For `a`, `b`,`c in R-{0}`, let `(a+b)/(1-ab)`, `b`, `(b+c)/(1-bc)` are in `A.P.` If `alpha`, `beta` are the roots of the quadratic equation
`2acx^(2)+2abcx+(a+c)=0`, then the value of `(1+alpha)(1+beta)` is

Options

A. `0`
B. `1`
C. `-1`
D. `2`

Answer

B

Explanation

`(b)` Given `(a+B)/(1-ab)`, `b`, `(b+c)/(1-bc)` are in `A.P.`
`impliesb-(a+b)/(1-ab)=(b+c)/(1-bc)-b`
`implies (-a(b^(2)+1))/(1-ab)=(c(b^(2)+1))/(1-bc)`
`impliesa+c=2abc`
Now, given quadratic equation is
`2acx^(2)+2abcx+2abc=0`
(Substituting `a+c=2abc` and then cancelling `2ac` )
`impliesx^(2)+bx+b=0`