### Related Formula For composite functions, g(f(x))$g(f(x))$ is determined by substituting the range of f(x)$f(x)$ into the appropriate domain intervals of g(y)$g(y)$: g(f(x)) = begincases f(x) & , & f(x) ge 0 \\ e^f(x) & , & f(x) < 0 endcases$g(f(x)) = \begin{cases} f(x) & , & f(x) \ge 0 \\ e^{f(x)} & , & f(x) < 0 \end{cases}$ ### Core Logic Let us analyze the definition of g(f(x))$g(f(x))$ branch-by-branch based on the domain of x$x$: 1. **Case 1: x le 0$x \le 0$** Here, f(x) = e^-x$f(x) = e^{-x}$. Since x le 0$x \le 0$, -x ge 0 implies e^-x ge 1 > 0$-x \ge 0 \implies e^{-x} \ge 1 > 0$. Since f(x) ge 0$f(x) \ge 0$, we use the upper branch of g(y)$g(y)$: g(f(x)) = f(x) = e^-x$g(f(x)) = f(x) = e^{-x}$ 2. **Case 2: x > 0$x > 0$** Here, f(x) = log_e x$f(x) = \log_e x$. - Subcase (a): If f(x) ge 0 implies log_e x ge 0 implies x ge 1$f(x) \ge 0 \implies \log_e x \ge 0 \implies x \ge 1$. Then, g(f(x)) = f(x) = log_e x$g(f(x)) = f(x) = \log_e x$. - Subcase (b): If f(x) < 0 implies log_e x < 0 implies 0 < x < 1$f(x) < 0 \implies \log_e x < 0 \implies 0 < x < 1$. Then, g(f(x)) = e^f(x) = e^log_e x = x$g(f(x)) = e^{f(x)} = e^{\log_e x} = x$. ### Step 1: Constructing the Composition Function Combining the branches obtained, the composite function is: g(f(x)) = begincases e^-x & , & x le 0 \\ x & , & 0 < x < 1 \\ log_e x & , & x ge 1 endcases$g(f(x)) = \begin{cases} e^{-x} & , & x \le 0 \\ x & , & 0 < x < 1 \\ \log_e x & , & x \ge 1 \end{cases}$

### Step 2: Injectivity and Surjectivity Analysis - **Injectivity (One-One Check):** Let's test two different inputs: x_1 = 0$x_1 = 0$ and x_2 = e$x_2 = e$. g(f(0)) = e^-0 = 1$g(f(0)) = e^{-0} = 1$ g(f(e)) = log_e e = 1$g(f(e)) = \log_e e = 1$ Since distinct inputs yield identical outputs (g(f(0)) = g(f(e)) = 1$g(f(0)) = g(f(e)) = 1$), the function is **many-one** (not one-one). - **Surjectivity (Onto Check):** Evaluating the range across the branches: - For x le 0$x \le 0$, e^-x in [1, infty)$e^{-x} \in [1, \infty)$. - For 0 < x < 1$0 < x < 1$, x in (0, 1)$x \in (0, 1)$. - For x ge 1$x \ge 1$, \log_e x in [0, infty)$\log_e x \in [0, \infty)$. The union of these sets gives the total range as [0, infty)$[0, \infty)$. Since the codomain is given as mathbbR$\mathbb{R}$, textRange neq textCodomain$\text{Range} \neq \text{Codomain}$, so the function is **into** (not onto). Therefore, the function is neither one-one nor onto. ### Pattern Recognition Sees: Piecewise branch composition. Shortcut: Sketching the graph quickly shows that a horizontal line at y=1$y=1$ intersects the function multiple times (not one-one) and no part of the graph goes below the x-axis (not onto). Trap: Always determine the range of the inner function first to select the correct branch of the outer function. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions