If the real part of the complex number (1 - cosθ + 2isinθ) ^ - 1 is (1)/(5) for θ ∈ ( 0, π ) , then the value of the integral ∫ _ 0 ^ θ sinxdx is equal to:

Correct answer: A. 1 Let z = (1)/(1 - cosθ + 2isinθ) ⇒ z = ((1 - cosθ) - 2isinθ)/((1 - cosθ + 2isinθ)(1 - cosθ - 2isinθ)) ⇒ z = 2 ^ 2 ((θ)/(2)) - 2isinθ ( 1 - cosθ ) ^ 2 + 4 ^ 2 θ Put, sinθ = 2sin(θ)/(2)cos(θ)/(2), 1 - cosθ = 2 ^ 2 (θ)/(2) ⇒ z = 2 ^ 2 ((θ)/(2)) - 4isin((θ)/(2)) ((θ)/(2)) 4 ^ 4 ((θ)/(2)) + 16 ^ 2 ((θ)/(2)) ^ 2 ((θ)/(2)) Hence, Re(z) = 2 ^ 2 ((θ)/(2)) 4 ^ 4 ((θ)/(2)) + 16 ^ 2 ((θ)/(2)) ^ 2 ((θ)/(2)) ⇒ 1 2 ^ 2 ((θ)/(2)) + 8 ^ 2 ((θ)/(2)) = (1)/(5) ⇒ 1 2 ^ 2 ((θ)/(2)) + 8 ^ 2 ((θ)/(2)) = (1)/(5) ⇒ ^ 2 ((θ)/(2)) + 4 ^ 2 ((θ)/(2)) = (5)/(2) ⇒ 1 - ^ 2 ((θ)/(2)) + 4 ^ 2 ((θ)/(2)) = (5)/(2) ⇒ 3 ^ 2 (θ)/(2) = (3)/(2) ⇒ ^ 2 ((θ)/(2)) = (1)/(2) ⇒ (θ)/(2) = nπ ± (π)/(4), n ∈ Z ⇒ θ = 2nπ ± (π)/(2) For θ ∈ ( 0, π ) , θ = (π)/(2) . Hence, ∴ ∫ _ 0 ^ θ sin θ = ∫ _ 0 ^ (π)/(2) sin θ = [- cosθ] _ 0 ^ (π)/(2) = - (0 - 1) = 1

JEE Main 202120 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 20 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2021 · 20 July · Evening Shift · Mathematics · Complex Number

If the real part of the complex number (1 - cosθ + 2isinθ) ^ - 1 is...

If the real part of the complex number

{\left(1 - cos\theta + 2isin\theta\right)}^{- 1}

\dfrac{1}{5}

for

\theta \in ( 0, \pi ) ,

then the value of the integral

{∫}_{0}^{\theta}sinxdx

is equal to:

Options

A $1$
B $2$
C $- 1$
D $0$

Correct answer

A. $1$

Step-by-step solution

Let

$z = \dfrac{1}{1 - cos\theta + 2isin\theta}$

$⇒ z = \dfrac{\left(1 - cos\theta\right) - 2isin\theta}{\left(1 - cos\theta + 2isin\theta\right)\left(1 - cos\theta - 2isin\theta\right)}$

$⇒ z = \dfrac{2\sin^{2}\left(\dfrac{\theta}{2}\right) - 2isin\theta}{( 1 - cos\theta{)}^{2} + 4\sin^{2}\theta}$

Put, $sin\theta = 2sin\dfrac{\theta}{2}cos\dfrac{\theta}{2}, 1 - cos\theta = 2\sin^{2}\dfrac{\theta}{2}$

$⇒ z = \dfrac{2\sin^{2}\left(\dfrac{\theta}{2}\right) - 4isin\left(\dfrac{\theta}{2}\right)\cos \left(\dfrac{\theta}{2}\right)}{4\sin^{4}\left(\dfrac{\theta}{2}\right) + 16\sin^{2}\left(\dfrac{\theta}{2}\right)\cos^{2}\left(\dfrac{\theta}{2}\right)}$

Hence,

$Re\left(z\right) = \dfrac{2\sin^{2}\left(\dfrac{\theta}{2}\right)}{4\sin^{4}\left(\dfrac{\theta}{2}\right) + 16\sin^{2}\left(\dfrac{\theta}{2}\right)\cos^{2}\left(\dfrac{\theta}{2}\right)}$

$⇒ \dfrac{1}{2\sin^{2}\left(\dfrac{\theta}{2}\right) + 8\cos^{2}\left(\dfrac{\theta}{2}\right)} = \dfrac{1}{5}$

$⇒ \sin^{2}\left(\dfrac{\theta}{2}\right) + 4\cos^{2}\left(\dfrac{\theta}{2}\right) = \dfrac{5}{2}$

$⇒ 1 - \cos^{2}\left(\dfrac{\theta}{2}\right) + 4\cos^{2}\left(\dfrac{\theta}{2}\right) = \dfrac{5}{2}$

$⇒ 3\cos^{2}\dfrac{\theta}{2} = \dfrac{3}{2}$

$⇒ \cos^{2}\left(\dfrac{\theta}{2}\right) = \dfrac{1}{2}$

$⇒ \dfrac{\theta}{2} = n\pi \pm \dfrac{\pi}{4}, n \in Z$

$⇒ \theta = 2n\pi \pm \dfrac{\pi}{2}$

For $\theta \in ( 0, \pi )$ , $\theta = \dfrac{\pi}{2}$ .

Hence,

$∴ {∫}_{0}^{\theta}sin\thetad\theta = {∫}_{0}^{\dfrac{\pi}{2}}sin\thetad\theta = {\left[- cos\theta\right]}_{0}^{\dfrac{\pi}{2}}$

$= - \left(0 - 1\right)$

$= 1$

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