Question:

Step 1: Let $f(x) = \cos x$ and $g(x) = \ln x$. The domain of $g(x) = \ln x$ is $(0, \infty)$, and its range is $(-\infty, \infty)$. The composite function is $f(g(x)) = \cos(\ln x)$. Since the range of $\ln x$ is $(-\infty, \infty)$, and the cosine function maps all real numbers to the interval $[-1, 1]$, the range of $f(g(x))$ is $[-1, 1]$. Step 2: Let $f(x) = \dfrac{2}{\pi}(\sin^{-1}x + \cos^{-1}x)$ and $g(x) = \text{sgn}(x^2 - x + 1)$. For $x \in [-1, 1]$, the identity $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ holds. Thus, for $x \in [-1, 1]$, $$f(x) = \dfrac{2}{\pi} \cdot \dfrac{\pi}{2} = 1$$ Consider $g(x) = \text{sgn}(x^2 - x + 1)$. The discriminant of the quadratic $x^2 - x + 1$ is $\Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3$. Since $\Delta < 0$ and the leading coefficient is positive, $x^2 - x + 1 > 0$ for all $x \in \mathbb{R}$. Therefore, $g(x) = \text{sgn}(x^2 - x + 1) = 1$ for all $x \in \mathbb{R}$. Now, consider the composite function $f(g(x))$. The domain of $f(g(x))$ is $\mathbb{R}$ because $g(x)=1$ for all $x \in \mathbb{R}$, and $1$ is within the domain of $f(x)$ (which is $[-1,1]$ for the simplified form). Since $g(x) = 1$ for all $x \in \mathbb{R}$, and $1 \in [-1, 1]$, we have $f(g(x)) = f(1) = 1$. Therefore, $f(g(x)) = 1$ for all $x \in \mathbb{R}$, and $g(x) = 1$ for all $x \in \mathbb{R}$. Thus, $f(g(x))$ and $g(x)$ are identical functions. Step 3: Let $f: \mathbb{R} \to [-2, 2]$ be defined by $f(x) = \dfrac{2x}{1+x^2}$. To determine if $f$ is injective, consider $f(x) = f(y)$. Alternatively, observe that for $x \neq 0$, $$f\left(\frac{1}{x}\right) = \dfrac{2(1/x)}{1+(1/x)^2} = \dfrac{2/x}{(x^2+1)/x^2} = \dfrac{2}{x} \cdot \dfrac{x^2}{x^2+1} = \dfrac{2x}{x^2+1} = f(x)$$ For example, $f(2) = \dfrac{2(2)}{1+2^2} = \dfrac{4}{5}$ and $f(1/2) = \dfrac{2(1/2)}{1+(1/2)^2} = \dfrac{1}{1+1/4} = \dfrac{1}{5/4} = \dfrac{4}{5}$. Since $f(2) = f(1/2)$ but $2 \neq 1/2$, the function $f$ is not injective. A bijective function must be both injective and surjective. Since $f$ is not injective, it is not bijective. Step 4: Let $f(x) = \sin^{-1}x$ and $g(x) = \cos x$. Consider $f(g(x)) = \sin^{-1}(\cos x)$. The domain of $f(g(x))$ is $\mathbb{R}$ since the range of $\cos x$ is $[-1, 1]$, which is the domain of $\sin^{-1}x$. To check for parity, evaluate $f(g(-x))$: $$f(g(-x)) = \sin^{-1}(\cos(-x)) = \sin^{-1}(\cos x)$$ Since $f(g(-x)) = f(g(x))$, the function $f(g(x))$ is an even function. The statement claims $f(g(x))$ is odd, which is false. Consider $g(f(x)) = \cos(\sin^{-1}x)$. The domain of $g(f(x))$ is $[-1, 1]$ since this is the domain of $\sin^{-1}x$. To check for parity, evaluate $g(f(-x))$: $$g(f(-x)) = \cos(\sin^{-1}(-x))$$ Since $\sin^{-1}(-x) = -\sin^{-1}x$, we have: $$g(f(-x)) = \cos(-\sin^{-1}x)$$ Since $\cos$ is an even function, $\cos(-\theta) = \cos(\theta)$. Thus, $g(f(-x)) = \cos(\sin^{-1}x) = g(f(x))$. Therefore, $g(f(x))$ is an even function. The statement claims $g(f(x))$ is even, which is true. However, since the first part of the statement (that $f(g(x))$ is odd) is false, the entire statement is incorrect.