Question

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

Answer 1

Step 1: Solve $\frac{x^2 + x + 2}{x^2 + 5x + 6} < 0$

Factorize the numerator and denominator:

$\frac{x^2 + x + 2}{x^2 + 5x + 6} = \frac{x^2 + x + 2}{(x + 2)(x + 3)}.$

Analyze the sign of the inequality $\frac{x^2 + x + 2}{(x + 2)(x + 3)} < 0$ using the critical points:

• The critical points are $x = -3$, $x = -2$, and the roots of $x^2 + x + 2 = 0$.
• Solve $x^2 + x + 2 = 0$ using the discriminant:
• The roots are complex, so $x^2 + x + 2 > 0$ for all $x \in \mathbb{R}$.

Thus, the inequality reduces to:

\frac{1}{(x + 2)(x + 3)} < 0\.

From the critical points $x = -3$ and $x = -2$, analyze the intervals:

• For $x \in (-\infty, -3)$, the product $(x + 2)(x + 3) > 0$.
• For $x \in (-3, -2)$, the product $(x + 2)(x + 3) < 0$.
• For $x \in (-2, \infty)$, the product $(x + 2)(x + 3) > 0$.

Therefore, the solution to $\frac{1}{(x+2)(x+3)} < 0$ is:

$x \in (-3, -2).$ $\hspace{20pt}(1)$

Step 2: Find the local minima of $f(x)$

The function is given as:

$f(x) = 1 + x(\lambda^2 - x^2).$

Find $f'(x)$:

$f'(x) = (\lambda^2 - x^2) + (-2x)x.$ $f'(x) = \lambda^2 - x^2 - 2x^2 = \lambda^2 - 3x^2.$

Set $f'(x) = 0$ to find the critical points:

$\lambda^2 - 3x^2 = 0.$ $x^2 = \frac{\lambda^2}{3}.$ $x = \pm \frac{\lambda}{\sqrt{3}}.$

From the critical points $x = \pm \frac{\lambda}{\sqrt{3}}$:

• At $x = \frac{\lambda}{\sqrt{3}}$, $f(x)$ achieves a local maximum.
• At $x = -\frac{\lambda}{\sqrt{3}}$, $f(x)$ achieves a local minimum.

Thus, the point of local minimum is:

$x = -\frac{\lambda}{\sqrt{3}}. \hspace{20pt}(2)$

Step 3: Condition for $x \in (-3, -2)$

From equation (1), $x \in (-3, -2)$. Substituting $x = -\frac{\lambda}{\sqrt{3}}$:

$-3 < -\frac{\lambda}{\sqrt{3}} < -2.$

Multiply through by $-1$ (reversing the inequality):

3 > \frac{\lambda}{\sqrt{3}} > 2\.

Multiply through by $\sqrt{3}$:

$3\sqrt{3} > \lambda > 2\sqrt{3}.$

Thus, the set of all positive $\lambda$ values is:

$\lambda \in (2\sqrt{3}, 3\sqrt{3}).$

Step 4: Compute $\alpha^2 + \beta^2$

From the interval $\lambda \in (2\sqrt{3}, 3\sqrt{3})$, we have:

• $\alpha = 2\sqrt{3}, \beta = 3\sqrt{3}.$

Compute $\alpha^2 + \beta^2$:

$\alpha^2 + \beta^2 = (2\sqrt{3})^2 + (3\sqrt{3})^2.$ $\alpha^2 + \beta^2 = 4(3) + 9(3) = 12 + 27 = 39.$

Final Answer is 39.