JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 28)
Let the set of all positive values of $$\lambda$$, for which the point of local minimum of the function $$(1+x(\lambda^2-x^2))$$ satisfies $$\frac{x^2+x+2}{x^2+5 x+6}<0$$, be $$(\alpha, \beta)$$. Then $$\alpha^2+\beta^2$$ is equal to _________.
Answer
39
Explanation
_9th_April_Morning_Shift_en_28_1.png)
$$\begin{aligned} & f(x)=1+x\left(\lambda^2-x^2\right) \\ & f(x)=-x^3+\left(\lambda^2 x+1\right) \\ & f^{\prime}(x)=-3 x^2+\lambda^2 \\ & x= \pm \frac{\lambda}{\sqrt{3}} \end{aligned}$$
$$-\frac{\lambda}{\sqrt{3}}$$ should satisfy the given condition
$$\begin{aligned} & \frac{x^2+x+2}{x^2+5 x+6}<6 \\ & \frac{1}{(x+2)(x+3)}<0 \end{aligned}$$
_9th_April_Morning_Shift_en_28_2.png)
$$\begin{aligned} &\begin{aligned} & x \in(-3,-2) \\ & -3<-\frac{\lambda}{\sqrt{3}}<-2 \\ & -3 \sqrt{3}<-\lambda<-2 \sqrt{3} \\ & 2 \sqrt{3}<\lambda<3 \sqrt{3} \\ & \left.\begin{array}{l} \alpha=2 \sqrt{3} \\ \beta=2 \sqrt{3} \end{array}\right\} \\ & (2 \sqrt{3})^2+(3 \sqrt{2})^2 \\ & 12+27 \\ \end{aligned}\\ &39 \end{aligned}$$
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