Let `f(x)=x^(2)-ax+b`, `'a'` is odd positive integar and the roots of the equation `f(x)=0` are two distinct prime numbers. If `a+b=35`, then the value of `f(10)=`

Options

A. `-8`
B. `-10`
C. `-4`
D. `0`

Explanation

To solve the problem step by step, we will analyze the given function and the conditions provided.

### Step 1: Understand the function and its roots
The function is given as:
\[ f(x) = x^2 - ax + b \]
where \( a \) is an odd positive integer, and the roots of the equation \( f(x) = 0 \) are two distinct prime numbers.

### Step 2: Use Vieta's formulas
According to Vieta's formulas, for a quadratic equation \( x^2 - (sum \ of \ roots)x + (product \ of \ roots) = 0 \):
- The sum of the roots \( \alpha + \beta = a \)
- The product of the roots \( \alpha \beta = b \)

### Step 3: Set up equations based on the problem
Since \( a + b = 35 \), we can express \( b \) in terms of \( a \):
\[ b = 35 - a \]

### Step 4: Substitute for \( b \)
From Vieta's formulas, we know:
\[ \alpha + \beta = a \]
\[ \alpha \beta = b = 35 - a \]

### Step 5: Analyze the roots
Given that \( \alpha \) and \( \beta \) are distinct prime numbers, we can assume one of the roots is 2 (the only even prime number) since the sum of two odd primes would be even, and \( a \) must be odd.

Let:
\[ \alpha = 2 \]
Then:
\[ \beta = a - 2 \]
Substituting this into the product equation:
\[ 2(a - 2) = 35 - a \]
Expanding this:
\[ 2a - 4 = 35 - a \]

### Step 6: Solve for \( a \)
Rearranging gives:
\[ 2a + a = 35 + 4 \]
\[ 3a = 39 \]
\[ a = 13 \]

### Step 7: Find \( b \)
Substituting \( a \) back into the equation for \( b \):
\[ b = 35 - a = 35 - 13 = 22 \]

### Step 8: Write the function
Now we can write the function:
\[ f(x) = x^2 - 13x + 22 \]

### Step 9: Calculate \( f(10) \)
Now we need to find \( f(10) \):
\[ f(10) = 10^2 - 13 \cdot 10 + 22 \]
Calculating this gives:
\[ f(10) = 100 - 130 + 22 \]
\[ f(10) = 100 - 130 + 22 = -8 \]

### Final Answer
Thus, the value of \( f(10) \) is:
\[ \boxed{-8} \]