
Write a range of $f\left( x \right)={{\sin }^{-1}}x$ other than $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
Answer
504.3k+ views
Hint: To find the range of given function other than the given range we will plot its graph and check its value in other points. Firstly we will draw a graph of a given function then we will see under what range its value falls. Then we will take a range accordingly and get our desired answer.
Complete step by step answer:
The function given to us is as follows:
$f\left( x \right)={{\sin }^{-1}}x$
The range is provided to us as follows:
$\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
We will find value of the function in this range as follows:
So at $x=-\dfrac{\pi }{2}$ we get the value as:
$\begin{align}
& f\left( \dfrac{\pi }{2} \right)={{\sin }^{-1}}\left( -\dfrac{\pi }{2} \right) \\
& \Rightarrow f\left( \dfrac{\pi }{2} \right)=-{{\sin }^{-1}}\left( \dfrac{\pi }{2} \right) \\
& \Rightarrow f\left( \dfrac{\pi }{2} \right)=-1 \\
\end{align}$
So at $x=\dfrac{\pi }{2}$ we get the value as:
$\begin{align}
& f\left( \dfrac{\pi }{2} \right)={{\sin }^{-1}}\left( \dfrac{\pi }{2} \right) \\
& \Rightarrow f\left( \dfrac{\pi }{2} \right)=1 \\
\end{align}$
So the value of function $f\left( x \right)={{\sin }^{-1}}x$ lies in $\left[ -1,1 \right]$
We get the graph of the function as below:
So we have to take a range that gives the value under the above graph.
So we can take the range as $\left[ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right]$
Hence range of $f\left( x \right)={{\sin }^{-1}}x$ other than $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ is $\left[ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right]$
Note: Trigonometric is a branch of mathematics that studies the relation between the side lengths and the angles of a triangle. There are six types of trigonometric functions which are sine, cosine, tangent, secant, cosecant and cotangent. As the six trigonometric functions are periodic in nature they are not injective and hence they are invertible by restricting the domain of the function. The graph of the inverse of the sine function is like a reflection over the line $y=x$ of the sine function. Sometimes we write the inverse function as $\arcsin \left( x \right)$ because the superscript $-1$ is not an exponent so to avoid any confusion a different notation can be used.
Complete step by step answer:
The function given to us is as follows:
$f\left( x \right)={{\sin }^{-1}}x$
The range is provided to us as follows:
$\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$
We will find value of the function in this range as follows:
So at $x=-\dfrac{\pi }{2}$ we get the value as:
$\begin{align}
& f\left( \dfrac{\pi }{2} \right)={{\sin }^{-1}}\left( -\dfrac{\pi }{2} \right) \\
& \Rightarrow f\left( \dfrac{\pi }{2} \right)=-{{\sin }^{-1}}\left( \dfrac{\pi }{2} \right) \\
& \Rightarrow f\left( \dfrac{\pi }{2} \right)=-1 \\
\end{align}$
So at $x=\dfrac{\pi }{2}$ we get the value as:
$\begin{align}
& f\left( \dfrac{\pi }{2} \right)={{\sin }^{-1}}\left( \dfrac{\pi }{2} \right) \\
& \Rightarrow f\left( \dfrac{\pi }{2} \right)=1 \\
\end{align}$
So the value of function $f\left( x \right)={{\sin }^{-1}}x$ lies in $\left[ -1,1 \right]$
We get the graph of the function as below:
So we have to take a range that gives the value under the above graph.
So we can take the range as $\left[ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right]$
Hence range of $f\left( x \right)={{\sin }^{-1}}x$ other than $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ is $\left[ \dfrac{\pi }{2},\dfrac{3\pi }{2} \right]$
Note: Trigonometric is a branch of mathematics that studies the relation between the side lengths and the angles of a triangle. There are six types of trigonometric functions which are sine, cosine, tangent, secant, cosecant and cotangent. As the six trigonometric functions are periodic in nature they are not injective and hence they are invertible by restricting the domain of the function. The graph of the inverse of the sine function is like a reflection over the line $y=x$ of the sine function. Sometimes we write the inverse function as $\arcsin \left( x \right)$ because the superscript $-1$ is not an exponent so to avoid any confusion a different notation can be used.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

When was the first election held in India a 194748 class 12 sst CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

December 10th of 1948 is an important day in the history class 12 sst CBSE

