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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If ∫ dx x^ 2 + x = |f(x)| + C (where C is the constant of...

If ∫dxx2+x=ln⁡∣f(x)∣+C∫ \dfrac{dx}{x^{2} + x} = \ln \left|f\left(x\right)\right| + C (where CC is the constant of integration), then the range of y=f(x),∀x∈R−{−1,0}y = f\left(x\right), \forall x \in \mathbb{R} - \left\{- 1,0\right\} is

Options

  1. AR−{1}\mathbb{R} - \left\{1\right\}
  2. BR−{0}\mathbb{R} - \left\{0\right\}
  3. CR−{0,1}\mathbb{R} - \left\{0,1\right\}
  4. DR−{0,−1}\mathbb{R} - \left\{0, - 1\right\}

Correct answer

C. R−{0,1}\mathbb{R} - \left\{0,1\right\}

Step-by-step solution

I=∫dxx(x+1)=∫dxx−∫dxx+1Ι = ∫ \dfrac{dx}{x\left(x + 1\right)} = ∫ \dfrac{dx}{x} - ∫ \dfrac{dx}{x + 1}
⇒I=ln⁡∣x∣−ln⁡∣x+1∣+C⇒ Ι = \ln \left|x\right| - \ln \left|x + 1\right| + C
=ln⁡∣xx+1∣+C= \ln \left|\dfrac{x}{x + 1}\right| + C
i.e. y=f(x)=xx+1y = f\left(x\right) = \dfrac{x}{x + 1}
⇒yx+y=x⇒ yx + y = x
⇒x=y1−y⇒ x = \dfrac{y}{1 - y}
⇒y∈R−{0,1}(asx≠0)⇒ y \in \mathbb{R} - \left\{0,1\right\} \left(asx \ne 0\right)
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