MathsCBSE Class 11 · Relations and Functions

Finding the Domain and Range of Real Functions (Class 11)

Q: What is the difference between codomain and range?

The codomain is the set the function is declared to map into; the range is the set of values actually taken.

Q: Why do we check the discriminant for the range?

Because writing y = f(x) as a quadratic in x, a real x exists only when the discriminant is non-negative.

Q: Is the greatest-integer function continuous?

No. It jumps at every integer, which is why its graph is a staircase. Continuity belongs to the study of limits in this course.

Q: Are logarithmic domains covered here?

The domain restriction for logarithms (argument greater than zero) belongs with the exponential and logarithmic functions in the chapter; this page focuses on roots, fractions, modulus and the greatest integer.

Updated 17 October 2026 · Fact-checked

The domain of a real function is the set of real x for which the formula gives a real value. Exclude x that make a denominator zero or a square-root expression negative. The range is the set of values f(x) takes; find it by solving y = f(x) for x and requiring x to be real, by completing the square, or by using the graph. Greatest-integer, modulus and signum functions have step or V-shaped graphs that make ranges easy to read.

Squash the graph onto the x-axis to read the domain and onto the y-axis to read the range; the root needs x − 2 ≥ 0.

Illustration drawn for QuestionBuddy.

Understand Domain and Range of Real Functions

A real function $f : A \to R$ with $A \subseteq R$ assigns to each $x \in A$ exactly one real number $f (x)$ . The domain is the set of allowed inputs; the range is the set of outputs.

Domain rules. Remove from $R$ every $x$ that causes:

a zero denominator (solve denominator $= 0$ );
a negative number under an even root: need radicand $\geq 0$ (or $> 0$ if the root is also in a denominator);
(logarithm: argument $> 0$ , covered with exponentials and logarithms).

Combine several restrictions by intersection. For $x^{2} - 5 x + 6$ : $(x - 2) (x - 3) \geq 0$ gives $x \leq 2$ or $x \geq 3$ . For $\frac{1}{∣ x ∣ - 2}$ : $∣ x ∣ - 2 > 0$ gives $∣ x ∣ > 2$ . Check whether simplifying a formula changes the domain: $\frac{x ^{2} - 4}{x - 2}$ equals $x + 2$ only for $x \neq = 2$ .

Range methods.
1. **Reverse (solve for $x$ ):** write $y = f (x)$ , solve for $x$ in terms of $y$ and require $x$ real and in the domain. For $y = \frac{x}{1 + x ^{2}}$ : $y x^{2} - x + y = 0$ needs discriminant $1 - 4 y^{2} \geq 0$ , so $- \frac{1}{2} \leq y \leq \frac{1}{2}$ .
2. Complete the square: $x^{2} - 4 x + 5 = (x - 2)^{2} + 1 \geq 1$ , so $\frac{1}{x ^{2} - 4 x + 5} \in (0, 1]$ .
3. Build from known pieces: start from the range of $x$ or of $∣ x ∣$ and apply the operations in order. $∣ x - 1∣ \geq 0$ , so $3 - ∣ x - 1∣ \leq 3$ .
4. Use graphs: the modulus function $∣ x ∣$ has range $[0, \infty)$ ; the signum function takes only $- 1, 0, 1$ ; the greatest-integer function $[x]$ has range $Z$ .

Greatest-integer function. $[x]$ is the largest integer not exceeding $x$ : $[2.5] = 2$ , $[- 2.5] = - 3$ . For non-integer $x$ , $[- x] = - [x] - 1$ ; for integer $x$ , $[- x] = - [x]$ . So $f (x) = [x] + [- x]$ is $0$ at integers and $- 1$ elsewhere, with range ${- 1, 0}$ . The function $x - [x]$ (the fractional part) has range $[0, 1)$ .

Signum and modulus quotients. $\frac{∣ x ∣}{x}$ is defined for $x \neq = 0$ and equals $1$ or $- 1$ ; its range is ${- 1, 1}$ , not an interval.

Always state the domain and range in set or interval notation, with brackets matching whether end values are attained.

