Let `'a'` be an integer. If there are `10` inegers satisfying the `((x-a)^(2)(x-2a))/((x-3a)(x-4a))le0` then

Options

A. `a` is prime number
B. `e^(a) lt pi^(a)`
C. `a^(2) - 2` is prime number
D. number of possible values of `a` is `2`

Explanation

To solve the inequality \(\frac{(x-a)^{2}(x-2a)}{(x-3a)(x-4a)} \leq 0\), we will analyze the expression step by step.

### Step 1: Identify the critical points
The critical points of the inequality occur when the numerator or denominator is equal to zero.

- **Numerator**: \((x-a)^{2}(x-2a) = 0\)
- \(x = a\) (double root)
- \(x = 2a\) (single root)

- **Denominator**: \((x-3a)(x-4a) = 0\)
- \(x = 3a\) (single root)
- \(x = 4a\) (single root)

Thus, the critical points are \(a\), \(2a\), \(3a\), and \(4a\).

### Step 2: Determine the intervals
The critical points divide the number line into intervals. We need to test the sign of the expression in each interval:
1. \((- \infty, a)\)
2. \((a, 2a)\)
3. \((2a, 3a)\)
4. \((3a, 4a)\)
5. \((4a, +\infty)\)

### Step 3: Test the intervals
We will test a point from each interval to determine where the expression is less than or equal to zero.

1. **Interval \((- \infty, a)\)**: Choose \(x = a - 1\)
\[ \frac{(a-1-a)^{2}(a-1-2a)}{(a-1-3a)(a-1-4a)} = \frac{(1)^{2}(-1)}{(-2)(-3)} = \frac{-1}{6} < 0 \]

2. **Interval \((a, 2a)\)**: Choose \(x = \frac{3a}{2}\)
\[ \frac{(\frac{3a}{2}-a)^{2}(\frac{3a}{2}-2a)}{(\frac{3a}{2}-3a)(\frac{3a}{2}-4a)} = \frac{(\frac{a}{2})^{2}(-\frac{a}{2})}{(-\frac{3a}{2})(-\frac{5a}{2})} = \frac{-\frac{a^{3}}{8}}{\frac{15a^{2}}{4}} < 0 \]

3. **Interval \((2a, 3a)\)**: Choose \(x = 2.5a\)
\[ \frac{(2.5a-a)^{2}(2.5a-2a)}{(2.5a-3a)(2.5a-4a)} = \frac{(1.5a)^{2}(0.5a)}{(-0.5a)(-1.5a)} = \frac{1.125a^{3}}{0.75a^{2}} > 0 \]

4. **Interval \((3a, 4a)\)**: Choose \(x = 3.5a\)
\[ \frac{(3.5a-a)^{2}(3.5a-2a)}{(3.5a-3a)(3.5a-4a)} = \frac{(2.5a)^{2}(1.5a)}{(0.5a)(-0.5a)} < 0 \]

5. **Interval \((4a, +\infty)\)**: Choose \(x = 5a\)
\[ \frac{(5a-a)^{2}(5a-2a)}{(5a-3a)(5a-4a)} = \frac{(4a)^{2}(3a)}{(2a)(a)} > 0 \]

### Step 4: Combine results
The expression is less than or equal to zero in the intervals:
- \((- \infty, a]\)
- \([2a, 3a]\)
- \((3a, 4a]\)

### Step 5: Count the integer solutions
To find the total number of integers satisfying the inequality, we need to count the integers in the intervals:
1. From \((- \infty, a]\) — Count integers from \(-\infty\) to \(a\).
2. From \([2a, 3a]\) — Count integers from \(2a\) to \(3a\).
3. From \((3a, 4a]\) — Count integers from \(3a\) to \(4a\).

### Step 6: Set up the equation
The total number of integers satisfying the inequality is given as \(10\):
\[ \text{Count from } (-\infty, a] + \text{Count from } [2a, 3a] + \text{Count from } (3a, 4] = 10 \]

### Step 7: Solve for \(a\)
Let’s denote:
- Count from \((- \infty, a]\) = \(a + 1\) (if \(a\) is positive)
- Count from \([2a, 3a]\) = \(a + 1\) (as there are \(a\) integers from \(2a\) to \(3a\))
- Count from \((3a, 4a]\) = \(a\) (as there are \(a\) integers from \(3a\) to \(4a\))

Thus, we have:
\[ (a + 1) + (a + 1) + a = 10 \]
\[ 3a + 2 = 10 \]
\[ 3a = 8 \implies a = \frac{8}{3} \]

Since \(a\) must be an integer, we check integer values around this to find suitable integers.

### Conclusion
After checking integer values, we find that \(a = 5\) or \(a = -2\) satisfy the conditions.