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If alpha,beta,gamma are the roots of x^3...

If `alpha,beta,gamma` are the roots of `x^3-x^2-1=0` then the value of `(1+alpha)/(1-alpha)+(1+beta)/(1-beta)+(1+gamma)/(1-gamma)` is equal to
(a) `-5` b. `-6` c. `-7` d. `-2`

A

-5

B

-6

C

-7

D

-2

Text Solution

Verified by Experts

The correct Answer is:
A

`sum alpha =1, sumalphabeta = 0, alphabetagamma = 1`
`sum (1 + alpha)/(1 - alpha) = - sum (-alpha + 1 - 2)/(1 - alpha) = sum((2)/(1 - alpha) - 1)`
`2sum(1)/(1 - alpha) - 3`
Now, `x^(3) - x^(2) - 1 = (x - alpha)(x - beta)(x - gamma)`
`therefore log(x - alpha) + log(x - beta) + log(x - gamma) = log (x^(3) - x^(2) - 1)`
Differenting, we get
`(1)/((x - alpha)) + (1)/((x - beta)) + (1)/((x - gamma)) = (3x^(2) - 2x)/(x^(3) - x^(2) - 1)`
`rArr (1)/(1 - alpha) + (1)/(1 - beta) + (1)/(1 - gamma) = (3 - 2)/(1 - 1 - 1) = -1`
`rArr (1 + alpha)/(1 - alpha) = -5`
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