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Quadratic Equation and Inequalities question

2021 · 25 Feb · Shift 2 · Q29
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  5. /2021 · 25 Feb · Shift 2 · Q29

Quadratic Equation and Inequalities question

2021 · 25 Feb · Shift 2 · Q29

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let α\alphaα and β\betaβ be the roots of x2 −-− 6x −-− 2 = 0. If an =α\alphaα n −β-\beta−β n for n ≥\ge≥ 1, then the value of a10−2a83a9{{{a_{10}} - 2{a_8}} \over {3{a_9}}}3a9​a10​−2a8​​ is :
  1. A
    3
  2. B
    2
  3. C
    4
  4. D
    1
View written solutionFree

Correct answer: B

  1. Given quadratic and its roots

The roots α,β\alpha, \betaα,β satisfy x2−6x−2=0.x^2-6x-2=0.x2−6x−2=0. So each root satisfies r2=6r+2.r^2=6r+2.r2=6r+2. Also, α+β=6,αβ=−2.\alpha+\beta=6, \qquad \alpha\beta=-2.α+β=6,αβ=−2.

We are given an=αn−βn,n≥1.a_n=\alpha^n-\beta^n, \qquad n\ge 1.an​=αn−βn,n≥1.

  1. Find a recurrence relation for ana_nan​

Since each root satisfies r2=6r+2r^2=6r+2r2=6r+2, multiplying by rn−2r^{n-2}rn−2 gives rn=6rn−1+2rn−2.r^n=6r^{n-1}+2r^{n-2}.rn=6rn−1+2rn−2. This is true for both r=αr=\alphar=α and r=βr=\betar=β.

Subtracting the two equations, αn−βn=6(αn−1−βn−1)+2(αn−2−βn−2).\alpha^n-\beta^n=6(\alpha^{n-1}-\beta^{n-1})+2(\alpha^{n-2}-\beta^{n-2}).αn−βn=6(αn−1−βn−1)+2(αn−2−βn−2). Hence, an=6an−1+2an−2.a_n=6a_{n-1}+2a_{n-2}.an​=6an−1​+2an−2​.

  1. Simplify the required expression

We need a10−2a83a9.\frac{a_{10}-2a_8}{3a_9}.3a9​a10​−2a8​​.

Using the recurrence for n=10n=10n=10, a10=6a9+2a8.a_{10}=6a_9+2a_8.a10​=6a9​+2a8​. Therefore, a10−2a8=(6a9+2a8)−2a8=6a9.a_{10}-2a_8=(6a_9+2a_8)-2a_8=6a_9.a10​−2a8​=(6a9​+2a8​)−2a8​=6a9​. So, a10−2a83a9=6a93a9=2.\frac{a_{10}-2a_8}{3a_9}=\frac{6a_9}{3a_9}=2.3a9​a10​−2a8​​=3a9​6a9​​=2.

  1. Check options

The value is 2.2.2. So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

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