In the expansion of ( x cos θ + 1 xsin θ ) ^ 16 , if l _ 1 is the least value of the term independent of x when (π)/(8) ≤ θ ≤ (π)/(4) and l _ 2 is the least value of the term independent of x when (π)/(16) ≤ θ ≤

Correct answer: B. 16: 1 T _ r + 1 = Cr 16 ((x)/(cos θ)) ^ 16 - r ((1)/(xsin θ)) ^ r for r = 8 term is free from 'x' T _ 9 = C8 16 1 ^ 8 θ ^ 8 θ T _ 9 = C8 16 2 ^ 8 ( 2θ) ^ 8 In θ ∈ [(π)/(8), (π)/(4)], l _ 1 = C8 16 2 ^ 8 ( ∵ Minimum value of l _ 1 at θ = (π)/(4) ) In θ ∈ [(π)/(16), (π)/(8)], l _ 2 = C8 16 2 ^ 8 ( 1 2 ) ^ 8 = C8 16 · 2 ^ 8 · 2 ^ 4 ( ∵ Minimum value of l _ 2 at θ = (π)/(8) ) l _ 2 l _ 1 = C8 16 . 2 ^ 8 2 ^ 4 C8 16 . 2 ^ 8 = 16

JEE Main 20209 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2020 · 9 January · Evening Shift · Mathematics · Binomial Theorem

In the expansion of ( x cos θ + 1 xsin θ ) ^ 16 , if l _ 1 is the...

In the expansion of

{\left(\dfrac{x}{\text{ cos } \theta} + \dfrac{1}{\text{ xsin } \theta}\right)}

^{16},

if

{l}_{1} is the least value of the term independent of

x

when

\dfrac{\pi}{8} \le \theta \le \dfrac{\pi}{4}

and

{l}_{2}

is the least value of the term independent of

x

when

\dfrac{\pi}{16} \le \theta \le \dfrac{\pi}{8},

then the ratio

{l}_{2}: {l}_{1}

is equal to:

Options

A $1: 8$
B $16: 1$
C $8: 1$
D $1: 16$

Correct answer

B. $16: 1$

Step-by-step solution

{T}_{r + 1} = Cr 16 {\left(\dfrac{x}{cos \theta}\right)}^{16 - r}{\left(\dfrac{1}{xsin \theta}\right)}^{r}

for

r = 8

term is free from

'x'

{T}_{9} = C8 16 \dfrac{1}{\sin^{8} \theta\cos^{8} \theta}

{T}_{9} = C8 16 \dfrac{{2}^{8}}{{\left(\sin 2\theta\right)}^{8}}

\theta \in \left[\dfrac{\pi}{8}, \dfrac{\pi}{4}\right], {l}_{1} = C8 16 {2}^{8}

(

∵

Minimum value of

{l}_{1}

\theta = \dfrac{\pi}{4}

)
In

\theta \in \left[\dfrac{\pi}{16}, \dfrac{\pi}{8}\right], {l}_{2} = C8 16 \dfrac{{2}^{8}}{{\left(\dfrac{1}{\sqrt{2}}\right)}^{8}} = C8 16 \cdot {2}^{8} \cdot {2}^{4}

(

∵

Minimum value of

{l}_{2}

\theta = \dfrac{\pi}{8}

)

\dfrac{{l}_{2}}{{l}_{1}} = \dfrac{C8 16 . {2}^{8}{2}^{4}}{C8 16 . {2}^{8}} = 16

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