Verified Solution • 2025–26 Session moderate

QJEEM24Ma4bdf: JEE Main 2024 Mathematics - Class 11 JEEM24Ma4bdf Solution: Consider the following two statements : Statement I: For any…

Consider the following two statements :

Statement I: For any two non-zero complex numbers z1,z2,(∣z1∣+∣z2∣)∣z1∣z1∣+z2∣z2∣∣≤2(∣z1∣+∣z2∣), and z_1, z_2,(|z_1|+|z_2|)\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) \text {, and }

Statement II : If x,y,zx, y, z are three distinct complex numbers and a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c} are three positive real numbers such that a∣y−z∣=b∣z−x∣=c∣x−y∣\frac{\mathrm{a}}{|y-z|}=\frac{\mathrm{b}}{|z-x|}=\frac{\mathrm{c}}{|x-y|} , then a2y−z+b2z−x+c2x−y=1\frac{\mathrm{a}^2}{y-z}+\frac{\mathrm{b}^2}{z-x}+\frac{\mathrm{c}^2}{x-y}=1 .

Between the above two statements,

Multiple Choice Options

  • A both Statement I and Statement II are incorrect.
  • B Statement I is correct but Statement II is incorrect.
  • C Statement I is incorrect but Statement II is correct.
  • D both Statement I and Statement II are correct.

Step-by-Step Verified Solution

amp;a∣y−z∣=b∣z−x∣=c∣x−y∣=λamp;⇒a2=λ2∣(y−z)∣2amp;b2=λ2∣(z−x)∣2amp;c2=λ2∣(x−y)∣2amp;a2(y−z‾)(y−z)(y−z)=a2(yˉ−zˉ)∣y−z∣2=a2(yˉ−zˉ)a2λ2=λ2(yˉ−zˉ)amp;⇒∑(a2y−z)=λ2(yˉ−zˉ+zˉ−xˉ+xˉ−yˉ)=0≠1\begin{aligned}& \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}=\lambda \\& \Rightarrow a^2=\lambda^2|(y-z)|^2 \\& b^2=\lambda^2|(z-x)|^2 \\& c^2=\lambda^2|(x-y)|^2 \\& \frac{a^2(\overline{y-z})}{(y-z)(y-z)}=\frac{a^2(\bar{y}-\bar{z})}{|y-z|^2}=\frac{a^2(\bar{y}-\bar{z})}{\frac{a^2}{\lambda^2}}=\lambda^2(\bar{y}-\bar{z}) \\& \Rightarrow \sum\left(\frac{a^2}{y-z}\right)=\lambda^2(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0 \neq 1 \\\end{aligned}

Statement I

amp;(∣z1∣+∣z2∣)∣z1∣z1∣+z2∣z2∣∣≤2(∣z1∣+∣z2∣)amp;⇒z1=∣z1∣eiθ1amp;z2=∣z2∣eiθ2amp;⇒z1∣z1∣=eiθ1amp;⇒z2∣z2∣=eiθ2amp;⇒∣eiθ1+eiθ2∣amp;=∣2+2cos⁡(θ1−θ2)∣amp;=∣2cos⁡(θ1−θ22)∣≤2\begin{aligned}& \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) \\& \Rightarrow \quad z_1=\left|z_1\right| e^{i \theta_1} \\& \quad z_2=\left|z_2\right| e^{i \theta_2} \\& \Rightarrow \frac{z_1}{\left|z_1\right|}=e^{i \theta_1} \\& \Rightarrow \frac{z_2}{\left|z_2\right|}=e^{i \theta_2} \\& \Rightarrow\left|e^{i \theta_1}+e^{i \theta_2}\right| \\& \quad=\left|\sqrt{2+2 \cos \left(\theta_1-\theta_2\right)}\right| \\& \quad=\left|2 \cos \left(\frac{\theta_1-\theta_2}{2}\right)\right| \leq 2\end{aligned}

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