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JEE Main 05-Apr-2024 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 05-Apr-2024 Paper - Shift 14 Marks
MCQ
Consider the following two statements:
Statement $I$ : For any two non $-$ zero complex numbers $z_1, z_2$ 
$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ 
Statement $II$ : If $x, y, z$ are three distinct complex numbers and $a, b, c$ are three positive real numbers
such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$ then
$\frac{a^2}{y-z}+\frac{b^2}{z-x}+\frac{c^2}{x-y}=1$
  • A
    both Statement I and Statement II are incorrect.
  • B
    Statement I is incorrect but Statement II is correct.
  • C
    Statement I is correct but Statement II is incorrect.
  • D
    both Statement I and Statement II are correct.
✓

Answer

Statement $I$ :
$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|$
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 $II$ :
$\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$
$\frac{a^2}{|y-z|^2}=\frac{b^2}{|z-x|^2}=\frac{c^2}{|x-y|^2}=\lambda$
$a^2=\lambda\left(|y-z|^2\right)=\lambda(y-z)(\bar{y}-\bar{z})$
$b^2=\lambda(z-x)(\bar{z}-\bar{x}) $ and $ c^2=\lambda(x-y)(\bar{x}-\bar{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

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