Maths Specialist CAS

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MATHS QUEST 11
TI-NSPIRE CAS CALCULATOR COMPANION
Advanced General
Mathematics
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VCE MATHEMATI CS UNI TS 1 & 2
PATRICK SCOBLE MARK BARNES RUTH BAKOGIANIS KYLIE BOUCHER
MARK DUNCAN TRACEY HERFT ROBYN WILLIAMS JENNIFER NOLAN GEOFF PHILLIPS
MATHS QUEST 11
TI-NSPIRE CAS CALCULATOR COMPANION
Advanced General
Mathematics
First published 2013 by
John Wiley & Sons Australia, Ltd
42 McDougall Street, Milton, Qld 4064
Typeset in 10/12 pt Times LT Std
© John Wiley & Sons Australia, Ltd 2013
The moral rights of the authors have been asserted.
ISBN: 978 1 118 31771 6
978 1 118 31768 6 (fexisaver)
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Contents
Introduction vi
CHAPTER 1
Number systems: real and complex 1
CHAPTER 2
Transformations 7
CHAPTER 3
Relations and functions 9
CHAPTER 4
Algebra 13
CHAPTER 5
Trigonometric ratios and their applications 23
CHAPTER 6
Sequences and series 27
CHAPTER 7
Variation 37
CHAPTER 8
Further algebra 45
CHAPTER 10
Linear and non-linear graphs 51
CHAPTER 11
Linear programming 61
CHAPTER 12
Coordinate geometry 65
CHAPTER 13
Vectors 67
CHAPTER 14
Statics of a particle 69
CHAPTER 15
Kinematics 71
CHAPTER 16
Geometry in two and three dimensions 75
Introduction
This booklet is designed as a companion to Maths Quest 11 Advanced General Mathematics Second
Edition.
It contains worked examples from the student text that have been re-worked using the TI-Nspire CX
CAS calculator with Operating System v3.
The content of this booklet will be updated online as new operating systems are released by Texas
Instruments.
The companion is designed to assist students and teachers in making decisions about the judicious use of
CAS technology in answering mathematical questions.
The calculator companion booklet is also available as a PDF fle on the eBookPLUS under the
preliminary section of Maths Quest 11 Advanced General Mathematics Second Edition.
vi Introduction
CHAPTER 1 • Number systems: real and complex 1
CHAPTER 1
Number systems: real
and complex
WORKED EXAMPLE 4
Express each of the following in the form
a
b
, where a ∈ Z and b ∈ Z \{0}.
a 0.6

b
0.23

c
0.41

d
2.1234

THINK WRITE
a 1 Write 0.6

in expanded form. a 0.6

= 0.666666… [1]
2 Multiply [1] by 10.
10 × 0.6

= 6.66666… [2]
3 Subtract [1] from [2].
9 × 0.6

= 6
0.6

=
6
9
4 State the simplest answer.
2
3
=
b 1 Write
0.23

in the expanded form. b 0.23

= 0.232323… [1]
2 Multiply [1] by 100.
100 × 0.23

= 23.232323… [2]
3 Subtract [1] from [2].
99 × 0.23

= 23
4 State the simplest answer. 0.23
23
99
=

c 1 Write
0.41

in the expanded form. c 0.41

= 0.41111… [1]
2 Multiply [1] by 10.
10 × 0.41

= 4.11111… [2]
3 Subtract [1] from [2].
9 × 0.41

= 3.7
4 State the simplest answer. 0.41
3.7
9
37
90
= =

d 1 Write
2.1234

in the expanded form. d 2.1234

= 2.1234234… [1]
2 Multiply [1] by 1000.
1000 × 2.1234

= 2123.423423… [2]
3 Subtract [1] from [2].
999 × 2.1234

= 2121.3
4 State the simplest answer.
2.1234
2121.3
999
21213
9999
2357
1111
=
=
=

