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:

Question

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:, 1 : 8, 16 : 1, 8 : 1, 1 : 16

MARKS App · Accepted Answer

T r + 1 = C r 16 x cos ⁡ θ 16 - r 1 x sin ⁡ θ r for r = 8 term is free from ' x ' T 9 = C 8 16 1 sin 8 ⁡ θ cos 8 ⁡ θ T 9 = C 8 16 2 8 sin ⁡ 2 θ 8 In θ ∈ π 8 , π 4 , l 1 = C 8 16 2 8 ( ∵ Minimum value of l 1 at θ = π 4 ) In θ ∈ π 16 , π 8 , l 2 = C 8 16 2 8 1 2 8 = C 8 16 · 2 8 · 2 4 ( ∵ Minimum value of l 2 at θ = π 8 ) l 2 l 1 = C 8 16 . 2 8 2 4 C 8 16 . 2 8 = 16

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