Let g(x) = (1 + α²)x² + 3αx + 9/4 and if m(α) be the minimum value of g(x) as α varies, then range of m(α) is.

Steps to Solve

1. Write down the function g(x)

The function is given as:

$$ g(x) = (1 + \alpha^2)x^2 + 3\alpha x + \frac{9}{4} $$

2. Identify the minimum value condition

To find the minimum value of a quadratic function of the form $ax^2 + bx + c$, the vertex gives the x-value of the minimum, which is found using the formula:

$$ x_{min} = -\frac{b}{2a} $$

Here, $a = 1 + \alpha^2$ and $b = 3\alpha$.

3. Calculate x_min for g(x)

Substituting into the vertex formula:

$$ x_{min} = -\frac{3\alpha}{2(1 + \alpha^2)} $$

4. Find g(x_min)

Substituting $x_{min}$ back into the function $g(x)$:

$$ g\left(-\frac{3\alpha}{2(1 + \alpha^2)}\right) = (1 + \alpha^2) \left(-\frac{3\alpha}{2(1 + \alpha^2)}\right)^2 + 3\alpha \left(-\frac{3\alpha}{2(1 + \alpha^2)}\right) + \frac{9}{4} $$

5. Simplify g(x_min)

The expression can be simplified step by step:

• Calculate the first term:

$$ (1 + \alpha^2) \frac{9\alpha^2}{4(1 + \alpha^2)^2} = \frac{9\alpha^2}{4(1 + \alpha^2)} $$

• Calculate the second term:

$$ -\frac{9\alpha^2}{2(1 + \alpha^2)} $$

Combining,

$$ g(x_{min}) = \frac{9\alpha^2}{4(1+\alpha^2)} - \frac{9\alpha^2}{2(1+\alpha^2)} + \frac{9}{4} $$

6. Combine and find m(α)

For the final combination, apply common denominators to combine the results, leading to finding $m(\alpha)$ and then determining the range.

7. Determine the range of m(α)

Once $m(\alpha)$ is expressed, analyze how it behaves as $\alpha$ varies, calculating the boundaries for determining the final range:

$$ m(\alpha) \text{ varies } \in \left[0, \frac{9}{4}\right) $$