JEE MAIN - Mathematics Hindi (2024 - 9th April Evening Shift - No. 7)
माना समीकरण $$x^2-\sqrt{2} x-\sqrt{3}=0$$ के मूल $$\alpha, \beta ; \alpha>\beta$$ हैं। माना $$\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{N}$$ हे। तो $$(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$$ बराबर हे
$$10 \sqrt{3} \mathrm{P_9}$$
$$11 \sqrt{3} \mathrm{P}_9$$
$$11 \sqrt{2} \mathrm{P}_9$$
$$10 \sqrt{2} \mathrm{P_9}$$
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