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If a x^2+(b-c)x+a-b-c=0 has unequal real...

If `a x^2+(b-c)x+a-b-c=0` has unequal real roots for all `c` in `R` , then

A

`blt0lta`

B

`altoltb`

C

`bltalt0`

D

`bgtagt0`

Text Solution

Verified by Experts

The correct Answer is:
C, D
We have `D=(b=c)^(2)-4a(a-b-c)gt0`
`impliesb^(2)+c^(2)-2bc-4a^(2)+4ab+4acgt0`
`impliesc^(2)+(4a-2b)c-4a^(2)+4ab+b^(2)gt0,AAc epsilonR`
Since `c epsionR` so we have
`(4a-2b)^(2)-4(-4a^(2)+4ab+b^(2))lt0`
`implies4a^(2)-4ab+b^(2)+4a^(2)-4ab-b^(2)lt0`
`impliesa(a-b)lt0`
If `agt0`, then `a-blt0`
i.e. `0ltaltb`
or `bgtagt0`
If `alt0`, then `a-bgt0`
i.e. `0gtagtb`
or `bltalt0`
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