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 value of $\frac{l_2}{l_1}$ is

• A. 8
• B. 32
• C. 16
• D. 64

Answer 1

Answer: (C)

Genral term $T_{r+1}={}^{16}{C}_r{\left(\frac{x}{\cos \theta }\right)}^{16-r}{\left(\frac{1}{x\sin \theta }\right)}^r$
$={}^{16}{C}_r\frac{1}{(\cos \theta {)}^{16-r}(\sin \theta {)}^r}\cdot x^{16-2r}$
If this term is independent of $x$, then $16-2r=0$
$\therefore$ The term independent of $x={}^{16}{C}_8\frac{1}{{\cos }^8\theta {\sin }^8\theta }$
$={}^{16}{C}_8\frac{2^8}{{\sin }^{8}2\theta };l_1={}^{16}{C}_8\frac{2^8}{{\sin }^8\frac{\pi }{2}}={}^{16}{C}_8{2}^8$
$l_2={}^{16}{C}_8\frac{2^8}{{\sin }^8\frac{\pi }{4}}={}^{16}{C}_8\cdot \frac{2^8}{{\left(\frac{1}{\sqrt{2}}\right)}^8}={}^{16}{C}_8{2}^{12}$
$\therefore \frac{l_2}{l_1}=\frac{2^{12}}{2^8}=2^4=16$