Consider the following two statements : Statement I : For any two non-zero complex numbers z 1 , z 2 ( z 1 + z 2 ) frac z 1 z 1 +frac z 2 z 2 leq 2 ( z 1 + z 2 ) and Statement II : If x , y , z are three distinct complex numbers and a , b , c are three positive real numbers such that a/ y-z =b/ z-x =c/ x-y , then frac a 2 y-z +frac b 2 z-x +frac c 2 x-y =1 Between the above two statements, In the light of the above statements, choose the correct answer from the options given below :

Statement I is correct and statement Il is incorrect.

Both statement I and statement Il are correct.

Statement I is incorrect and statement Il is correct.

Both statements 1 and statements ll are incorrect.

Mathematics · Complex Numbers

JEE Main 2024 — 5 April, Shift 1 — Question 14

Consider the following two statements :

Statement I : For any two non-zero complex numbers $\mathrm{z}_{1}, \mathrm{z}_{2}$ $\left(\left|z_{1}\right|+\left|z_{2}\right|\right)\left|\frac{z_{1}}{\left|z_{1}\right|}+\frac{z_{2}}{\left|z_{2}\right|}\right| \leq 2\left(\left|z_{1}\right|+\left|z_{2}\right|\right)$ and

Statement II : If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three distinct complex numbers and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are three positive real numbers such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$ , then $\frac{a^{2}}{y-z}+\frac{b^{2}}{z-x}+\frac{c^{2}}{x-y}=1$ Between the above two statements,

In the light of the above statements, choose the correct answer from the options given below :

Option A:
Both statement I and statement Il are correct.
Option B:
Statement I is correct and statement Il is incorrect.
Correct
Option C:
Statement I is incorrect and statement Il is correct.
Option D:
Both statements 1 and statements ll are incorrect.

Answer: B

Step-by-step solution

\left(\left|z_{1}\right|+\left|z_{2}\right|\right)\left|\frac{z_{1}}{\left|z_{1}\right|}+\frac{z_{2}}{\left|z_{2}\right|}\right|

Since $\left|\frac{z_{1}}{\left|z_{1}\right|}+\frac{z_{2}}{\left|z_{2}\right|}\right| \leq\left|\frac{z_{1}}{\left|z_{1}\right|}\right|+\left|\frac{z_{2}}{\left|z_{2}\right|}\right|$

$\left|\frac{z_{1}}{\left|z_{1}\right|}+\frac{z_{2}}{\left|z_{2}\right|}\right| \leq \frac{\left|z_{1}\right|}{\left|z_{1}\right|}+\frac{\left|z_{2}\right|}{\left|z_{2}\right|}$

$\left|\frac{z_{1}}{\left|z_{1}\right|}+\frac{z_{2}}{\left|z_{2}\right|}\right| \leq 2$ $\left(\left|z_{1}\right|+\left|z_{2}\right|\right)$

$\left(\left|\frac{z_{1}}{\left|z_{1}\right|}+\frac{z_{2}}{\left|z_{2}\right|}\right|\right) \leq 2\left(\left|z_{1}\right|+\left|z_{2}\right|\right)$

$\therefore$ statement $I$ is correct

For Statement II :

$\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$

$\frac{a^{2}}{|y-z|^{2}}=\frac{b^{2}}{|z-x|^{2}}=\frac{c^{2}}{|x-y|^{2}}=\lambda$

$a^{2}=\lambda\left(|y-z|^{2}\right)=\lambda(y-z)(\bar{y}-\bar{z})$ , $\mathrm{b}^{2}=\lambda(\mathrm{z}-\mathrm{x})(\overline{\mathrm{z}}-\overline{\mathrm{x}})$ and $\mathrm{c}^{2}=\lambda(\mathrm{x}-\mathrm{y})(\overline{\mathrm{x}}-\overline{\mathrm{y}})$

$\frac{a^{2}}{y-z}+\frac{b^{2}}{z-x}+\frac{c^{2}}{x-y}=\lambda(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0$

Statement II is false

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Mathematics
Chapter: Complex Numbers
Topic: Properties of Complex Numbers