Question:

Let f(x, y) = ln(1 + x2 + y2 ) for (x, y) ∈ R2\R^2. Define
P=2fx2(0,0) Q=2fxy(0,0) R=2fyx(0,0) S=2fy2(0,0)\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

Updated On: Oct 1, 2024
  • PS − QR > 0 and P < 0
  • PS − QR > 0 and P > 0
  • PS − QR < 0 and P > 0
  • PS − QR < 0 and P < 0
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The Correct Option is B

Solution and Explanation

The correct option is (B) : PS − QR > 0 and P > 0.
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