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Question

If x4+1x6+1 dx=tan1f(x)23tan1g(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 roots
Let I=x4+1x6+1 dx
I=(x2+1)22x2(x2+1)(x4x2+1) dx
I=(1+1x2)(x21+1x2) dx2x2(x3)2+1 dxI=(1+1x2)(x1x)2+1 dx2x2(x3)2+1 dx
Put x1x=t(1+1x2)dx=dt
And x3=u3x2 dx=du
I=tan1t23tan1u+CI=tan1(x1x)23tan1(x3)+C
f(x)=x1x, g(x)=x3
Clearly, both f(x) and g(x) are odd functions.

Also, f(x)=1+1x20
and f(x)>0 x(0,)
g(x)=3x20
Therefore, both f(x) and g(x) are monotonic.

f(x)=g(x)x1x=x31=1x2+x2
We know that,
1x2+x22
Hence no solution.

f(x)g(x) dx=x1xx3=1x+13x3+C

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