JEE Main 2025 8 April Evening Shift
Exam JEE Main
Year 2025
Date 8 April
Shift Evening Shift
Subject Mathematics
Chapter Matrices JEE Main 2025 · 8 April · Evening Shift · Mathematics · Matrices
Let α be a solution of x^2+x+1=0 , and for some a and b in ℝ,[...
Let
α \alpha α be a solution of
x 2 + x + 1 = 0 x^2+x+1=0 x 2 + x + 1 = 0 , and for some
a a a and
b b b in
\mathbb{R},\left[ 4 a b \begin{array}{lll}4 & \mathrm{a} & \mathrm{b}\end{array} 4 a b \right]\left[ 1 16 13 − 1 − 1 2 − 2 − 14 − 8 \begin{array}{ccc}1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8\end{array} 1 − 1 − 2 16 − 1 − 14 13 2 − 8 \right]=\left[ 0 0 0 \begin{array}{ccc}0 & 0 & 0\end{array} 0 0 0 \right] . If
4 α 4 \frac{4}{\alpha^4} α 4 4 + m α a + n α b = 3 +\frac{\mathrm{m}}{\alpha^{\mathrm{a}}}+\frac{\mathrm{n}}{\alpha^{\mathrm{b}}}=3 + α a m + α b n = 3 , then
m + n \mathrm{m}+\mathrm{n} m + n is equal to
Step-by-step solution
x^2+x+1=0
< b r / > <br/> < b r / > \alpha
i s r o o t < b r / > is root<br/> i sroo t < b r / > < b r / > ∴ α 2 + α + 1 = 0 ⇒ α = ω as ω 2 [cube root of unity] < b r / > \begin{aligned}<br/>& \therefore \alpha^2+\alpha+1=0 \\
& \Rightarrow \alpha=\omega \text{ as } \omega^2 \text{ [cube root of unity] }<br/>\end{aligned} < b r / > ∴ α 2 + α + 1 = 0 ⇒ α = ω as ω 2 [cube root of unity] < b r / > <br/>also<br/>\left. \begin{array}{l}<br/>{\left[\begin{array}{ll}<br/>4-a-2 b & 64-a-14 b<br/>\end{array} 52+2 a-8 b\right.} \end{array}
\right] = [ 0 0 0 ] ∴ a + 2 b = 4 a + 14 b = 64 ⇒ 12 b = 60 ⇒ b = 5 ⇒ a = − 6 ∴ 4 α 4 + m α − 6 + n α 5 = 3 ⇒ 4 ω + m 1 + n ω 2 = 3 ⇒ 4 ω 2 + m + n ω = 3 \begin{aligned} & =\left[\begin{array}{ll}0 & 0&0\end{array}\right] \\ \therefore & a+2 b=4 \\ & a+14 b=64 \\ \Rightarrow & 12 b=60 \Rightarrow b=5 \\ \Rightarrow & a=-6 \\ \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{aligned} ∴ ⇒ ⇒ ∴ ⇒ ⇒ = [ 0 0 0 ] a + 2 b = 4 a + 14 b = 64 12 b = 60 ⇒ b = 5 a = − 6 α 4 4 + α − 6 m + α 5 n = 3 ω 4 + 1 m + ω 2 n = 3 4 ω 2 + m + nω = 3 < b r / > ⇒ 4 ( − 1 2 − 3 2 i ) + m + n ( − 1 2 + 3 2 i ) = 3 ∴ − 2 + m − n 2 = 3.... ( 1 ) − 4 3 2 + n 3 2 = 0 ∴ n = 4 m = 7 ∴ m + n = 11 < b r / > \begin{aligned}<br/>& \Rightarrow 4\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)+\mathrm{m}+\mathrm{n}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)=3 \\
& \therefore-2+\mathrm{m}-\frac{\mathrm{n}}{2}=3....(1) \\
& & \frac{-4 \sqrt{3}}{2}+\frac{\mathrm{n} \sqrt{3}}{2}=0 \\
& \therefore \mathrm{n}=4 \\
& \mathrm{~m}=7 \\
& \therefore \mathrm{~m}+\mathrm{n}=11<br/>\end{aligned} < b r / > ⇒ 4 ( − 2 1 − 2 3 i ) + m + n ( − 2 1 + 2 3 i ) = 3 ∴ − 2 + m − 2 n = 3.... ( 1 ) ∴ n = 4 m = 7 ∴ m + n = 11 < b r / > 2 − 4 3 + 2 n 3 = 0
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