
Range of ${{\sin }^{-1}}x-{{\cos }^{-1}}x$ is:
A) $\left[ -\dfrac{3\pi }{2},\dfrac{\pi }{2} \right]$
B) $\left[ -\dfrac{5\pi }{3},\dfrac{\pi }{3} \right]$
C) $\left[ -\dfrac{3\pi }{2},\pi \right]$
D) $\left[ 0,\pi \right]$
Answer
555.9k+ views
Hint: First, start with the range of the inverse of sin is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and of cos is $\left[ 0,\pi \right]$. After that add and subtract ${{\cos }^{-1}}x$ to get positive signs of ${{\cos }^{-1}}x$. As, we know that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$, substitute $\dfrac{\pi }{2}$ in function. After that substitute the range of ${{\cos }^{-1}}x$ in the function and solve it further to get the range of the function.
Complete step-by-step answer:
Given:- $f\left( x \right)={{\sin }^{-1}}x-{{\cos }^{-1}}x$
As we know that the range of ${{\sin }^{-1}}x$ is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and the range of ${{\cos }^{-1}}x$ is $\left[ 0,\pi \right]$.
Now, add and subtract ${{\cos }^{-1}}x$ in the function,
$\Rightarrow$$f\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x-{{\cos }^{-1}}x-{{\cos }^{-1}}x$
As we know that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$. Substitute $\dfrac{\pi }{2}$ in place of ${{\sin }^{-1}}x+{{\cos }^{-1}}x$,
$\Rightarrow$$f\left( x \right)=\dfrac{\pi }{2}-2{{\cos }^{-1}}x$
Substitute the range of ${{\cos }^{-1}}x$ to find the range of the function,
$\Rightarrow$$f\left( x \right)\in \left[ \dfrac{\pi }{2}-2\left( \pi \right),\dfrac{\pi }{2}-2\left( 0 \right) \right]$
Open the bracket and multiply the terms,
$f\left( x \right)\in \left[ \dfrac{\pi }{2}-2\pi ,\dfrac{\pi }{2}-0 \right]$
Subtract the terms,
$\Rightarrow$$f\left( x \right)\in \left[ -\dfrac{3\pi }{2},\dfrac{\pi }{2} \right]$
Thus, the range of the function $f\left( x \right)$ is $\left[ -\dfrac{3\pi }{2},\dfrac{\pi }{2} \right]$.
Hence, option (C) is the correct answer.
Note: A function is a relation for which each value from the set of the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.
The range of a function is the set of all possible outputs for the function.
Inverse trigonometric functions are also called “Arc Functions” since, for a given value of trigonometric functions, they produce the length of arc needed to obtain that particular value. The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent.
The domain and range of inverse trigonometry functions are: -
Complete step-by-step answer:
Given:- $f\left( x \right)={{\sin }^{-1}}x-{{\cos }^{-1}}x$
As we know that the range of ${{\sin }^{-1}}x$ is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and the range of ${{\cos }^{-1}}x$ is $\left[ 0,\pi \right]$.
Now, add and subtract ${{\cos }^{-1}}x$ in the function,
$\Rightarrow$$f\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x-{{\cos }^{-1}}x-{{\cos }^{-1}}x$
As we know that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$. Substitute $\dfrac{\pi }{2}$ in place of ${{\sin }^{-1}}x+{{\cos }^{-1}}x$,
$\Rightarrow$$f\left( x \right)=\dfrac{\pi }{2}-2{{\cos }^{-1}}x$
Substitute the range of ${{\cos }^{-1}}x$ to find the range of the function,
$\Rightarrow$$f\left( x \right)\in \left[ \dfrac{\pi }{2}-2\left( \pi \right),\dfrac{\pi }{2}-2\left( 0 \right) \right]$
Open the bracket and multiply the terms,
$f\left( x \right)\in \left[ \dfrac{\pi }{2}-2\pi ,\dfrac{\pi }{2}-0 \right]$
Subtract the terms,
$\Rightarrow$$f\left( x \right)\in \left[ -\dfrac{3\pi }{2},\dfrac{\pi }{2} \right]$
Thus, the range of the function $f\left( x \right)$ is $\left[ -\dfrac{3\pi }{2},\dfrac{\pi }{2} \right]$.
Hence, option (C) is the correct answer.
Note: A function is a relation for which each value from the set of the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.
The range of a function is the set of all possible outputs for the function.
Inverse trigonometric functions are also called “Arc Functions” since, for a given value of trigonometric functions, they produce the length of arc needed to obtain that particular value. The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent.
The domain and range of inverse trigonometry functions are: -
| Function | Domain | Range |
| ${{\sin }^{-1}}x$ | [-1, 1] | $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ |
| ${{\cos }^{-1}}x$ | [-1, 1] | $\left[ 0,\pi \right]$ |
| ${{\tan }^{-1}}x$ | For all real numbers | $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ |
| ${{\cot }^{-1}}x$ | For all real numbers | $\left( 0,\pi \right)$ |
| ${{\sec }^{-1}}x$ | $\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$ | $\left[ 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi \right]$ |
| ${{\operatorname{cosec}}^{-1}}x$ | $\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$ | $\left[ -\dfrac{\pi }{2},0 \right)\cup \left( 0,\dfrac{\pi }{2} \right]$. |
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

If overrightarrow a overrightarrow b overrightarrow class 12 maths CBSE

If a b and c are unit coplanar vectors then left 2a class 12 maths CBSE

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

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

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Give 10 examples of unisexual and bisexual flowers

