JEE Main 2022 · 29 June · morning · Single correct · +4 / −1

If the constant term in the expansion of
( 3x³ - 2x² + 5/x⁵ )¹⁰ is 2ᵏ.l, where l is an odd integer, then the value of k is equal to:

  1. (a) 6
  2. (b) 7
  3. (c) 8
  4. (d) 9
Answer and solution

Answer (d) 9

Solution

Maths sir · 4 steps

Make the powers of x cancel, then count every factor of 2, including those in the coefficient.

  1. General term. Take 3x³ a times, -2x² b times and 5/x⁵ c times, with a+b+c=10. The general term is 10!/(a! b! c!) 3ᵃ(-2)ᵇ5ᶜ x³ᵃ⁺²ᵇ⁻⁵ᶜ.
  2. Make x vanish. For a constant term, 3a+2b-5c=0. Substitute c=10-a-b to get 8a+7b=50. The only solution in non-negative integers is a=1, b=6, which gives c=3.

    10!/(1! 6! 3!)× 3×(-2)⁶× 5³ = 840× 3× 64× 125

  3. Count the 2s. 840 = 2³× 105 and 64 = 2⁶, while 3 and 125 are odd. So the constant term is 2⁹×(105× 3× 125) = 2⁹× 39375.
  4. So. Here l = 39375 is odd, so k = 9, which is option (d).

The idea to keep. For a multinomial constant term, solve two conditions together: the powers add to n, and the powers of x add to zero. Then factor the full coefficient.

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 2022.