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 B (−1,0)Let z=α is the real root of the equation α3+(3+2i)α+(−1+ia)=0 ⇒(α3+3α−1)+(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=0⇒α=−a2 so using value of α we get, −a38−3a2−1=0⇒a3+12a+8=0 Now, Let f(x)=a3+12a+8 f′(a)=3a2+12>0 ∀ a∈R ∴f is an increasing function ∀ x∈R 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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