If the roots of the equation, `x^3 + px^2+qx-1 = 0` form an increasing` G.P.` where` p` and `q `are real,then

Options

A. `p +q = 0`
B. `pin (-3, oo)`
C. one of the roots is untiy
D. one roots is smaller than 1 and one roots is greater than 1

Answer

1,3,4

Explanation

Let the roots be `a//r,a, ar ` where `a gt 0, r gt 1`. Now,
`a//r + a + ar = - p" "(1)`
`a(a//r) + a(ar) + a(ar) + (ar) (a//r) =q" "(2)`
`(a//r) (a) (ar) = q " "(3)`
`rArr a^(3) - 1`
`rArr a= 1`
Hence, (3) is correct,. From (1), putting a=1, we get
`-p= 3 gt 0 " "(because r +(1)/(r) gt 2)`
`rArr p lt - 3`
Hence, (2) is not correct. Also,
`1//r + 1 + r = - q" "(4)`
From (2), putting a= 1, we get
`1//r + r 1= q" "(5)`
`"^(-)p = q rArr p+q = 0`
Hence,(1) correct . Now, as `r gt 1`
`a//r = 1// r lt 1`
and `ar = r gt 1`
Hence, (4) is correct.