Contents
Number systems: real
and complex 1
2 Maths Quest 11 Advanced General Mathematics
Note: The CAS calculator can
perform some of these calculations
for you. On a Calculator page, press:
• MENU b
• 2: Number 2
• 2: Approximate to fraction 2
Complete the entry line as:
0.6666666666666666666666¢
approxFraction(5E-14)
Then press ENTER ·.
WORKED EXAMPLE 13
Expand and simplify the following where possible.
a 7( 18 3) − b 2 3( 10 5 3) −
−
c ( 5 3 6)(2 3 2) + −
THINK WRITE
a,
b
&
c
1 On a Calculator page, complete the entry
lines as:
7( 18 3) −
2 3( 10 5 3) − −
( 5 3 6)(2 3 2) + −
Press ENTER · after each entry.
Note: The calculator must be set in either
Real or Auto modes.
2 Write the answers. a
b
c
3( 2 1) 7 −
30 2 30 −
(2 3 2) 5 6 3 18 2 − − +
CHAPTER 1 • Number systems: real and complex 3
WORKED EXAMPLE 16
If z
1
= 2 − 3i and z
2
=
−
3 + 4i, fnd
a z
1
+ z
2
b z
1
− z
2
c 3z
1
− 4z
2
.
THINK WRITE
a,
b
&
c
1 On a Calculator page, press:
• MENU b
• 1: Actions 1
• 1: Defne 1
Complete the entry lines as:
Defne z1 = 2 − 3i
Defne z2 = 3 + 4i
z1 + z2
z1 − z2
3z1 − 4z2
Press ENTER · after each entry.
2 Write the answers. a
b
c
z
1
+ z
2
=
−
1 + i
z
1
− z
2
= 5 − 7i
3z
1
− 4z
2
= 18 − 25i
WORKED EXAMPLE 17
Simplify
a 2i(2 − 3i) b (2 − 3i)(
−
3 + 4i).
THINK WRITE
a Expand the brackets. a 2i(2 − 3i) = 4i − 6i
2
= 6 + 4i
b Expand the brackets as for binomial
expansion and simplify.
b (2 − 3i)(
−
3 + 4i) =
−
6 + 8i + 9i − 12i
2
= 6 + 17i
Alternatively, on a Calculator page,
complete the entry lines as:
2i × (2 − 3i)
(2 − 3i) × (
−
3 + 4i)
Press ENTER · after each entry.
4 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 18
If z
1
= 2 + 3i and z
2
=
−
4 − 5i, fnd
a z z
1 2
+ b z z
1 2
+ c z z
1 2
d
z z
1 2 .
THINK WRITE
On a Calculator page, complete the entry
lines as:
Defne z1 = 2 + 3i
Defne z2 =
−
4 − 5i
conj(z1) + conj(z2)
conj(z1 + z2)
conj(z1) × conj(z2)
conj(z1 × z2)
Press ENTER · after each entry.
Note: ‘conj’ can be typed directly onto the
screen or can be found by pressing:
• MENU b
• 2: Number 2
• 9: Complex Number Tools 9
• 1: Complex Conjugate 1
WORKED EXAMPLE 19
Express each of the following in a + bi form.
a
i 4
2
−
b
i
i
3
3
− 4
c (3 − 2i)
−1
d
i
i
2 3
2
−
+
THINK WRITE
a,
b,
c
&
d
On a Calculator page, complete the entry
lines as:
i 4
2
−
i
i
3 4
3
−
(3 − 2i)
−
1
i
i
2 3
2
−
+
Press ENTER · after each entry.
CHAPTER 1 • Number systems: real and complex 5
WORKED EXAMPLE 21
Factorise each of the following quadratic expressions over C.
a 2z
2
+ 6z b 2z
2
− 6 c 2z
2
+ 3
THINK WRITE
a,
b
&
c
Ensure the calculator is set in
Rectangular mode. On a Calculator
page, press:
• MENU b
• 3: Algebra 3
• C: Complex C
• 2: Factor 2
Complete the entry lines as:
cFactor(2z
2
+ 6z, z)
cFactor(2z
2
− 6, z)
cFactor(2z
2
+ 3, z)
Press ENTER · after each entry.
WORKED EXAMPLE 23
Solve the following using the formula for the solution of a quadratic equation.
a 2z
2
+ 4z + 5 = 0 b 2iz
2
+ 4z − 5i = 0
THINK WRITE
a
&
b
Ensure the calculator is set in
Rectangular mode. On a Calculator
page, press:
• MENU b
• 3: Algebra 3
• C: Complex C
• 1: Solve 1
Complete the entry lines as:
cSolve(2z
2
+ 4z + 5 = 0, z)
cSolve(2iz
2
+ 4z − 5i = 0, z)
Press ENTER · after each entry.
CHAPTER 2 • Transformations 7
CHAPTER 2
Transformations
WORKED EXAMPLE 4
Find the image rule for each of the following, given the original rule and translation.
a y = x, T−
2,
−
3
b y = 2x
2
, T−

