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Question

# If ∫x4+1x6+1 dx=tan−1f(x)−23tan−1g(x)+C, then

A
both f(x) and g(x) are odd function
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B
f(x) is monotonic function in (0,)
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C
f(x)=g(x) has no real roots
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D
f(x)g(x) dx=1x13x3+C
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Solution

## The correct option is C f(x)=g(x) has no real rootsLet I=∫x4+1x6+1 dx ⇒I=∫(x2+1)2−2x2(x2+1)(x4−x2+1) dx ⇒I=∫(1+1x2)(x2−1+1x2) dx−2∫x2(x3)2+1 dx⇒I=∫(1+1x2)(x−1x)2+1 dx−2∫x2(x3)2+1 dx Put x−1x=t⇒(1+1x2)dx=dt And x3=u⇒3x2 dx=du ∴I=tan−1t−23tan−1u+C⇒I=tan−1(x−1x)−23tan−1(x3)+C ∴f(x)=x−1x, g(x)=x3 Clearly, both f(x) and g(x) are odd functions. Also, f′(x)=1+1x2≠0 and f′(x)>0 ∀ x∈(0,∞) g′(x)=3x2≥0 Therefore, both f(x) and g(x) are monotonic. f(x)=g(x)⇒x−1x=x3⇒1=1x2+x2 We know that, 1x2+x2≥2 Hence no solution. ∫f(x)g(x) dx=x−1xx3=−1x+13x3+C

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