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Question

If α,β are the roots of the equation x2+αx+β=0 such that αβ and ||xβ|α|<1, then

A
inequality is satisfied by exactly three integral values of x
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B
inequality is satisfied by all values of xε(4,2)
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C
roots of the equation are opposite in sign
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D
All of the above
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Solution

The correct option is D All of the above
α+β=α and αβ=βα=1

β=2

(α,β)=(1,2)

||x+2|1|<1

1<|x+2|1<1

Let us inequality on left side first

1<|x+2|1

|x+2|>0x(,)

Right inequality ;

|x+2|1<1

|x+2|<2

2<x+2<2

x(4,0)

Combining both we get x(4,0)

(3,2,1) three integers satisfy the equation and all values between (4,0) satisfy the equation
α,β are of different sign

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