The number of integers satisfying the inequation `|x-1|le[(sqrt(2)+1)^(6)+(sqrt(2)-1)^(6)]` where `[.]` denotes greatest integer function is greater than and equal to :

Options

A. `198`
B. `396`
C. `397`
D. `398`

Explanation

To solve the problem, we need to find the number of integers satisfying the inequality:

\[ |x - 1| \leq \left\lfloor (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 \right\rfloor \]

### Step 1: Calculate \((\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6\)

Using the binomial theorem, we can expand both terms:

1. **Expansion of \((\sqrt{2} + 1)^6\)**:
\[ (\sqrt{2} + 1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (1)^k \]

2. **Expansion of \((\sqrt{2} - 1)^6\)**:
\[ (\sqrt{2} - 1)^6 = \sum_{k=0}^{6} \binom{6}{k} (\sqrt{2})^{6-k} (-1)^k \]

### Step 2: Combine the expansions

When we add these two expansions, the terms with odd powers of \((\sqrt{2})\) will cancel out, and we will be left with only the even powers:

\[ (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 = 2 \left( \binom{6}{0} (\sqrt{2})^6 + \binom{6}{2} (\sqrt{2})^4 + \binom{6}{4} (\sqrt{2})^2 + \binom{6}{6} (1)^0 \right) \]

### Step 3: Calculate the coefficients

1. **Calculate each term**:
- \(\binom{6}{0} = 1\)
- \(\binom{6}{2} = 15\)
- \(\binom{6}{4} = 15\)
- \(\binom{6}{6} = 1\)

2. **Substituting values**:
\[ = 2 \left( 1 \cdot 2^3 + 15 \cdot 2^2 + 15 \cdot 2 + 1 \right) \]
\[ = 2 \left( 8 + 60 + 30 + 1 \right) = 2 \cdot 99 = 198 \]

### Step 4: Apply the greatest integer function

Now we apply the greatest integer function:
\[ \left\lfloor 198 \right\rfloor = 198 \]

### Step 5: Solve the inequality

Now we need to solve the inequality:
\[ |x - 1| \leq 198 \]

This can be rewritten as:
\[ -198 \leq x - 1 \leq 198 \]

### Step 6: Rearranging the inequality

Adding 1 to all parts of the inequality gives:
\[ -197 \leq x \leq 199 \]

### Step 7: Count the integers

The integers satisfying this inequality are:
\[ -197, -196, \ldots, 0, 1, \ldots, 199 \]

To count the total number of integers:
- From \(-197\) to \(0\) there are \(198\) integers.
- From \(1\) to \(199\) there are \(199\) integers.

Thus, the total number of integers is:
\[ 198 + 199 = 397 \]

### Final Answer

The number of integers satisfying the given inequality is:
\[ \boxed{397} \]