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
637.5k+ 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
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry 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

