Let α > 0, β > 0 be such that
α³ + β² = 4. If the maximum value of the term independent of x in
the binomial expansion of ( α x^(1/9) + β x^(- 1/6) )¹⁰ is 10k,
then k is equal to :

Solution

Maths sir · 4 steps

Find the r that kills the power of x, then the term becomes 210(α³β²)². AM-GM finishes it.

1. General term. The (r+1)th term is C(10, r)α¹⁰⁻ʳβʳ x^((10-r)/9-r/6). For the term independent of x, set the power to zero: 2(10-r) = 3r, so r = 4.

   T₅ = C(10, 4)α⁶β⁴ = 210(α³β²)²

2. Maximise the product. Apply AM-GM to the positive numbers α³ and β²: α³β² ≤ ((α³+β²)/2)² = 2² = 4. Equality holds when α³ = β² = 2, so the maximum is reached.
3. Largest term. The maximum of the term is 210 × 4² = 210 × 16 = 3360.
4. So. 10k = 3360, which gives k = 336, option (b).

The idea to keep. When the coefficient depends on parameters tied by a fixed sum, rewrite it as a power of a product of those parameters. AM-GM on that sum then gives the maximum at once.

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