If number of integral values of x satisfying the inequality `((x)/(100))^(7logx-log^(2)x-6)ge10^(12)` are `alpha` then

Options

A. `alpha` is divisible by 5
B. `alpha` is divisible by 2
C. sum of digits of `alpha` is a composite number
D. sum of digits of `alpha` is a prime number

Explanation

To solve the inequality \(\left(\frac{x}{100}\right)^{7 \log x - \log^2 x - 6} \geq 10^{12}\), we will follow these steps:

### Step 1: Rewrite the Inequality
We start by taking the logarithm of both sides of the inequality. This gives us:
\[ 7 \log x - \log^2 x - 6 \cdot \log\left(\frac{x}{100}\right) \geq 12 \]
We can rewrite \(\log\left(\frac{x}{100}\right)\) as \(\log x - \log 100 = \log x - 2\). Thus, the inequality becomes:
\[ 7 \log x - \log^2 x - 6(\log x - 2) \geq 12 \]

### Step 2: Simplify the Inequality
Now we simplify the expression:
\[ 7 \log x - \log^2 x - 6 \log x + 12 \geq 12 \]
This simplifies to:
\[ \log x - \log^2 x \geq 0 \]

### Step 3: Rearranging the Terms
Rearranging gives us:
\[ -\log^2 x + \log x \geq 0 \]
Factoring out \(-\log x\):
\[ -\log x(\log x - 1) \geq 0 \]

### Step 4: Analyzing the Factors
The product \(-\log x(\log x - 1) \geq 0\) implies that either:
1. \(-\log x \geq 0\) and \(\log x - 1 \leq 0\)
2. \(-\log x \leq 0\) and \(\log x - 1 \geq 0\)

From the first case, we have:
\[ \log x \leq 0 \quad \text{(which means } x \leq 1\text{)} \]
And:
\[ \log x \leq 1 \quad \text{(which means } x \leq 10\text{)} \]
This gives us \(x \in (0, 1]\).

From the second case:
\[ \log x \geq 0 \quad \text{(which means } x \geq 1\text{)} \]
And:
\[ \log x \geq 1 \quad \text{(which means } x \geq 10\text{)} \]
This gives us \(x \in [10, \infty)\).

### Step 5: Combine the Intervals
Combining the intervals, we have:
\[ x \in (0, 1] \cup [10, \infty) \]

### Step 6: Count Integral Values
Now we count the integral values of \(x\):
- From \(0 < x \leq 1\), the only integral value is \(x = 1\).
- From \(x \geq 10\), the integral values are \(10, 11, 12, \ldots\) up to \(99,999\).

The number of integers from \(10\) to \(99,999\) is:
\[ 99,999 - 10 + 1 = 99,990 \]

### Total Integral Values
Adding the integral values from both intervals:
\[ 1 + 99,990 = 99,991 \]

Thus, the number of integral values of \(x\) satisfying the inequality is:
\[ \alpha = 99,991 \]

### Final Answer
The value of \(\alpha\) is \(99,991\).

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