Domain and Range of Real Functions: formulas to remember

Denominator rule: $f (x) = p (x) / q (x) n ee d s q (x) \neq = 0$; Exclude the zeros of q.
Root rule: $\sqrt g (x) n ee d s g (x) \geq 0$; If also in a denominator: g(x) > 0.
Range by discriminant: $y = f (x) a s a q u a d r a t i c in x : d i scr iminan t \geq 0$; Gives the allowed y values.
Greatest integer: $[- x] = - [x] - 1 f or n o n - in t e g er x$; And [−x] = −[x] for integers.
Completing the square: $x^{2} + b x + c = (x + b /2)^{2} + c - b^{2} /4$; Gives the minimum value.

How to solve Domain and Range of Real Functions questions

Use these steps for domain and range.

1List every restriction: zero denominators, negative radicands, and any modulus or logarithm conditions.
2Solve each restriction as an inequality and intersect them to get the domain.
3For the range, choose a method: solve $y = f (x)$ for $x$ , complete the square, or build up from simple pieces.
4Write the condition for $x$ to be real and in the domain, and solve it for $y$ .
5State the range with correct brackets and check one or two values (the extremes) by substitution.
6If a formula simplifies, still use the original domain.

Domain and Range of Real Functions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Domain and range in a fencing problem

Suppose a gardener has 40 m of fencing for a rectangular plot of width $x$ metres, so its length is $20 - x$ and its area is $A (x) = x (20 - x)$ .

State the sensible domain and the range of $A$ .

Try it first, then show the solutionSolution

Both sides must be positive: $x > 0$ and $20 - x > 0$ , so the domain is $(0, 20)$ .
$A (x) = 100 - (x - 10)^{2} \leq 100$ , with equality at $x = 10$ .
As $x \to 0$ or $20$ , $A \to 0$ but never reaches it, so the range is $(0, 100]$ .

Answer: Domain $(0, 20)$ ; range $(0, 100]$ m².

Memory hook: The domain is what the situation allows; the range is what comes out.

Greatest-integer pricing

Suppose a car park charges Rs 20 for every hour or part of an hour. For a stay of $t$ hours the charge is $20 \times (- [- t])$ rupees.

Find the charge for a stay of 2.3 hours and explain why $- [- t]$ rounds up.

Try it first, then show the solutionSolution

$- [- 2.3] = - (- 3) = 3$ .
Charge $= 20 \times 3 =$ Rs 60.
$[- t]$ rounds $- t$ down, so $- [- t]$ rounds $t$ up to the next whole hour.

Answer: Rs 60.

Memory hook: Minus, floor of minus: the ceiling.

Tricks to remember Domain and Range of Real Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Bottom zero, root negative
Two things ruin a real-valued function: dividing by zero and rooting a negative.
Use it when: Domain.
Square then compare
Completing the square exposes the least or greatest value.
Use it when: Range of quadratics.
Solve for x
If you can solve y = f(x) for x, the range is the set of y that allow real x.
Use it when: Range of rational functions.
Floor of minus
[−x] = −[x] − 1 unless x is an integer.
Use it when: Greatest-integer simplifications.

Domain and Range of Real Functions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Domain with a root · 2 marks · Very short answerBoard level

Find the domain of $f (x) = x^{2} - 5 x + 6$ .

Try it first, then show the solutionSolution

Need $x^{2} - 5 x + 6 \geq 0$ , i.e. $(x - 2) (x - 3) \geq 0$ . (1 mark)
This holds for $x \leq 2$ or $x \geq 3$ ; domain $(- \infty, 2] \cup [3, \infty)$ . (1 mark)

Answer: $(- \infty, 2] \cup [3, \infty)$

Example 2: Domain and range together · 3 marks · Short answerBoard level

Find the domain and range of $f (x) = 2 x - x^{2}$ .

