Question:

Let the functions \( f: \mathbb{R} \to \mathbb{R} \) and \( g: \mathbb{R}^2 \to \mathbb{R} \) be given by \[ f(x_1, x_2) = x_1^2 + x_2^2 - 2x_1x_2, \quad g(x_1, x_2) = 2x_1^2 + 2x_2^2 - x_1x_2. \] Consider the following statements: 
S1: For every compact subset \( K \) of \( \mathbb{R} \), \( f^{-1}(K) \) is compact. 
S2: For every compact subset \( K \) of \( \mathbb{R} \), \( g^{-1}(K) \) is compact. Then, which one of the following is correct?

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For continuous functions, the preimage of a compact set is compact. However, not all continuous functions guarantee this property for their inverse images, so analyze the specific behavior of the function.
Updated On: Apr 9, 2025
  • S1 is TRUE and S2 is FALSE
  • S2 is TRUE and S1 is FALSE
  • both S1 and S2 are TRUE
  • neither S1 nor S2 is TRUE
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The Correct Option is B

Solution and Explanation

To apply the inverse function theorem, we need to check the determinant of the Jacobian matrix of \( F \). The inverse function theorem fails where the Jacobian determinant is zero.

The Jacobian matrix of \( F \) is:
\[ J_F(x, y) = \begin{pmatrix} \frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial y} \\ \frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2 - 3y^2 - 3 & -6xy \\ 6xy - 3y^2 & 3x^2 - 3 \end{pmatrix} \]

We compute the determinant of \( J_F(x, y) \):
\[ \det(J_F(x, y)) = (3x^2 - 3y^2 - 3)(3x^2 - 3) - (-6xy)(6xy - 3y^2) \]

By simplifying the determinant expression, we find that the inverse function theorem fails at exactly two points. Therefore, the correct answer is:

\[ \boxed{\text{(C) not applicable at exactly two points of } \mathbb{R}^2.} \]
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