Paragraph: Let p, q be integers and let α, β be the roots of the equation, x^ 2 -x-1=0 , where α ≠ β . For n=0,1,2, , let a_ n =p α^ n +q β^ n FACT: If a and b are rational numbers and a+b 5 =0 , then a=0=b . Question

Correct answer: C. 12 α ^ 2 = α + 1 ⇒ α ^ 4 = ( α ^ 2 ) ^ 2 ⇒ α ^ 4 = ( α + 1 ) ^ 2 ⇒ α ^ 4 = α ^ 2 + 2α + 1 ⇒ α ^ 4 = α + 1 + 2α + 1 ⇒ α ^ 4 = 3α + 2 Alsoα + β = 1 ∴ a _ 4 = 28 ⇒ p α ^ 4 + q β ^ 4 = p(3α + 2) + q(3β + 2) = 28 ⇒ p(3α + 2) + q(3 - 3α + 2) = 28 ⇒ α(3p - 3q) + 2p + 5q = 28 (as α ∈ Q ^ c ) (as it is root of quadratic equation) ⇒ p = q, 2p + 5q = 28 ⇒ p = q = 4 ∴ p + 2q = 12

JEE Advanced 2017Paper 2

Exam: JEE Advanced
Year: 2017
Paper: Paper 2
Subject: Mathematics
Chapter: Quadratic Equation

JEE Advanced 2017 · Paper 2 · Mathematics · Quadratic Equation

Paragraph: Let p, q be integers and let α, β be the roots of the...

Paragraph: Let

p, q

be integers and let

\alpha, \beta

be the roots of the equation,

x^{2}-x-1=0

, where

\alpha \neq \beta

. For

n=0,1,2, \ldots

, let

a_{n}=p \alpha^{n}+q \beta^{n}

FACT: If

a

and

b

are rational numbers and

a+b \sqrt{5}=0

, then

a=0=b

.
Question: If

a_{4}=28

, then

p+2 q=

Options

A $14$
B $7$
C $12$
D $21$

Correct answer

C. $12$

Step-by-step solution

{\alpha}^{2} = \alpha + 1

<br />

⇒ {\alpha}^{4} = {( {\alpha}^{2} )}^{2}

<br />

⇒ {\alpha}^{4} = {( \alpha + 1 )}^{2}

<br />

⇒ {\alpha}^{4} = {\alpha}^{2} + 2\alpha + 1

<br />

⇒ {\alpha}^{4} = \alpha + 1 + 2\alpha + 1

<br />

⇒ {\alpha}^{4} = 3\alpha + 2

<br /> <p>Also\alpha + \beta = 1

$∴ {a}_{4} = 28 ⇒ p{\alpha}^{4} + q{\beta}^{4} = p\left(3\alpha + 2\right) + q\left(3\beta + 2\right) = 28$

$⇒ p\left(3\alpha + 2\right) + q\left(3 - 3\alpha + 2\right) = 28$

$⇒ \alpha\left(3p - 3q\right) + 2p + 5q = 28$ (as $\alpha \in {Q}^{c}$ ) (as it is root of quadratic equation)

$⇒ p = q, 2p + 5q = 28 ⇒ p = q = 4$

$∴ p + 2q = 12$

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