The real numbers `x_1, x_2, x_3` satisfying the equation `x^3-x^2+b x+gamma=0` are in A.P. Find the intervals in which `beta and gamma` lie.

Options

A. ` (-infty, (1)/(3))`
B. `(-infty, - (1)/(3))`
C. `((1)/(3) , infty)`
D. ` (-(1)/(3), infty)`

Answer

1

Explanation

From the question, the real roots of `x^(3) - x^(2) + betax + y = 0` are `x_(1),x_(2),x_(3)` and the they are in A.P.As `x_(1),x_(2),x_(3)` are in A.P., let `x_(1) = a-d, x_(2) = a, x_(3) = a + d`. Now,
`x_(1) + x_(2) + x_(3) = -(-1)/(1) = 1`
`rArr a-d + a+d = 1`
` rArr a= (1)/(3)" "(1)`
`x_(1),x_(2)+x_(x_(2)x_(3) + x_(3)+x_(1) = (beta)/(1) = beta`
`rArr (a-d)a+a(a+d) + (a+d) (a-d) = beta" "(2)`
`x_(1)x_(2)x_(3)= -(gamma)/(1) = gamma`
`rArr (a-d)a(a+d) = - gamma`
Form (1) and (2), we get
`3a^(2) -d^(2) = beta `
`rArr 3 (1)/(9) - d^(2) = beta , so beta = (1)/(3) - d^(2) lt (1)/(3)`
From (1) and (3) , we get
`(1)/(3) ((1)/(9) - d^(2)) = - gamma`
`rArr gamma = (1)/(3) ( d^(2)-(1)/(9)) gt (1)/(3) (-(1)/(9)) = - (1)/(27)`
` gamma in (-(1)/(27), + infty)`