Question

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 1

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.