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Question

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

A
(2,1)
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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)
Let z=α is the real root of the equation
α3+(3+2i)α+(1+ia)=0
(α3+3α1)+(2α+a)i=0
As α is a real root of the equation, the real & the imaginary part of the equation must be zero.
a+2α=0α=a2
so using value of α we get, a383a21=0a3+12a+8=0

Now, Let f(x)=a3+12a+8
f(a)=3a2+12>0 aR
f is an increasing function xR
f(2)=(2)3+12(2)+8<0
f(1)=(1)3+12(1)+8<0
f(0)=(0)3+12(0)+8>0
f(1)=(1)3+12(1)+8>0
f(2)=(2)3+12(2)+8>0

a lies in the interval (1,0)

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