Let $$f(x)=1x6+x4$$ and F be a antiderivative of f such that $$F(1)=$$π$$4+23Statement-1$$ $$F13=$$π$$6Statement-2$$: $$F(x)=$$$$tan$$−$$1$$⁡$$x+13x$$−$$1x3$$

Correct option is 3STATEMENT-1 is True, STATEMENT-2 is False

Questions

Let $f (x) = \frac{1}{x^{6} + x^{4}}$ and $F$ be a antiderivative
of $f$ such that $F (1) = \frac{π}{4} + \frac{2}{3}$
Statement-1 $F (\frac{1}{\sqrt{3}}) = \frac{π}{6}$
Statement-2: $F (x) = \tan^{- 1} x + \frac{1}{3 x} - \frac{1}{x^{3}}$

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STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation forSTATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

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detailed solution

Correct option is C

f(x)=1x4x2+1=1x4−1x2+1x2+1So F(x)=−13x3+1x+tan−1⁡x+CF(1)=23+π4+Cso C=0F(x)=tan−1⁡x+1x−13x3⇒F13=π6.

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Let f(x)=1x6+x4 and F be a antiderivative of f such that F(1)=π4+23Statement-1 F13=π6Statement-2: F(x)=tan−1⁡x+13x−1x3

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free study material

Let $f (x) = \frac{1}{x^{6} + x^{4}}$ and $F$ be a antiderivative
of $f$ such that $F (1) = \frac{π}{4} + \frac{2}{3}$
Statement-1 $F (\frac{1}{\sqrt{3}}) = \frac{π}{6}$
Statement-2: $F (x) = \tan^{- 1} x + \frac{1}{3 x} - \frac{1}{x^{3}}$