Jee Mains

Question 33 — JEE (Main) 24 Jan 2023 (S I)

Log In Register Lost Password
Viewing 3 posts - 1 through 3 (of 3 total)
  • Author
    Posts
  • 7 October, 2026 at 11:17 pm #11832
    Om Sharma
    Keymaster

    Let $\mathrm{p}, \mathrm{q} \in \mathrm{R}$ and $(1-\sqrt{3} \mathrm{i})^{200}=2^{199}(\mathrm{p}+\mathrm{iq}), \mathrm{i}=\sqrt{-1}$. Then $\mathrm{p}+\mathrm{q}+\mathrm{q}^{2}$ and $\mathrm{p}-\mathrm{q}+\mathrm{q}^{2}$ are roots of the equation. [JEE (Main) 24 Jan 2023 S I]
    (A) $\mathrm{x}^{2}-4 \mathrm{x}+1=0$
    (B) $x^{2}+4 x-1=0$
    (C) $x^{2}+4 x+1=0$
    (D) $x^{2}-4 x-1=0$

    7 October, 2026 at 11:17 pm #11833
    Om Sharma
    Keymaster

    Solution

    $(1-\sqrt{3} \mathrm{i})^{200}=2^{199}(\mathrm{p}+\mathrm{iq}) \Rightarrow 2^{200}\left(\cos \frac{\pi}{3}-\mathrm{i} \sin \frac{\pi}{3}\right)^{200}=2^{199}(\mathrm{p}+\mathrm{iq})$

    \[ \begin{aligned} & 2\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)=p+i q \Rightarrow p=-1, q=-\sqrt{3} \\ & \alpha=p+q+q^{2}=2-\sqrt{3}, \beta=p-q+q^{2}=2+\sqrt{3} \\ & \alpha+\beta=4 \Rightarrow \alpha \cdot \beta=1 \text { equation } x^{2}-4 x+1=0 \end{aligned} \]
    7 October, 2026 at 11:17 pm #11834
    Om Sharma
    Keymaster

    Answer

    A

  • Author
    Posts
Log In Register Lost Password
Viewing 3 posts - 1 through 3 (of 3 total)
  • You must be logged in to reply to this topic.