Function F Analysis

The question describes a function \( f : (0, 1) \rightarrow (0, 1) \) which is differentiable and surjective, with the condition that \( f'(x) \ne 0 \) for every \( x \in (0, 1) \). A new function \( F : (0, 1) \times (0, 1) \rightarrow (0, 1) \times (0, 1) \) is defined as \( F(x, y) = (f(x), f(y)) \).

Let's analyze the properties of the function \( f \).

• Since \( f \) is differentiable on \( (0, 1) \) and \( f'(x) \ne 0 \) for all \( x \in (0, 1) \), by the Intermediate Value Theorem for Derivatives (Darboux's Theorem), \( f'(x) \) must maintain the same sign throughout the interval \( (0, 1) \).
• If \( f'(x) > 0 \) for all \( x \), then \( f \) is strictly increasing.
• If \( f'(x) < 0 \) for all \( x \), then \( f \) is strictly decreasing.
• A strictly monotonic function is always injective (one-to-one). Thus, \( f \) is injective.
• Since \( f \) is both injective and surjective, it is a bijection from \( (0, 1) \) to \( (0, 1) \). This means for every \( z' \in (0, 1) \), there exists a unique \( z \in (0, 1) \) such that \( f(z) = z' \).

Analyzing Option 1: F is injective

The function is \( F(x, y) = (f(x), f(y)) \). To check if \( F \) is injective, we assume \( F(x_1, y_1) = F(x_2, y_2) \) for some \( (x_1, y_1), (x_2, y_2) \in (0, 1) \times (0, 1) \) and see if this implies \( (x_1, y_1) = (x_2, y_2) \).

If \( F(x_1, y_1) = F(x_2, y_2) \), then \( (f(x_1), f(y_1)) = (f(x_2), f(y_2)) \).

This equality of ordered pairs means:

• \( f(x_1) = f(x_2) \)
• \( f(y_1) = f(y_2) \)

Since \( f \) is injective (as established above), \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), and \( f(y_1) = f(y_2) \) implies \( y_1 = y_2 \).

Therefore, \( (x_1, y_1) = (x_2, y_2) \). This shows that \( F \) is injective.

Option 1 is true.

Analyzing Option 2: f is increasing

We know that \( f'(x) \ne 0 \) for all \( x \in (0, 1) \). This means \( f'(x) \) is either strictly positive throughout \( (0, 1) \) or strictly negative throughout \( (0, 1) \). If \( f'(x) > 0 \), \( f \) is increasing. If \( f'(x) < 0 \), \( f \) is decreasing.

The condition \( f'(x) \ne 0 \) does not guarantee that \( f'(x) > 0 \). For example, the function \( f(x) = 1-x \) defined on \( (0, 1) \) is differentiable, surjective from \( (0, 1) \) to \( (0, 1) \), and \( f'(x) = -1 \ne 0 \) for all \( x \). However, this function is decreasing.

Therefore, \( f \) is not necessarily increasing. It could be decreasing.

Option 2 is false.

Analyzing Option 3: F is bijective

The statement says that for every \( (x', y') \in (0, 1) \times (0, 1) \), there exists a unique \( (x, y) \in (0, 1) \times (0, 1) \) such that \( F(x, y) = (x', y') \). This is the definition of \( F \) being a bijection (both injective and surjective).

We already showed in the analysis of Option 1 that \( F \) is injective because \( f \) is injective.

Now let's check surjectivity. For any \( (x', y') \in (0, 1) \times (0, 1) \), we need to find \( (x, y) \in (0, 1) \times (0, 1) \) such that \( F(x, y) = (x', y') \).

\( F(x, y) = (f(x), f(y)) = (x', y') \).

This means we need \( f(x) = x' \) and \( f(y) = y' \). Since \( f \) is surjective from \( (0, 1) \) to \( (0, 1) \), for any \( x' \in (0, 1) \), there exists an \( x \in (0, 1) \) such that \( f(x) = x' \). Similarly, for any \( y' \in (0, 1) \), there exists a \( y \in (0, 1) \) such that \( f(y) = y' \).

Furthermore, since \( f \) is injective, the \( x \) and \( y \) found are unique for the given \( x' \) and \( y' \).

Thus, for every \( (x', y') \in (0, 1) \times (0, 1) \), there exists a unique \( (x, y) \in (0, 1) \times (0, 1) \) such that \( F(x, y) = (x', y') \). This confirms that \( F \) is a bijection.

Option 3 is true.

Analyzing Option 4: Total derivative DF(x, y) is invertible

The function \( F(x, y) \) is a multivariable function from \( \mathbb{R}^2 \) to \( \mathbb{R}^2 \), given by \( F(x, y) = (F_1(x, y), F_2(x, y)) = (f(x), f(y)) \). The total derivative of \( F \) at \( (x, y) \) is represented by the Jacobian matrix:

Let's compute the partial derivatives:

• \( \frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x} (f(x)) = f'(x) \)
• \( \frac{\partial F_1}{\partial y} = \frac{\partial}{\partial y} (f(x)) = 0 \) (since \( f(x) \) does not depend on \( y \))
• \( \frac{\partial F_2}{\partial x} = \frac{\partial}{\partial x} (f(y)) = 0 \) (since \( f(y) \) does not depend on \( x \))
• \( \frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y} (f(y)) = f'(y) \)

So, the Jacobian matrix is:

A matrix is invertible if and only if its determinant is non-zero.

The determinant of \( DF(x, y) \) is:

The problem states that \( f'(x) \ne 0 \) for every \( x \in (0, 1) \). Since \( (x, y) \in (0, 1) \times (0, 1) \), both \( x \) and \( y \) are in \( (0, 1) \).

Therefore, \( f'(x) \ne 0 \) and \( f'(y) \ne 0 \).

The product of two non-zero numbers is non-zero. So, \( f'(x) f'(y) \ne 0 \).

Thus, \( \det(DF(x, y)) \ne 0 \) for all \( (x, y) \in (0, 1) \times (0, 1) \).

This means the total derivative \( DF(x, y) \) is invertible for all \( (x, y) \in (0, 1) \times (0, 1) \).

Option 4 is true.

Based on the analysis, options 1, 3, and 4 are true statements.