Question

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

$T_{r+1}{=}^{16}{C}_r{\left(\frac{x}{cos\theta }\right)}^{16-r}{\left(\frac{1}{xsin\theta }\right)}^r$
${=}^{16}{C}_r{\left(x\right)}^{16-2r}\times \frac{1}{{\left(cos\theta \right)}^{16-r}{\left(sin\theta \right)}^r}$
For independent of $x;16-2r=0\Rightarrow r=8$
$\Rightarrow T_9{=}^{16}{C}_8\frac{1}{co{s}^8\theta si{n}^8\theta }$
${=}^{16}{C}_8\frac{2^8}{{\left(sin2\theta \right)}^8}$
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 ${\theta }_2=\frac{\pi }{8}$
$\frac{{\ell }_2}{{\ell }_1}=\frac{{\left(sin2{\theta }_1\right)}^8}{{\left(sin2{\theta }_2\right)}^8}={\left(\sqrt{2}\right)}^8=\frac{16}{1}$