Functions: Domain and Range Restrictions - Worksheets, Questions and Revision

10 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 4)Functions: Domain and Range Restrictions revision guide

This topic is chapter 8 of IGCSE Maths Practice Book 1.

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2.7 Functions: Domain and Range Restrictions

EDEXCEL 4MA1 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The function f(x) = 5 has domain all real numbers. State the value of f(2) for this constant function and explain why there is no domain restriction in this context.

(Total for Question 1 is 1 mark)

Consider g(x) = 3/(x - 4). State the single value of x that must be excluded from the domain because it makes the expression undefined.

(Total for Question 2 is 1 mark)

h(x) = √x + 2. State the restriction on x so that h(x) is real, and give the domain in inequality form.

(Total for Question 3 is 1 mark)

The function r(x) = √4 - x. State the domain in inequality form and give one example value of x inside the domain and one outside.

(Total for Question 4 is 2 marks)

A mapping diagram shows function T: { -1, 0, 1, 2 } -> R with arrows T(-1) = 4, T(0) = 1, T(1) = 0, T(2) = -3. State the domain and range of T using set notation.

(Total for Question 5 is 3 marks)

For the function p(x) = (x² - 1)/(x + 1), state any value(s) excluded from the domain and give the domain in set notation.

(Total for Question 6 is 2 marks)

Given s(x) = 2/(x² - 9), find all x excluded from the domain and give the domain in interval notation.

(Total for Question 7 is 2 marks)

Consider the linear function L(x) = -2x + 5 with domain x in [0, 3]. Find the range of L over this restricted domain and give it in inequality form.

(Total for Question 8 is 3 marks)

Let Q(x) = x² - 4x + 3. The domain of Q is restricted to 1 ≤ x ≤ 4 for this question. Find the range of Q on this restricted domain by locating its vertex and evaluating Q at the vertex and at the domain endpoints. Give the range in inequality form.

(Total for Question 9 is 8 marks)

Consider the rational function R(x) = (2x + 1)/(x - 2). (a) State all values excluded from the domain because they make the denominator zero. (b) For the domain x > 2, find the range of R. You should show algebraic reasoning: consider behaviour as x -> 2+ and x -> infinity, and check whether R takes any finite minimum or maximum on the domain x > 2. Give the final range in inequality or interval notation.

(Total for Question 10 is 12 marks)

Mark scheme · 2.7 Functions: Domain and Range Restrictions

Question 1

B1 f(2) = 5 cao
Answer: f(2) = 5

Question 2

B1 x = 4 cao
Answer: x = 4

Question 3

B1 x ≥ -2 or domain x ≥ -2 cao
Answer: x ≥ -2

Question 4

M1 states domain 4 - x ≥ 0, leading to x ≤ 4
A1 gives correct examples, e.g. x = 0 inside and x = 5 outside, or similar, cao
Answer: Domain x ≤ 4; example inside x = 0, outside x = 5

Question 5

M1 states domain as { -1, 0, 1, 2 }
A1 states range as {4, 1, 0, -3} cao
B1 both domain and range given in correct set notation
Answer: Domain = {-1, 0, 1, 2}; Range = {4, 1, 0, -3}

Question 6

M1 identifies x = -1 as a value making denominator zero
A1 domain = {x : x in R, x ≠ -1} or written as (-infinity, -1) union (-1, infinity) cao
Answer: Excluded x = -1; domain {x : x in R, x ≠ -1}

Question 7

M1 solves x² - 9 = 0 to get x = 3 and x = -3
A1 domain = (-infinity, -3) union (-3, 3) union (3, infinity) cao
Answer: Excluded x = -3 and x = 3; domain (-infinity, -3) U (-3, 3) U (3, infinity)

Question 8

M1 evaluates L at the endpoints x = 0 and x = 3, giving 5 and -1
A1 identifies the max and min correctly and orders them for the range, e.g. -1 ≤ L(x) ≤ 5
B1 gives the range in correct inequality form, -1 ≤ L(x) ≤ 5 cao
Answer: -1 ≤ L(x) ≤ 5

Question 9

M1 completes the square or uses vertex formula to find x-coordinate of vertex x_v = 2
M1 evaluates Q at the vertex to find Q(2) = -1
M1 evaluates Q at the endpoints Q(1) = 0 and Q(4) = 3
M1 compares values and identifies minimum and maximum correctly, showing minimum -1 at x = 2 and maximum 3 at x = 4
A1 states the range in correct inequality form, -1 ≤ Q(x) ≤ 3 cao
B1 explicit statement that the vertex lies inside the domain and is used to find the minimum, oe
A1 correct numerical substitution at endpoints and vertex with no arithmetic error, cao
B1 final answer presented clearly as the set of possible outputs, -1 ≤ Q(x) ≤ 3
Answer: -1 ≤ Q(x) ≤ 3

Question 10

M1 states x = 2 is excluded from the domain because denominator x - 2 = 0
M1 rearranges R(x) into form 2 + 5/(x - 2) or equivalent algebraic manipulation, showing R(x) = 2 + 5/(x - 2)
M1 uses form to consider limits: as x -> 2+ then 5/(x - 2) -> +infinity so R -> +infinity, showing no upper bound
M1 as x -> +infinity, 5/(x - 2) -> 0 so R -> 2, indicating a horizontal asymptote at y = 2
M1 checks monotonic behaviour on x > 2, for example differentiating R or noting 5/(x - 2) is positive and decreases on x > 2, so R decreases from +infinity toward 2
A1 concludes that R(x) takes every value greater than 2 on x > 2, so range is (2, infinity) cao
B1 correct statement that y = 2 is not attained for finite x > 2, since 5/(x - 2) would have to be 0 which only occurs as x -> infinity
A1 final presentation of domain exclusion and range for x > 2, domain x ≠ 2 and range (2, infinity), cao
M1 alternative valid reasoning accepted: solves R(x) = y for x and shows for x > 2 this yields y > 2, oe
A1 correct algebraic manipulation in alternative route, e.g. solving y = (2x + 1)/(x - 2) leading to x = (2 + 2y)/(y - 2), and concluding y > 2 for x > 2, cao
B1 clear statement of excluded value(s) in part (a) and clear linking words in part (b) showing consideration of endpoints and limits
A1 final boxed answer: Domain excludes x = 2; Range for x > 2 is (2, infinity) cao
Answer: Excluded: x = 2. For domain x > 2, range is (2, infinity).

Functions: Domain and Range Restrictions - Worksheets, Questions and Revision

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2.7 Functions: Domain and Range Restrictions

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