Functions: Additional Mathematics Tingkatan 4 questions

Section A: objective questions

11 marks

Choose the correct answer. Each question carries 1 mark.

A function from set X to set Y is a relation in which each element of X is mapped to

every element of Y
at least one element of Y
one and only one element of Y
no element of Y

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Each object has exactly one image. Mapping to more than one element (one-to-many) is not a function.

For the teacher: Remembering, low difficulty, standard 1.1.1

Textbook p. 2

Diagram 1 shows a relation from set P to set Q. Why is this relation not a function?

Diagram 1

The object 1 has two images
The image c has two objects
Set P has more elements than set Q
Element b has no image

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1 is mapped to both a and b, so it has two images. Two objects sharing one image (2 and 3 to c) is allowed.

For the teacher: Understanding, low difficulty, standard 1.1.1

Textbook p. 3

Diagram 2 shows the graph of a relation. Which statement is correct?

A curve opening to the right, symmetrical about the x-axis, starting at x = 0.4 — Diagram 2

It is a function because it is a smooth curve
It is a function because it is symmetrical
It is not a function because it passes through the x-axis
It is not a function because a vertical line cuts it at two points

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For each x greater than 0.4 there are two values of y, so a vertical line cuts the graph twice (vertical line test).

For the teacher: Understanding, low difficulty, standard 1.1.1

Textbook p. 4

Given f(x) = 2x² − 3x + 1, find f(−2).

3
15
−13
11

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f(−2) = 2(4) − 3(−2) + 1 = 8 + 6 + 1 = 15. 3 comes from forgetting that −3 × (−2) is positive.

For the teacher: Applying, low difficulty, standard 1.1.3

Textbook p. 9

Diagram 3 shows the function f(x) = x². State the range of f.

Diagram 3

{0, 1, 4}
{0, 1, 2, 4, 5}
{−2, −1, 0, 1, 2}
{1, 4}

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The images are 4, 1, 0, 1, 4, so the range is {0, 1, 4}. {0, 1, 2, 4, 5} is the codomain, and {−2, −1, 0, 1, 2} is the domain.

For the teacher: Understanding, low difficulty, standard 1.1.2

Textbook p. 7

Given f(x) = |3x − 1|, find the values of x such that f(x) = 8.

x = 3 and x = 73
x = −3 and x = 73
x = 3 and x = −73
x = 3 only

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3x − 1 = 8 gives x = 3. 3x − 1 = −8 gives x = −7/3. An absolute value equation has two cases.

For the teacher: Applying, medium difficulty, standard 1.1.3

Textbook p. 10

Given f : x → 2x + 3 and g : x → x², find fg(x).

(2x + 3)²
4x² + 9
2x² + 9
2x² + 3

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fg(x) = f[g(x)] = f(x²) = 2x² + 3. (2x + 3)² is gf(x).

For the teacher: Applying, low difficulty, standard 1.2.2

Textbook p. 14

Given f(x) = x + 43, find f⁻¹(x).

x − 43
3x + 4
3x − 4
3x + 4

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Let y = (x + 4)/3. Then x = 3y − 4, so f⁻¹(x) = 3x − 4. The last option is the reciprocal, which f⁻¹ does not mean.

For the teacher: Applying, medium difficulty, standard 1.3.3

Textbook p. 28

The function f is defined by f(x) = 3x − 2, x ≠ 2. Find f⁻¹(x).

3x + 2
3x + 2
x − 23
3x − 2

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y = 3/(x − 2), so x − 2 = 3/y and x = 3/y + 2. Hence f⁻¹(x) = 3/x + 2, x ≠ 0.

For the teacher: Applying, high difficulty, standard 1.3.3

Textbook p. 28

Which function, defined for all real numbers x, does not have an inverse function?

f(x) = 2x + 1
f(x) = x³
f(x) = 1 − x
f(x) = x² + 1

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x² + 1 is many-to-one: f(2) = f(−2) = 5, so a horizontal line cuts its graph twice. The other three are one-to-one.

For the teacher: Understanding, medium difficulty, standard 1.3.2

Textbook p. 24

The function f is defined by f(x) = x² − 4 for 0 ≤ x ≤ 3. State the domain of f⁻¹.

0 ≤ x ≤ 3
−4 ≤ x ≤ 5
−4 ≤ x ≤ 3
0 ≤ x ≤ 5

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The domain of f⁻¹ is the range of f. f(0) = −4 and f(3) = 5, and f increases between them, so −4 ≤ x ≤ 5.

For the teacher: Applying, high difficulty, standard 1.3.2

Textbook p. 26

Section B: structured questions

34 marks

Answer all questions. Show your working.

Diagram 4 shows a relation from set P to set Q.

Diagram 4

[1 mark] (a)Is the relation a function?

YesNo

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Yes

Each object has exactly one image. Element 4 of set Q has no object, which is allowed.

