if for nonzero `x`, `af(x) + bf(1/x) =1/x-5`, where `a!=b` then f(2) =

Options

A. `(3(2b+3a))/(2(a^(2)-b^(2))`
B. `(3(2b-3a))/(2(a^(2)-b^(2))`
C. `(3(3a-2b))/(2(a^(2)-b^(2))`
D. `(6)/(a+b)`

Answer

B

Explanation

We have, `a f(x)+bf((1)/(x))=(1)/(x)-5" "`…(i)
On replacing x by `(1)/(x)`, we get
`af((1)/(x))+bf(x)=x-5`
`implies bf(x)+af((1)/(x))=x-5" "`……(ii)
Multiplying (i) by a (ii) by b and then subtracting, we get
`(a^(2)-b^(2))f(x)=((a)/(x)-bx)-5(a-b)`
`implies f(x)=(1)/(a^(2)-b^(2))((1)/(x)-bx)-(5)/(a+b)`
`implies f(2)=(3(2b-3a))/(2(a^(2)-b^(2))`