In the expansion of ( x/(cos θ) + 1/(x sin θ) )¹⁶, if ℓ ₁ is the least value of the term independent of x when π/8 ≤ θ ≤ π/4 and ℓ ₂ is the least value of the term independent of x when π/16 ≤ θ ≤ π/8, then the ratio ℓ ₂ : ℓ ₁ is equal to :

Solution

Maths sir · 4 steps

Find the constant term, then turn sin θ cos θ into 1/2 sin 2θ.

1. General term. Tᵣ₊₁ = C(16, r)(x/(cos θ))¹⁶⁻ʳ(1/(x sin θ))ʳ. The power of x is 16 - 2r, which is zero when r = 8.

   T₉ = C(16, 8)/(cos⁸ θ sin⁸ θ) = (C(16, 8) 2⁸)/(sin⁸ 2θ)

2. Least value. T₉ is least when sin 2θ is greatest. For π/8 ≤ θ ≤ π/4, 2θ runs from π/4 to π/2. The greatest value of sin 2θ is 1, so ℓ₁ = C(16, 8) 2⁸.
3. Second range. For π/16 ≤ θ ≤ π/8, 2θ runs from π/8 to π/4. The greatest value of sin 2θ is 1/√2, and (1/√2)⁸ = 1/16. So ℓ₂ = C(16, 8) 2⁸ × 16.
4. So. ℓ₂ : ℓ₁ = 16 : 1, which is option (b).

The idea to keep. When sin θ and cos θ appear with equal powers, combine them into sin 2θ. A quantity like 1/(sinⁿ 2θ) is least where sin 2θ is largest on the given interval.

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