if the equation x^2+(a-b)x+(1-a-b)=0 has real and unequal roots for all the real values of b then

A) a<1 B) a>1 C) - 1<a<1 D) - infinity < a < +infinity

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Solution

Discriminant of above equation is – (a-b)² – 4(1-a-b) For unequal roots D>0 => a² + b² -2ab – 4 + 4a + 4b > 0 b² -2b(a -2) + a² -4 +4a > 0 now this is also quadratic in b and it is given this should hold for all b which means D for this equation should be <0 So that graph will always be above x axis and inequality will always be true (-2(a-2))² – 4(a² -4 +4a) <0 a² + 4 -4a -a² +4 -4a < 0 -8a +8<0 a>1

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