If Ι = ∫ tan^ - 1 (e^ x ) e^ x + e^ - x dx = [tan^ - 1 (f(x))]2 2 + C (where C is the constant of integration), then the range of y = f(x) x ∈ ℝ is

Correct answer: C. (0, ∞) Given integral = ∫ tan^ - 1 (e^ x ) 1 + e^ 2x e^ x dx Let, tan^ - 1 (e^ x ) = t ⇒ 1 1 + e^ 2x ⋅ e^ x dx = dt ⇒ Ι = ∫ tdt = t^ 2 2 + C = [tan^ - 1 ( e^ x )] ^ 2 2 + C Hence, f( x) = e^ x ∈ (0, ∞)

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If Ι = ∫ tan^ - 1 (e^ x ) e^ x + e^ - x dx = [tan^ - 1 (f(x))]2 2 +...

Ι = ∫ \dfrac{tan^{- 1}\left(e^{x}\right)}{e^{x} + e^{- x}}dx = \dfrac{\left[tan^{- 1}\left(f\left(x\right)\right)\right]2}{2} + C

(where

C

is the constant of integration), then the range of

y = f\left(x\right) \forall x \in \mathbb{R}

Options

A $\left(- \infty, \infty\right)$
B $\left[0, \infty\right)$
C $\left(0, \infty\right)$
D $\left(- \infty, 0\right)$

Correct answer

C. $\left(0, \infty\right)$

Step-by-step solution

Given integral

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

Let,

tan^{- 1}\left(e^{x}\right) = t

⇒ \dfrac{1}{1 + e^{2x}} ⋅ e^{x}dx = dt

⇒ Ι = ∫ tdt =\dfrac{t^{2}}{2} + C

= \dfrac{{\left[tan^{- 1} \left( e^{x}\right)\right]}^{2}}{2} + C

Hence, f\left( x\right)

= e^{x} \in \left(0,

\infty\right)

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