If the constant term in the expansion of (⁵√3/x+(2 x)/∛5)¹², x ≠ 0, is α × 2⁸ × ⁵√3, then 25 α is equal to :

Solution

Maths sir · 4 steps

Set the power of x to zero, then split 3^(6/5) as 3 × 3^(1/5) to read off α.

1. General term. Tᵣ₊₁ = C(12, r)(3^(1/5)/x)¹²⁻ʳ(2x/5^(1/3))ʳ. The power of x is -(12-r) + r = 2r - 12.
2. Constant term. Set 2r - 12 = 0, so r = 6. This term has 3^(6/5), 2⁶ and 5^(-6/3) = 1/25.

   T₇ = C(12, 6) × 3^(6/5) × 2⁶/25 = 924 × 3 × 3^(1/5) × 2⁶/25

3. Match with given form. Set this equal to α × 2⁸ × 3^(1/5). The 3^(1/5) cancels, and 2⁶/2⁸ = 1/4, so α = (924 × 3)/(25 × 4) = 2772/100 = 27.72.
4. So. 25α = 2772/4 = 693, which is option (c).

The idea to keep. For a constant term, write the power of x in Tᵣ₊₁ and set it to zero. Split a power like 3^(6/5) into 3 × 3^(1/5) before you compare with the given form.

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