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 value of $\frac{l_{2}}{l_{1}}$ is

• A. 8
• B. 32
• C. 16
• D. 64

Answer 1

Answer: (C)

Genral term $T_{r+1}={ }^{16} C_{r}\left(\frac{x}{\cos \theta}\right)^{16-r}\left(\frac{1}{x \sin \theta}\right)^{r}$
$={ }^{16} C_{r} \frac{1}{(\cos \theta)^{16-r}(\sin \theta)^{r}} \cdot x^{16-2 r}$
If this term is independent of $x$, then $16-2 r=0$
$\therefore $ The term independent of $x={ }^{16} C_{8} \frac{1}{\cos ^{8} \theta \sin ^{8} \theta}$
$={ }^{16} C_{8} \frac{2^{8}}{\sin ^{8} 2 \theta} ; \quad l_{1}={ }^{16} C_{8} \frac{2^{8}}{\sin ^{8} \frac{\pi}{2}}={ }^{16} C_{8} 2^{8} $
$l_{2}={ }^{16} C_{8} \frac{2^{8}}{\sin ^{8} \frac{\pi}{4}}={ }^{16} C_{8} \cdot \frac{2^{8}}{\left(\frac{1}{\sqrt{2}}\right)^{8}}={ }^{16} C_{8} 2^{12}$
$\therefore \frac{l_{2}}{l_{1}}=\frac{2^{12}}{2^{8}}=2^{4}=16$