If `a f(x+1)+b f(1/(x+1))=x,x !=-1,a != b,`then `f(2)` is equal to

Options

A. A.
`(2a+b)/(2(a^(2)-b^(2)))`
B. B. `(a)/(a^(2)-b^(2))`
C. C. `(a+2b)/(a^(2)-b^(2))`
D. D. none of these

Answer

A

Explanation

`af(x+1)+bf((1)/(x+1))=(x+1)-1 " (1)" `
Replacing `x+1" by " (1)/(x+1),` we get
`af((1)/(x+1))+bf(x+1)=(1)/(x+1)-1 " (2)" `
`Eq. (1)xx a-Eq.(2) xx b`
`implies (a^(2)-b^(2))f(x+1)=a(x+1)-a-(b)/(x+1)+b`
Putting `x=1, (a^(2)-b^(2))f(2)=2a-a-(b)/(2)+b=a+(b)/(2)=(2a+b)/(2)`