Key Concepts and Formulas

• Absolute Value Inequalities: $|x| < a \iff -a < x < a$ and $|x| > a \iff x < -a \text{ or } x > a$.
• Solving Rational Inequalities: Analyze the signs of the numerator and denominator separately. A fraction is non-negative if both numerator and denominator are positive or both are negative. Remember to exclude values where the denominator is zero.
• Quadratic Formula: The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

1. Analyzing Set T: $x^{2}-7|x|+9 \leq 0$

• Explanation: We want to find all integers $x$ that satisfy the given inequality. Since $x^2 = |x|^2$, we can substitute $y = |x|$ to simplify the inequality.
• Step 1.1: Substitute $|x|$ Let $y = |x|$. The inequality becomes $y^2 - 7y + 9 \leq 0$.
• Step 1.2: Find roots of the quadratic equation We find the roots of $y^2 - 7y + 9 = 0$ using the quadratic formula: $y = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(9)}}{2(1)}$ $y = \frac{7 \pm \sqrt{49 - 36}}{2}$ $y = \frac{7 \pm \sqrt{13}}{2}$ The roots are $y_1 = \frac{7 - \sqrt{13}}{2}$ and $y_2 = \frac{7 + \sqrt{13}}{2}$.
• Step 1.3: Determine the interval for $y$ Since the quadratic $y^2 - 7y + 9$ has a positive leading coefficient (1), its parabola opens upwards. For $y^2 - 7y + 9 \leq 0$, $y$ must lie between the roots: $\frac{7 - \sqrt{13}}{2} \leq y \leq \frac{7 + \sqrt{13}}{2}$
• Step 1.4: Substitute back $y = |x|$ and approximate values We know that $3 < \sqrt{13} < 4$, so $3.5 < \sqrt{13} < 3.7$ (approximately $\sqrt{13} \approx 3.6$). So, the interval becomes: $\frac{7 - 3.6}{2} \leq |x| \leq \frac{7 + 3.6}{2}$ $\frac{3.4}{2} \leq |x| \leq \frac{10.6}{2}$ $1.7 \leq |x| \leq 5.3$
• Step 1.5: Identify integer values for $x$ Since $x \in \mathbb{Z}$ (integers), $|x|$ must also be an integer. The possible integer values for $|x|$ in the interval $[1.7, 5.3]$ are $2, 3, 4, 5$. If $|x|=2 \Rightarrow x = \pm 2$ If $|x|=3 \Rightarrow x = \pm 3$ If $|x|=4 \Rightarrow x = \pm 4$ If $|x|=5 \Rightarrow x = \pm 5$ Therefore, the set $T$ is $T = \{-5, -4, -3, -2, 2, 3, 4, 5\}$.

2. Analyzing Set S: $\frac{|x+3|-1}{|x|-2} \geq 0$

• Explanation: We need to find values of $x$ for which the given expression is non-negative, considering the specified domain $x \in [-6,3]$ and excluding $x=-2, 2$. A fraction is non-negative if both numerator and denominator are positive, or both are negative. The denominator cannot be zero.

• Step 2.1: Analyze the Numerator: $|x+3|-1$ $|x+3|-1 \geq 0 \Rightarrow |x+3| \geq 1$. This means $x+3 \geq 1$ or $x+3 \leq -1$. $x \geq -2$ or $x \leq -4$. So, the numerator is non-negative for $x \in (-\infty, -4] \cup [-2, \infty)$.

• Step 2.2: Analyze the Denominator: $|x|-2$ $|x|-2 > 0 \Rightarrow |x| > 2$. (Denominator cannot be zero) This means $x > 2$ or $x < -2$. So, the denominator is positive for $x \in (-\infty, -2) \cup (2, \infty)$.

• Step 2.3: Combine conditions for $\frac{\text{Numerator}}{\text{Denominator}} \geq 0$ We need (Numerator $\geq 0$ AND Denominator $> 0$) OR (Numerator $\leq 0$ AND Denominator $< 0$).

