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The range of real number alpha for which...

The range of real number `alpha` for which the equation `z + alpha|z-1| + 2i = 0` has a solution is :

A

`[-(sqrt(5))/(2),(sqrt(5))/(2)]`

B

`[-(sqrt(3))/(2),(sqrt(3))/(2)]`

C

`[0, (sqrt(5))/(2)]`

D

`[-oo, (-sqrt(5))/(2)]uu [(sqrt(5))/(2), oo]`

Text Solution

Verified by Experts

The correct Answer is:
A
Let `z = x + iy`
We have `z + alpha |z-1|+2i=0`
`x+iy+alpha sqrt((x-1)^(2)+y^(2))+2i=0`
Equating real and imaginary parts.
`y+2=0` and `x + alpha sqrt((x-1)^(2)+y^(2))=0`
`because " " x^(2)=alpha^(2)(x^(2)-2x+5) " " (because y = -2)`
or `(1-alpha^(2))x^(2)+2alpha^(2)x-5alpha^(2)=0`
Since x is real
`therefore D = B^(2)-4AC ge 4alpha^(4)+20 alpha^(2)(1-alpha^(2))ge 0`
`rArr -4alpha^()4)+5alpha^(2)ge 0`
`rArr 4alpha^(2)(alpha^(2)-(5)/(4))le 0`
`rArr - (sqrt(5))/(2)le alpha le (sqrt(5))/(2)`
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