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Question

If for a non-zero x, af(x)+bf(1x)=1x5, where ab, then 21f(x)dx=

A
1a2+b2(alog2+5a+7b2)
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B
1a2b2(alog25a+7b2)
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C
1a2+b2(alog2+5a7b2)
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D
none of these
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Solution

The correct option is A 1a2b2(alog25a+7b2)
Given af(x)+bf(1x)=1x5 ...(1)
Replacing x1x in (1), we get
af(1x)+bf(x)=x5 ...(2)
On multiplying equation (1) by a and equation (2) by b respectively
a2f(x)+abf(1x)=a(1x5) ...(3)
abf(1x)+b2f(x)=b(x5) ...(4)
On subtracting (4) from (3)
(a2b2)f(x)=axbx5a+5bf(x)=1(a2b2)(axbx5a+5b)21f(x)dx=1(a2b2)21(axbx5a+5b)dx=1(a2b2)[alog|x|b2x25(ab)x]21=1(a2b2)[alog22b10(ab)alog1+b2+5(ab)]=1(a2b2)[alog25a+72b]

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