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
b>a>0
answer is C.
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Detailed Solution
We have, D=(b−c)2−4a(a−b−c)>0⇒ b2+c2−2bc−4a2+4ab+4ac>0⇒ c2+(4a−2b)c−4a2+4ab+b2>0,∀c∈RSince, c∈R, so we have (4a−2b)2−4−4a2+4ab+b2<0⇒ 4a2−4ab+b2+4a2−4ab−b2<0⇒ a(a−b)<0If a>0, then a−b<0i.e. 0a>0If a<0, then a−b>0i.e. 0>a>bor b