Mathematics · 2020
JEE Main · 9 January 2020, Shift 2 · Q56
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…
In the expansion of $\displaystyle \left(\frac{x}{\cos \theta}+\frac{1}{x \sin \theta}\right)^{16}$, if $\displaystyle l_1$ is the least value of the term independent of $\displaystyle x$ when $\displaystyle \frac{\pi}{8} \leq \theta \leq \frac{\pi}{4}$ and $\displaystyle l_2$ is the least value of the term independent of $\displaystyle x$ when $\displaystyle \frac{\pi}{16} \leq \theta \leq \frac{\pi}{8}$, then the ratio $\displaystyle l_2: l_1$ is equal to :
Official answer
(3)
$\displaystyle 16$ : $\displaystyle 1$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.