Try it first, then show the solutionSolution

Domain: $2 x - x^{2} \geq 0$ , i.e. $x (2 - x) \geq 0$ , so $0 \leq x \leq 2$ . (1 mark)
Complete the square: $2 x - x^{2} = 1 - (x - 1)^{2}$ . For $x \in [0, 2]$ , $(x - 1)^{2} \in [0, 1]$ , so $2 x - x^{2} \in [0, 1]$ . (1 mark)
Hence $f (x) \in [0, 1]$ ; $0$ is attained at $x = 0, 2$ and $1$ at $x = 1$ . Range $[0, 1]$ . (1 mark)

Answer: Domain $[0, 2]$ ; range $[0, 1]$

Example 3: Root in a denominator · 3 marks · Short answerHarder board level

Find the domain and range of $f (x) = \frac{1}{∣ x ∣ - 2}$ .

Try it first, then show the solutionSolution

The root is in the denominator, so $∣ x ∣ - 2 > 0$ , i.e. $∣ x ∣ > 2$ : $x < - 2$ or $x > 2$ . (1 mark)
Domain: $(- \infty, - 2) \cup (2, \infty)$ . (½ mark)
$∣ x ∣ - 2$ takes all values in $(0, \infty)$ , so its square root does too, and the reciprocal takes all values in $(0, \infty)$ . (1 mark)
Range $(0, \infty)$ . (½ mark)

Answer: Domain $(- \infty, - 2) \cup (2, \infty)$ ; range $(0, \infty)$

Example 4: Range by the discriminant · 4 marks · Short answerHarder board level

Find the range of $f (x) = \frac{x}{1 + x ^{2}}$ .

Try it first, then show the solutionSolution

Let $y = \frac{x}{1 + x ^{2}}$ . Then $y x^{2} - x + y = 0$ . (1 mark)
For $y = 0$ , $x = 0$ works. For $y \neq = 0$ , $x$ is real only if the discriminant $1 - 4 y^{2} \geq 0$ . (1 mark)
So $y^{2} \leq \frac{1}{4}$ , i.e. $- \frac{1}{2} \leq y \leq \frac{1}{2}$ . (1 mark)
Both ends are attained, at $x = \pm 1$ . Range $[- \frac{1}{2}, \frac{1}{2}]$ . (1 mark)

Answer: $[- \frac{1}{2}, \frac{1}{2}]$

Example 5: Greatest-integer function · 5 marks · Long answerHOTS / competency

Find the domain and range of $f (x) = [x] + [- x]$ , and evaluate $f (2.5)$ and $f (3)$ .

Try it first, then show the solutionSolution

Domain: $[x]$ and $[- x]$ exist for every real $x$ , so the domain is $R$ . (1 mark)
If $x$ is an integer, $[- x] = - x = - [x]$ , so $f (x) = 0$ . (1 mark)
If $x$ is not an integer, $[- x] = - [x] - 1$ , so $f (x) = - 1$ . (1 mark)
Range $= {- 1, 0}$ , a set of two values, not an interval. (1 mark)
$f (2.5) = 2 + (- 3) = - 1$ ; $f (3) = 3 + (- 3) = 0$ . (1 mark)

Answer: Domain $R$ ; range ${- 1, 0}$ ; $f (2.5) = - 1$ , $f (3) = 0$

Where students lose marks in Domain and Range of Real Functions

The slips examiners see most often, and what to write instead.

Mistake: Simplifying before finding the domain
Cancelling a factor hides a restriction.
Wrong working
Domain of $\frac{x ^{2} - 4}{x - 2}$ taken as $R$ after writing $x + 2$ .
Right working
Domain $R - {2}$ .
Do this: Find the domain from the original formula.
Usually costs: The domain mark.
Mistake: Using ≥ when the root is in a denominator
The radicand cannot be zero there.
Wrong working
$\frac{1}{x - 1}$ : $x \geq 1$ .
Right working
$x > 1$ .
Do this: Check whether the radical sits below a fraction bar.
Usually costs: 1 mark.
Mistake: Writing a range as an interval when it is a set of values
Step functions take only certain values.
Wrong working
Range of $[x] + [- x]$ written as $[- 1, 0]$ .
Right working
${- 1, 0}$ .
Do this: Evaluate at integers and non-integers separately.
Usually costs: The range mark.
Mistake: Wrong bracket at an end point
Attained and unattained values are treated alike.
Wrong working
Range of $\frac{1}{x ^{2} - 4 x + 5}$ written as $(0, 1)$ .
Right working
$(0, 1]$ , since $1$ is attained at $x = 2$ .
Do this: Test whether the end value is reached.
Usually costs: 1 mark.

