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
b
a<0
c
b
d
None of these
answer is C.
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Detailed Solution
We have discriminant D=(b−c)2−4a(a−b−c)>0⇒b2+c2−2bc−4a2+4ab+4ac>0⇒ c2+(4a−2b)c−4a2+4ab+b2>0 for all c∈R Discriminant of above expression in c must be negative. ⇒(4a−2b)2−4−4a2+4ab+b2<0⇒ 4a2−4ab+b2+4a2−4ab−b2<0⇒ a(a−b)<0⇒ a<0 and a−b>0 or a>0 and a−b<0⇒ba>0