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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If I = ∫ x^ 5 + 1 x + 1 dx= f(x) + c and f(0) = 0 (where, c is the...

If I=∫x5+1x+1dx=f(x)+cI = ∫ \dfrac{x^{5} + 1}{x + 1}dx= f\left(x\right) + c and f(0)=0f\left(0\right) = 0 (where, cc is the constant of integration), then

Options

  1. Af(1)>1f\left(1\right) > 1
  2. B0<f(1)<10 < f\left(1\right) < 1
  3. Cf(1)=1f\left(1\right) = 1
  4. Df(1)>2f\left(1\right) > 2

Correct answer

B. 0<f(1)<10 < f\left(1\right) < 1

Step-by-step solution

The given integral is
I=∫(x+1)(x4−x3+x2−x+1)x+1dxI = ∫ \dfrac{\left(x + 1\right)\left(x^{4} - x^{3} + x^{2} - x + 1\right)}{x + 1}dx
=∫(x4−x3+x2−x+1)dx= ∫ \left(x^{4} - x^{3} + x^{2} - x + 1\right)dx
=x55−x44+x33−x22+x+c= \dfrac{x5}{5} - \dfrac{x4}{4} + \dfrac{x3}{3} - \dfrac{x2}{2} + x + c
∴f(x)=x55−x44+x33−x22+x∴ f\left(x\right) = \dfrac{x^{5}}{5} - \dfrac{x^{4}}{4} + \dfrac{x^{3}}{3} - \dfrac{x^{2}}{2} + x
∴f(1)=15−14+13−12+1∴ f\left(1\right) = \dfrac{1}{5} - \dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{2} + 1
=1−120−16=4760= 1 - \dfrac{1}{20} - \dfrac{1}{6} = \dfrac{47}{60}
⇒0<f(1)<1⇒ 0 < f\left(1\right) < 1
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