Let `P(x)` be a polynomial of degree `11` such that `P(x) = (1)/(x + 1)` for `x = 0, 1, 2, 3,"…."11`. The value of `P(12)` is.

Options

A. `(1)/(13)`
B. `1`
C. `0`
D. `2`

Explanation

To solve the problem, we need to find the value of the polynomial \( P(12) \) given that \( P(x) = \frac{1}{x + 1} \) for \( x = 0, 1, 2, \ldots, 11 \).

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
The polynomial \( P(x) \) is of degree 11, and it is defined such that \( P(k) = \frac{1}{k + 1} \) for \( k = 0, 1, 2, \ldots, 11 \).

2. **Constructing a New Function**:
We define a new function \( Q(x) = P(x)(x + 1) - 1 \). This function \( Q(x) \) will also be a polynomial of degree 11.

3. **Finding Roots of \( Q(x) \)**:
Since \( P(k) = \frac{1}{k + 1} \) for \( k = 0, 1, 2, \ldots, 11 \), we can substitute these values into \( Q(x) \):
\[ Q(k) = P(k)(k + 1) - 1 = 0 \quad \text{for } k = 0, 1, 2, \ldots, 11 \]
Thus, \( Q(x) \) has roots at \( x = 0, 1, 2, \ldots, 11 \).

4. **Expressing \( Q(x) \)**:
Since \( Q(x) \) is a polynomial of degree 11 with roots at \( 0, 1, 2, \ldots, 11 \), we can express it as:
\[ Q(x) = a(x)(x - 1)(x - 2) \cdots (x - 11) \]
for some constant \( a \).

5. **Finding the Constant \( a \)**:
To find \( a \), we can evaluate \( Q(-1) \):
\[ Q(-1) = P(-1)(-1 + 1) - 1 = P(-1)(0) - 1 = -1 \]
Now substituting \( x = -1 \) into the expression for \( Q(x) \):
\[ Q(-1) = a(-1)(-2)(-3) \cdots (-12) = a(-1)^{12}(12!) = a(12!) \]
Setting these equal gives:
\[ a(12!) = -1 \implies a = -\frac{1}{12!} \]

6. **Final Expression for \( Q(x) \)**:
Thus, we have:
\[ Q(x) = -\frac{1}{12!} x(x - 1)(x - 2) \cdots (x - 11) \]

7. **Finding \( P(12) \)**:
Now we need to find \( P(12) \):
\[ Q(12) = P(12)(12 + 1) - 1 \]
Therefore,
\[ Q(12) = P(12)(13) - 1 \]
We can calculate \( Q(12) \):
\[ Q(12) = -\frac{1}{12!} (12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) = -\frac{12!}{12!} = -1 \]
Setting this equal gives:
\[ -1 = P(12)(13) - 1 \implies P(12)(13) = 0 \]
Thus,
\[ P(12) = 0 \]

### Conclusion:
The value of \( P(12) \) is \( \boxed{0} \).