JEE Main 2020 · 8 January · evening · Single correct · +4 / −1

If α and β be the coefficients of x⁴ and x² respectively in the expansion of
( x + √(x² - 1) )⁶ + ( x - √(x² - 1) )⁶, then

  1. (a) α + β = 60
  2. (b) α - β = 60
  3. (c) α + β = -30
  4. (d) α - β = -132
Answer and solution

Answer (d) α - β = -132

Solution

Maths sir · 4 steps

Add the two expansions and the odd powers of the root cancel, so only even powers of √(x²-1) remain.

  1. Pair the expansions. Let y = √(x²-1). In (x+y)⁶ + (x-y)⁶ the terms with odd powers of y cancel and the even ones double, so the sum is 2[x⁶ + 15x⁴y² + 15x²y⁴ + y⁶].
  2. Remove the root. Put y² = x² - 1 into each term and expand. Every term becomes a plain polynomial in x.
    TermExpandedx⁴ partx² part
    x⁶x⁶00
    15x⁴y²15x⁶ - 15x⁴-150
    15x²y⁴15x⁶ - 30x⁴ + 15x²-3015
    y⁶x⁶ - 3x⁴ + 3x² - 1-33
    Total inside bracket32x⁶ - 48x⁴ + 18x² - 1-4818
  3. Double it. The whole expression is 64x⁶ - 96x⁴ + 36x² - 2. So α = -96 and β = 36.
  4. So. α - β = -96 - 36 = -132, which is option (d). For comparison, α + β = -60, so (a) and (c) are both wrong.

The idea to keep. Whenever you see (a+b)ⁿ + (a-b)ⁿ, write only the even-power terms of b and multiply by 2. Do not forget that factor of 2 at the end.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2020.