In the expansion of $\left(\frac{x}{\cos \theta}+\frac{1}{x \sin \theta}\right)^{16}$, if $I_{1}$ is the least value of the term independent of $x$ when $\frac{\pi}{16} \leq \theta \leq \frac{\pi}{8}$, and $I_{2}$ is the least value of the term independent of $x$ when $\frac{\pi}{8} \leq \theta \leq \frac{\pi}{4}$, then the ratio $I_{2}: I_{1}$ is equal to:

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Binomial Theorem

In the expansion of

(\frac{x}{c o s θ} + \frac{1}{x s i n θ})^{16}

, if

I_{1}

is the least value of the term independent of

x

when

\frac{π}{16} \leq θ \leq \frac{π}{8}

, and

I_{2}

is the least value of the term independent of

x

when

\frac{π}{8} \leq θ \leq \frac{π}{4}

, then the ratio

I_{2} : I_{1}

is equal to:

Answer: (A)

The term free of

x

is the middle term, where the powers balance at

r = 8

, and it equals

(8 16) \frac{1}{sin ^{8} θ cos ^{8} θ} = (8 16) \frac{2 ^{8}}{sin ^{8} 2 θ}

. This is smallest where

sin 2 θ

is largest. On

\frac{π}{16} \leq θ \leq \frac{π}{8}

the angle

2 θ

rises to

\frac{π}{4}

, where

sin 2 θ = \frac{1}{2}

, giving

I_{1} = (8 16) 2^{8} \cdot 2^{4}

. On

\frac{π}{8} \leq θ \leq \frac{π}{4}

the angle

2 θ

reaches

\frac{π}{2}

, where

sin 2 θ = 1

, giving

I_{2} = (8 16) 2^{8}

. Dividing,

I_{2} : I_{1} = 1 : 2^{4} = 1 : 16

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option