Question
Find the domain of the following function:$\text{f(x)}=\sin^{-1}\text{x}+\sin^{-1}2\text{x}$

Answer

Let f(x) = g(x) + h(x), where
Therefore the domain of f(x) is given by intersection of the domain of g(x) and h(x)
The domain of g(x) is [-1, 1]
The domain of h(x) is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$
Therefore, the intersection of g(x) and h(x) is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$
Hence, the domain is $\Big[-\frac{1}{2},\frac{1}{2}\Big]$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

In the following cases, find the distance of each of the given points from the corresponding given plane.
Point: $(3, -2, 1)$
Plane: $2x - y + 2z + 3 = 0$
Write a vector in the direction of vector $5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ which has magnitude of 8 unit.
If u, v and w are functions of x, then show that 
$\frac{d}{d x}(u . v . w)=\frac{d u}{d x} v . w+u . \frac{d v}{d x} \cdot w+u \cdot v \frac{d w}{d x}$ 
in two ways - first by repeated application of product rule, second by logarithmic differentiation.
vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are such that $|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=\frac{2}{3}$ and $\big(\vec{\text{a}}\times\vec{\text{b}}\big)$ is a unit vector. write the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Evaluate the integral in Exercise:
$\int_{0}^{1}\frac{\text{x}}{\text{x}^{2}+1}\text{dx}$
Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are given by:
$\text{a}_\text{ij}=\frac{|2\text{i}-3\text{j}|}{2}$
Show that $f(x)=(x-1) e^x+1$ is an increasing function for all $x > 0$.
Find $\lambda $ and $\mu $ if $\left( {2\hat i + 6\hat j + 27\hat k} \right) \times \left( {\hat i + \lambda \hat j + u\hat k} \right) = \overrightarrow 0 $
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{25\pi}{6}\Big)$
Write a vector satisfying $\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1.$