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 \( \frac{1}{F(10)} \) given that \( F(r) = \frac{1}{r} \) for \( r = 1, 2, 3, \ldots, 9 \).

### Step-by-Step Solution:

1. **Define the Polynomial**:
We know that \( F(r) = \frac{1}{r} \) for \( r = 1, 2, 3, \ldots, 9 \). We can rearrange this to form a new function:
\[ R \cdot F(R) - 1 = 0 \]
This implies that \( R \cdot F(R) - 1 \) is a polynomial of degree 8 with roots at \( r = 1, 2, 3, \ldots, 9 \).

2. **Construct a New Function**:
Let \( G(R) = R \cdot F(R) - 1 \). Since \( G(R) \) is a polynomial of degree 8, we can express it as:
\[ G(R) = K(R - 1)(R - 2)(R - 3)(R - 4)(R - 5)(R - 6)(R - 7)(R - 8)(R - 9) \]
where \( K \) is a constant.

3. **Determine the Constant \( K \)**:
To find \( K \), we can substitute \( R = 0 \):
\[ G(0) = 0 \cdot F(0) - 1 = -1 \]
On the right side, substituting \( R = 0 \):
\[ G(0) = K(-1)(-2)(-3)(-4)(-5)(-6)(-7)(-8)(-9) = K \cdot (-1)^9 \cdot 9! = -K \cdot 9! \]
Setting both sides equal gives:
\[ -1 = -K \cdot 9! \implies K = \frac{1}{9!} \]

4. **Find \( F(10) \)**:
Now we can find \( F(10) \):
\[ G(10) = K(10 - 1)(10 - 2)(10 - 3)(10 - 4)(10 - 5)(10 - 6)(10 - 7)(10 - 8)(10 - 9) \]
Substituting \( K \):
\[ G(10) = \frac{1}{9!} \cdot (9)(8)(7)(6)(5)(4)(3)(2)(1) = \frac{9!}{9!} = 1 \]
Thus, we have:
\[ 10 \cdot F(10) - 1 = 1 \implies 10 \cdot F(10) = 2 \implies F(10) = \frac{2}{10} = \frac{1}{5} \]

5. **Calculate \( \frac{1}{F(10)} \)**:
Finally, we need to find \( \frac{1}{F(10)} \):
\[ \frac{1}{F(10)} = \frac{1}{\frac{1}{5}} = 5 \]

### Final Answer:
\[ \frac{1}{F(10)} = 5 \]