If `z^3+3+2i(z+(-1+i a)=0` has on ereal roots, then the value of `a` lies in the interval `(a in R)` `(-2,1)` b. `(-1,0)` c. `(0,1)` d. `(-2,3)`

Options

A. `(2,-1)`
B. `(-1,0)`
C. `(0,1)`
D. `(-2,3)`

Answer

A, B, D

Explanation

Let `z = alpha` be real root. Then,
`alpha^(3) + (3+ 2i)alpha+(-1+ia) = 0`
` rArr (alpha^(3) + 3alpha-1) + i(a+ 2alpha) = 0`
` rArr alpha^(3) + 3alpha-1 = 0 and alpha = - a//2`
` rArr (a^(3))/(8) + (3alpha)/(2) - 1 =0`
`rArr a^(3) + 12a + 8 = 0`
Let `f(a) = a^(3) + 12a + 8`
` f'(a) = 3a^(2) + 12 gt 0`
So, f(a) cuts x-axis only once.
Now `f(-1) gt 0,f(0)gt 0`
`f(-2) gt0,f(1) gt 0 and f(3) gt 0`
Hence, `a in (-1,0) or a in (-2,1) or a in (-2,3)`