Question

The domain of ${{\sin }^{-1}}\left[ x \right]$, where $\left[ x \right]$ is the greatest integer function, given by: -
(a) [-1, 1]
(b) [-1, 2)
(c) {-1, 0, 1}
(d) None of these

Answer 1

Consider the range of ${{\sin }^{-1}}\left( a \right)$ as: - $\dfrac{-\pi }{2}\le {{\sin }^{-1}}a\le \dfrac{\pi }{2}$. Take sine function in the above relation to get rid of the sine inverse function. Now, replace ‘a’ with [x] and consider the two cases of inequality. Use the definition of greatest integer function to find the set of values of x in both the cases. Take their intersection to get the answer.

Complete step-by-step solution:
We have been asked to find the domain of the function, ${{\sin }^{-1}}\left[ x \right]$. That means we have to find the values of ‘x’ for which the function is defined.
Now, we know that the range of ${{\sin }^{-1}}\left( a \right)$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$.
Therefore, mathematically it can be written as: -
$\Rightarrow \dfrac{-\pi }{2}\le {{\sin }^{-1}}a\le \dfrac{\pi }{2}$
Taking sine function in all the term, we get,
$\Rightarrow \sin \left( \dfrac{-\pi }{2} \right)\le \sin \left( {{\sin }^{-1}}a \right)\le \sin \left( \dfrac{\pi }{2} \right)$
Applying the formula: - $\sin \left( {{\sin }^{-1}}a \right)=a$, we get,
$\Rightarrow -1\le a\le 1$
Substituting, a = [x], we get,
$\Rightarrow -1\le \left[ x \right]\le 1$
Here, $\left[ x \right]$ is called the greatest integer function.
Let us see, what is the greatest integer function?
So, a greatest integer function, denoted by $\left[ x \right]$, takes a real number as input and outputs the nearest integer which is equal to less than the number. For example: -

& \Rightarrow x=0.1\Rightarrow \left[ x \right]=0 \\\ & \Rightarrow x=-0.2\Rightarrow \left[ x \right]=-1 \\\ & \Rightarrow x=-9.2\Rightarrow \left[ x \right]=-10 \\\ & \Rightarrow x=2\Rightarrow \left[ x \right]=2 \\\ \end{aligned}$$ Clearly, we can see that if x is an integer then $$\left[ x \right]$$ will give the same integer. If x is not an integer then $$\left[ x \right]$$ will give an integer less than that real number. Now, let us come to the condition we have found: - $$\Rightarrow -1\le \left[ x \right]\le 1$$ (i) Case (i): - Consider, $$\left[ x \right]\ge -1$$, For $$\left[ x \right]$$ to be greater than or equal to -1, the value of x can be any real number from -1 to $$\infty $$. $$\Rightarrow x\in \left[ -1,\infty \right)$$ (ii) Case (ii): - Consider, $$\left[ x \right]\le 1$$, For $$\left[ x \right]$$ to be less than or equal to 1, the value of x can be any real number between $$-\infty $$ and 2. $$\Rightarrow x\in \left( -\infty ,2 \right)$$ Since, we have to satisfy both the cases, therefore, we have to consider the intersection of the two sets of values of x. $$\begin{aligned} & \Rightarrow x\in \left[ -1,\infty \right)\cap \left( -\infty ,2 \right) \\\ & \Rightarrow x\in \left[ -1,2 \right) \\\ \end{aligned}$$ **Hence, option (b) is the correct answer.** **Note:** One may note that in case (i) we cannot consider the values of x less than -1 because if we even deviate a little from -1, the value of $$\left[ x \right]$$ will be -2 which should not be our answer. Also, note that in case (ii), we cannot consider the value of x equal to 2 or greater than 2. If we consider x = 2, then $$\left[ x \right]$$ will become 2 which contradicts the condition, $$\left[ x \right]\le 1$$.