Let I = ∫ dx 1 + 3si n ^ 2 x = (1)/(2) ^ - 1 (2f(x)) + C (where, C is the constant of integration). If f((π)/(4)) = 1, then the fundamental period of y = f(x) is

Correct answer: B. π Dividing the numerator and denominator by ^ 2 x, we get I = ∫ ^ 2 x ^ 2 x + 3 ^ 2 x dx = ∫ ^ 2 x 1 + 4 ^ 2 x dx Let, tan x = t ⇒ ^ 2 xdx = dt ∴ I = ∫ dt 1 + 4 t ^ 2 = (1)/(2)ta n ^ - 1 (2t) + C I = (1)/(2)ta n ^ - 1 (2tan x) + C ⇒ f(x) = tan x Hence, the fundamental period is ‘π’

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

Let I = ∫ dx 1 + 3si n ^ 2 x = (1)/(2) ^ - 1 (2f(x)) + C (where, C...

Let

I = ∫ \dfrac{dx}{1 + 3si{n}^{2}x} =

\dfrac{1}{2}\tan^{- 1}\left(2f\left(x\right)\right) + C

(where,

C

is the constant of integration). If

f\left(\dfrac{\pi}{4}\right) = 1,

then the fundamental period of

y = f\left(x\right)

Options

A $\dfrac{\pi}{4}$
B $\pi$
C $2\pi$
D $\dfrac{\pi}{6}$

Correct answer

B. $\pi$

Step-by-step solution

Dividing the numerator and denominator by

\cos^{2}x,

we get

I = ∫ \dfrac{\sec^{2}x}{\sec^{2}x + 3\tan^{2}x}dx

= ∫ \dfrac{\sec^{2}x}{1 + 4\tan^{2}x}dx

Let,

tan x = t

⇒ \sec^{2}xdx = dt

∴ I = ∫ \dfrac{dt}{1 + 4{t}^{2}}

= \dfrac{1}{2}ta{n}^{- 1}\left(2t\right) + C

I = \dfrac{1}{2}ta{n}^{- 1}\left(2tan x\right) + C ⇒ f\left(x\right) = tan x

Hence, the fundamental period is

‘\pi’

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