Question

Find the domain of $f\left( x \right)={{\cos }^{-1}}2x+{{\sin }^{-1}}x$ .

Answer 1

Hint: In this question we are given the function as an inverse trigonometric function. To solve the question first we need to find the domain of ${{\sin }^{-1}}x$ and ${{\cos }^{-1}}2x$ with the help of inverse trigonometric function in which ${{\sin }^{-1}}$ and ${{\cos }^{-1}}$ lies between -1 and 1.

Complete step-by-step answer:
We know that the value of ${{\sin }^{-1}}\theta $ always lies between -1 and 1 for any angle $\theta $ . Therefore, ${{\sin }^{-1}}\theta $ will be defined on the domain -1 and 1.
i.e. $-1\le \theta \le 1$ ……(1)
Now we will find the value of ${{\sin }^{-1}}x$ by substituting $\theta $ as $x$ in equation (1)
i.e, $-1\le \theta \le 1$
$\Rightarrow -1\le x\le 1$
Therefore, ${{\sin }^{-1}}x$ lies between -1 and 1 . So, the domain of ${{\sin }^{-1}}x$ is $\left[ -1,1 \right]$ ……(2)
We also know that the value of ${{\cos }^{-1}}\theta $ always lies between -1 and 1
Same as equation (1)
Now we will find the value of ${{\cos }^{-1}}2x$ by substituting $\theta $ as $2x$ in equation (1)
i.e. $-1\le \theta \le 1$
$\Rightarrow -1\le 2x\le 1$
By dividing $'2'$ throughout the above inequality, we get –
$\Rightarrow -\dfrac{1}{2}\le x\le \dfrac{1}{2}$
Therefore, ${{\cos }^{-1}}2x$ lies between \[-\dfrac{1}{2}\] and $\dfrac{1}{2}$ So, the domain of ${{\cos }^{-1}}2x$ is $\left[ -\dfrac{1}{2},\dfrac{1}{2} \right]$ …..(3)
Now, find the value of $f\left( x \right)={{\cos }^{-1}}2x+{{\sin }^{-1}}x$ .By substituting equation (2) and (3) . in
${{\cos }^{-1}}2x+{{\sin }^{-1}}x$ , we get –
\[{{\cos }^{-1}}2x+{{\sin }^{-1}}x=\left[ -\dfrac{1}{2},\dfrac{1}{2} \right]\cap \left[ -1,1 \right]\]
\[=\left[ -\dfrac{1}{2},\dfrac{1}{2} \right]\].

Note-: Students should know what is the domain and range of trigonometric functions. The domain of a function is the specific set of values that the independent variable in a function can take on. The range is the resulting value that the dependent variable can have as x varies throughout the domain.
They should not get confused between domain and range of inverse trigonometric functions.