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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

Let \(\alpha, \beta ; \alpha>\beta\), be the roots of the equation \(x^2-\sqrt{2} x-\sqrt{3}=0\). Let \(P_n=\alpha^n-\beta^n, n \in N\). Then \((11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}\) is equal to :

  1. A  \(10 \sqrt{2} \mathrm{P}_9\)
  2. B  \(10 \sqrt{3} \mathrm{P}_9\)
  3. C  \(11 \sqrt{2} \mathrm{P}_9\)
  4. D  \(11 \sqrt{3} \mathrm{P}_9\)
Verified Solution

Answer & Solution

Correct Answer

(B)  \(10 \sqrt{3} \mathrm{P}_9\)

Step-by-step Solution

Detailed explanation

\( x^2-\sqrt{2 x}-\sqrt{3}=0\left\langle_\beta^\alpha\right. \) \( \alpha^{n+2}-\sqrt{2} \alpha^{n+1}-\sqrt{3} \alpha^n=0 \) \( \text { and } \beta^{n+2}-\sqrt{2} \beta^{n+1}-\sqrt{3} \beta^n=0\) Subtracting…
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