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  • In the expansion of <img alt="\left ( \frac{x}{cos\theta }+\frac{1}{x\sin \theta } \right )^{16}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%20%28%20%5Cfrac%7Bx%7D%7Bcos%5Ctheta%20%7D&plus;%5Cfrac%7B1%7D%7Bx%5Csin%20%5Ctheta%2

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}\leq \theta \leq \frac{\pi }{4} and  L_{2}  is the least value of the term independent of x when \frac{\pi }{16}\leq \theta \leq \frac{\pi }{8}, then the ratio L_{2}:L_{1}  is equal to : 
Option: 1 16:1
Option: 2 8:1
Option: 3 1:8
Option: 4 1:16
 

Answers (1)

best_answer

General Term of Binomial Expansion\left(T_{r+1}\right)^{\mathrm{th}} \text { term is called as general term in }(x+y)^{n}\;\text{and general term is given by}

\mathrm{T}_{\mathrm{r}+1}=^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{\;x}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}

Term independent of x: It means term containing x0,

 

Now,

\\\mathrm{T}_{\mathrm{r}+1}=^{16} \mathrm{C}_{\mathrm{r}}\left(\frac{\mathrm{x}}{\cos \theta}\right)^{16-\mathrm{r}}\left(\frac{1}{\mathrm{x} \sin \theta}\right)^{\mathrm{r}}\\\text{for r = 8 term is free from 'x' }\\\begin{aligned} &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{1}{\sin ^{8} \theta \cos ^{8} \theta}\\ &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}\\ &\text { in } \theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right], L_{1}=^{16} \mathrm{C}_{8} 2^{8} \end{aligned}

\\\because \text{Min value of L}_1\;\text{at }\theta=\pi/4\\\text { in } \theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right], L_{2}=16 \mathrm{C}_{8} \frac{2^{8}}{\left(\frac{1}{\sqrt{2}}\right)^{8}}=^{16} \mathrm{C}_{8} \cdot 2^{8} \cdot 2^{4}\\\\\because \text{Min value of L}_2\;\text{at }\theta=\pi/8\\\frac{L_{2}}{L_{1}}=\frac{16 \mathrm{C}_{8} \cdot 2^{8} 2^{4}}{^{16} \mathrm{C}_{8} \cdot 2^{8}}=16

Correct option 1

Posted by

avinash.dongre

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