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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If the integral ∫ x^ 4 + x^ 2 + 1 x^ 2 - x + 1 dx = f(x) + C,...

If the integral ∫x4+x2+1x2−x+1dx=f(x)+C,∫ \dfrac{x^{4} + x^{2} + 1}{x^{2} - x + 1}dx = f\left(x\right) + C, (where CC is the constant of integration and x∈Rx \in \mathbb{R}), then the minimum value of f^{'} \left(x\right) is&

Options

  1. A11
  2. B14\dfrac{1}{4}
  3. C34\dfrac{3}{4}
  4. D22

Correct answer

C. 34\dfrac{3}{4}

Step-by-step solution

As x4+x2+1=(x2+1)2−x2=(x2+1+x)(x2+1−x)x^{4} + x^{2} + 1 = { \left(x^{2} + 1\right)}^{2} - x^{2} = \left(x^{2} + 1 + x\right)\left(x^{2} + 1 - x\right)
Thus, ∫x4+x2+1x2−x+1dx=∫∫ \dfrac{x^{4} + x^{2} + 1}{x^{2} - x + 1}dx = ∫\left(x^{2} + x + 1\right)dx<br/>i.e.,<br/>i.e.,f^{'}\left(x\right)= x^{2} + x + 1 = (x+12)2+34{\left(x + \dfrac{1}{2}\right)}^{2} + \dfrac{3}{4} \ge 34\dfrac{3}{4}
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