Mathematics - Functions Question with Solution | TestHub

We know that, ${\cos }^{-1}x$ is defined when $-1⩽x⩽1$

So, ${\cos }^{-1}\left(\frac{x^2-5x+6}{x^2-9}\right)$ is defined when

$-1⩽\frac{x^2-5x+6}{x^2-9}⩽1$ and $x^2-9\ne 0$

 $\Rightarrow -1\le \frac{\left(x-3\right)\left(x-2\right)}{\left(x-3\right)\left(x+3\right)}\le 1$ and $x\ne \pm 3$

$\Rightarrow -1\le \frac{x-2}{x+3}\le 1$ and $x\ne \pm 3$, $\Rightarrow -1\le \frac{x+3-5}{x+3}\le 1$ and $x\ne \pm 3$

$\Rightarrow -1\le 1-\frac{5}{x+3}\le 1$ and $x\ne \pm 3$, 

$\Rightarrow -2\le -\frac{5}{x+3}⩽0$ and $x\ne \pm 3$

$\Rightarrow 0\le \frac{5}{x+3}⩽2$ and $x\ne \pm 3$, $\Rightarrow x+3\ge \frac{5}{2}$ and $x\ne \pm 3$

$\Rightarrow x\ge \frac{-1}{2}$ and $x\ne \pm 3$

$\therefore x\in \left[\frac{-1}{2},\infty \right)-\left\{3\right\} \cdots \left(1\right)$

Now, we know that $\log \left(x\right)$ is defined for $x>0$.

So, $\log \left(x^2-3x+2\right)$ is defined for $x^2-3x+2>0$, $\Rightarrow \left(x-1\right)\left(x-2\right)>0$

$\Rightarrow x>2$ or $x<1 \cdots \left(2\right)$

from equations $\left(1\right)$ & $\left(2\right)$

$\therefore x\in \left[-\frac{1}{2},1\right)\cup \left(2,\infty \right)-\left\{3\right\}$