Key Concepts and Formulas

• Composite Function $f(g(x))$: For two functions $f$ and $g$, the composite function $f(g(x))$ is obtained by substituting the function $g(x)$ into the function $f(x)$. The domain of $f(g(x))$ consists of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.
• One-one (Injective) Function: A function $h: A \rightarrow B$ is one-one if for every $x_1, x_2 \in A$, $h(x_1) = h(x_2)$ implies $x_1 = x_2$. Graphically, any horizontal line intersects the graph of a one-one function at most once.
• Onto (Surjective) Function: A function $h: A \rightarrow B$ is onto if for every $y \in B$, there exists at least one $x \in A$ such that $h(x) = y$. This means the range of the function must be equal to its codomain.

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Step-by-Step Solution

Step 1: Define the functions $f(x)$ and $g(x)$ in piecewise form.

We are given $f(x) = |x-1|$ and $g(x) = \begin{cases} \mathrm{e}^x, & x \geq 0 \\ x+1, & x \leq 0 \end{cases}$.

The function $f(x) = |x-1|$ can be written as: $f(x) = \begin{cases} x-1, & \text{if } x-1 \geq 0 \implies x \geq 1 \\ -(x-1), & \text{if } x-1 < 0 \implies x < 1 \end{cases}$ So, $f(x) = \begin{cases} x-1, & \text{if } x \geq 1 \\ 1-x, & \text{if } x < 1 \end{cases}$ The domain of $f$ is $\mathbf{R}$, and its range is $[0, \infty)$.

The function $g(x)$ is already given in piecewise form: $g(x) = \begin{cases} \mathrm{e}^x, & \text{if } x \geq 0 \\ x+1, & \text{if } x \leq 0 \end{cases}$ The domain of $g$ is $\mathbf{R}$.

Let's analyze the range of $g(x)$:

• For $x \geq 0$, $g(x) = \mathrm{e}^x$. As $x$ ranges from $0$ to $\infty$, $\mathrm{e}^x$ ranges from $\mathrm{e}^0 = 1$ to $\infty$. So, the range for this part is $[1, \infty)$.
• For $x \leq 0$, $g(x) = x+1$. As $x$ ranges from $-\infty$ to $0$, $x+1$ ranges from $-\infty$ to $1$. So, the range for this part is $(-\infty, 1]$.

Combining both parts, the range of $g(x)$ is $(-\infty, 1] \cup [1, \infty) = \mathbf{R}$.

Step 2: Determine the composite function $f(g(x))$.

The composite function $f(g(x))$ means we substitute $g(x)$ into $f(x)$. We need to consider the two cases for $g(x)$ based on its definition and apply the definition of $f$ accordingly.

Case 1: $x \geq 0$. In this case, $g(x) = \mathrm{e}^x$. Now we need to apply $f$ to $g(x) = \mathrm{e}^x$. We must check whether $g(x) \geq 1$ or $g(x) < 1$. Since $x \geq 0$, $\mathrm{e}^x \geq \mathrm{e}^0 = 1$. So, for $x \geq 0$, $g(x) = \mathrm{e}^x \geq 1$. Therefore, we use the first case of $f(x)$, where $f(y) = y-1$ for $y \geq 1$. Substituting $y = g(x) = \mathrm{e}^x$: $f(g(x)) = f(\mathrm{e}^x) = \mathrm{e}^x - 1$, for $x \geq 0$.

Case 2: $x \leq 0$. In this case, $g(x) = x+1$. Now we need to apply $f$ to $g(x) = x+1$. We must check whether $g(x) \geq 1$ or $g(x) < 1$. For $x \leq 0$, $x+1 \leq 0+1 = 1$. So, for $x \leq 0$, $g(x) = x+1 \leq 1$. This means we need to consider two sub-cases for $g(x) = x+1$: when $g(x) \geq 1$ and when $g(x) < 1$. However, we already established that for $x \leq 0$, $g(x) \leq 1$. So, if $g(x) = 1$, which happens when $x+1=1 \implies x=0$, we use $f(1) = 1-1 = 0$. If $g(x) < 1$, which happens when $x+1 < 1 \implies x < 0$, we use the second case of $f(x)$, where $f(y) = 1-y$ for $y < 1$. Substituting $y = g(x) = x+1$: $f(g(x)) = f(x+1) = 1 - (x+1) = 1 - x - 1 = -x$, for $x < 0$.

