Let the set of all positive values of lambda, | vedclass

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(C) First,solve the inequality $\frac{x^2+x+2}{x^2+5x+6} < 0$.Since the numerator $x^2+x+2$ has a discriminant $D = 1^2 - 4(1)(2) = -7 < 0$,it is alwa

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$39$

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(C) First,solve the inequality $\frac{x^2+x+2}{x^2+5x+6} Since the numerator $x^2+x+2$ has a discriminant $D = 1^2 - 4(1)(2) = -7 Thus,the inequality reduces to $\frac{1}{(x+2)(x+3)} This gives $x \in (-3, -2)$.Next,consider the function $f(x) = 1 + x\lambda^2 - x^3$.Find the derivative: $f'(x) = \lambda^2 - 3x^2$.Set $f'(x) = 0$ to find critical points: $3x^2 = \lambda^2 \Rightarrow x = \pm \frac{\lambda}{\sqrt{3}}$.Using the second derivative test: $f''(x) = -6x$.For local minimum,$f''(x) > 0$,so $-6x > 0 \Rightarrow x Thus,the point of local minimum is $x = -\frac{\lambda}{\sqrt{3}}$.Given that this point must satisfy $x \in (-3, -2)$,we have:$-3 Multiplying by $-1$ reverses the inequality:$2 $2\sqrt{3} So,$\alpha = 2\sqrt{3}$ and $\beta = 3\sqrt{3}$.Then $\alpha^2 + \beta^2 = (2\sqrt{3})^2 + (3\sqrt{3})^2 = 12 + 27 = 39$.

Let the set of all positive values of $\lambda$,for which the point of local minimum of the function $f(x) = 1 + x(\lambda^2 - x^2)$ satisfies $\frac{x^2+x+2}{x^2+5x+6} < 0$,be $(\alpha, \beta)$. Then $\alpha^2 + \beta^2$ is equal to:

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