If the integral I = ∫ 2x^ 2 4 + x^ 2 dx = 2x - f(x) + c , where f (2) = π, then the minimum value of y = f(x) x ∈ [-2,2] is (where, c is the constant of integration)

Correct answer: B. - π I = 2∫ ( x^ 2 + 4 - 4 x^ 2 + 4 )dx = 2∫ (1 - 4 x^ 2 + 4 )dx = 2(x - (4)/(2) ^ - 1 ((x)/(2))) + c = 2x - 4 ^ - 1 ((x)/(2)) + c ∴ f( x) = 4 ^ - 1 ((x)/(2)) So, min ( f( x)) = 4 ^ - 1 (-1) = - π

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If the integral I = ∫ 2x^ 2 4 + x^ 2 dx = 2x - f(x) + c , where f...

If the integral

I = ∫ \dfrac{2x^{2}}{4 + x^{2}}dx = 2x - f\left(x\right) + c

, where

f \left(2\right) = \pi,

then the minimum value of

y = f\left(x\right) \forall x \in \left[-2,2\right]

is (where,

c is the constant of integration)

Options

A $0$
B $- \pi$
C $2\pi$
D $- 4\pi$

Correct answer

B. $- \pi$

Step-by-step solution

I = 2∫ \left(\dfrac{x^{2} + 4 - 4}{x^{2} + 4}\right)dx

<br />

= 2∫ \left(1 - \dfrac{4}{x^{2} + 4}\right)dx

<br />

= 2\left(x - \dfrac{4}{2}\tan^{- 1} \left(\dfrac{x}{2}\right)\right) + c

<br />

= 2x - 4\tan^{- 1} \left(\dfrac{x}{2}\right) +

c

∴ f\left(

x\right) = 4\tan^{- 1} \left(\dfrac{x}{2}\right)
So,

min \left(

f\left(

x\right)\right) = 4\tan^{- 1} \left(-1\right) = - \pi

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