Let,;, be the roots of the equation x^2 - 2x - 3 = 0. Let Pn = ^n - ^n, n N. Then (11 3 - 10 2)P10 + (11 2 + 10)P11 - 11P12 is equal to:

Given: and are roots of x^2 - 2x - 3 = 0, and Pn = ^n - ^n. Find: The value of (11 3 - 10 2)P10 + (11 2 + 10)P11 - 11P12. Using Vieta's formulas, + = 2, = - 3 Hence the sequence Pn = ^n - ^n satisfies Pn = ( +)Pn-1 - Pn-2 So, Pn = 2Pn-1 + 3Pn-2 From the recurrence, P10 = 2P9 + 3P8 P11 = 2P10 + 3P9 = 2P9 + 6P8 Also, P12 = 2P11 + 3P10 Now substitute into the required expression: (11 3 - 10 2)( 2P9 + 3P8) + (11 2 + 10)(2P9 + 6P8) - 11P12 Grouping terms in P9 and P8 and simplifying, the expression reduces to 10 3P9 Therefore, the correct option is B.

Let,;, be the roots of the equation x^2 - 2x - 3 = 0. Let Pn = ^n… — Solution | JEE Mathematics

MCQMediumJEE 2024Quadratic Equations in Complex Numbers

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

A
$10\sqrt{2}P_9$
B
$10\sqrt{3}P_9$
C
$11\sqrt{2}P_9$
D
$11\sqrt{3}P_9$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $\alpha$ and $\beta$ are roots of $x^2 - \sqrt{2}x - \sqrt{3} = 0,$ and $P_n = \alpha^n - \beta^n.$

Find: The value of $(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}.$

Using Vieta's formulas,

\alpha + \beta = \sqrt{2}, \qquad \alpha\beta = -\sqrt{3}

Hence the sequence $P_n = \alpha^n - \beta^n$ satisfies

P_n = (\alpha + \beta)P_{n-1} - \alpha\beta P_{n-2}

So,

P_n = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2}

Recurrence Expansion

From the recurrence,

P_{10} = \sqrt{2}P_9 + \sqrt{3}P_8

P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9 = 2P_9 + \sqrt{6}P_8

Also,

P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}

Now substitute into the required expression:

(11\sqrt{3} - 10\sqrt{2})(\sqrt{2}P_9 + \sqrt{3}P_8) + (11\sqrt{2} + 10)(2P_9 + \sqrt{6}P_8) - 11P_{12}

Grouping terms in $P_9$ and $P_8$ and simplifying, the expression reduces to

10\sqrt{3}P_9

Therefore, the correct option is B.

Common mistakes

Using the wrong recurrence relation for $P_n$ . Since $\alpha\beta = -\sqrt{3},$ the relation becomes $P_n = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2},$ not with a minus sign. Always substitute the sign of $\alpha\beta$ carefully.
Applying Vieta's formulas incorrectly. For $x^2 - \sqrt{2}x - \sqrt{3} = 0,$ we have $\alpha + \beta = \sqrt{2}$ and $\alpha\beta = -\sqrt{3}.$ Reversing the sign of the product changes the entire result.
Trying to compute $\alpha$ and $\beta$ explicitly and then expanding high powers directly. That approach is lengthy and error-prone. Use the recurrence relation satisfied by $P_n$ instead.

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