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 $\frac{\pi}{8} \le\,\theta\,\le\, \frac{\pi}{4}$ and $l_2$ is the least value of the term independent of $x$ when $\frac{\pi}{16} \le\,\theta\,\le\, \frac{\pi}{8},$ then the ratio $l_{2} : l_{1}$ is equal to :

• A. $16 : 1$
• B. $8 : 1$
• C. $1 : 8$
• D. $1 : 16$

Answer 1

Answer: (A)

$T_{r+1} = ^{16}C_{r} \left(\frac{x}{cos\,\theta}\right)^{16-r}\left(\frac{1}{x\,sin\,\theta }\right)^{r}$
$= ^{16}C_{r} \left(x\right)^{16-2r}\times \frac{1}{\left(cos\,\theta \right)^{16-r}\left(sin\,\theta \right)^{r}}$
For independent of $x; 16 - 2r = 0 \Rightarrow r = 8$
$\Rightarrow T_{9} = ^{16}C_{8} \frac{1}{cos^{8}\,\theta \,sin^{8}\,\theta }$
$= ^{16}C_{8} \frac{2^{8}}{\left(sin\,2\theta\right)^{8} }$
for $\theta\in \left[\frac{\pi}{8}, \frac{\pi}{4}\right]\ell_{1}$ is least for $\theta_{1} = \frac{\pi }{4}$
for $\theta \in \left[\frac{\pi }{16}, \frac{\pi }{8}\right]\ell _{2}$ is least for $\theta _{2} = \frac{\pi }{8}$
$\frac{\ell _{2}}{\ell _{1}} = \frac{\left(sin\,2\theta_{1} \right)^{8}}{\left(sin\,2\theta_{2} \right)^{8}} = \left(\sqrt{2}\right)^{8} = \frac{16}{1}$