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Q.

If ax2+(b−c)x+a−b−c=0 has unequal real roots for all c∈R, then

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a

b < 0 < a

b

a < 0 < b

c

b < a < 0

d

b > a > 0

answer is C.

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Detailed Solution

We have, D=(b−c)2−4a(a−b−c)>0 or  b2+c2−2bc−4a2+4ab+4ac>0 or  c2+(4a−2b)c−4a2+4ab+b2>0 for all c∈RDiscriminant of the above expression in c must be negative.Hence,    (4a−2b)2−4−4a2+4ab+b2<0 or     4a2−4ab+b2+4a2−4ab−b2<0 or     a(a−b)<0⇒     a<0 and a−b>0 or a>0 and a−b<0⇒     ba>0
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