JEE Main 2021 · 31 August · morning · Numerical answer · +4 / −0

If ( 3⁶/4⁴ )k is the term, independent of x, in the binomial expansion of ( x/4 - 12/x² )¹², then k is equal to ___________.

Answer and solution

Answer 55

Solution

Maths sir · 4 steps

Write the general term, collect the powers of x, and set the total power to zero.

  1. General term. With a = x/4 and b = -12/x², the term Tᵣ₊₁ = C(12, r)(x/4)¹²⁻ʳ(-12/x²)ʳ. The power of x in it is (12-r) - 2r = 12 - 3r.
  2. Find r. Independent of x means 12 - 3r = 0, so r = 4. The term you need is T₅, and (-12)⁴ is positive.

    T₅ = C(12, 4)12⁴/4⁸ = 495 × (3⁴ × 4⁴)/4⁸ = (495 × 3⁴)/4⁴

  3. Match the form. Set 3⁶/4⁴k = (495 × 3⁴)/4⁴. The 4⁴ cancels, leaving 3⁶ k = 495 × 3⁴, so k = 495/3² = 495/9.
  4. So. k = 55.

The idea to keep. For a term independent of x, write the power of x in Tᵣ₊₁ as a function of r and set it to zero. Split numbers like 12⁴ into prime powers to match the given form.

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 2021.