 3]$

$\Rightarrow \bar{z} = 3 - i y$

$z_{1} = \frac{2 i y + (9 + y^{2})}{2 - 3 (3 + i y) + 5 (3 - i y)}$

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

$= \frac{9 + 9 i y + y^{2} - 2 y^{2} + i y^{3} + 2 i y}{8 (1 + y^{2})}$

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

$= \frac{1}{8} [\frac{10 - (1 + y^{2})}{1 + y^{2}}] = \frac{1}{8} [\frac{10}{1 + y^{2}} - 1]$

Thus, $1 + y^{2} \in [1, \infty) \Rightarrow \frac{1}{1 + y^{2}} \in (0, 1]$

$\Rightarrow \frac{10}{1 + y^{2}} \in (0, 10] \Rightarrow \frac{10}{1 + y^{2}} - 1 \in (- 1, 9]$

$∴ Re (z_{1}) \in (\frac{- 1}{8}, \frac{9}{8}]$ ; $α = \frac{- 1}{8}, β = \frac{9}{8}$

$∴$ $24 (β - α) = 24 (\frac{9}{8} + \frac{1}{8}) = 30$

Want Us to Email you About Special Offers & Updates?