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Question

If z3 + (3 + 2i)z + (-1 + ia) = 0 has one real root , then value lies in interval of

A
(-z , -1)
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B
(-1, 0)
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C
(0, 1)
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D
(1, 2)
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Solution

The correct option is C (-1, 0)
z3+(3+2i)z+(1+ia)=0
Say z=α is the real root of the above equation.Hence substituting
α3+(3+2i)α+(1+2a)=0(α3+3α1)+i(a+2α)=0
equating we get
α3+3α1=0a+2α=0α=a2
Putting α=a2 in α3+3α1=0 we get
(a2)3+3(a2)1=0a383a21=0a3+12a+8=0
Consider the function f(a)=a3+12a+8
Now let's check for the sign of the function at points (2,1,0,1,3)
f(0)=8>0f(1)=112+8=5<0f(2)=824+8<0f(1)=1+12+8>0f(3)=27+36+8>0

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