Let A denotes the set of values of x for which `(x+2)/(x-4) le0` and B denotes the set of values of x for which `x^2-ax-4 le 0`. If B is the subset of A then a cannot take integral value (a) 0, (b) 1 (c) 2 (d) 3

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem step by step, we need to analyze the sets A and B defined by the inequalities given in the question.

### Step 1: Determine the set A

We start with the inequality:
\[ \frac{x + 2}{x - 4} \leq 0 \]

To find the values of \( x \) that satisfy this inequality, we need to identify the critical points where the expression is equal to zero or undefined.

1. **Find the zeros of the numerator**:
\[ x + 2 = 0 \implies x = -2 \]

2. **Find the zeros of the denominator**:
\[ x - 4 = 0 \implies x = 4 \]

3. **Determine the intervals**: The critical points divide the number line into intervals:
- \( (-\infty, -2) \)
- \( (-2, 4) \)
- \( (4, \infty) \)

4. **Test the intervals**:
- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ \frac{-1}{-7} > 0 \quad \text{(not in A)} \]
- For \( -2 < x < 4 \) (e.g., \( x = 0 \)):
\[ \frac{2}{-4} < 0 \quad \text{(in A)} \]
- For \( x > 4 \) (e.g., \( x = 5 \)):
\[ \frac{7}{1} > 0 \quad \text{(not in A)} \]

5. **Include the endpoints**: Since the inequality is less than or equal to zero, we include \( x = -2 \) (where the expression equals zero) but exclude \( x = 4 \) (where the expression is undefined).

Thus, the set \( A \) is:
\[ A = [-2, 4) \]

### Step 2: Determine the set B

Next, we consider the inequality:
\[ x^2 - ax - 4 \leq 0 \]

This is a quadratic inequality. The roots of the corresponding equation \( x^2 - ax - 4 = 0 \) can be found using the quadratic formula:
\[ x = \frac{a \pm \sqrt{a^2 + 16}}{2} \]

Let the roots be \( r_1 \) and \( r_2 \). The quadratic will be less than or equal to zero between its roots if \( r_1 \) and \( r_2 \) are real and distinct.

### Step 3: Conditions for B to be a subset of A

For \( B \) to be a subset of \( A \), the roots \( r_1 \) and \( r_2 \) must lie within the interval \( [-2, 4) \).

1. **Condition on the roots**:
- The roots must satisfy:
\[ -2 \leq r_1 < r_2 < 4 \]

2. **Finding the conditions**:
- The roots \( r_1 \) and \( r_2 \) can be expressed as:
\[ r_1 = \frac{a - \sqrt{a^2 + 16}}{2}, \quad r_2 = \frac{a + \sqrt{a^2 + 16}}{2} \]

3. **Set conditions**:
- For \( r_1 \geq -2 \):
\[ \frac{a - \sqrt{a^2 + 16}}{2} \geq -2 \implies a - \sqrt{a^2 + 16} \geq -4 \implies a + 4 \geq \sqrt{a^2 + 16} \]
Squaring both sides (valid since both sides are non-negative):
\[ a^2 + 8a + 16 \geq a^2 + 16 \implies 8a \geq 0 \implies a \geq 0 \]

- For \( r_2 < 4 \):
\[ \frac{a + \sqrt{a^2 + 16}}{2} < 4 \implies a + \sqrt{a^2 + 16} < 8 \implies \sqrt{a^2 + 16} < 8 - a \]
Squaring both sides:
\[ a^2 + 16 < (8 - a)^2 \implies a^2 + 16 < 64 - 16a + a^2 \]
Simplifying gives:
\[ 16a < 48 \implies a < 3 \]

### Step 4: Conclusion

From the conditions derived, we have:
\[ 0 \leq a < 3 \]
This means that \( a \) can take integral values of 0, 1, and 2, but cannot take the integral value of 3.

Thus, the answer is:
\[ \text{(d) 3} \]