Let `f(x) = x^(2) + px +q`. The value of(p, q) so that f(1) =3 is an extreme value of f on [0, 2] is

Options

A. `(-2,2)`
B. `(1,4)`
C. `(-2,4)`
D. `(-2,3)`

Answer

To solve the problem, we need to find the values of \( p \) and \( q \) such that \( f(1) = 3 \) is an extreme value of the function \( f(x) = x^2 + px + q \) on the interval \([0, 2]\).

### Step 1: Set up the function and conditions
We have:
\[ f(x) = x^2 + px + q \]
We need to ensure that:
1. \( f(1) = 3 \)
2. \( f'(1) = 0 \) (since \( f(1) \) is an extreme value)

### Step 2: Calculate \( f(1) \)
Substituting \( x = 1 \) into the function:
\[ f(1) = 1^2 + p(1) + q = 1 + p + q \]
Setting this equal to 3:
\[ 1 + p + q = 3 \]
This simplifies to:
\[ p + q = 2 \quad \text{(Equation 1)} \]

### Step 3: Calculate the derivative \( f'(x) \)
The derivative of \( f(x) \) is:
\[ f'(x) = 2x + p \]
Now, substituting \( x = 1 \):
\[ f'(1) = 2(1) + p = 2 + p \]
Setting this equal to 0 for the extreme value condition:
\[ 2 + p = 0 \]
This simplifies to:
\[ p = -2 \quad \text{(Equation 2)} \]

### Step 4: Substitute \( p \) into Equation 1
Now, substitute \( p = -2 \) into Equation 1:
\[ -2 + q = 2 \]
Solving for \( q \):
\[ q = 2 + 2 = 4 \quad \text{(Equation 3)} \]

### Step 5: Final values of \( p \) and \( q \)
Thus, the values of \( p \) and \( q \) are:
\[ p = -2, \quad q = 4 \]

### Conclusion
The final answer is:
\[ (p, q) = (-2, 4) \]