[3 marks] (b)State the domain, the codomain and the range.

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Domain = {2, 4, 6} [1]1
Codomain = {1, 2, 3, 4} [1]1
Range = {1, 2, 3} [1]1

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 1.1.2

Textbook p. 8

Diagram 5 shows part of the graph of f(x) = |3x − 6| for the domain 0 ≤ x ≤ 4.

V shaped graph of f(x) = |3x − 6| from (0, t) down to (2, 0) and up to (4, 6) — Diagram 5

[1 mark] (a)Find the value of t.

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t = |3(0) − 6| = 6 [1]1

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[2 marks] (b)State the range of f for the given domain.

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f(2) = 0 is the least value / f(0) = f(4) = 6 is the greatest [1]1
0 ≤ f(x) ≤ 6 [1]1

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[2 marks] (c)Find the range of values of x such that f(x) ≤ 3.

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−3 ≤ 3x − 6 ≤ 3 [1]1
1 ≤ x ≤ 3 [1]1

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For the teacher: Applying, medium difficulty, standard 1.1.2

Textbook p. 8

Two functions are defined by f : x → 3x − 1 and g : x → x² + 2.

[2 marks] (a)Find fg(x).

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f(x² + 2) = 3(x² + 2) − 1 [1]1
= 3x² + 5 [1]1

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[2 marks] (b)Find gf(x).

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g(3x − 1) = (3x − 1)² + 2 [1]1
= 9x² − 6x + 3 [1]1

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[2 marks] (c)Find the values of x for which fg(x) = 32.

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3x² + 5 = 32, so x² = 9 [1]1
x = 3 or x = −3 [1]1

Check: 3(9) + 5 = 32.

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For the teacher: Applying, low difficulty, standard 1.2.2

Textbook p. 14

The function f is defined by f : x → x + 3.

[2 marks] (a)Given fg : x → 2x² − x + 5, find g(x).

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g(x) + 3 = 2x² − x + 5 [1]1
g(x) = 2x² − x + 2 [1]1

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[3 marks] (b)Given gf : x → x² + 2x − 3, find g(x).

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Let y = x + 3, so x = y − 3 [1]1
g(y) = (y − 3)² + 2(y − 3) − 3 = y² − 4y [1]1
g(x) = x² − 4x [1]1

Check: g(f(x)) = (x + 3)² − 4(x + 3) = x² + 2x − 3.

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For the teacher: Analysing, high difficulty, standard 1.2.4

Textbook p. 16

An oil spill is circular. Its radius is r = g(t) = 5t metres after t minutes, and its area is A = f(r) = πr² m².

[2 marks] (a)Express A as a function of t.

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A = f[g(t)] = π(5t)² [1]1
= 25πt² m² [1]1

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[2 marks] (b)Find the time when the area is 900π m².

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25πt² = 900π, so t² = 36 [1]1
t = 6 minutes [1]1

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For the teacher: Applying, low difficulty, standard 1.2.5

Textbook p. 17

A function f is defined by f(x) = 2x + 3x − 1, x ≠ 1.

[3 marks] (a)Find f⁻¹(x).

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y(x − 1) = 2x + 3, so xy − y = 2x + 3 [1]1
x(y − 2) = y + 3 [1]1
f⁻¹(x) = (x + 3)/(x − 2) [1]1

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[1 mark] (b)Find f⁻¹(4).

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(4 + 3)/(4 − 2) = 7/2 [1]1

Check: f(7/2) = 10 ÷ (5/2) = 4.

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[1 mark] (c)State the value of x for which f⁻¹(x) is not defined.

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x = 2 [1]1

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For the teacher: Applying, medium difficulty, standard 1.3.3

Textbook p. 28

Diagram 6 shows the graph of f(x) = x² − 1 for −2 ≤ x ≤ 2. Hana says that f has an inverse function.

Graph of f(x) = x² − 1 for −2 to 2 with a horizontal line cutting it at two points — Diagram 6

[1 mark] (a)Is Hana correct? Explain using the horizontal line test.

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No. A horizontal line cuts the graph at two points, so f is not one-to-one [1]1

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[1 mark] (b)The domain is changed to 0 ≤ x ≤ 3. Explain why f now has an inverse function.

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Only the right half of the curve is used, so every horizontal line cuts it at most once and f is one-to-one [1]1

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[3 marks] (c)For the domain 0 ≤ x ≤ 3, find f⁻¹(x) and state its domain.

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y = x² − 1, x = √(y + 1) (x ≥ 0) [1]1
f⁻¹(x) = √(x + 1) [1]1
Domain: −1 ≤ x ≤ 8 [1]1

f(0) = −1 and f(3) = 8.

Tick each point you got.

For the teacher: Evaluating, medium difficulty, standard 1.3.2

Textbook p. 24

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Section B: structured questions

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18