If $\forall \lambda \in R$, equation $ax^{2}+\left ( b-\lambda \right )x+\left ( a-b-\lambda \right )=0,a\neq 0,a,b\in R$ has real roots, then which of the following may be true?

For the equation to have real roots for all real lambda, the discriminant must be non-negative, giving (b - lambda)^2 - 4a(a - b - lambda) >= 0. This simplifies to lambda^2 + (4a - 2b)lambda + (b^2 - 4a^2 + 4ab) >= 0. For this quadratic in lambda to be non-negative for all lambda, its discriminant must be less than or equal to 0, so (4a - 2b)^2 - 4(b^2 - 4a^2 + 4ab) <= 0. This yields 16a(a - b) <= 0, meaning a and (a - b) have opposite signs or a = 0, but since a cannot be zero, we get a(a - b) < 0. If a > 0, then a < b, and if a < 0, then b <= a < 0, making the latter a valid condition.

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