Let the set of all positive values of λ, for which the point of local minimum of the function (1+x (λ^2-x^2 ) ) satisfies x^2+x+2 x^2+5 x+6 < 0, be…

$\begin{aligned} & \frac{x^2+x+2}{x^2+5 x+6} < 0 \\ & \Rightarrow \frac{1}{(x+2)(x+3)} < 0\end{aligned}$​x2+5x+6x2+x+2​<0⇒(x+2)(x+3)1​<0​

$\begin{aligned}
& x \in(-3,-2) \ldots \ldots \ldots . .(1) \
& f(x)=1+x\left(\lambda^2-x^2\right)
\end{aligned}$

Finding local minima $f^{\prime}(x)=\left(\lambda^2-x^2\right)+(-2 x) \cdot x$f′(x)=(λ2−x2)+(−2x)⋅x Put $\mathrm{f}^{\prime}(\mathrm{x})=0$f′(x)=0 $\begin{aligned}
& \Rightarrow \lambda^2=3 x^2 \
& \Rightarrow x= \pm \frac{\lambda}{\sqrt{3}}
\end{aligned} We want local min\Rightarrow \mathrm{x}=\frac{-\lambda}{\sqrt{3}}$from (1)$from(1)x \in(-3,-2)$$\begin{aligned} & -3 < \frac{-\lambda}{\sqrt{3}} < -2 \ & 3 \sqrt{3}>\lambda>2 \sqrt{3} \ & \alpha=2 \sqrt{3}, \beta=3 \sqrt{3} \ & \alpha^2+\beta^2=12+27=39\end{aligned}$