Question

In the expansion $\begin{array}{l}{\left(\frac{x}{\cos \theta }+\frac{1}{x\sin \theta }\right)}^{16}\end{array}$if $l_1$ is the least value of the term independent of $x$ when $\frac{\mathrm{\pi }}{8}\le \theta \le \frac{\mathrm{\pi }}{4}$ and $l_2$ is the least value of the term independent of $x$ when $\frac{\mathrm{\pi }}{16}\le \theta \le \frac{\mathrm{\pi }}{8}$, 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

$16:1$

Determine the ratio $l_2:l_1$

Step 1: Calculating $I_1$ and $I_2$

We know $r^{th}$ term in the expansion of ${(a+b)}^n$ is $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$ then,

$\Rightarrow \begin{array}{l}{T}_{r+1}={}^{16}{C}_r{\left(\frac{x}{\cos \theta }\right)}^{16-r}{\left(\frac{1}{x\sin \theta }\right)}^r\end{array}.....\left(i\right)$

For the independent term of $x$, put $16-2r=0$

$\therefore r=8$

$$\begin{gathered} \Rightarrow T_{8+1}=\begin{array}{l}{}^{16}{C}_8{\left(\frac{1}{\sin \theta \cos \theta }\right)}^8\end{array} \\ ={}^{16}{C}_8{2}^8{\left(\frac{1}{\sin 2\theta }\right)}^8;[\because \sin 2\theta =2\sin \theta \cos \theta ] \end{gathered}$$

$l_1={}^{16}{C}_8{2}^8$in $\frac{\mathrm{\pi }}{8}\le \theta \le \frac{\mathrm{\pi }}{4};\left[\because \sin 2\theta =1\text{is least in}\frac{\mathrm{\pi }}{8}\le \theta \le \frac{\mathrm{\pi }}{4}\right]$

$$\begin{gathered} \begin{array}{l}\Rightarrow l_2={}^{16}{C}_8\frac{2^8}{{\left(\frac{1}{\sqrt{2}}\right)}^8}\end{array};\left[\because \sin 2\theta =\frac{1}{\sqrt{2}}in\frac{\mathrm{\pi }}{16}\le \theta \le \frac{\mathrm{\pi }}{8}\right] \\ {=}^{16}{C}_8{2}^{12} \end{gathered}$$

Step 3: Calculating the ratio

$$\begin{gathered} \Rightarrow \frac{l_2}{l_1}=\frac{{}^{16}{C}_8{2}^{12}}{{}^{16}{C}_8{2}^8} \\ =\frac{16}{1} \end{gathered}$$
Hence, option A is the correct answer.