Question

Let $f\left( x \right) = {\sin ^{ - 1}}x$ and $g\left( x \right) = \dfrac{{{x^2} - x - 2}}{{2{x^2} - x - 6}}$. If $g\left( 2 \right) = \mathop {\lim }\limits_{x \to 2} g\left( x \right)$, then the domain of the function $fog$ is
1. $( - \infty , - 2] \cup \left[ {\dfrac{{ - 4}}{3},\infty } \right]$
2. $( - \infty , - 1] \cup \left[ {2,\infty } \right)$
3. $( - \infty , - 2] \cup \left[ { - 1,\infty } \right]$
4. $( - \infty ,2] \cup \left[ {\dfrac{{ - 3}}{2},\infty } \right)$

Answer 1

Hint:The process of combining two or more functions into a single function is known as function composition. The output of one function inside the parenthesis becomes the input of the outside function in the function composition. The range is the set of possible values for the function. To solve this question, use the range of $f\left( x \right)$.

Formula Used:
$fog = f\left( {g\left( x \right)} \right)$

Complete step by step Solution:
$f\left( x \right) = {\sin ^{ - 1}}x$and $g\left( x \right) = \dfrac{{{x^2} - x - 2}}{{2{x^2} - x - 6}}$
Here, $g\left( 2 \right) = \mathop {\lim }\limits_{x \to 2} g\left( x \right)$
$ = \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - x - 2}}{{2{x^2} - x - 6}}$
$ = \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 2x + x - 2}}{{2{x^2} - 4x + 3x - 6}}$
$ = \mathop {\lim }\limits_{x \to 2} \dfrac{{x\left( {x - 2} \right) + 1\left( {x - 2} \right)}}{{2x\left( {x - 2} \right) + 3\left( {x - 2} \right)}}$
$ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x + 1} \right)\left( {x - 2} \right)}}{{\left( {2x + 3} \right)\left( {x - 2} \right)}}$
$ = \dfrac{3}{7}$
As we know that, Domain of $fog$ is ${\sin ^{ - 1}}\left( {g\left( x \right)} \right)$
$ \Rightarrow \left| {g\left( x \right)} \right| \leqslant 1$ because the range of $\sin x$ is $\left[ { - 1,1} \right]$
Since $\left| {g\left( x \right)} \right| \leqslant 1$
$\therefore - 1 \leqslant g\left( x \right) \leqslant 1$
$ - 1 \leqslant \dfrac{{{x^2} - x - 2}}{{2{x^2} - x - 6}} \leqslant 1$
$ - 1 \leqslant \dfrac{{\left( {x + 1} \right)\left( {x - 2} \right)}}{{\left( {2x + 3} \right)\left( {x - 2} \right)}} \leqslant 1$
$ - 1 \leqslant \dfrac{{\left( {x + 1} \right)}}{{\left( {2x + 3} \right)}} \leqslant 1$
$ \Rightarrow \dfrac{{\left( {x + 1} \right)}}{{\left( {2x + 3} \right)}} \geqslant - 1,\dfrac{{\left( {x + 1} \right)}}{{\left( {2x + 3} \right)}} \leqslant 1$
$\dfrac{{\left( {x + 1} \right)}}{{\left( {2x + 3} \right)}} + 1 \geqslant 0,\dfrac{{\left( {x + 1} \right)}}{{\left( {2x + 3} \right)}} - 1 \leqslant 0$
$\dfrac{{\left( {x + 1} \right) + \left( {2x + 3} \right)}}{{\left( {2x + 3} \right)}} \geqslant 0,\dfrac{{\left( {x + 1} \right) - \left( {2x + 3} \right)}}{{\left( {2x + 3} \right)}} \leqslant 0$
$3x + 4 \geqslant 0, - x - 2 \leqslant 0$
$x \geqslant \dfrac{{ - 4}}{3},x \leqslant - 2$
$x \in \left[ {\dfrac{{ - 4}}{3},\infty } \right),x \in \left( { - \infty , - 2} \right]$
$ \Rightarrow x \in \left( { - \infty , - 2} \right] \cup \left[ {\dfrac{{ - 4}}{3},\infty } \right)$

Hence, the correct option is 1.

Note:The key concept involved in solving this problem is a good knowledge of the domain. Students must remember that a function's domain is the set of numbers that can be entered into a given function. Basically, the set of $x - $ values that you can put into any of the given equations or functions.