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In the expansion of ( x cos⁡θ + 1 x sin⁡θ )^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 (xcos⁡θ+1xsin⁡θ)16,{(\frac{x}{\mathrm{cos}⁡\theta }+\frac{1}{x\mathrm{sin}⁡\theta })}^{16},(cos⁡θx​+xsin⁡θ1​)16, if l1{l}_{1}l1​ is the least value of the term independent of xxx when π8≤θ≤π4\frac{\pi }{8}\leq \theta \leq \frac{\pi }{4}8π​≤θ≤4π​ and l2{l}_{2}l2​ is the least value of the term independent of xxx when π16≤θ≤π8,\frac{\pi }{16}\leq \theta \leq \frac{\pi }{8},16π​≤θ≤8π​, then the ratio l2:l1{l}_{2}:{l}_{1}l2​:l1​ is equal to:

Official previous-year question

Held on 9 Jan 2020 · Verified 6 Jul 2026.

Options

  1. A

    1:81:81:8

  2. B

    16:116:116:1

  3. C

    8:18:18:1

  4. D

    1:161:161:16

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