4, 5
c y = f (x), T

h, k
THINK WRITE
a 1 State the image equations. a x′ = x − 2
y′ = y − 3
2
Find x and y in terms of x′ and y′. x = x′ + 2
y = y′ + 3
3
Substitute into y = x. y = x
⇒ y′ + 3 = x′ + 2
⇒ y′ = x′ − 1
4 Express the answer without using the
primes.
Given y = x under translation T−
2,
−
3
, the equation
of the image (or image rule) is y = x − 1.
5 Note: The effect of the
transformation can be illustrated on a
CAS calculator.
To do this, open a Graphs page.
Complete the entry lines as:
f 1(x) = x
f 2(x) = x − 1
Press ENTER · after each entry.
b 1 State the image equations. b x′ = x − 4
y′ = y + 5
2
Find x and y in terms of x′ and y′. x = x′ + 4
y = y′ − 5
3
Substitute into y = 2x
2
. y = 2x
2
⇒ y′ − 5 = 2(x′ + 4)
2
⇒ y′ = 2(x′ + 4)
2
+ 5
4 Express the answer without using the
primes.
Note: In frst form of the answer,
the turning point is (
−
4, 5), which
was the answer expected as
(0, 0) → (
−
4, 5).
Given y = 2x
2
under translation T−
4, 5
, the equation
of the image (or image rule) is
y = 2(x

