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JEE Main 8-April-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 8-April-2025 Paper - Shift 24 Marks
MCQ
Let $\alpha$ be a solution of $x^{2}+x+1=0$, and for some $a$ and $b$ in
$\mathbb{R},\left[\begin{array}{lll}4 & \mathrm{a} & \mathrm{b}\end{array}\right]\left[\begin{array}{ccc}1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$. If $\frac{4}{\alpha^{4}}$ $+\frac{\mathrm{m}}{\alpha^{\mathrm{a}}}+\frac{\mathrm{n}}{\alpha^{\mathrm{b}}}=3$, then $\mathrm{m}+\mathrm{n}$ is equal to __________
  • A
    3
  • ✓
    11
  • C
    7
  • D
    8
✓

Answer

Correct option: B.
11
(B) 11
$x^{2}+x+1=0$
$\alpha$ is root
$\therefore \alpha^{2}+\alpha+1=0$
$\Rightarrow \alpha=\omega$ as $\omega^{2}$ [cube root of unity]
also
$\begin{array}{l}{\left[\begin{array}{lll}4-a-2 b & 64-a-14 b & 52+2 a-8 b\end{array}\right]} \\ \quad=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right] \\ \therefore a+2 b=4\end{array}$
$\begin{aligned} & a+14 b=64 \\ \Rightarrow & 12 b=60 \Rightarrow b=5 \\ \Rightarrow & a=-6\end{aligned}$
$\begin{array}{l}\therefore \frac{4}{\alpha^4}+\frac{m}{\alpha^6}+\frac{n}{\alpha^5}=3 \\ \Rightarrow \frac{4}{\omega}+\frac{m}{1}+\frac{n}{\omega^2}=3 \\ \Rightarrow 4 \omega^2+m+n \omega=3\end{array}$
$ \begin{array}{l} \Rightarrow 4\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)+m+n\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=3 \\ \therefore-2+m-\frac{n}{2}=3 \quad \ldots(1) \\ \& \frac{-4 \sqrt{3}}{2}+\frac{n \sqrt{3}}{2}=0 \\ \therefore n=4 \\ m=7 \\ \therefore m+n=11 \end{array} $

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