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
A
`0`
B
`1`
C
`-1`
D
`2`
Text Solution
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The correct Answer is:
B
`(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`