+ 4)
2
+ 5
or y = 2x
2
+ 16x + 37
Contents
Transformations 7
8 Maths Quest 11 Advanced General Mathematics
5 Note: The effect of the transformation
can be illustrated on a CAS calculator.
To do this, open a Graphs page.
Complete the entry lines as:
f 1(x) = 2x
2
f 2(x) = 2(x + 4)
2
+ 5
Press ENTER · after each entry.
c 1 State the image equations. c x′ = x + h
y′ = y + k
2
Find x and y in terms of x′ and y′. x = x′ − h
y = y′ − k
3
Substitute into y = f (x). y′ − k = f (x′ − h)
⇒ y′ = f (x′ − h) + k
4 Express the answer without using the
primes.
Given y = f (x) under translation T
h, k
, the equation
of the image (or image rule) is y = f (x − h) + k.
CHAPTER 3 • Relations and functions 9
CHAPTER 3
Relations and functions
WORKED EXAMPLE 3
Find the range for the following functions.
a f: R
+
→ R, f (x) = 4x − 1
b f: R → R, f (x) =
−
x
2
− 4x + 5
c f: R → R, f (x) = 2
x
− 1
THINK WRITE
a 1
f (x) = 4x − 1 is linear. The domain
is x ∈ R
+
or x ∈ (0, ∞).
a When x = 0, f (0) = 4(0) − 1
=
−
1
2
f (0) =
−
1, but (0,
−
1) is not included
and therefore this lower end of the
range must be represented using a
round bracket. State the range.
The range: y ∈ (
−
1, ∞).
b 1
f (x) =
−
x
2
− 4x + 5 is an inverted
parabola over the set of real
numbers. Use x
b
a 2
=
−
to
determine the x-value of the
turning point, as this can be used to
indicate the maximum y-value of
the graph.
b =
=
− −
−
−
x
( 4)
2
2
2 Substitute this x-value into f (x) to
determine the maximum y-value.
f (
−
2) =
−
4 + 8 + 5 = 9
3 State the range. The range: y ∈ (
−
∞, 9].
c 1
f (x) = 2
x
− 1 is an exponential
graph over the set of real numbers.
As x →
−
∞, 2
x
→
−
1, y =
−
1 is an
asymptote. Use a CAS calculator
to draw this graph.
c
2 Use the graph and the information
described above to state the range.
The range is: y ∈ (
−
1, ∞).
Contents
Relations and functions 9
10 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 4
If f: R → R, f (x) = 2x
2
− 4x + 1, fnd
a f (x
2
) b f (2x + 1)
THINK WRITE
a To fnd f (x
2
), substitute x
2
for x and
simplify.
a f (x) = 2x
2
− 4x + 1
f (x
2
) = 2(x
2
)
2
− 4(x
2
) + 1
= 2x
4
− 4x
2
+ 1
b To fnd f (2x + 1), substitute 2x + 1 for
x and simplify.
b f (x) = 2x
2
− 4x + 1
f (2x + 1) = 2(2x + 1)
2
− 4(2x + 1) + 1
= 2[4x
2
+ 4x + 1] − 8x − 4 + 1
= 8x
2
+ 8x + 2 − 8x − 4 + 1
= 8x
2
− 1
Alternatively, on a Calculator page,
complete the entry lines as:
Defne f (x) = 2x
2
− 4x + 1
f (x
2
)
f (2x + 1)
Press ENTER · after each entry.
WORKED EXAMPLE 6
If f: (
−
∞,
−
1] → R, f (x) = x
2
+ 2x + 2, fnd the domain, range and rule of f
−
1
(x), and sketch the
graphs of f and f
−
1
on the same set of axes.
THINK WRITE
1 First, determine if the inverse function,
f
−
1
(x), exists. Since f (x) = x
2
+ 2x + 2 is
an upright parabola, it is necessary to
locate the turning point using =
−
x
b
a 2
.
= =
−
−
x
2
2
1
Since the turning point occurs at x =
−
1, and the
domain is x ∈ (
−
∞,
−
1], f (x) is a 1–1 function
and f
−
1
(x) exists.
2 Determine the range of f (x). The domain
is x ∈ (
−
∞,
−
1], so substitute the end value
of x to determine the range (this value is
