# In the expansion of

Question:

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 of the term independent of $x$ when

$\frac{\pi}{16} \leq \theta \leq \frac{\pi}{8}$, then the ratio $l_{2}: l_{1}$ is equal to :

1. (1) $1: 8$

2. (2) $16: 1$

3. (3) $8: 1$

4. (4) $1: 16$

Correct Option: , 2

Solution:

General term of the given expansion

$T_{r+1}={ }^{16} C_{r}\left(\frac{x}{\sin \theta}\right)^{16-r}\left(\frac{1}{x \cos \theta}\right)^{r}$

For $r=8$ term is free from ' $x$ '

$T_{9}={ }^{16} C_{8} \frac{1}{\sin ^{8} \theta \cos ^{8} \theta}$

$T_{9}={ }^{16} C_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}$

When $\theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right]$, then least value of the term

independent of $x$,
$l_{1}={ }^{16} C_{8} 2^{8} \quad\left[\because\right.$ min. value of $l_{1}$ at $\left.\theta=\pi / 4\right]$

When $\theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right]$, then least value of the term

independent of $x$,

$l_{2}={ }^{16} C_{8}=\frac{2^{8}}{\left(\frac{1}{\sqrt{2}}\right)^{8}}={ }^{16} C_{8} \cdot 2^{8} \cdot 2^{4}$

$\left[\because\right.$ min. value of $l_{2}$ at $\left.\theta=\pi / 8\right]$

Now, $\frac{l_{2}}{l_{1}}=\frac{{ }^{16} C_{8} \cdot 2^{8} \cdot 2^{4}}{{ }^{16} C_{8} \cdot 2^{8}}=16$