Question

In the expansion of `(x/cosθ + 1/sinθ)^16`. If l₁ is the least value of the term independent of x when `π/8 ≤ θ ≤ π/4` and l₂ is the least value of the term independent of x when `π/16 ≤ θ ≤ π/8`, then the ratio l₂ : l₁ is equal to ______.

Options

• 1 : 8

• 16 : 1

• 8 : 1

• 1 : 16

Answer 1

In the expansion of `(x/cosθ + 1/sinθ)^16`. If l₁ is the least value of the term independent of x when `π/8 ≤ θ ≤ π/4` and l₂ is the least value of the term independent of x when `π/16 ≤ θ ≤ π/8`, then the ratio l₂ : l₁ is equal to 16 : 1.

Explanation:

General term of the given expansion

T_r+1 = `""^16C_r(x/(cosθ))^(16 - r) (1/(xsinθ))^r`

For r = 8 term is free from ‘x’

T₉ = `""^16C_8 1/(sin^8θcos^8θ)`; T₉ = `""^16C_8 2^8/(sin2θ)^8`

When `θ ∈ [π/8, π/4]`, then least value of the term independent of x,

l₁ = ¹⁶C₈ 2⁸  ...[∵ min value of l₁ at θ = `π/4`]

When `θ ∈ [π/16, π/8]`, then least value of the term independent of x,

l₂ = `""^16C_8 2^8/(1/sqrt(2))^8`  

= ¹⁶C₈.2⁸.2⁴   ...[∵ min value of l₂ at θ = `π/8`]

Now, `l_2/l_1 = (""^16C_8. 2^8. 2^4)/(""^16C_8. 2^8)` = 16