Question

Assertion :Let $z_1$ & $z_2$ are two distinct points in the argand plane such that $a|z_1|=b|z_2|$ where $a,bϵR$.
The expression $\frac{a{z}_1}{b{z}_2}+\frac{b{z}_2}{a{z}_1}$ is a point on the line segment [-2, 2] of the real axis. Reason: When arg $\left(z_1\right)=\theta$ & arg $\left(z_2\right)=\theta +\alpha$, then $\frac{a{z}_1}{b{z}_2}+\frac{b{z}_2}{a{z}_1}=e^{i\alpha }+e^{-i\alpha }=2cos\alpha$

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true (R) is false.
• D. (A) is false (R) is true.

Answer 1

The correct option is A Both (A) & (R) are individually true & (R) is correct explanation of (A).

Given $a|z_1|=b|z_2|$

$\Rightarrow a{r}_1=b{r}_2$

Now $\frac{a{z}_1}{b{z}_2}+\frac{b{z}_2}{a{z}_1}=\frac{a}{b}\frac{r_1{e}^{i\theta }}{b{r}_2{e}^{i(\theta +\alpha )}}+\frac{b}{a}\frac{r_2}{r_1}\frac{e^{i(\theta +\alpha )}}{e^{i\theta }}$

$=\frac{e^{i\theta }}{e^{i(\theta +\alpha )}}+\frac{e^{i(\theta +\alpha )}}{e^{i\theta }}=e^{-i\alpha }+e^{i\alpha }$ = $2cos\alpha$

$\Rightarrow -2\le 2cos\alpha \le 2$

Hence Assertion (A) & Reason (R) both are true & Reason (R) is correct explanation of Assertion (A)

Note : Reason (R) is solution of Assertion (A).