Find the domain and range of a function $f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}$.
A) $\mathbb{R},\{ - 1,1\} $
B) $\mathbb{R} - \{ 3\} ,\{ - 1,1\} $
C) ${\mathbb{R}^{\text{T}}},\mathbb{R}$
D) None of these
Answer
638.4k+ views
Hint:
Domain of the function is the set of all values taken by $x$ and range is the set of all values taken by $f(x)$. Denominator of a fraction cannot be zero. The modulus function $\left| x \right|$ takes the value $x$ and $ - x$ when $x > 0$ and $x < 0$ respectively.
Useful formula:
The function $f(x) = \left| x \right|$ takes the value $x$ if $x > 0$ and $ - x$ if $x < 0$.
Complete step by step solution:
The given function is $f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}$.
Let $f$ be a function defined from the set $A$ to the set $B$.
Then $A$ is called the domain of the function and contains all possible values $x$ can take.
Also $B$ is called the co-domain of the set.
Then the set of all images of the function, which will be a subset of the co-domain, is called the range of the function.
Now consider the function given.
$f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}$
To find the domain let us check what all values $x$ can take here.
We know that division by zero is not defined.
So the denominator of a function cannot be zero.
This gives,
$x - 3 \ne 0$
Adding $3$ on both sides we get,
$x \ne 3$
So the only value which could not be taken by $x$ is $3$.
This gives the domain is the set of all real numbers except three, that is $\mathbb{R} - \{ 3\} $.
Now the range is the set of all values taken by $f(x)$.
We have $f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}$
Consider $\left| {x - 3} \right|$.
We know that $f(x) = \left| x \right|$ takes the value $x$ if $x > 0$ and $ - x$ if $x < 0$.
So we have,
$\left| {x - 3} \right| = x - 3$ if $x - 3 > 0$ and $\left| {x - 3} \right| = - (x - 3)$ if $x - 3 < 0$
$\left| {x - 3} \right| = x - 3$ if $x > 3$ and $\left| {x - 3} \right| = - (x - 3)$ if $x < 3$
If $\left| {x - 3} \right| = x - 3$, then $\dfrac{{\left| {x - 3} \right|}}{{x - 3}} = \dfrac{{x - 3}}{{x - 3}} = 1$
And if $\left| {x - 3} \right| = - (x - 3)$, then $\dfrac{{\left| {x - 3} \right|}}{{x - 3}} = \dfrac{{ - (x - 3)}}{{x - 3}} = - 1$
That is,
$f(x) = 1$ if $x > 3$ and $f(x) = - 1$ if $x < 3$.
So $f(x)$ takes two values $1$ and $ - 1$.
This gives the range of the function is $\{ - 1,1\} $.
Therefore the answer is option B.
Note:
When a function is defined, its domain and co-domain are also mentioned. The domain and co-domain need not be different as in this case. They may be the same as well.
Domain of the function is the set of all values taken by $x$ and range is the set of all values taken by $f(x)$. Denominator of a fraction cannot be zero. The modulus function $\left| x \right|$ takes the value $x$ and $ - x$ when $x > 0$ and $x < 0$ respectively.
Useful formula:
The function $f(x) = \left| x \right|$ takes the value $x$ if $x > 0$ and $ - x$ if $x < 0$.
Complete step by step solution:
The given function is $f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}$.
Let $f$ be a function defined from the set $A$ to the set $B$.
Then $A$ is called the domain of the function and contains all possible values $x$ can take.
Also $B$ is called the co-domain of the set.
Then the set of all images of the function, which will be a subset of the co-domain, is called the range of the function.
Now consider the function given.
$f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}$
To find the domain let us check what all values $x$ can take here.
We know that division by zero is not defined.
So the denominator of a function cannot be zero.
This gives,
$x - 3 \ne 0$
Adding $3$ on both sides we get,
$x \ne 3$
So the only value which could not be taken by $x$ is $3$.
This gives the domain is the set of all real numbers except three, that is $\mathbb{R} - \{ 3\} $.
Now the range is the set of all values taken by $f(x)$.
We have $f(x) = \dfrac{{\left| {x - 3} \right|}}{{x - 3}}$
Consider $\left| {x - 3} \right|$.
We know that $f(x) = \left| x \right|$ takes the value $x$ if $x > 0$ and $ - x$ if $x < 0$.
So we have,
$\left| {x - 3} \right| = x - 3$ if $x - 3 > 0$ and $\left| {x - 3} \right| = - (x - 3)$ if $x - 3 < 0$
$\left| {x - 3} \right| = x - 3$ if $x > 3$ and $\left| {x - 3} \right| = - (x - 3)$ if $x < 3$
If $\left| {x - 3} \right| = x - 3$, then $\dfrac{{\left| {x - 3} \right|}}{{x - 3}} = \dfrac{{x - 3}}{{x - 3}} = 1$
And if $\left| {x - 3} \right| = - (x - 3)$, then $\dfrac{{\left| {x - 3} \right|}}{{x - 3}} = \dfrac{{ - (x - 3)}}{{x - 3}} = - 1$
That is,
$f(x) = 1$ if $x > 3$ and $f(x) = - 1$ if $x < 3$.
So $f(x)$ takes two values $1$ and $ - 1$.
This gives the range of the function is $\{ - 1,1\} $.
Therefore the answer is option B.
Note:
When a function is defined, its domain and co-domain are also mentioned. The domain and co-domain need not be different as in this case. They may be the same as well.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

