Question

Let f(x, y) = ln(1 + x² + y² ) for (x, y) ∈ \(\R^2\). Define
\(\begin{matrix} P=\frac{∂^2f}{∂x^2}|_{(0,0)} &  Q=\frac{∂^2f}{∂x∂y}|_{(0,0)} \\  R=\frac{∂^2f}{∂y∂x}|_{(0,0)} &  S=\frac{∂^2f}{∂y^2}|_{(0,0)} \end{matrix}\)
Then

• A. PS − QR > 0 and P < 0
• B. PS − QR > 0 and P > 0
• C. PS − QR < 0 and P > 0
• D. PS − QR < 0 and P < 0

Answer 1

The Correct Option is B

To solve the given problem involving the function \(f(x, y) = \ln(1 + x^2 + y^2)\), we need to find the second derivatives at the point \((0,0)\) and determine the expressions for \(PS - QR\) and whether \(P > 0\) or \(P < 0\).b

1. First, compute the partial derivatives:
   • Compute the first order partial derivative with respect to \(x\):
   • Compute the second order partial derivative with respect to \(x\):
   • Evaluate at \((0,0)\):
2. Next, compute the mixed second order partial derivatives:
   • Compute the mixed derivative with respect to \(x\) and \(y\) (which is symmetric hence \(Q = R\)):
   • Evaluate at \((0,0)\):
3. Finally, compute the partial derivative with respect to \(y^2\):
   • Compute the second partial derivative with respect to \(y\):
   • Evaluate at \((0,0)\):

Now, evaluate the expression \(PS - QR\):

\[PS - QR = (2)(2) - (0)(0) = 4\]

Since \(4 > 0\) and \(P = 2 > 0\), the correct statement is:

• PS − QR > 0 and P > 0

This concludes that the correct answer is: PS − QR > 0 and P > 0.