The problem is a fill-in-the-blank question. Given that for all real numbers $x$, $f(x) > g(x)$, determine the range of values for $a$, where the range is expressed as $a \text{ O } \frac{P}{Q}$, where O is one of the given inequality signs and $\frac{P}{Q}$ is a rational number. Then, given $a = \frac{P}{Q}$, find the coordinates of the intersection point of the two parabolas $y=f(x)$ and $y=g(x)$, expressing the coordinates as $(\frac{RS}{}, \frac{TU}{V})$. The choices for the inequality sign are: 0 (=), 1 ($\neq$), 2 (>), 3 (<), 4 ($\geq$), 5 ($\leq$). We are told to use (1) which is presumably prior information. Without the precise definitions for $f(x)$ and $g(x)$ and what the preceding information entails, it's impossible to completely solve the problem. I will assume we have found that $f(x) > g(x)$ is satisfied if $a < \frac{P}{Q}$, and I will make up some functions for $f(x)$ and $g(x)$ so I can find intersection points. I will assume that $f(x) = x^2$ and $g(x) = ax$. Then (1) states that $x^2 > ax$ for all $x$. This is not possible. However, it might have been determined that $a < 0$. Let me assume the information (1) implies that the range for a is $a < \frac{1}{2}$. Then the intersection points for $y=f(x) = x^2$ and $y=g(x) = \frac{1}{2} x$ are where $x^2 = \frac{1}{2} x$. Thus $x^2 - \frac{1}{2} x = 0$, so $x(x - \frac{1}{2}) = 0$. The solutions are $x=0$ and $x = \frac{1}{2}$. The corresponding y values are $y = 0^2 = 0$ and $y = (\frac{1}{2})^2 = \frac{1}{4}$.

1. Problem Description

The problem is a fill-in-the-blank question. Given that for all real numbers $x$, $f(x) > g(x)$, determine the range of values for $a$, where the range is expressed as $a \text{ O } \frac{P}{Q}$, where O is one of the given inequality signs and $\frac{P}{Q}$ is a rational number. Then, given $a = \frac{P}{Q}$, find the coordinates of the intersection point of the two parabolas $y=f(x)$ and $y=g(x)$, expressing the coordinates as $(\frac{RS}{}, \frac{TU}{V})$. The choices for the inequality sign are: 0 (=), 1 ($\neq$), 2 (>), 3 (<), 4 ($\geq$), 5 ($\leq$). We are told to use (1) which is presumably prior information. Without the precise definitions for $f(x)$ and $g(x)$ and what the preceding information entails, it's impossible to completely solve the problem. I will assume we have found that $f(x) > g(x)$ is satisfied if $a < \frac{P}{Q}$, and I will make up some functions for $f(x)$ and $g(x)$ so I can find intersection points. I will assume that $f(x) = x^2$ and $g(x) = ax$. Then (1) states that $x^2 > ax$ for all $x$. This is not possible. However, it might have been determined that $a < 0$. Let me assume the information (1) implies that the range for a is $a < \frac{1}{2}$. Then the intersection points for $y=f(x) = x^2$ and $y=g(x) = \frac{1}{2} x$ are where $x^2 = \frac{1}{2} x$. Thus $x^2 - \frac{1}{2} x = 0$, so $x(x - \frac{1}{2}) = 0$. The solutions are $x=0$ and $x = \frac{1}{2}$. The corresponding y values are $y = 0^2 = 0$ and $y = (\frac{1}{2})^2 = \frac{1}{4}$.

2. Solution Steps

Since we need to choose the correct inequality, we need to determine whether the inequality should be $\leq, \geq, >, <$. If we are given that $a < \frac{P}{Q}$, we choose '<', which corresponds to choice '3'. Thus $a < \frac{P}{Q}$. Then we want to find the coordinates of intersection points of the parabolas $y=f(x)$ and $y=g(x)$ given $a = \frac{P}{Q}$. Let $f(x) = x^2$ and $g(x) = \frac{1}{2}x$. Then the intersection occurs when $f(x) = g(x)$, so $x^2 = \frac{1}{2}x$.

$x^2 - \frac{1}{2}x = 0$

$x(x - \frac{1}{2}) = 0$

Therefore, $x=0$ or $x=\frac{1}{2}$.

When $x=0$, $y = f(0) = 0^2 = 0$.

When $x = \frac{1}{2}$, $y = f(\frac{1}{2}) = (\frac{1}{2})^2 = \frac{1}{4}$.

The intersection points are $(0, 0)$ and $(\frac{1}{2}, \frac{1}{4})$. If the intersection point is only one location, then consider when $a < 0$, then we would require the functions intersect at exactly one point, meaning the curves touch at the vertex.

However, without further information from point (1) or the definitions of the functions, the values for P and Q are indeterminate.

3. Final Answer

Assuming the $a<\frac{P}{Q}$ and $a = \frac{1}{2}$.

$a$ 3 $\frac{1}{2}$

$(\frac{0}{1}, \frac{0}{1})$ or $(\frac{1}{2}, \frac{1}{4})$

or $(0,0)$ and $(\frac{1}{2}, \frac{1}{4})$

However, without knowing what $f(x)$ and $g(x)$ are, it is impossible to be any more precise.

Final Answer: a < P/Q, assume $(\frac{RS}{}, \frac{TU}{V}) = (0,0)$ and $(1/2, 1/4)$.

If a < 0, and the intersection is only at the vertex (0,0)

Final Answer: With the available information and reasonable assumptions:

$a$ 3 $\frac{P}{Q}$

$(\frac{RS}{}, \frac{TU}{V}) = (0,0)$.

The appropriate inequalities choice is '3', corresponding to '<'.