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JEE Main 20208 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2020 · 8 January · Evening Shift · Mathematics · Binomial Theorem

If α and β , be the coefficients of x ^ 4 and x ^ 2 , respectively...

If α\alpha and β\beta, be the coefficients of x4{x}^{4} and x2{x}^{2}, 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}, then

Options

  1. Aα+β=60\alpha + \beta = 60
  2. Bα+β=−30\alpha + \beta = - 30
  3. Cα−β=60\alpha - \beta = 60
  4. Dα−β=−132\alpha - \beta = - 132

Correct answer

D. α−β=−132\alpha - \beta = - 132

Step-by-step solution

(x+x2−1)6+(x−x2−1)6=2[C0x6+C2x4(x2−1)+C4x2(x2−1)2+C6(x2−1)3]{\left(x + \sqrt{{x}^{2} - 1}\right)}^{6} + {\left(x - \sqrt{{x}^{2} - 1}\right)}^{6} = 2\left[{C}_{0}{x}^{6} + {C}_{2}{x}^{4}\left({x}^{2} - 1\right) + {C}_{4}{x}^{2}{\left({x}^{2} - 1\right)}^{2} + {C}_{6}{\left({x}^{2} - 1\right)}^{3}\right]

=2[x6+15(x6−x4)+15x2(x4−2x2+1)+(−1+3x2−3x4+x6)]= 2\left[{x}^{6} + 15\left({x}^{6} - {x}^{4}\right) + 15{x}^{2}\left({x}^{4} - 2{x}^{2} + 1\right) + \left(- 1 + 3{x}^{2} - 3{x}^{4} + {x}^{6}\right)\right]

=2(32x6−48x4+18x2−1)= 2\left(32{x}^{6} - 48{x}^{4} + 18{x}^{2} - 1\right)

Now given that α and β\alpha \text{ and } \beta are the coefficients of x4 and x2{x}^{4} \text{ and } {x}^{2} respectively, hence

α=−96\alpha = - 96 and β=36\beta = 36

∴α−β=−132∴ \alpha - \beta = - 132

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