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(A) The given equation is $(3 + \alpha) x^2 + (6 + 2 \alpha) x + (5 + 2 \alpha) = 0$.For the roots to be complex,the discriminant $D$ must be less tha

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$-2$

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(A) The given equation is $(3 + \alpha) x^2 + (6 + 2 \alpha) x + (5 + 2 \alpha) = 0$.For the roots to be complex,the discriminant $D$ must be less than $0$.$D = (2 \alpha + 6)^2 - 4 (\alpha + 3) (2 \alpha + 5) $4(\alpha + 3)^2 - 4(\alpha + 3)(2 \alpha + 5) Dividing by $4$: $(\alpha + 3)(\alpha + 3 - 2 \alpha - 5) $(\alpha + 3)(-\alpha - 2)  0$.This implies $\alpha  -2$.Comparing with $\alpha > L$ or $\alpha The condition $L y^2 + M y + r For this to hold for all $y \in R$,the coefficient of $y^2$ must be negative (which is $-2 $D = (-3)^2 - 4(-2)(r) $9 + 8 r The greatest integer $[r]$ not exceeding $r$ is $[-1.125] = -2$.

Let $[r]$ denote the greatest integer not exceeding $r$. The roots of the equation $3 x^2 + 6 x + 5 + \alpha (x^2 + 2 x + 2) = 0$ are complex numbers whenever $\alpha > L$ or $\alpha < M$. If $(L - M)$ is minimum,then the greatest value of $[r]$ such that $L y^2 + M y + r < 0$ for all $y \in R$ is:

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