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Question

If for non-zero x, af (x) + bf (1x)=1x25 where a b then 21 f(x) dx is

A
1a2b2[a(ln225)+47b2]
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B
1a2b2[a(2ln225)+47b2]
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C
1a2b2[a(2ln225)+47b2]
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D
1a2b2[a(ln225)47b2]
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Solution

The correct option is A 1a2b2[a(ln225)+47b2]
Given that

a f(x) +bf (1x)=1x25 ...(1)

Replacing x with 1x

a.f (1x)+b.f(x)=x25 ...(2)

Now [(1) × a - (2) × b]

(a2b2)f(x)=a(1x25)b(x25)

f(x)=1a2b2[a(1x25)b(x25)]

Integrating

21f(x)dx=1a2b2{a(logx25x)b(x2225x)}21

=1a2b2[{a(log250)b(50250)}{a(025)b(1225)}]

= 1a2b2[a(log225)+472b]


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