Key Concepts and Formulas

1. Discriminant of a Quadratic Equation: For a quadratic equation $ax^2 + bx + c = 0$ ($a \neq 0$), the roots are real if and only if the discriminant $D = b^2 - 4ac \ge 0$.
2. Solving Quadratic Inequalities: To solve an inequality like $m^2 - 2m - 15 \ge 0$, find the roots of the corresponding equation ($m^2 - 2m - 15 = 0$). These roots partition the number line into intervals. Test a value from each interval or use the parabola's shape (upward if the leading coefficient is positive) to determine where the inequality holds.
3. Set Operations:
   • Intersection ($A \cap B$): Elements present in both A and B.
   • Union ($A \cup B$): All elements from A and B combined.
   • Set Difference ($A - B$): Elements present in A but not in B.
   • Set Difference ($B - A$): Elements present in B but not in A.

Step-by-Step Solution

The problem requires us to define set A based on the nature of roots of a quadratic equation and then evaluate given set operations with set B.

Step 1: Determine the set A

Set A contains all real numbers $m$ for which the roots of the quadratic equation $x^2 – (m + 1)x + m + 4 = 0$ are real. For the roots to be real, the discriminant $D$ must be non-negative ($D \ge 0$). The coefficients are $a = 1$, $b = -(m+1)$, and $c = m+4$. The discriminant is: $D = b^2 - 4ac = (-(m+1))^2 - 4(1)(m+4)$ $D = (m+1)^2 - 4(m+4)$ $D = (m^2 + 2m + 1) - (4m + 16)$ $D = m^2 + 2m + 1 - 4m - 16$ $D = m^2 - 2m - 15$ For real roots, we need $D \ge 0$: $m^2 - 2m - 15 \ge 0$ To solve this quadratic inequality, we find the roots of $m^2 - 2m - 15 = 0$. Factoring the quadratic, we get: $(m - 5)(m + 3) = 0$ The roots are $m = 5$ and $m = -3$. Since the coefficient of $m^2$ is positive (1), the parabola $y = m^2 - 2m - 15$ opens upwards. Thus, the inequality $m^2 - 2m - 15 \ge 0$ is satisfied when $m$ is outside or at the roots. So, $m \le -3$ or $m \ge 5$. Therefore, set A is given by the union of two intervals: $A = (-\infty, -3] \cup [5, \infty)$

Step 2: Identify set B

Set B is given as: $B = [-3, 5)$ This means $m$ is greater than or equal to -3 and strictly less than 5.

Step 3: Visualize sets A and B on a number line

It is helpful to visualize these sets on a number line to perform the set operations accurately. Set A: ---]-------[--- -3 5 Set B: ---[-------)--- -3 5

Step 4: Evaluate each option

We will now evaluate each of the given options using the sets A and B.

• Option (A): $A \cap B = \{-3\}$ The intersection $A \cap B$ consists of elements common to both A and B. Looking at the number line, the only point that belongs to both $A = (-\infty, -3] \cup [5, \infty)$ and $B = [-3, 5)$ is $m = -3$. Therefore, $A \cap B = \{-3\}$. This statement is TRUE.

• Option (B): $B - A = (-3, 5)$ The set difference $B - A$ consists of elements in B that are not in A. Set B is $[-3, 5)$. Set A contains numbers $\le -3$ and $\ge 5$. When we remove elements of A from B:

  • The element $-3$ is in B and also in A, so it is removed.
  • The elements in the interval $(-3, 5)$ are in B. Are they in A? No, because A only has numbers $\le -3$ or $\ge 5$. So, these elements remain. Thus, $B - A = (-3, 5)$. This statement is TRUE.

• Option (C): $A \cup B = \mathbb{R}$ The union $A \cup B$ consists of all elements that are in A or in B or in both. $A = (-\infty, -3] \cup [5, \infty)$ and $B = [-3, 5)$. Combining $(-\infty, -3]$ and $[-3, 5)$ gives $(-\infty, 5)$. Now, combining $(-\infty, 5)$ with the remaining part of A, which is $[5, \infty)$, covers all real numbers. $(-\infty, 5) \cup [5, \infty) = \mathbb{R}$. This statement is TRUE.

• Option (D): $A - B = (-\infty, -3) \cup (5, \infty)$ The set difference $A - B$ consists of elements in A that are not in B. Set A is $(-\infty, -3] \cup [5, \infty)$. Set B is $[-3, 5)$. Let's consider the parts of A:

  • For $(-\infty, -3]$: Elements in this interval that are not in B. The interval $(-\infty, -3)$ is not in B. The element $-3$ is in A, but it is also in B, so it's removed from A. This leaves $(-\infty, -3)$.
  • For $[5, \infty)$: Elements in this interval that are not in B. The interval $(5, \infty)$ is not in B. The element $5$ is in A. Is it in B? No, because B is $[-3, 5)$ which strictly excludes 5. So, $5$ is in A and not in B. This leaves $[5, \infty)$. Combining these, we get: $A - B = (-\infty, -3) \cup [5, \infty)$. The option states $A - B = (-\infty, -3) \cup (5, \infty)$. Our result includes the element $5$, while the option's result excludes $5$. Therefore, the statement in option (D) is NOT TRUE.

Common Mistakes & Tips

• Interval Endpoints: Be extremely careful with inclusive ([ or ]) and exclusive (( or )) endpoints when performing set operations. A single endpoint difference can make a statement false.
• Quadratic Inequality Signs: Correctly interpreting the inequality sign ($\ge$ or $\le$) and the direction of the parabola is crucial for defining set A.
• Set Difference Logic: For $A-B$, focus on elements that are exclusively in A. If an element is in both A and B, it must be removed from A.

Summary

We first determined set A by applying the discriminant condition for real roots of the given quadratic equation, resulting in $A = (-\infty, -3] \cup [5, \infty)$. Set B was given as $B = [-3, 5)$. By carefully analyzing the intersection, union, and differences between these two sets, we found that options (A), (B), and (C) are true statements, while option (D) is false because it incorrectly excludes the element 5 from the set difference $A - B$.

The final answer is $\boxed{D}$.