Sign In

JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 12)

Let $$S_{1}=\left\{z_{1} \in \mathbf{C}:\left|z_{1}-3\right|=\frac{1}{2}\right\}$$ and $$S_{2}=\left\{z_{2} \in \mathbf{C}:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\}$$. Then, for $$z_{1} \in S_{1}$$ and $$z_{2} \in S_{2}$$, the least value of $$\left|z_{2}-z_{1}\right|$$ is :
0
$$\frac{1}{2}$$
$$\frac{3}{2}$$
$$\frac{5}{2}$$

Explanation

$$\because$$ $${\left| {{Z_2} + |{Z_2} - 1|} \right|^2} = {\left| {{Z_2} - |{Z_2} + 1|} \right|^2}$$

$$ \Rightarrow \left( {{Z_2} + |{Z_2} - 1|} \right)\left( {{{\overline Z }_2} + |{Z_2} - 1|} \right) = \left( {{Z_2} - |{Z_2} + 1|} \right)\left( {{{\overline Z }_2} - |{Z_2} + 1|} \right)$$

$$ \Rightarrow {Z_2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) + {\overline Z _2}\left( {|{Z_2} - 1| + |{Z_2} + 1|} \right) = |{Z_2} + 1{|^2} - |{Z_2} - 1{|^2}$$

$$ \Rightarrow \left( {{Z_2} + {{\overline Z }_2}} \right)\left( {|{Z_2} + 1| + |{Z_2} - 1|} \right) = 2\left( {{Z_2} + {{\overline Z }_2}} \right)$$

$$\Rightarrow$$ Either $${Z_2} + {\overline Z _2} = 0$$ or $$|{Z_2} + 1| + |{Z_2} - 1| = 2$$

So, Z2 lies on imaginary axis or on real axis within $$[ - 1,1]$$

Also $$|{Z_1} - 3| = {1 \over 2} \Rightarrow {Z_1}$$ lies on the circle having center 3 and radius $${1 \over 2}$$.

JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Complex Numbers Question 62 English Explanation

Clearly $$|{Z_1} - {Z_2}{|_{\min }} = {3 \over 2}$$

Comments (0)

Advertisement