at the turning point).
x =
−
1
⇒ f (
−
1) = 1 − 2 + 2
= 1
The point (
−
1, 1) is the minimum point on the
graph.
3
The domain of f (x) = the range of f
−
1
(x).
The range of f (x) = the domain of f
−
1
(x).
State the domain and range of f
−
1
(x).
Domain f (x): x ∈ (
−
∞,
−
1]
⇒ range of f
−
1
(x) is y ∈ (
−
∞,
−
1]
Range of f (x): y ∈ [1, ∞)
⇒ Domain of f
−
1
(x) is [1, ∞)
CHAPTER 3 • Relations and functions 11
4
To determine the rule of f
−
1
(x), let
f (x) = y and interchange x and y.
Then make y the subject.
Let y = x
2
+ 2x + 2
⇒ y = (x + 1)
2
+ 1
Interchange x and y
⇒ x = (y + 1)
2
+ 1
⇒ x − 1 = (y + 1)
2
⇒ = ± − − y x 1 1
5
Since f (x) = x
2
+ 2x + 2, and x ∈ (
−
∞,
−
1]
(left side of the parabola), then f
−
1
should
be = − −
−
y x 1 1.
Fully defne the rule for the inverse
function.
∞ → = − −
− − −
f R f x x : [1,  ) ,  ( ) 1 1
1 1
6 Sketch the graphs over the required
domains, showing the line y = x.
−1
y
x
−2
−3
−4
−5
1
2
3
5
4
−2 −3 −4 −5 1
0
2 3 4 5
y = x
−1
f
−
1
(x) =
−
1−
f(x) = x
2
+ 2x + 2
√x − 1
7 To view the graph and its inverse on a
CAS calculator, open a new Graphs page.
Complete the function entry lines as:
f 1(x) = x
2
+ 2x + 2 | x ≤
−
1
= − −
−
f x x 2( ) 1 1 | x ≥
−
1
f 3(x) = x
Press ENTER · after each entry.
CHAPTER 4 • Algebra 13
CHAPTER 4
Algebra
WORKED EXAMPLE 9
Transpose each of the following formulas to make the pronumerals indicated
in brackets the subject.
a A =
4
3
π r
2
(r) b =
−
P
ab ac
d
a ( ) c = − m pq rs s ( )
THINK WRITE
a 1 Write the equation. a A =
4
3
πr
2
2 Multiply both sides of the equation
by 3.
3 × A =
4
3
πr
2
× 3
3A = 4πr
2
3
Divide both sides by 4π.
π
π
π
π
=
=
A r
A
r
3
4
4
4
3
4
2
2
4 Take the square root of both sides.
Note: From an algebraic point of view
we should write ± in front of the root.
However, since r represents a physical
quantity (radius of a sphere in this
case), it can take only positive values.
π
π
=
=
A
r
r
A
3
4
3
4
2
b 1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 1: Solve 1
Complete the entry line as:
solve(p = (a × b − a × c) ÷ d, a)
Then press ENTER ·.
b
2 Write the answer. =
−
a
dP
b c
Note: Capital P should be used in the answer.
Contents
Algebra 13
14 Maths Quest 11 Advanced General Mathematics
c 1 Write the equation. c = − m pq rs
2
The inverse of x is x
2
so square
both sides.
m
2
= pq − rs
3 Subtract pq from both sides. m
2
− pq = pq − rs − pq
m
2
− pq =
−
rs
4
Divide both sides by
−
r.
−
=
=
−
−
−
−
−
m pq
r
rs
r
s
m pq
r
2
2
5 Multiply the numerator and
denominator by
−
1 (optional).
=
−
s
pq m
r
2
WORKED EXAMPLE 12
Solve for x in 2(x + 5) = 3(2x − 6).
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3:Algebra 3
• 1:Solve 1
Complete the entry line as:
solve(2(x + 5) = 3(2x − 6),x)
Then press ENTER ·.
2 Write the answer. Solving 2(x + 5) = 3(2x − 6) for x, gives x = 7.
CHAPTER 4 • Algebra 15
WORKED EXAMPLE 13
Find the value of x that will make the following a true statement:
+
= −
x x 2
3
5
2
.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3:Algebra 3
• 1:Solve 1
Complete the entry line as:
solve
+
= −






