The number of integral solution of `(x^(2)(3x-4)^(3)(x-2)^(4))/(In(2x-1)(x-5)^(5)(2x-7)^(6))le 0` is :

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To solve the inequality

\[ \frac{x^2 (3x - 4)^3 (x - 2)^4}{\ln(2x - 1)(x - 5)^5 (2x - 7)^6} \leq 0, \]

we will follow these steps:

### Step 1: Analyze the numerator and denominator
The numerator is \(x^2 (3x - 4)^3 (x - 2)^4\).

- \(x^2\) is always non-negative (0 or positive).
- \((3x - 4)^3\) can be negative or positive depending on the value of \(x\).
- \((x - 2)^4\) is always non-negative.

Thus, the numerator is non-negative when \(x^2\) and \((x - 2)^4\) are positive, and \((3x - 4)^3\) is negative.

The denominator is \(\ln(2x - 1)(x - 5)^5 (2x - 7)^6\).

- \(\ln(2x - 1)\) is defined for \(2x - 1 > 0\) or \(x > \frac{1}{2}\).
- \((x - 5)^5\) is zero when \(x = 5\) and changes sign at this point.
- \((2x - 7)^6\) is always positive (even power).

### Step 2: Find critical points
1. Set \(3x - 4 = 0\) to find when the numerator changes sign:
\[ 3x - 4 = 0 \implies x = \frac{4}{3} \approx 1.33. \]

2. The denominator has critical points at:
- \(x = 1\) (where \(\ln(2x - 1) = 0\)),
- \(x = 5\) (where \((x - 5)^5 = 0\)),
- \(x = \frac{7}{2} = 3.5\) (where \((2x - 7)^6 = 0\)).

### Step 3: Determine intervals
The critical points divide the number line into intervals:
- \((-\infty, \frac{1}{2})\)
- \((\frac{1}{2}, \frac{4}{3})\)
- \((\frac{4}{3}, 1)\)
- \((1, \frac{7}{2})\)
- \((\frac{7}{2}, 5)\)
- \((5, \infty)\)

### Step 4: Test each interval
We will test each interval to see where the expression is less than or equal to zero.

1. **Interval \((-\infty, \frac{1}{2})\)**: Not valid since \(\ln(2x - 1)\) is undefined.
2. **Interval \((\frac{1}{2}, \frac{4}{3})\)**:
- Choose \(x = 1\):
\[ \text{Numerator: } (3(1) - 4)^3 < 0 \quad \text{Denominator: } \text{positive} \quad \Rightarrow \text{Negative}. \]
3. **Interval \((\frac{4}{3}, 1)\)**: Not valid since \(x = 1\) is excluded.
4. **Interval \((1, \frac{7}{2})\)**:
- Choose \(x = 2\):
\[ \text{Numerator: } (3(2) - 4)^3 > 0 \quad \text{Denominator: } \text{positive} \quad \Rightarrow \text{Positive}. \]
5. **Interval \((\frac{7}{2}, 5)\)**:
- Choose \(x = 4\):
\[ \text{Numerator: } (3(4) - 4)^3 > 0 \quad \text{Denominator: } \text{positive} \quad \Rightarrow \text{Positive}. \]
6. **Interval \((5, \infty)\)**:
- Choose \(x = 6\):
\[ \text{Numerator: } (3(6) - 4)^3 > 0 \quad \text{Denominator: } \text{positive} \quad \Rightarrow \text{Positive}. \]

### Step 5: Identify valid solutions
From the analysis, the only interval where the expression is less than or equal to zero is \((\frac{1}{2}, \frac{4}{3})\). The integer solutions in this interval are:
- \(x = 1\) (not included),
- \(x = 2\) (included),
- \(x = 3\) (included).

Thus, the integral solutions are \(2\) and \(3\), totaling **two integral solutions**.

### Final Answer
The number of integral solutions is **2**.