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
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
\(\begin{matrix} P=\frac{∂^2f}{∂x^2}|_{(0,0)} & Q=\frac{∂^2f}{∂x∂y}|_{(0,0)} \\ R=\frac{∂^2f}{∂y∂x}|_{(0,0)} & S=\frac{∂^2f}{∂y^2}|_{(0,0)} \end{matrix}\)
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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