Let's combine these results. We need to be careful at $x=0$. For $x=0$: $g(0) = \mathrm{e}^0 = 1$ (using the $x \geq 0$ case for $g$). $f(g(0)) = f(1) = |1-1| = 0$. From Case 1 ($x \geq 0$), $f(g(x)) = \mathrm{e}^x - 1$. At $x=0$, this gives $\mathrm{e}^0 - 1 = 1 - 1 = 0$. From Case 2 ($x \leq 0$), if we consider $x=0$, $g(0) = 0+1 = 1$ (using the $x \leq 0$ case for $g$). $f(g(0)) = f(1) = |1-1| = 0$. If we use $f(g(x)) = -x$ for $x < 0$, at $x=0$ this would give $0$. So, the composite function $f(g(x))$ is: $f(g(x)) = \begin{cases} \mathrm{e}^x - 1, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$

Step 3: Check if $f(g(x))$ is one-one.

To check for one-one property, we see if distinct inputs map to distinct outputs.

Consider the interval $x \geq 0$, where $f(g(x)) = \mathrm{e}^x - 1$. The function $\mathrm{e}^x$ is strictly increasing for all $x$. Therefore, $\mathrm{e}^x - 1$ is also strictly increasing for $x \geq 0$. This means that for any $x_1, x_2 \geq 0$, if $x_1 \neq x_2$, then $\mathrm{e}^{x_1} - 1 \neq \mathrm{e}^{x_2} - 1$. So, the function is one-one on $[0, \infty)$.

Consider the interval $x < 0$, where $f(g(x)) = -x$. The function $-x$ is strictly decreasing for all $x$. Therefore, for any $x_1, x_2 < 0$, if $x_1 \neq x_2$, then $-x_1 \neq -x_2$. So, the function is one-one on $(-\infty, 0)$.

Now we need to check if there are any overlaps in the outputs between the two intervals. For $x \geq 0$, the range of $f(g(x)) = \mathrm{e}^x - 1$ is $[\mathrm{e}^0 - 1, \infty) = [0, \infty)$. For $x < 0$, the range of $f(g(x)) = -x$ is $(0, \infty)$ (since $x$ approaches 0 from the negative side, $-x$ approaches 0 from the positive side, and as $x \to -\infty$, $-x \to \infty$).

Let's check if different inputs from different intervals can map to the same output. Consider an output $y$. If $y \geq 0$, we can find an $x \geq 0$ such that $\mathrm{e}^x - 1 = y \implies \mathrm{e}^x = y+1 \implies x = \ln(y+1)$. This $x$ is valid if $y+1 \geq 1 \implies y \geq 0$. If $y > 0$, we can find an $x < 0$ such that $-x = y \implies x = -y$. This $x$ is valid if $-y < 0 \implies y > 0$.

Let's test some values. If $x_1 = 1$ (from $x \geq 0$), $f(g(1)) = \mathrm{e}^1 - 1 = e-1$. If $x_2 = -(e-1)$ (from $x < 0$), $f(g(-(e-1))) = -(-(e-1)) = e-1$. Since $1 \neq -(e-1)$ (as $e-1 \approx 1.718 > 0$), but $f(g(1)) = f(g(-(e-1)))$, the function $f(g(x))$ is not one-one.

Step 4: Check if $f(g(x))$ is onto.

The codomain of $f(g(x))$ is $\mathbf{R}$. We need to see if the range of $f(g(x))$ covers all real numbers.

From Step 3, we found the range of $f(g(x))$ for $x \geq 0$ is $[0, \infty)$. And the range of $f(g(x))$ for $x < 0$ is $(0, \infty)$.

The total range of $f(g(x))$ is the union of these two ranges: $[0, \infty) \cup (0, \infty) = [0, \infty)$.

Since the range of $f(g(x))$ is $[0, \infty)$, and the codomain is $\mathbf{R}$, the range is not equal to the codomain. Specifically, negative real numbers are not in the range of $f(g(x))$. Therefore, the function $f(g(x))$ is not onto.

Step 5: Conclude the properties of $f(g(x))$.

We have determined that $f(g(x))$ is neither one-one nor onto.

This corresponds to option (A).

Common Mistakes & Tips

• Careful with piecewise function composition: When composing piecewise functions, ensure you correctly identify the conditions on the inner function's output that determine which piece of the outer function to apply.
• Check endpoints of intervals: Pay close attention to the values of the composite function at the boundaries of the piecewise intervals. This is crucial for determining continuity and the exact range.
• Range of $g(x)$: Before composing, understanding the range of the inner function $g(x)$ is helpful in determining the domain for the outer function $f$ and subsequently the range of the composite function.

Summary

To find the composite function $f(g(x))$, we first defined both $f(x)$ and $g(x)$ in piecewise forms. We then substituted $g(x)$ into $f(x)$, considering the different definitions of $g(x)$ and the conditions under which $f(x)$ is defined. This resulted in a piecewise definition for $f(g(x))$. We then analyzed the one-one and onto properties of $f(g(x))$ by examining its behavior over its domain and determining its range. We found that $f(g(x))$ is not one-one because different inputs can lead to the same output, and it is not onto because its range does not cover all real numbers.

The final answer is \boxed{A}.