In the expantion of \left(\left(\right. \frac{x}{cosθ} + \frac{1}{xsinθ} \left.\right)\right)^{1 / 6}, if l1 is the least value of the term independent of x when \frac{\pi}{8} \leq \theta \leq \frac{\pi}{4} and l2 is the least value of the term independent of x when \frac{\pi}{16} \leq \theta \leq \frac{\pi}{8}, then the ratio l2 : l1 is equal to :

Engineering

Mathematics

Introduction to Binomial Theorem

Question

In the expantion of ${(\frac{x}{cosθ} + \frac{1}{xsinθ})}^{1 / 6}$ , if l₁ is the least value of the term independent of x when $\frac{π}{8} \leq θ \leq \frac{π}{4}$ and l₂ is the least value of the term independent of x when $\frac{π}{16} \leq θ \leq \frac{π}{8}$ , then the ratio l₂ : l₁ is equal to :

1 : 16

16 : 1

8 : 1

1 : 8

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$T_{r + 1} =^{16} C_{r} {(\frac{x}{cosθ})}^{16 - r} {(\frac{1}{xsinθ})}^{r}$

for r = 8 term is free from 'x'

$T_{9} =^{16} C_{8} \frac{1}{\sin^{8} {θcos}^{8} θ}$

$T_{9} =^{16} C_{8} \frac{2^{8}}{{(\sin 2 θ)}^{8}}$

$in θ \in [\frac{π}{8}, \frac{π}{4}], L_{1} =^{16} C_{8} 2^{8}$ $∵$ {Min value of L₁ at θ = π/4}

$in θ \in [\frac{π}{16}, \frac{π}{8}], L_{2} =^{16} C_{8} \frac{2^{8}}{{(\frac{1}{\sqrt{2}})}^{8}} =^{16} C_{8} . 2^{8} . 2^{4}$ { $∵$ min value of L₂ at θ = π/8]

$\frac{L_{2}}{L_{1}} = \frac{^{16} C_{8} . 2^{8} 2^{4}}{^{16} C_{8} . 2^{8}} = 16$