Let R be the set of all real numbers and let f be a function from R to R such that
`f(X)+(X+1/2)f(l-X)=1,`
for all `xinR." Then " 2f(0)+3f(1)` is equal to-

Options

A. 2
B. 0
C. -2
D. -4

Answer

C

Explanation

`f(x)+(x+1/2)f(1-x)=1`
`f(l-x)+(1-x+1/2)f(1-(1-x))=1`
`f(1-x)+(3/2-x)f(x)=1`
`(1-f(x))/(x+1/2)+(3/2-x)f(x)=1`
`1-f(x)+(3/2x-x^(2)+3/4-x/2)f(x)=x+1/2`
`f(x)(x-x^(2)-1/4)=x-1/2`
`f(x)(4x-4x^(2)-1)=4x-2`
`f(x)=(-2+4x)/(4x-4x^(2)-1)`
`2f(0)+3f(1)=2((-2+0)/(0-0-1))+3((-2+4)/(4-4-1))`
`=+4+(3(+2))/-1=+4-6=-2`
Alternate
Put x = 0
`f(0)+1/2f(1)=1rArr2f(0)+f(1)=2`
put x = 1
`f(1)+3/2f(0)=1rArr2f(1)+3f(0)=2`
solving above `f(0)=2" and "f(1)=-2`
`:. 2f(0)+3f(1)=4-6=-2`