Let `A = {x:x in R " and " x^(2) -5x + 4 le 0}` and `B={x:x in R " and " x^(2)-12x + 45 gt 0}`
then which of the following is not true?

Options

A. `A cap B = A`
B. `A cup B = R`
C. `A cap B = phi`
D. `A sube B`

Answer

To solve the problem, we need to determine the sets A and B based on the given inequalities and then analyze the statements to find which one is not true.

### Step 1: Determine Set A
Set A is defined as:
\[ A = \{ x \in \mathbb{R} : x^2 - 5x + 4 \leq 0 \} \]

First, we need to solve the inequality \( x^2 - 5x + 4 \leq 0 \).

1. **Factor the quadratic expression:**
\[ x^2 - 5x + 4 = (x - 1)(x - 4) \]

2. **Set the expression to zero to find the critical points:**
\[ (x - 1)(x - 4) = 0 \implies x = 1 \quad \text{and} \quad x = 4 \]

3. **Determine the intervals:**
The critical points divide the number line into intervals: \( (-\infty, 1) \), \( (1, 4) \), and \( (4, \infty) \).

4. **Test each interval:**
- For \( x < 1 \) (e.g., \( x = 0 \)): \( (0 - 1)(0 - 4) = 4 > 0 \)
- For \( 1 < x < 4 \) (e.g., \( x = 2 \)): \( (2 - 1)(2 - 4) = -2 < 0 \)
- For \( x > 4 \) (e.g., \( x = 5 \)): \( (5 - 1)(5 - 4) = 4 > 0 \)

5. **Conclusion for Set A:**
The inequality \( (x - 1)(x - 4) \leq 0 \) holds for \( x \) in the interval:
\[ A = [1, 4] \]

### Step 2: Determine Set B
Set B is defined as:
\[ B = \{ x \in \mathbb{R} : x^2 - 12x + 45 > 0 \} \]

1. **Factor the quadratic expression:**
\[ x^2 - 12x + 45 = (x - 6)(x - 6) \]

2. **Set the expression to zero to find the critical points:**
\[ (x - 6)(x - 6) = 0 \implies x = 6 \]

3. **Determine the intervals:**
The critical point divides the number line into intervals: \( (-\infty, 6) \) and \( (6, \infty) \).

4. **Test each interval:**
- For \( x < 6 \) (e.g., \( x = 0 \)): \( (0 - 6)(0 - 6) = 36 > 0 \)
- For \( x > 6 \) (e.g., \( x = 7 \)): \( (7 - 6)(7 - 6) = 1 > 0 \)

5. **Conclusion for Set B:**
The inequality \( (x - 6)(x - 6) > 0 \) holds for:
\[ B = (-\infty, 6) \cup (6, \infty) \]

### Step 3: Analyze the Statements
Now we need to analyze the statements regarding sets A and B.

1. **Intersection \( A \cap B \):**
- \( A = [1, 4] \)
- \( B = (-\infty, 6) \cup (6, \infty) \)
- The intersection \( A \cap B = [1, 4] \) (since all elements in A are less than 6).

2. **Union \( A \cup B \):**
- The union \( A \cup B = (-\infty, 6) \cup (6, \infty) \) (since all elements in A are included in B).

3. **Check the statements:**
- **Statement 1:** \( A \cap B = A \) (True)
- **Statement 2:** \( A \cup B = \mathbb{R} \) (True)
- **Statement 3:** \( A \cap B = \emptyset \) (Not True, since \( A \cap B = [1, 4] \))
- **Statement 4:** \( A \subseteq B \) (True, since all elements of A are less than 6)

### Conclusion
The statement that is **not true** is:
\[ A \cap B = \emptyset \]