Question

Let $f\left(x\right)=\cos \left({\cos }^{-1}\left(\frac{2}{\left[x\right]-1}\right)\right)$ where $[\cdot ]$ is the greatest integer function, then

• A. Range of $f\left(x\right)=[-1,1]-\left\{0\right\}$
• B. Domain of $f\left(x\right)=R-\{0,1\}$
• C. Domain of $f\left(x\right)$ excludes $3$ integral values of $x$
• D. Range of $f\left(x\right)=\{\frac{2}{k}:k\in Z-\{-1,0,1\left\}\right\}$

Answer 1

Answer: (C), (D)

Range of $f\left(x\right)=\{\frac{2}{k}:k\in Z-\{-1,0,1\left\}\right\}$

$f\left(x\right)=\cos \left({\cos }^{-1}\left(\frac{2}{\left[x\right]-1}\right)\right)$
For $f\left(x\right)$ to be defined,
$-1\le \frac{2}{\left[x\right]-1}\le 1\Rightarrow \frac{2}{\left[x\right]-1}+1\ge 0\text{and}\frac{2}{\left[x\right]-1}-1\le 0\Rightarrow \frac{\left[x\right]+1}{\left[x\right]-1}\ge 0\text{and}-\frac{\left[x\right]-3}{\left[x\right]-1}\le 0\Rightarrow \frac{\left[x\right]+1}{\left[x\right]-1}\ge 0\text{and}\frac{\left[x\right]-3}{\left[x\right]-1}\ge 0\Rightarrow \left[x\right]\in (-\infty ,-1]\cup [3,\infty )\Rightarrow x\in (-\infty ,0)\cup [3,\infty )$
$\therefore$ Domain of $f\left(x\right)$ excludes $3$ integral values i.e. $\{0,1,2\}$

Now,
$f\left(x\right)=\cos \left({\cos }^{-1}\frac{2}{\left[x\right]-1}\right)$
$\Rightarrow y=\frac{2}{\left[x\right]-1}$
As $\left[x\right]\in Z-\{0,1,2\}$
$\Rightarrow \left[x\right]-1\in Z-\{-1,0,1\}$
$\therefore$ Range of $f\left(x\right)$ is
$\{\frac{2}{k}:k\in Z-\{-1,0,1\left\}\right\}$