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 first identify the critical points where the expression is equal to zero or undefined.

1. **Numerator**: \(x + 2 = 0 \Rightarrow x = -2\)
2. **Denominator**: \(x - 4 = 0 \Rightarrow x = 4\)

Next, we analyze the sign of the expression in the intervals defined by these points: \((- \infty, -2)\), \((-2, 4)\), and \((4, \infty)\).

- For \(x < -2\) (e.g., \(x = -3\)):
\[ \frac{-3 + 2}{-3 - 4} = \frac{-1}{-7} > 0 \quad (\text{not included}) \]

- For \(-2 < x < 4\) (e.g., \(x = 0\)):
\[ \frac{0 + 2}{0 - 4} = \frac{2}{-4} < 0 \quad (\text{included}) \]

- For \(x > 4\) (e.g., \(x = 5\)):
\[ \frac{5 + 2}{5 - 4} = \frac{7}{1} > 0 \quad (\text{not included}) \]

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

### Step 2: Determine the set B
Next, we analyze the set B defined by the inequality:
\[ x^2 - ax - 4 \leq 0 \]
This is a quadratic inequality. For this quadratic to be less than or equal to zero, it must have real roots and the roots must lie within the interval defined by A.

1. **Finding the roots**:
The roots of the quadratic \(x^2 - ax - 4 = 0\) can be found using the quadratic formula:
\[ x = \frac{-(-a) \pm \sqrt{(-a)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{a \pm \sqrt{a^2 + 16}}{2} \]

Let the roots be \(\alpha = \frac{a - \sqrt{a^2 + 16}}{2}\) and \(\beta = \frac{a + \sqrt{a^2 + 16}}{2}\).

2. **Condition for B to be a subset of A**:
For \(B\) to be a subset of \(A\), both roots \(\alpha\) and \(\beta\) must lie within the interval \([-2, 4)\).

- **Condition 1**: \(\alpha \geq -2\)
\[ \frac{a - \sqrt{a^2 + 16}}{2} \geq -2 \Rightarrow a - \sqrt{a^2 + 16} \geq -4 \Rightarrow a + 4 \geq \sqrt{a^2 + 16} \]
Squaring both sides (valid since both sides are positive):
\[ (a + 4)^2 \geq a^2 + 16 \Rightarrow a^2 + 8a + 16 \geq a^2 + 16 \Rightarrow 8a \geq 0 \Rightarrow a \geq 0 \]

- **Condition 2**: \(\beta < 4\)
\[ \frac{a + \sqrt{a^2 + 16}}{2} < 4 \Rightarrow a + \sqrt{a^2 + 16} < 8 \Rightarrow \sqrt{a^2 + 16} < 8 - a \]
Squaring both sides:
\[ a^2 + 16 < (8 - a)^2 \Rightarrow a^2 + 16 < 64 - 16a + a^2 \Rightarrow 16a < 48 \Rightarrow a < 3 \]

### Step 3: Combine the conditions
From the conditions derived:
- \(a \geq 0\)
- \(a < 3\)

Thus, the valid range for \(a\) is:
\[ 0 \leq a < 3 \]

### Step 4: Identify the integral values of \(a\)
The integral values that \(a\) can take within this range are \(0, 1, 2\). The only integer that \(a\) cannot take is \(3\) since \(a < 3\).

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