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# What is the domain of the function $f\left( x \right) = {\sin ^{ - 1}}\left( {{{\log }_2}\left( {\dfrac{{{x^2}}}{2}} \right)} \right)$ ?(A) $\left[ { - 2, - 1} \right]$ (B) $\left[ {1,2} \right]$ (C) $\left[ { - 2, - 1} \right] \cup \left[ {1,2} \right]$ (D) Not defined

Last updated date: 20th Jun 2024
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Hint:Try to solve the problem from the outermost function, i.e. inverse sine function. Use the domain of inverse sine function to form an inequality. Now further solve this inequality with the use of the definition of the logarithmic function.

Let’s first try to understand the question properly. Here we are given with a composite function which is a combination of Sine inverse function, logarithmic function and polynomial function. And we have to find the domain of the function$f\left( x \right)$.
For that, we need to know the definition of the domain of the function. The domain of a function is the set of all possible inputs for the function. So, for a function $g\left( x \right)$, the set of all the possible values of $'x'$ will be the function’s domain.
The best way to solve this problem is to go from outside to inside.
The domain of inverse sine function is $\left[ { - 1,1} \right]$, that means whatever is inside the scope of ${\sin ^{ - 1}}()$ must attain a value in the closed interval $\left[ { - 1,1} \right]$.
$\Rightarrow {\log _2}\left( {\dfrac{{{x^2}}}{2}} \right) \in \left[ { - 1,1} \right] \Rightarrow - 1 \leqslant {\log _2}\left( {\dfrac{{{x^2}}}{2}} \right) \text{and}\, {\log _2}\left( {\dfrac{{{x^2}}}{2}} \right) \leqslant 1 \Rightarrow - 1 \leqslant {\log _2}\left( {\dfrac{{{x^2}}}{2}} \right) \leqslant 1$
The above inequality is just showing that the insides of the inverse sine function should be greater than equal to $- 1$ and smaller than equal to$1$ .
But we know that by the definition of the logarithmic function: ${\log _a}b = y \Rightarrow {a^y} = b$ and the domain of the logarithmic function is positive integers which also satisfy the conditions here.
So, we can write that: ${2^{ - 1}} \leqslant \dfrac{{{x^2}}}{2} \text{and}\, \dfrac{{{x^2}}}{2} \leqslant {2^1} \Rightarrow {2^{ - 1}} \leqslant \dfrac{{{x^2}}}{2} \leqslant {2^1}$
We can carefully evaluate the above inequality
$\Rightarrow \dfrac{1}{2} \leqslant \dfrac{{{x^2}}}{2} \leqslant 2 \Rightarrow \dfrac{1}{2} \times 2 \leqslant {x^2} \leqslant 2 \times 2 \Rightarrow 1 \leqslant {x^2} \leqslant 4$
Now, for ${x^2} = 1 \Rightarrow x = \pm 1$ and for ${x^2} = 4 \Rightarrow x = \pm 2$
The above inequality can be represented as: $1 \leqslant {x^2} \leqslant 4 \Rightarrow - 1 \leqslant x \leqslant - 2$ and $1 \leqslant x \leqslant 2$
Therefore, for the function, $f\left( x \right)$ we have $x \in \left[ { - 2, - 1} \right] \cup \left[ {1,2} \right]$ as its domain.

So, the correct answer is “Option C”.

Note:Be careful with the signs while transformations in the inequality. The interval closed using square braces $'\left[ {\,} \right]'$ symbolizes for both sides of the closed interval it means starting and end of the values are also included.Students should remember domain and range of all trigonometric and inverse trigonometric functions for solving these types of problems.