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MCQMediumJEE 2024Nature of Roots & Formation of Equations

JEE Mathematics 2024 Question with Solution

Let α,β;α>β,\alpha, \beta; \alpha > \beta, be the roots of the equation x2−2x−3=0.x^2 - \sqrt{2}x - \sqrt{3} = 0. Let Pn=αn−βn,n∈N.P_n = \alpha^n - \beta^n, n \in N. Then (113−102)P10+(112+10)P11−11P12(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12} is equal to:

  • A

    102P910\sqrt{2}P_9

  • B

    103P910\sqrt{3}P_9

  • C

    112P911\sqrt{2}P_9

  • D

    113P911\sqrt{3}P_9

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: α\alpha and β\beta are roots of x2−2x−3=0,x^2 - \sqrt{2}x - \sqrt{3} = 0, and Pn=αn−βn.P_n = \alpha^n - \beta^n. Find: The value of (113−102)P10+(112+10)P11−11P12.(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}.

Using Vieta's formulas,

α+β=2,αβ=−3\alpha + \beta = \sqrt{2}, \qquad \alpha\beta = -\sqrt{3}

Since Pn=αn−βn,P_n = \alpha^n - \beta^n, it satisfies the recurrence relation

Pn=(α+β)Pn−1−αβPn−2P_n = (\alpha + \beta)P_{n-1} - \alpha\beta P_{n-2}

So,

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

Therefore,

P10=2P9+3P8P_{10} = \sqrt{2}P_9 + \sqrt{3}P_8 P11=2P10+3P9=2P9+6P8P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9 = 2P_9 + \sqrt{6}P_8 P12=2P11+3P10P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}

Substituting into the required expression,

(113−102)(2P9+3P8)+(112+10)(2P9+6P8)−11P12(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}

On simplifying and combining terms back in terms of P9P_9 and P8,P_8, the expression reduces to

103P910\sqrt{3}P_9

Therefore, the correct option is B, that is 103P910\sqrt{3}P_9.

Recurrence Relation Approach

Given: α,β\alpha, \beta are roots of x2−2x−3=0.x^2 - \sqrt{2}x - \sqrt{3} = 0. Also, Pn=αn−βn.P_n = \alpha^n - \beta^n. Find: The value of the given linear combination of P10,P11,P12.P_{10}, P_{11}, P_{12}.

From the quadratic equation,

α+β=2,αβ=−3\alpha + \beta = \sqrt{2}, \qquad \alpha\beta = -\sqrt{3}

Hence,

Pn=(α+β)Pn−1−αβPn−2=2Pn−1+3Pn−2P_n = (\alpha + \beta)P_{n-1} - \alpha\beta P_{n-2} = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2}

Now write the successive terms:

P10=2P9+3P8P_{10} = \sqrt{2}P_9 + \sqrt{3}P_8 P11=2P10+3P9P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9 P12=2P11+3P10P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}

Substituting these relations into

(113−102)P10+(112+10)P11−11P12(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}

and simplifying gives

103P910\sqrt{3}P_9

Thus, the required value is 103P910\sqrt{3}P_9.

Common mistakes

  • Using the recurrence with the wrong sign for αβ.\alpha\beta. Since αβ=−3,\alpha\beta = -\sqrt{3}, the relation becomes Pn=2Pn−1+3Pn−2,P_n = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2}, not with a minus sign. Always substitute the product carefully.

  • Applying Vieta's formulas incorrectly to the quadratic equation. For x2−2x−3=0,x^2 - \sqrt{2}x - \sqrt{3} = 0, the sum is 2\sqrt{2} and the product is −3.-\sqrt{3}. A sign error here changes every later step.

  • Trying to find explicit values of α\alpha and β\beta and then expanding large powers. That is unnecessary and inefficient. Use the recurrence relation for PnP_n directly to reduce the expression.

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