In the expansion of x cos ⁡ θ + 1 x sin ⁡ θ 16 , if l 1 is the least value of the term…

In the expansion of xcos⁡θ+1xsin⁡θ16, if l1 is the least value of the term independent of x when π8≤θ≤π4 and l2 is the least value of the term independent of x when π16≤θ≤π8, then the ratio l2:l1 is equal to:
  1. 1:8
  2. 16:1
  3. 8:1
  4. 1:16

Solution

Tr+1=Cr16xcos⁡θ16-r1xsin⁡θr
for r=8 term is free from 'x'
T9=C8161sin8⁡θcos8⁡θ
T9=C81628sin⁡2θ8
In θ∈π8,π4,l1=C81628 (∵ Minimum value of l1 at θ=π4)
In θ∈π16,π8,l2=C81628128=C816·28·24 (∵Minimum value of l2 at θ=π8)
l2l1=C816.2824C816.28=16

Asked in: JEE Main 2020 (09 Jan Shift 2)

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