Question:

Step 1: Analyze $f(x) = \ln(\tan(\pi[x])) + |x^2+2x-3|$. For $\ln(\tan(\pi[x]))$ to be defined, $\tan(\pi[x])$ must be positive. However, for any integer value of $[x]$, $\tan(\pi[x]) = 0$. The logarithm $\ln(0)$ is undefined. Therefore, the domain $D_f$ is empty. A function with an empty domain cannot be surjective. Step 2: Analyze $g(x) = \dfrac{x^2+2x-3}{x-1}$. The function can be simplified as $g(x) = \dfrac{(x+3)(x-1)}{x-1}$. The domain of $g(x)$ is $D_g = \mathbb{R} \setminus \{1\}$. For $x \in D_g$, $g(x) = x+3$. The value $g(1)$ is undefined. If $x=1$ were in the domain, $g(1)$ would be $1+3=4$. Since $x=1$ is excluded from the domain, the value $4$ is not attained by $g(x)$. Thus, the range of $g(x)$ is $\mathbb{R} \setminus \{4\}$. The codomain is given as $\mathbb{R}$. Since the range $\mathbb{R} \setminus \{4\}$ is not equal to the codomain $\mathbb{R}$, the function $g$ is not surjective. Step 3: Analyze $h(x) = \ln\left(\dfrac{1-x}{1+x}\right)$. For $h(x)$ to be defined, the argument of the logarithm must be positive: $\dfrac{1-x}{1+x} > 0$. This inequality holds when $x \in (-1, 1)$. Thus, the domain $D_h = (-1, 1)$. To find the range, consider the behavior of $h(x)$ at the boundaries of its domain: As $x \to -1^+$, $\dfrac{1-x}{1+x} \to \dfrac{1-(-1)}{0^+} = \dfrac{2}{0^+} \to +\infty$. Therefore, $h(x) \to \ln(+\infty) \to +\infty$. As $x \to 1^-$, $\dfrac{1-x}{1+x} \to \dfrac{0^+}{1+1} = \dfrac{0^+}{2} \to 0^+$. Therefore, $h(x) \to \ln(0^+) \to -\infty$. Since $h(x)$ is a continuous function on its domain $(-1, 1)$ and its values span from $-\infty$ to $+\infty$, the range of $h(x)$ is $\mathbb{R}$. The codomain is given as $\mathbb{R}$. Since the range equals the codomain, $h$ is surjective. Step 4: Analyze $k(x) = \sqrt{[x]+[-x]+1} + \sqrt{\{x\}+\{-x\}+1}$. We use the properties of the greatest integer function and fractional part function: 1. If $x$ is an integer ($x \in \mathbb{Z}$): $[x]+[-x] = x+(-x) = 0$. $\{x\}+\{-x\} = 0+0 = 0$. In this case, $k(x) = \sqrt{0+1} + \sqrt{0+1} = \sqrt{1} + \sqrt{1} = 1+1 = 2$. 2. If $x$ is not an integer ($x \notin \mathbb{Z}$): $[x]+[-x] = -1$. $\{x\}+\{-x\} = 1$. In this case, $k(x) = \sqrt{-1+1} + \sqrt{1+1} = \sqrt{0} + \sqrt{2} = 0+\sqrt{2} = \sqrt{2}$. The range of $k(x)$ is $\{\sqrt{2}, 2\}$. The codomain is given as $\mathbb{R}^+ = (0, \infty)$. Since the range $\{\sqrt{2}, 2\}$ is not equal to the codomain $\mathbb{R}^+$, the function $k$ is not surjective.