Question

Find range and domain
\[f\left( x \right)=\text{ si}{{\text{n}}^{-1}}\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}+2} \right)\]
\[A.\left[ 0,\dfrac{\pi }{2} \right]\]
\[B.\left( 0,\dfrac{\pi }{3} \right)\]
\[C.\left[ 0,\dfrac{\pi }{4} \right)\]
\[D.\text{ None of these}\]

Answer 1

Hint: Here, we are going to find the domain and range of the given function, by taking the identity function which is \[\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}+2} \right)\], hence we are about to find the domain the function. After that, range is solved based on the definition related to sine function in the range.

Complete step by step solution:
Let us solve the given problem,
\[f\left( x \right)=\text{ si}{{\text{n}}^{-1}}\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}+2} \right)\]
We know that the inverse sine function is only defined for \[-1\le x\le 1\], hence we know that our function is only defined when \[\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}+2} \right)\] obeys that same interval.
Substituting the given function in \[-1\le x\le 1\], we get
\[-1\le \left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}+2} \right)\ge 1\]
Let us consider the case 1 and case 2,
Case 1:\[\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}+2} \right)\ge 1\]
Now taking the \[{{x}^{2}}+2\]on right-hand side, then we get
\[\Rightarrow {{x}^{2}}+1\ge -\left( {{x}^{2}}+2 \right)\]
Separating the \[{{x}^{2}}\]terms and constant terms,
 \[\Rightarrow {{x}^{2}}+{{x}^{2}}\ge -2-1\]
Adding the terms, we get as
\[\Rightarrow 2{{x}^{2}}\ge -3\]
Therefore, \[\Rightarrow {{x}^{2}}\ge \dfrac{-3}{2}\].

Case 2:
 \[\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}+2} \right)\le 1\]
\[\Rightarrow {{x}^{2}}+1\le \left( {{x}^{2}}+2 \right)\]
Thus, for the above inequality system, there are no values of x that make the equation true.
Hence no solution.

Range of the function normally ranges from \[-\dfrac{\pi }{2}\]to \[\dfrac{\pi }{2}\], since but negative values are produced only by negative inputs to the function, the \[\left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}+2} \right)\] makes it so that we get only positive (0) values.
Range of the function will be:\[\left\{ y:0\le y\le \dfrac{\pi }{2} \right\}\].
The correct answer is option(A).

Note: Whenever we face such questions the key concept is to be clear about the definitions of domain and range. Do not mistake the closed brackets for range and domain here, keep in mind we have just the value that occurs before that number where the function is undefined but we have to exclude the value where the function is undefined.