If the integral ∫ x 4 + x 2 + 1 x 2 - x + 1 d x = f x + C , (where C is the constant of integration and x ∈ R ), then the minimum value of f ' x is

Question

If the integral ∫ x 4 + x 2 + 1 x 2 - x + 1 d x = f x + C , (where C is the constant of integration and x ∈ R ), then the minimum value of f ' x is, 1, 1 4, 3 4, 2

MARKS App · Accepted Answer

As x 4 + x 2 + 1 = x 2 + 1 2 - x 2 = x 2 + 1 + x x 2 + 1 - x Thus, ∫ x 4 + x 2 + 1 x 2 - x + 1 d x = ∫ x 2 + x + 1 d x i.e., f ' x = x 2 + x + 1 = x + 1 2 2 + 3 4 ≥ 3 4

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