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