If $$\alpha = \lim_{x \to \pi/4} \frac{\tan^3 x - \tan x}{\cos x + \frac{\pi}{4}}$$ and $$\beta = \lim_{x \to 0} \cos x^{\cot x}$$ are the roots of the equation, $$ax^2 + bx - 4 = 0$$, then the ordered pair $$a, b$$ is:

The limit is given by:

$$\alpha = \lim_{x \to \pi/4} \frac{\tan^3 x - \tan x}{\cos(x + \pi/4)}$$

Substituting $$x = \pi/4$$ gives the indeterminate form $$\frac{1-1}{0} = \frac{0}{0}$$. We can apply L'Hôpital's Rule:

$$\alpha = \lim_{x \to \pi/4} \frac{3\tan^2 x \sec^2 x - \sec^2 x}{-\sin(x + \pi/4)}$$

Now, substitute $$x = \pi/4$$:

$$\alpha = \frac{3(1)^2(\sqrt{2})^2 - (\sqrt{2})^2}{-\sin(\pi/4 + \pi/4)} = \frac{3(2) - 2}{-\sin(\pi/2)} = \frac{6 - 2}{-1} = -4$$

The limit is given by:

$$\beta = \lim_{x \to 0} (\cos x)^{\cot x}$$

This is of the form $$1^\infty$$. We use the formula $$\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[f(x) - 1]}$$:

$$\beta = e^{\lim_{x \to 0} \cot x (\cos x - 1)} = e^{\lim_{x \to 0} \frac{\cos x - 1}{\tan x}}$$

Applying L'Hôpital's Rule to the exponent:

$$\lim_{x \to 0} \frac{-\sin x}{\sec^2 x} = \frac{0}{1} = 0$$

Thus, $$\beta = e^0 = 1$$.

The roots of the equation $$ax^2 + bx - 4 = 0$$ are $$\alpha = -4$$ and $$\beta = 1$$.

Using the properties of roots:

• Product of roots: $$\alpha \beta = \frac{-4}{a}$$ $$(-4)(1) = \frac{-4}{a} \implies -4 = \frac{-4}{a} \implies a = 1$$
• Sum of roots: $$\alpha + \beta = -\frac{b}{a}$$ $$-4 + 1 = -\frac{b}{1} \implies -3 = -b \implies b = 3$$

The values are $$a = 1$$ and $$b = 3$$. The ordered pair $$(a, b)$$ is (1, 3).

Correct Option: C