Mathematics - Functions Question with Solution | TestHub

For $f(x)$, the argument of ${\sin }^{-1}$ must be in $[-1,1]$.

So, $-1\le \frac{x^2+1}{2x}\le 1$. For $x>0$, this implies $(x-1{)}^2\le 0$, which means $x=1$. For $x<0$, let $x=-t$ ($t>0$), then $-1\le \frac{t^2+1}{-2t}\le 1\Rightarrow -1\le -\frac{t^2+1}{2t}\le 1\Rightarrow -1\le \frac{t^2+1}{2t}\le 1$, which implies $t=1$, so $x=-1$. Thus, the domain of $f(x)$ is $\{-1,1\}$ (A is correct). For $x=1$, $f(1)={\sin }^{-1}(1)=\frac{\pi }{2}$. For $x=-1$, $f(-1)={\sin }^{-1}(-1)=-\frac{\pi }{2}$. So, the range of $f(x)$ is $\left\{-\frac{\pi }{2},\frac{\pi }{2}\right\}$ (B is correct). For $g(x)$, the argument of ${\cos }^{-1}$ must be in $[-1,1]$.

So, $-1\le \frac{x^2-1}{x^2+1}\le 1$. Since $x^2+1>0$, this is equivalent to $-(x^2+1)\le x^2-1\le x^2+1$. Both inequalities $2x^2\ge 0$ and $-1\le 1$ are true for all $x\in \mathbb{R}$. Thus, the domain of $g(x)$ is $\mathbb{R}$ (D is correct). Let $t=\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}$. Since $x^2\ge 0$, we have $x^2+1\ge 1$, so $0<\frac{1}{x^2+1}\le 1$. This implies $0<\frac{2}{x^2+1}\le 2$, so $-2\le -\frac{2}{x^2+1}<0$. Therefore, $-1\le 1-\frac{2}{x^2+1}<1$. So, $t\in [-1,1)$. The range of $g(x)$ is ${\cos }^{-1}([-1,1))$, which is $(0,\pi ]$. Thus, C is incorrect.