Let f (x) be a polynomial of degree 8 such that `F(r)=1/r, r=1,2,3,…,8,9,` then `(1)/(F(10)) =`

Answer

To solve the problem, we need to find \( \frac{1}{F(10)} \) for the polynomial \( F(x) \) of degree 8, given that \( F(r) = \frac{1}{r} \) for \( r = 1, 2, 3, \ldots, 9 \).

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
Since \( F(x) \) is a polynomial of degree 8, we can express it in the form:
\[ F(x) = k(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(x - 8) + \frac{1}{x} \]
Here, \( k \) is a constant that we need to determine.

2. **Finding the Constant \( k \)**:
We know that \( F(r) = \frac{1}{r} \) for \( r = 1, 2, \ldots, 9 \). This means:
\[ F(r) - \frac{1}{r} = 0 \]
Thus, we can define a new function:
\[ f(x) = xF(x) - 1 \]
This function \( f(x) \) is a polynomial of degree 9 because \( F(x) \) is of degree 8.

3. **Identifying the Roots of \( f(x) \)**:
Since \( F(r) = \frac{1}{r} \) for \( r = 1, 2, \ldots, 9 \), we have:
\[ f(1) = 0, \quad f(2) = 0, \quad \ldots, \quad f(9) = 0 \]
Therefore, the roots of \( f(x) \) are \( 1, 2, 3, \ldots, 9 \). Thus, we can express \( f(x) \) as:
\[ f(x) = k(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(x - 8)(x - 9) \]

4. **Finding the Value of \( k \)**:
To find \( k \), we can evaluate \( f(0) \):
\[ f(0) = 0F(0) - 1 = -1 \]
Substituting \( x = 0 \) into our expression for \( f(x) \):
\[ f(0) = k(-1)(-2)(-3)(-4)(-5)(-6)(-7)(-8)(-9) = k(-9!) = -1 \]
Thus,
\[ k = \frac{1}{9!} \]

5. **Substituting \( k \) back into \( F(x) \)**:
Now we can write:
\[ F(x) = \frac{1}{9!}(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(x - 8)(x - 9) + \frac{1}{x} \]

6. **Calculating \( F(10) \)**:
Now we substitute \( x = 10 \):
\[ F(10) = \frac{1}{9!}(10 - 1)(10 - 2)(10 - 3)(10 - 4)(10 - 5)(10 - 6)(10 - 7)(10 - 8)(10 - 9) + \frac{1}{10} \]
This simplifies to:
\[ F(10) = \frac{1}{9!}(9)(8)(7)(6)(5)(4)(3)(2)(1) + \frac{1}{10} \]
\[ = \frac{9!}{9!} + \frac{1}{10} = 1 + \frac{1}{10} = \frac{10}{10} + \frac{1}{10} = \frac{11}{10} \]

7. **Finding \( \frac{1}{F(10)} \)**:
Finally, we calculate:
\[ \frac{1}{F(10)} = \frac{10}{11} \]

### Final Answer:
\[ \frac{1}{F(10)} = \frac{10}{11} \]