Question

Let \[ f(x, y) = x^4 + y^4 - 2x^2 + 4xy - 2y^2 + \alpha \] be a real valued function. Then, which one of the following statements is TRUE for all \( \alpha \)?

• A. \( (0, 0) \) is not a stationary point of \( f \)
• B. \( f \) has a local maxima at \( (0, 0) \)
• C. \( f \) has a local minima at \( (0, 0) \)
• D. \( f \) has a saddle point at \( (0, 0) \)

Hint:

When the discriminant \( D = 0 \) at a stationary point, it indicates that the point is a saddle point, which is neither a maximum nor a minimum.

Answer 1

The Correct Option is D

We are asked to determine the type of point at \( (0, 0) \) for the given function. To solve this, we will calculate the first and second derivatives of the function and analyze the point \( (0, 0) \).

Step 1: Find the first derivatives. 
The first partial derivatives of \( f(x, y) \) are: \[ \frac{\partial f}{\partial x} = 4x^3 - 4x + 4y, \] \[ \frac{\partial f}{\partial y} = 4y^3 + 4x - 4y. \] At \( (0, 0) \), both derivatives are zero: \[ \frac{\partial f}{\partial x} (0, 0) = 0, \frac{\partial f}{\partial y} (0, 0) = 0. \] This confirms that \( (0, 0) \) is a stationary point.

Step 2: Find the second derivatives. 
Next, we calculate the second partial derivatives: \[ \frac{\partial^2 f}{\partial x^2} = 12x^2 - 4, \] \[ \frac{\partial^2 f}{\partial y^2} = 12y^2 - 4, \] \[ \frac{\partial^2 f}{\partial x \partial y} = 4. \] At \( (0, 0) \), the second derivatives are: \[ \frac{\partial^2 f}{\partial x^2} (0, 0) = -4, \frac{\partial^2 f}{\partial y^2} (0, 0) = -4, \frac{\partial^2 f}{\partial x \partial y} (0, 0) = 4. \]

Step 3: Analyze the discriminant. 
The discriminant \( D \) is given by: \[ D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2. \] Substituting the values at \( (0, 0) \): \[ D = (-4)(-4) - (4)^2 = 16 - 16 = 0. \] Since \( D = 0 \), the point \( (0, 0) \) is a saddle point.

Final Answer: (D) \( f \) has a saddle point at \( (0, 0) \)