If f(y) = `2y^(2)` + by + c and f(0) = 3 and f(2) = 1, then the value of f(1) is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the value of \( f(1) \) given the function \( f(y) = 2y^2 + by + c \) and the conditions \( f(0) = 3 \) and \( f(2) = 1 \).

### Step-by-Step Solution:

1. **Use the condition \( f(0) = 3 \)**:
\[ f(0) = 2(0)^2 + b(0) + c = c \]
Since \( f(0) = 3 \), we have:
\[ c = 3 \]

2. **Use the condition \( f(2) = 1 \)**:
\[ f(2) = 2(2)^2 + b(2) + c \]
Substituting \( c = 3 \):
\[ f(2) = 2(4) + 2b + 3 = 8 + 2b + 3 \]
Setting this equal to 1:
\[ 8 + 2b + 3 = 1 \]
Simplifying:
\[ 11 + 2b = 1 \]
\[ 2b = 1 - 11 \]
\[ 2b = -10 \]
\[ b = -5 \]

3. **Now we have the values of \( b \) and \( c \)**:
\[ b = -5, \quad c = 3 \]
Therefore, the function becomes:
\[ f(y) = 2y^2 - 5y + 3 \]

4. **Calculate \( f(1) \)**:
\[ f(1) = 2(1)^2 - 5(1) + 3 \]
Simplifying:
\[ f(1) = 2 - 5 + 3 \]
\[ f(1) = 0 \]

Thus, the value of \( f(1) \) is \( 0 \).