Question 13

Explanation

Given functions $f(x) = \frac{x-a}{x+a}$ for $x \neq -a$ and $g(x) = bx^2 - x + 2$.

The question asks whether $\frac{g(-2)}{f(-2)} > 0$.

Calculate Function Values

First, calculate the values of $g(-2)$ and $f(-2)$.

For $g(-2)$, substitute $x = -2$ into function $g(x)$:

For $f(-2)$, substitute $x = -2$ into function $f(x)$:

Therefore, the ratio we are looking for is:

We can simplify by factoring the numerator:

For $\frac{g(-2)}{f(-2)} > 0$, we need $\frac{4(b + 1)(2-a)}{2+a} > 0$.

Since $4$ is always positive, we need to analyze the sign of $\frac{(b + 1)(2-a)}{2+a}$.

Statement Analysis

Analysis of Statement $(1)$: $a < 5$ and $b < 5$

From this statement, we only know that both $a$ and $b$ are less than $5$. However, there is no definite information about the signs of the factors involved.

Let's check some possibilities:

1. If $b = 0$, then $b + 1 = 1 > 0$
2. If $b = -2$, then $b + 1 = -1 < 0$
3. For $2-a$, if $a = 1$, then $2-a = 1 > 0$
4. For $2-a$, if $a = 3$, then $2-a = -1 < 0$
5. For $2+a$, if $a = 0$, then $2+a = 2 > 0$
6. For $2+a$, if $a = -3$, then $2+a = -1 < 0$

Since the signs of these factors can vary with many possible combinations, we cannot determine whether $\frac{g(-2)}{f(-2)} > 0$.

Therefore, statement $(1)$ ALONE is not sufficient.

Analysis of Statement $(2)$: $a > 2$ and $b > 0$

From this statement, let's analyze the sign of each factor:

1. Since $b > 0$, then $b + 1 > 1 > 0$ (positive)
2. Since $a > 2$, then $2 - a < 0$ (negative)
3. Since $a > 2$, then $2 + a > 4 > 0$ (positive)

Thus, we have:

Since the result is negative, $\frac{g(-2)}{f(-2)} < 0$, so the answer is no, $\frac{g(-2)}{f(-2)}$ is not greater than $0$.

Therefore, statement $(2)$ ALONE is sufficient to answer the question definitively (the answer is no).