Analyzing Bijective Functions on Set A = (-1, 1)

The question asks us to determine which of the given functions, \(f(x) = x|x|\) and \(g(x) = \cos(\pi x)\), are bijective functions when considered as functions mapping from set \(A\) to itself.

First, let's clarify the set \(A\). The notation \(A = \{x \in \mathbb{R}: -1 > x < 1\}\) is unconventional. We will interpret it as the standard open interval \(A = \{x \in \mathbb{R}: -1 < x < 1\}\), which is \(A = (-1, 1)\).

A function is bijective if and only if it is both injective (one-to-one) and surjective (onto).

• Injective Function: A function \(f\) is injective if \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\) for all \(x_1, x_2\) in the domain.
• Surjective Function: A function \(f: A \to A\) is surjective if its range is equal to its codomain (\(A\)). In other words, for every element \(y\) in the codomain \(A\), there exists at least one element \(x\) in the domain \(A\) such that \(f(x) = y\).

We need to check these properties for both \(f(x)\) and \(g(x)\) on the domain and codomain \(A = (-1, 1)\).

Function 1: Investigating \(f(x) = x|x|\)

We can define \(f(x)\) piecewise:

• If \(x \ge 0\), then \(|x| = x\), so \(f(x) = x \cdot x = x^2\).
• If \(x < 0\), then \(|x| = -x\), so \(f(x) = x \cdot (-x) = -x^2\).

Thus, \(f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0 \end{cases}\).

1. Injectivity Check for \(f(x)\) on \(A = (-1, 1)\):

Assume \(f(x_1) = f(x_2)\) for \(x_1, x_2 \in (-1, 1)\).

• Case 1: \(x_1, x_2 \ge 0\). Then \(f(x_1) = x_1^2\) and \(f(x_2) = x_2^2\). If \(x_1^2 = x_2^2\), then \(x_1 = x_2\) since \(x_1, x_2\) are non-negative.
• Case 2: \(x_1, x_2 < 0\). Then \(f(x_1) = -x_1^2\) and \(f(x_2) = -x_2^2\). If \(-x_1^2 = -x_2^2\), then \(x_1^2 = x_2^2\). Since \(x_1, x_2\) are negative, \(x_1 = x_2\).
• Case 3: \(x_1 \ge 0\) and \(x_2 < 0\). Then \(f(x_1) = x_1^2\) and \(f(x_2) = -x_2^2\). If \(x_1^2 = -x_2^2\), then \(x_1^2 + x_2^2 = 0\). Since squares are non-negative, this only holds if \(x_1 = 0\) and \(x_2 = 0\). However, we assumed \(x_2 < 0\). Thus, this case implies \(x_1=0\), and for \(f(x_1)=f(x_2)\), we must have \(f(0)=0\). If \(x_2 < 0\), then \(f(x_2) = -x_2^2\). Setting \(-x_2^2 = 0\) implies \(x_2=0\), contradicting \(x_2 < 0\). So, this case cannot lead to \(f(x_1)=f(x_2)\) unless \(x_1=x_2=0\).

In all valid scenarios, \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). Therefore, \(f(x)\) is injective on \(A = (-1, 1)\).

2. Surjectivity Check for \(f(x)\) on \(A = (-1, 1)\):

We need to find the range of \(f(x)\) for \(x \in (-1, 1)\).

• If \(x \in [0, 1)\), then \(f(x) = x^2\). The values range from \(0^2 = 0\) to values approaching \(1^2 = 1\). So, \(f(x) \in [0, 1)\).
• If \(x \in (-1, 0)\), then \(f(x) = -x^2\). The values range from values approaching \(-(0)^2 = 0\) down to values approaching \(-(-1)^2 = -1\). So, \(f(x) \in (-1, 0)\).

Combining these, the range of \(f(x)\) on \((-1, 1)\) is \((-1, 0) \cup [0, 1) = (-1, 1)\). This is exactly the codomain \(A\). Therefore, \(f(x)\) is surjective on \(A = (-1, 1)\).

Conclusion for \(f(x)\): Since \(f(x) = x|x|\) is both injective and surjective on \(A = (-1, 1)\), it is a bijective function.

Function 2: Investigating \(g(x) = \cos(\pi x)\)

We need to check if \(g(x) = \cos(\pi x)\) is bijective from \(A = (-1, 1)\) to \(A = (-1, 1)\).

1. Injectivity Check for \(g(x)\) on \(A = (-1, 1)\):

Let's test if different inputs can give the same output.

• Consider \(x_1 = 0.5\). Then \(g(0.5) = \cos(\pi \cdot 0.5) = \cos(\pi/2) = 0\).
• Consider \(x_2 = -0.5\). Then \(g(-0.5) = \cos(\pi \cdot (-0.5)) = \cos(-\pi/2) = 0\).

We have \(x_1 = 0.5 \ne -0.5 = x_2\), but \(g(0.5) = g(-0.5) = 0\). Therefore, \(g(x)\) is not injective on \(A = (-1, 1)\).

2. Surjectivity Check for \(g(x)\) on \(A = (-1, 1)\):

Let's determine the range of \(g(x) = \cos(\pi x)\) for \(x \in (-1, 1)\).

• The input to the cosine function is \(\pi x\). As \(x\) ranges from \(-1\) to \(1\) (exclusive), \(\pi x\) ranges from \(-\pi\) to \(\pi\) (exclusive).
• The cosine function \(\cos(\theta)\) has a maximum value of \(1\) (at \(\theta = 0\)) and approaches \(-1\) as \(\theta\) approaches \(\pi\) or \(-\pi\).
• So, for \(\theta = \pi x\) where \(x \in (-1, 1)\), the range of \(\cos(\theta)\) is \((\cos(\pi), \cos(0)] = (-1, 1]\).

The range of \(g(x)\) is \((-1, 1]\). However, the codomain is \(A = (-1, 1)\). Since the range \((-1, 1]\) is not equal to the codomain \((-1, 1)\) (because \(1\) is in the range but not in the codomain), the function \(g(x)\) is not surjective.

Conclusion for \(g(x)\): Since \(g(x) = \cos(\pi x)\) is neither injective nor surjective on \(A = (-1, 1)\), it is not a bijective function.

Final Conclusion

Based on the analysis:

• \(f(x) = x|x|\) is bijective from \(A = (-1, 1)\) to itself.
• \(g(x) = \cos(\pi x)\) is not bijective from \(A = (-1, 1)\) to itself.

Therefore, only the first function is bijective.