For $a, b \in R$, the equation $x^2+(a-b) x+(1-a-b)=0$ has unequal roots for all values of b . Then

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Professor PC Thomas Classes · Answer

Discriminant $\Delta = (a-b)^2 - 4(1-a-b) > 0$ for all $b$. Quadratic in $b$: $b^2 - 2(a-2)b + (a^2+4a-4) > 0 \implies D' < 0 \implies a > 1$.