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The question provides several conditions about a function \( z = f(x, y) \):

1. The second partial derivative with respect to \( y \) is:
   \[
   \frac{\partial^2 z}{\partial y^2} = 2
   \]
2. It also gives specific information about the function at \( y = 1 \):
   \[
   f(x, 1) = x + 2
   \]
   \[
   f_y(x, 1) = x + 1
   \]
   
The task is to determine which statement (A, B, C, or D) is **incorrect**.

### Analysis:

- From the condition \( \frac{\partial^2 z}{\partial y^2} = 2 \), we know that \( f_{yy} = 2 \). This means the second partial derivative of \( f \) with respect to \( y \) is constant and equal to 2.
- We can integrate \( f_{yy} = 2 \) with respect to \( y \) to find \( f_y(x, y) \):
  \[
  f_y(x, y) = 2y + g(x)
  \]
  where \( g(x) \) is an arbitrary function of \( x \).
  
- Using the condition \( f_y(x, 1) = x + 1 \), we substitute \( y = 1 \) into the expression for \( f_y(x, y) \):
  \[
  2(1) + g(x) = x + 1 \implies 2 + g(x) = x + 1 \implies g(x) = x - 1
  \]
  Thus, \( f_y(x, y) = 2y + (x - 1) \).

- To find \( f(x, y) \), integrate \( f_y(x, y) \) with respect to \( y \):
  \[
  f(x, y) = \int \left( 2y + (x - 1) \right) \, dy = y^2 + (x - 1)y + h(x)
  \]
  where \( h(x) \) is another arbitrary function of \( x \).

- Using the condition \( f(x, 1) = x + 2 \), we substitute \( y = 1 \) into the expression for \( f(x, y) \):
  \[
  1^2 + (x - 1)(1) + h(x) = x + 2 \implies 1 + (x - 1) + h(x) = x + 2
  \]
  Simplifying:
  \[
  1 + x - 1 + h(x) = x + 2 \implies h(x) = 2
  \]
  Thus, the function \( f(x, y) \) is:
  \[
  f(x, y) = y^2 + (x - 1)y + 2
  \]

### Now, let's verify the statements:

- **A**: \( f_x(1, 1) = 1 \):
  \[
  f_x(x, y) = (x - 1)y + 0 \implies f_x(1, 1) = (1 - 1)(1) = 0
  \]
  So, this is **incorrect**.

- **B**: \( f_{yy}(1, 1) = 0 \):
  \[
  f_{yy}(x, y) = 2 \quad \text{(constant for all \( x \) and \( y \))}.
  \]
  So, \( f_{yy}(1, 1) = 2 \), not 0. This is **incorrect**.

- **C**: \( f(x, y) = y^2 + (x - 1)y + 2 \) is the correct form of the function we derived, so this is **correct**.

- **D**: \( f_{xy}(1, 1) = 1 \):
  \[
  f_{xy}(x, y) = 1 \quad \text{(since \( f_x(x, y) = (x - 1)y \))},
  \]
  So, \( f_{xy}(1, 1) = 1 \) is **correct**.

### Conclusion:
The **incorrect** statements are **A** and **B**.

Would you like a detailed breakdown of any part, or do you have any questions?

Here are five related questions to explore further:

1. How do you compute partial derivatives for more complex functions?
2. What is the geometric interpretation of second partial derivatives like \( f_{yy} \)?
3. How can you determine boundary conditions for partial differential equations (PDEs)?
4. What role does integration play in solving partial derivatives?
5. How can you check if a given function is differentiable across a domain?

**Tip**: When solving partial derivatives, always cross-check the conditions and boundary values to avoid mistakes in applying them.

若函数 z = f(x, y) 满足 ∂²z/∂y² = 2. 且 f(x, 1) = x + 2, f_y(x, 1) = x + 1, 则下列说法不正确的是?

Solve a multivariable calculus problem involving partial derivatives. Learn how to compute second-order partial derivatives and integrate with respect to a variable to find the solution for f(x, y).

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