If `f(x)` is a polynomial of degree n such that `f(0)=0 , f(x)=1/2,....,f(n)=n/(n+1)`, ,then the value of `f (n+ 1)` is

Options

A. 1, when n is even
B. `n/(n+2)`, when n is odd
C. 1, when n is odd
D. `n/(n+2),` when n is even

Answer

To solve the problem, we need to find the value of \( f(n + 1) \) given the conditions on the polynomial \( f(x) \). Let's go through the solution step by step.

### Step 1: Define the polynomial \( f(x) \)
We know that \( f(x) \) is a polynomial of degree \( n \) and it satisfies the following conditions:
- \( f(0) = 0 \)
- \( f(k) = \frac{k}{k + 1} \) for \( k = 1, 2, \ldots, n \)

### Step 2: Construct a new polynomial \( q(x) \)
We define a new polynomial \( q(x) \) as follows:
\[ q(x) = (x + 1)f(x) - x \]
This polynomial \( q(x) \) will also be of degree \( n + 1 \) since \( f(x) \) is of degree \( n \).

### Step 3: Evaluate \( q(x) \) at specific points
Now, we evaluate \( q(x) \) at the points \( x = 0, 1, 2, \ldots, n \):
- For \( x = 0 \):
\[ q(0) = (0 + 1)f(0) - 0 = 1 \cdot 0 - 0 = 0 \]
- For \( x = k \) where \( k = 1, 2, \ldots, n \):
\[ q(k) = (k + 1)f(k) - k = (k + 1) \cdot \frac{k}{k + 1} - k = k - k = 0 \]

### Step 4: Roots of \( q(x) \)
From the evaluations, we see that \( q(x) \) has roots at \( x = 0, 1, 2, \ldots, n \). Therefore, we can express \( q(x) \) as:
\[ q(x) = a \cdot x(x - 1)(x - 2) \cdots (x - n) \]
for some constant \( a \).

### Step 5: Determine the constant \( a \)
To find \( a \), we can evaluate \( q(-1) \):
\[ q(-1) = (-1 + 1)f(-1) - (-1) = 0 + 1 = 1 \]
Now substituting \( x = -1 \) into the polynomial form:
\[ q(-1) = a \cdot (-1)(-2)(-3) \cdots (-n - 1) = a \cdot (-1)^{n + 1} \cdot (n + 1)! \]
Setting this equal to 1 gives:
\[ a \cdot (-1)^{n + 1} \cdot (n + 1)! = 1 \implies a = \frac{(-1)^{n + 1}}{(n + 1)!} \]

### Step 6: Write the polynomial \( q(x) \)
Now we have:
\[ q(x) = \frac{(-1)^{n + 1}}{(n + 1)!} \cdot x(x - 1)(x - 2) \cdots (x - n) \]

### Step 7: Evaluate \( q(n + 1) \)
Now we can find \( q(n + 1) \):
\[ q(n + 1) = \frac{(-1)^{n + 1}}{(n + 1)!} \cdot (n + 1)(n)(n - 1) \cdots (1) = \frac{(-1)^{n + 1}}{(n + 1)!} \cdot (n + 1)! \]
This simplifies to:
\[ q(n + 1) = (-1)^{n + 1} \]

### Step 8: Relate \( q(n + 1) \) back to \( f(n + 1) \)
Using the relationship \( q(n + 1) = (n + 2)f(n + 1) - (n + 1) \), we get:
\[ (-1)^{n + 1} = (n + 2)f(n + 1) - (n + 1) \]
Rearranging gives:
\[ (n + 2)f(n + 1) = (-1)^{n + 1} + (n + 1) \]
Thus:
\[ f(n + 1) = \frac{(-1)^{n + 1} + (n + 1)}{n + 2} \]

### Step 9: Conclusion
Now we can conclude:
- If \( n \) is even, \( (-1)^{n + 1} = -1 \) and \( f(n + 1) = \frac{-1 + (n + 1)}{n + 2} = \frac{n}{n + 2} \).
- If \( n \) is odd, \( (-1)^{n + 1} = 1 \) and \( f(n + 1) = \frac{1 + (n + 1)}{n + 2} = 1 \).

Thus, the value of \( f(n + 1) \) is:
- \( f(n + 1) = 1 \) if \( n \) is odd.
- \( f(n + 1) = \frac{n}{n + 2} \) if \( n \) is even.