Question

The number of elements in the domain of the function $f\left(x\right)={\sin }^{-1}\left(\frac{x^2-2x}{3}\right)+\sqrt{\left[x\right]+[-x]}$ where $[.]$ denotes greatest integer function, are

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

$f\left(x\right)={\sin }^{-1}\left(\frac{x^2-2x}{3}\right)+\sqrt{\left[x\right]+[-x]}$

${\sin }^{-1}\left(\frac{x^2-2x}{3}\right)$ is defined for

$-1\underset{–}{<}\frac{x^2-2x}{3}\underset{–}{<}1\Rightarrow -3\le x^2-2x\le 3\Rightarrow -3\le {(x-1)}^2-1\le 3\Rightarrow -2\le {(x-1)}^2\le 4\Rightarrow 0\le {(x-1)}^2\le 4\Rightarrow -2\le x-1\le 2\Rightarrow -1\le x\le 3$

And $\sqrt{\left[x\right]+[-x]}$ is defined only when $\left[x\right]+[-x]>0$

Which is always true for $x\in I$

Otherwise,

If $\left[x\right]=x-f$ where $f$ is the fractional part

then $[-x]=-x-1+f$

$\left[x\right]+[-x]=-1$

Hence, the domain is only integer values of $x$

Hence, the values of $x$ satisfying both the conditions are

$\Rightarrow x=-1,0,1,2,3$

$\therefore$ the total number of elements in the domain is $5$.