Question

In the expansion of ${(\frac{x}{\cos \theta }+\frac{1}{x\sin \theta })}^{16}$, 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

The correct option is A $16:1$

$T_{r+1}={}^{16}{C}_r{\left(\frac{x}{\cos \theta }\right)}^{16-r}{\left(\frac{1}{x\sin \theta }\right)}^r$
For term independent of $x$,
$16-2r=0\Rightarrow r=8$
$T_9={}^{16}{C}_8{\left(\frac{1}{\sin \theta \cos \theta }\right)}^8=16C_8{2}^8{\left(\frac{1}{\sin 2\theta }\right)}^8$
$l_1={}^{16}{C}_8{2}^8\text{at}\theta =\frac{\pi }{4}$
$l_2={}^{16}{C}_8\frac{2^8}{{\left(\frac{1}{\sqrt{2}}\right)}^8}={}^{16}{C}_8{2}^{12}\text{at}\theta =\frac{\pi }{8}$
$\frac{l_2}{l_1}=16:1$