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Q.

The range of real number 'α'  for which the equation Zα|z1|+2i=0  has a solution

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a

[132,132]

b

[52,52]

c

(,52][52,)

d

[0,152]

answer is A.

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Detailed Solution

z=x+iyx+iy+2i=2|z1|=α|(x1)+iy|x+i(y+2)=α(x1)2+y2y+2=0x=α(x1)2+4x2=α2(x22x+5)x2(α21)2α2x+5α2=0xR,Δ0b24ac04α2(54α2)0

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