  Case A: Numerator $\geq 0$ AND Denominator $> 0$ $(x \in (-\infty, -4] \cup [-2, \infty))$ AND $(x \in (-\infty, -2) \cup (2, \infty))$ Let's find the intersection:

  1. Intersection of $(-\infty, -4]$ and $((-\infty, -2) \cup (2, \infty))$ is $(-\infty, -4]$.
  2. Intersection of $[-2, \infty)$ and $((-\infty, -2) \cup (2, \infty))$ is $(2, \infty)$. So, Case A yields $x \in (-\infty, -4] \cup (2, \infty)$.

  Case B: Numerator $\leq 0$ AND Denominator $< 0$ Numerator $\leq 0 \Rightarrow |x+3| \leq 1 \Rightarrow -1 \leq x+3 \leq 1 \Rightarrow -4 \leq x \leq -2$ Denominator $< 0 \Rightarrow |x| < 2 \Rightarrow -2 < x < 2$ The intersection of these two intervals is $(-2 < x < 2) \cap [-4, -2] = \emptyset$

  Combining Case A and Case B, the solution to the inequality $\frac{|x+3|-1}{|x|-2} \geq 0$ is $x \in (-\infty, -4] \cup (2, \infty)$.

• Step 2.4: Apply Domain Restrictions for S The set $S$ is defined for $x \in [-6, 3]$ and $x \neq -2, 2$. Intersection of $( (-\infty, -4] \cup (2, \infty) )$ with $[-6, 3]$:

  1. Intersection of $(-\infty, -4]$ and $[-6, 3]$ is $[-6, -4]$.
  2. Intersection of $(2, \infty)$ and $[-6, 3]$ is $(2, 3]$. So, before excluding points, we have $x \in [-6, -4] \cup (2, 3]$. Now, we apply the exclusion $x \neq -2, 2$. Neither $-2$ nor $2$ are present in $[-6, -4] \cup (2, 3]$. Therefore, the set $S = [-6, -4] \cup (2, 3]$.

3. Finding the Intersection $S \cap T$

• Explanation: We need to find the elements that are common to both set $S$ and set $T$. Since $T$ contains only integers, we will check which of the integers in $T$ fall within the intervals defined by $S$.

• Step 3.1: List elements of T and intervals of S $T = \{-5, -4, -3, -2, 2, 3, 4, 5\}$ $S = [-6, -4] \cup (2, 3]$

• Step 3.2: Check each element of T against S

  • Is $-5 \in S$? Yes, $-5 \in [-6, -4]$.
  • Is $-4 \in S$? Yes, $-4 \in [-6, -4]$.
  • Is $-3 \in S$? No, $-3$ is not in $[-6, -4]$ and not in $(2, 3]$.
  • Is $-2 \in S$? No, $-2$ is explicitly excluded from $S$.
  • Is $2 \in S$? No, $2$ is explicitly excluded from $S$.
  • Is $3 \in S$? Yes, $3 \in (2, 3]$.
  • Is $4 \in S$? No, $4$ is not in $[-6, -4]$ and not in $(2, 3]$.
  • Is $5 \in S$? No, $5$ is not in $[-6, -4]$ and not in $(2, 3]$.

• Step 3.3: Form the intersection set The elements common to both $S$ and $T$ are $\{-5, -4, 3\}$. So, $S \cap T = \{-5, -4, 3\}$.

• Step 3.4: Count the number of elements The number of elements in $S \cap T$ is 3.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when solving inequalities, especially when dealing with absolute values.
• Denominator Restrictions: Always remember that the denominator of a fraction cannot be zero. Exclude such values from the solution set.
• Domain Restrictions: Pay close attention to the domain specified for each set (e.g., $x \in [-6, 3]$, $x \in \mathbb{Z}$, exclusions like $x \neq -2, 2$). These restrictions are critical for the final solution.

Summary

To solve this problem, we first analyzed set $T$ by substituting $y=|x|$ and solving the resulting quadratic inequality. This gave us the integer values in $T$. Next, we analyzed set $S$ by considering the cases where the numerator and denominator of the given fraction were both positive or both negative, while respecting the given domain. Finally, we found the intersection of the two sets, $S \cap T$, which contains the elements $\{-5, -4, 3\}$. Therefore, the number of elements in $S \cap T$ is 3.

Final Answer

The final answer is \boxed{3}, which corresponds to option (D).