Exam tips

List the restrictions (denominator, root, modulus) one per line before intersecting them.
State the domain in interval notation with correct brackets.
For ranges, name the method you use: discriminant, completing the square or graph.
Check whether end values are attained and choose the bracket accordingly.
For [x]-type functions, test an integer and a non-integer value.

Practice questions on Domain and Range of Real Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The domain of $f (x) = \frac{1}{x ^{2} - 9}$ is:
(a) $R$
(b) $R - {3}$
(c) $R - {- 3, 3}$
(d) $R - {9}$
Hint
Solve $x^{2} - 9 = 0$ .
Answer
(c). $x = \pm 3$ must be excluded.
Q2 · 1 mark · MCQHarder board level
The range of $f (x) = 3 - ∣ x - 1∣$ is:
(a) $(- \infty, 3]$
(b) $[3, \infty)$
(c) $(- \infty, 3)$
(d) $R$
Hint
$∣ x - 1∣ \geq 0$ .
Answer
(a). $∣ x - 1∣$ takes all values $\geq 0$ , so $3 - ∣ x - 1∣ \leq 3$ , with $3$ attained at $x = 1$ .
Q3 · 3 marks · Short answerHarder board level
Find the domain of $f (x) = 4 - x + \frac{1}{x - 1}$ .
Hint
Both parts must be defined; the second also forbids $x = 1$ .
Answer
$4 - x \geq 0 \Rightarrow x \leq 4$ ; $x - 1 > 0 \Rightarrow x > 1$ . Domain $(1, 4]$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The range of $f (x) = \frac{∣ x ∣}{x}$ is ${- 1, 1}$ .
Reason (R): $\frac{∣ x ∣}{x} = 1$ for $x > 0$ and $- 1$ for $x < 0$ , and it is not defined at $x = 0$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Consider positive and negative $x$ separately.
Answer
(a). The function takes only the values $1$ and $- 1$ .
Q5 · 4 marks · Short answerHOTS / competency
Find the domain and range of (i) $f (x) = \frac{x ^{2} - 4}{x - 2}$ , and (ii) $g (x) = \frac{1}{x ^{2} - 4 x + 5}$ .
Hint
Simplify (i) but keep the original domain; complete the square for (ii).
Answer
(i) Domain $R - {2}$ . For $x \neq = 2$ , $f (x) = x + 2$ , which takes every real value except $4$ ; range $R - {4}$ . (ii) $x^{2} - 4 x + 5 = (x - 2)^{2} + 1 \geq 1$ , never zero: domain $R$ . So $0 < g (x) \leq 1$ ; range $(0, 1]$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the domain condition for √g(x)?Show answer
g(x) ≥ 0.
2.What is the range of |x|?Show answer
[0, ∞).
3.What is [−2.5]?Show answer
−3.
4.What is the range of x − [x]?Show answer
[0, 1).
5.What are the possible values of |x|/x?Show answer
−1 and 1 (x not zero).

Domain and Range of Real Functions: frequently asked questions

What is the difference between codomain and range?

The codomain is the set the function is declared to map into; the range is the set of values actually taken.

Why do we check the discriminant for the range?

Because writing y = f(x) as a quadratic in x, a real x exists only when the discriminant is non-negative.

Is the greatest-integer function continuous?

No. It jumps at every integer, which is why its graph is a staircase. Continuity belongs to the study of limits in this course.

Are logarithmic domains covered here?

The domain restriction for logarithms (argument greater than zero) belongs with the exponential and logarithmic functions in the chapter; this page focuses on roots, fractions, modulus and the greatest integer.