JEE PYQ: Binomial Theorem Question 34
Question 34 - 2020 (09 Jan Shift 2)
In the expansion of $\left(\frac{x}{\cos\theta} + \frac{1}{x\sin\theta}\right)^{16}$, if $l_1$ is the least value of the term independent of $x$ when $\frac{\pi}{8} \leq \theta \leq \frac{\pi}{4}$ and $l_2$ is the least value when $\frac{\pi}{16} \leq \theta \leq \frac{\pi}{8}$, then the ratio $l_2 : l_1$ is equal to:
(1) $1 : 8$ (2) $16 : 1$ (3) $8 : 1$ (4) $1 : 16$
Show Answer
Answer: (2)
Solution
$T_9 = {}^{16}C_8 \frac{2^8}{(\sin 2\theta)^8}$. $l_2/l_1 = 16$.
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