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QJEEM24Madb8f: JEE Main 2024 Mathematics - Class 11 JEEM24Madb8f Solution: Let be the roots of the equation such that . Let be integers…

Let α,β\alpha, \beta be the roots of the equation x2−6x+3=0x^2-\sqrt{6} x+3=0 such that \operatorname{Im}(\alpha)>\operatorname{Im}(\beta) . Let a,ba, b be integers not divisible by 3 and nn be a natural number such that α99β+α98=3n(a+ib),i=−1\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^n(a+i b), i=\sqrt{-1} . Then n+a+bn+a+b is equal to __________.

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JEE Main 2024 (Online) 29th January Evening Shift Mathematics - Complex Numbers Question 23 English Explanation

amp;x=6±i62=62(1±i)amp;α=3(eiπ4),β=3(e−iπ4)amp;∴α99β+α98=α98(αβ+1)amp;=α98(α+β)β=349(ei99π4)×2amp;=349(−1+i)amp;=3n(a+ib)amp;∴n=49,a=−1,b=1amp;∴n+a+b=49−1+1=49\begin{aligned}& x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i) \\& \alpha=\sqrt{3}\left(e^{i \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(e^{-i \frac{\pi}{4}}\right) \\& \therefore \frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}\left(\frac{\alpha}{\beta}+1\right) \\& =\frac{\alpha^{98}(\alpha+\beta)}{\beta}=3^{49}\left(e^{i 99 \frac{\pi}{4}}\right) \times \sqrt{2} \\& =3^{49}(-1+i) \\& =3^n(a+i b) \\& \therefore n=49, a=-1, b=1 \\& \therefore n+a+b=49-1+1=49\end{aligned}

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