Let I = ∫ dx (cos x - sin x) ^ 2 = (1)/(f(x)) + C (where, C is the constant of integration). If f((π)/(3)) = 1 - 3 , then the number of solution(s) of f(x) = 2020 in x ∈ ((π)/(2), π) is/are

Correct answer: 1 I = ∫ ^ 2 x (1 - tan x) ^ 2 dx Let tan x = t ⇒ ^ 2 xdx = dt ⇒ I = ∫ dt (1 - t) ^ 2 = ∫ (1 - t) ^ - 2 dt ⇒ I = (1)/(1 - t) + C = (1)/(1 - tan x) + C ∴ f(x) = 1 - tan x If f(x) = 2020 ⇒ 1 - tan x = 2020 ⇒ tan x = - 2019 Which happens once in x ∈ ((π)/(2), π)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

Let I = ∫ dx (cos x - sin x) ^ 2 = (1)/(f(x)) + C (where, C is the...

Let

I = ∫ \dfrac{dx}{{\left(cos x - sin x\right)}^{2}}

= \dfrac{1}{f\left(x\right)} + C

(where,

C

is the constant of integration). If

f\left(\dfrac{\pi}{3}\right) = 1 - \sqrt{3}

, then the number of solution(s)

of f\left(x\right) = 2020

x \in \left(\dfrac{\pi}{2}, \pi\right)

is/are

Correct answer

Step-by-step solution

I = ∫ \dfrac{\sec^{2} x}{{\left(1 - tan x\right)}^{2}}dx

Let

tan x = t

⇒ \sec^{2} xdx = dt

⇒ I = ∫ \dfrac{dt}{{\left(1 - t\right)}^{2}} = ∫ {\left(1 - t\right)}^{- 2}dt

⇒ I = \dfrac{1}{1 - t} + C

= \dfrac{1}{1 - tan x} + C

∴ f\left(x\right) = 1 - tan x

f\left(x\right) = 2020 ⇒ 1 - tan x = 2020

⇒ tan x = - 2019

Which happens once in

x \in \left(\dfrac{\pi}{2}, \pi\right)

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