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Question

Let f be a real-valued function and satisfies f(x)+f(y)=1x+1y x,yR{0}. If 323(f(x))5f(x)1(f(x))4dx=12ln(2α3β), then

A
αβ
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B
α+β=17
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C
tan1αβ=sec1(α5)
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D
(αβ) is not a prime number
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Solution

The correct option is C tan1αβ=sec1(α5)
f(x)+f(y)=1x+1y x,yR{0}
Put x=y, we get
f(x)=1x

Now, I=32⎜ ⎜ ⎜3x51x11x4⎟ ⎟ ⎟dx

On multiplying and dividing by 1x2,
I=32⎜ ⎜ ⎜3x71x31x21x6⎟ ⎟ ⎟dx

Put 1x21x6=t
dt=(2x3+6x7)dx

I=12807291564dtt

=12[lnt]807291564

=12ln(21037)
On comparing with 12ln(2α3β),
α=10,β=7

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