Domain of $$\sin^{-1}x$$

Step 1: Find the Domain of $$f(x)$$ (Set $$S_1$$)

For the function $$\sin^{-1}(y)$$ to be well-defined, its inner argument must lie strictly within the closed interval $$[-1, 1]$$:

$$-1 \le \frac{x^2 - 3x + 2}{x^2 + 1} \le 1$$

Since the denominator $$x^2 + 1$$ is strictly positive for all real numbers $$x$$, we can safely multiply the entire inequality by $$(x^2 + 1)$$ without flipping the inequality signs:

$$-(x^2 + 1) \le x^2 - 3x + 2 \le x^2 + 1$$

This breaks down into two simple inequalities:

1. Left-hand side:

   $$-x^2 - 1 \le x^2 - 3x + 2 \implies 2x^2 - 3x + 3 \ge 0$$

   Let's check the discriminant ($$D = b^2 - 4ac$$) of this quadratic equation:

   $$D = (-3)^2 - 4(2)(3) = 9 - 24 = -15 < 0$$

   Since the leading coefficient is positive ($$2 > 0$$) and $$D < 0$$, this quadratic expression is always positive for all real numbers.

   Thus, $$x \in \mathbb{R}$$.

2. Right-hand side:

   $$x^2 - 3x + 2 \le x^2 + 1 \implies -3x + 2 \le 1$$  $$-3x \le -1 \implies x \ge \frac{1}{2}$$

Taking the intersection of both parts, we get the domain $$S_1$$:

$$S_1 = \left[ \frac{1}{2}, \infty \right)$$

Step 2: Find the Domain of $$g(x)$$ (Set $$S_2$$)

Similarly, for $$\cos^{-1}(y)$$, the argument must satisfy:

$$-1 \le \frac{2|x| - 3}{5} \le 1$$

Multiply the entire inequality by $$5$$:

$$-5 \le 2|x| - 3 \le 5$$

Add $$3$$ to all parts:

$$-2 \le 2|x| \le 8$$

Divide by $$2$$:

$$-1 \le |x| \le 4$$

Since the absolute value $$|x|$$ is always greater than or equal to $$0$$, the left bound $$-1 \le |x|$$ is trivially true for all real numbers. We only need to solve the right bound:

$$|x| \le 4 \implies -4 \le x \le 4$$

Thus, the domain $$S_2$$ is:

$$S_2 = [-4, 4]$$

Step 3: Find the Intersection Set $$S_1 \cap S_2$$

Now, we find the common region shared by both domains on the number line:

$$S_1 \cap S_2 = \left[ \frac{1}{2}, \infty \right) \cap [-4, 4] = \left[ \frac{1}{2}, 4 \right]$$

Step 4: Count the Integer Values ($$N$$)

We simply list the whole integers that lie securely within the interval $$\left[\frac{1}{2}, 4\right]$$:

$$\text{Integers} = \{1, 2, 3, 4\}$$

Counting them up, we find exactly $$4$$ integer values.

Therefore, $$N = 4$$.

Correct Option: B