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Binomial Theorem question

2020 · 8 Jan · Shift 2 · Q31
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  5. /2020 · 8 Jan · Shift 2 · Q31

Binomial Theorem question

2020 · 8 Jan · Shift 2 · Q31

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If α\alphaα and β\betaβ be the coefficients of x4 and x2 respectively in the expansion of (x+x2−1)6+(x−x2−1)6{\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}(x+x2−1​)6+(x−x2−1​)6, then
  1. A
    α+β=60\alpha + \beta = 60α+β=60
  2. B
    α−β=60\alpha - \beta = 60α−β=60
  3. C
    α+β=−30\alpha + \beta = -30α+β=−30
  4. D
    α−β=−132\alpha - \beta = -132α−β=−132
View written solutionFree

Correct answer: D

  1. Let a=x+x2−1,b=x−x2−1.a=x+\sqrt{x^2-1},\qquad b=x-\sqrt{x^2-1}.a=x+x2−1​,b=x−x2−1​. Then ab=(x+x2−1)(x−x2−1)=x2−(x2−1)=1.ab=(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})=x^2-(x^2-1)=1.ab=(x+x2−1​)(x−x2−1​)=x2−(x2−1)=1. So b=1ab=\dfrac1ab=a1​ and the given expression is a6+b6=a6+a−6.a^6+b^6=a^6+a^{-6}.a6+b6=a6+a−6.

  2. Use the identity: a6+b6=(a3+b3)2−2(ab)3.a^6+b^6=(a^3+b^3)^2-2(ab)^3.a6+b6=(a3+b3)2−2(ab)3. Since ab=1ab=1ab=1, a6+b6=(a3+b3)2−2.a^6+b^6=(a^3+b^3)^2-2.a6+b6=(a3+b3)2−2. So first compute a3+b3a^3+b^3a3+b3.

  3. Now a+b=2x,ab=1.a+b=2x,\qquad ab=1.a+b=2x,ab=1. Hence a3+b3=(a+b)3−3ab(a+b).a^3+b^3=(a+b)^3-3ab(a+b).a3+b3=(a+b)3−3ab(a+b). Substituting, a3+b3=(2x)3−3(1)(2x)=8x3−6x.a^3+b^3=(2x)^3-3(1)(2x)=8x^3-6x.a3+b3=(2x)3−3(1)(2x)=8x3−6x.

  4. Therefore, a6+b6=(8x3−6x)2−2.a^6+b^6=(8x^3-6x)^2-2.a6+b6=(8x3−6x)2−2. Expand: (8x3−6x)2=64x6−96x4+36x2.(8x^3-6x)^2=64x^6-96x^4+36x^2.(8x3−6x)2=64x6−96x4+36x2. Thus a6+b6=64x6−96x4+36x2−2.a^6+b^6=64x^6-96x^4+36x^2-2.a6+b6=64x6−96x4+36x2−2.

  5. From this expansion:

  • coefficient of x4x^4x4 is α=−96,\alpha=-96,α=−96,
  • coefficient of x2x^2x2 is β=36.\beta=36.β=36.
  1. Now check the options: α+β=−96+36=−60≠60,\alpha+\beta=-96+36=-60\neq 60,α+β=−96+36=−60=60, so A is false.

α−β=−96−36=−132,\alpha-\beta=-96-36=-132,α−β=−96−36=−132, so D is true.

Also, α+β=−60≠−30,\alpha+\beta=-60\neq -30,α+β=−60=−30, so C is false, and α−β=−132≠60,\alpha-\beta=-132\neq 60,α−β=−132=60, so B is false.

Therefore the correct option is D.\boxed{D}.D​.

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