JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 25)
Let $$\alpha, \beta$$ be the roots of the equation $$x^2-\sqrt{6} x+3=0$$ such that $$\operatorname{Im}(\alpha)>\operatorname{Im}(\beta)$$. Let $$a, b$$ be integers not divisible by 3 and $$n$$ be a natural number such that $$\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^n(a+i b), i=\sqrt{-1}$$. Then $$n+a+b$$ is equal to __________.
Answer
49
Explanation
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$$\begin{aligned} & x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i) \\ & \alpha=\sqrt{3}\left(e^{i \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(e^{-i \frac{\pi}{4}}\right) \\ & \therefore \frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}\left(\frac{\alpha}{\beta}+1\right) \\ & =\frac{\alpha^{98}(\alpha+\beta)}{\beta}=3^{49}\left(e^{i 99 \frac{\pi}{4}}\right) \times \sqrt{2} \\ & =3^{49}(-1+i) \\ & =3^n(a+i b) \\ & \therefore n=49, a=-1, b=1 \\ & \therefore n+a+b=49-1+1=49 \end{aligned}$$
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