For two positive real numbers a and b such that 1/a² + 1/b³ = 4, then minimum value of the constant term in the expansion of ( ax^(1/8) + bx^(- 1/12) )¹⁰ is :

Solution

Maths sir · 4 steps

Find which term has no x, then treat 1/a² and 1/b³ as two numbers with a fixed sum.

1. General term. The (r+1)th term is C(10, r) a¹⁰⁻ʳ bʳ x^((10-r)/8 - r/12). The constant term is the one where this power of x is zero.
2. Find r. (10-r)/8 = r/12 gives 120 - 12r = 8r, so r = 6. The constant term is C(10, 6) a⁴ b⁶ = 210 a⁴ b⁶.
3. Rewrite the condition. Let p = 1/a² and q = 1/b³. Then p + q = 4, and a⁴ b⁶ = 1/(p² q²), so the constant term is smallest when pq is largest.

   (p+q)/2 ≥ √pq ⇒ pq ≤ 4 ⇒ a⁴ b⁶ = 1/(pq)² ≥ 1/16

4. So. Equality holds at p = q = 2, so the minimum constant term is 210 × 1/16 = 105/8, which is option (c).

The idea to keep. For a min or max under a sum condition, give names to the pieces in that sum. If the target is a product of those pieces, AM-GM on their fixed sum settles it.

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