In the expansion of ( x cosθ + 1 x sinθ )^16, if _1 is the least value of the term independent of x when π 8≤ θ ≤ π 4 and _2 is the least value of…
JEE Main 2020 — Mathematics Algebra
2020mcqmedium
In the expansion of (cosθx+xsinθ1)16, if l1 is the least value of the term independent of x when 8π≤θ≤4π and l2 is the least value of the term independent of x when 16π≤θ≤8π, then the ratio l2:l1 is equal to:
Official previous-year question
Held on 9 Jan 2020 · Verified 6 Jul 2026.
Options
A
1:8
B
16:1
C
8:1
D
1:16
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Solution
Tr+1=Cr16(cosθx)16−r(xsinθ1)r
for r=8 term is free from ′x′
T9=C816sin8θcos8θ1
T9=C816(sin2θ)828
In θ∈[8π,4π],l1=C81628 (∵ Minimum value of l1 at θ=4π)
In θ∈[16π,8π],l2=C816(21)828=C816⋅28⋅24 (∵Minimum value of l2 at θ=8π)
l1l2=C816.28C816.2824=16
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