JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 27)
Let $$\alpha, \beta$$ be roots of $$x^2+\sqrt{2} x-8=0$$. If $$\mathrm{U}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$$, then $$\frac{\mathrm{U}_{10}+\sqrt{2} \mathrm{U}_9}{2 \mathrm{U}_8}$$ is equal to ________.
Answer
4
Explanation
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$$\Rightarrow \alpha^2+\sqrt{2} \alpha=8$$
$$\begin{aligned} \alpha+\beta=\sqrt{2}, \quad \alpha \beta=-8, & \Rightarrow \alpha+\sqrt{2}=\frac{8}{\alpha} \\ & \Rightarrow \beta+\sqrt{2}=\frac{8}{\beta} \end{aligned}$$
$$\begin{aligned} & \frac{U_{10}+\sqrt{2} U_9}{2 U_8}=\frac{\alpha^{10}+\beta^{10}+\sqrt{2} \alpha^9+\sqrt{2} \beta^9}{2 \alpha^8+2 \beta^8} \\ & =\frac{\alpha^9(\alpha+\sqrt{2})+\beta^9(\beta+\sqrt{2})}{2\left(\alpha^8+\beta^8\right)} \\ & =\frac{\alpha^9 \cdot\left(\frac{8}{\alpha}\right)+\beta^9\left(\frac{8}{\beta}\right)}{2\left(\alpha^8+\beta^8\right)}=\frac{8}{2}=4 \end{aligned}$$
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