x x
x
2
3
5
2
,
Then press ENTER ·.
2 Write the answer. Solving
+
= −
x x 2
3
5
2
for x, gives = x
26
5
or 5
1
5
.
WORKED EXAMPLE 14
Solve the following equation for x: + =
− x x x
2 3
2
1
1
.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3:Algebra 3
• 1:Solve 1
Complete the entry line as:
solve + =
−






x x x
x
2 3
2
1
1
, 
Then press ENTER ·.
2 Write the answer. Solving + =
− x x x
2 3
2
1
1
for x, gives = x
7
5
or 1
2
5
.
16 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 15
Solve the following pair of simultaneous equations graphically:
a x + 2y = 4 b y + 3x = 17
x − y = 1 2x − 3y = 4
THINK WRITE
a 1 Rule up a set of axes. Label the
origin and the x and y axes.
a (See graph at step 7 on page 107.)
2 Find the x-intercept for the equation
x + 2y = 4, by making y = 0.
x-intercept: when y = 0,
x + 2y = 4
x + 2 × 0 = 4
x = 4
The x-intercept is at (4, 0).
3 Find the y-intercept for the equation
x + 2y = 4, by making x = 0. Divide
both sides of the equation by 2.
y-intercept: when x = 0,
x + 2y = 4
0 + 2y = 4
2y = 4
y = 2
The y-intercept is at (0, 2).
4 Plot the points on graph paper and join
them with the straight line. Label the
graph.
(Refer to the graph at step 7.)
5 Find the x-intercept for the equation
x – y = 1, by making y = 0.
x-intercept: when y = 0,
x − y = 1
x − 0 = 1
x = 1
The x-intercept is at (1, 0).
6 Find the y-intercept for the equation
x − y = 1, by making x = 0. Multiply
both sides of the equation by
−
1.
y-intercept: when x = 0,
x − y = 1
0 − y = 1

−
y = 1

−
y ×
−
1 = 1 ×
−
1
y =
−
1
The y-intercept is at (0,
−
1).
7 Plot the points on graph paper and join
them with the straight line. Label the
graph.
1
1
−1
2
2 4
(2, 1)
y
x
x – y = 1
x + 2y = 4
0
8 From the graph, read the coordinates of
the point of intersection.
The point of intersection between the two graphs
is (2, 1).
9 Verify the answer by substituting the
point of intersection into the original
equations.
Substitute x = 2 and y = 1 into x + 2y = 4.
LHS = 2 + 2 × 1 RHS = 4
= 2 + 2
= 4
LHS = RHS
Substitute x = 2 and y = 1 into x − y = 1
LHS = 2 − 1 RHS = 1
= 1
LHS = RHS
In both cases LHS = RHS; therefore, the solution
set (2, 1) is correct.
CHAPTER 4 • Algebra 17
b 1 Rearrange both equations to make y
the subject. To do this, on a
Calculator page, complete the entry
lines as:
solve(y + 3x = 17, y)
solve(2x − 3y = 4, y)
Press ENTER · after each entry.
b
2 On a Graphs page, complete the
function entry lines as:
=
−
f x
x
1( )
2( 2)
3
f 2(x) = 17 − 3x
Press ENTER · after each entry.
3 To fnd the point of intersection,
complete the following steps.
Press:
• MENU b
• 8:Geometry 8
• 1:Points & Lines 1
• 3:Intersection Point(s) 3.
Click on each line.
The point of intersection will appear.
ENTER ·.
4 Write the answer. The point of intersection between the two graphs
is (5, 2).
18 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 18
Solve the following simultaneous equations.
2x + 3y = 4
3x + 2y = 10
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3:Algebra 3
• 1:Solve 1
Then complete the entry line as:
solve(2x + 3y = 4 and 3x + 2y = 10, x).
Then press ENTER ·
Note: The term ‘and’ can be found in the
catalogue k or can be typed.
2 Answer the question. Solution: x =
22
5
, y =
− 8
5
or
−






,
22
5
 
8
5
.
CHAPTER 4 • Algebra 19
WORKED EXAMPLE 21
A train (denoted as train 1) leaves station A and moves in the direction of station B with an
average speed of 60 km/h. Half an hour later another train (denoted as train 2) leaves station A
and moves in the direction of the frst train with an average speed of 70 km/h. Find:
a the time needed for the second train to catch up with the frst train
b the distance of both trains from station A at that time.
THINK WRITE
1 Defne the variables.
Note: Since the frst train left half an hour
earlier, the time taken for it to reach the
meeting point will be x + 0.5.
Let x = the time taken for train 2 to reach train 1.
Therefore, the travelling time, t, for each train is:
Train 1: t
1
= x + 0.5
Train 2: t
2
= x
2 Write the speed of each train. Train 1: v
1
= 60
Train 2: v
2
= 70
3 Write the distance travelled by each of
the trains from station A to the point of
the meeting.
(Distance = speed × time.)
Train 1: d
1
= 60(x + 0.5)
Train 2: d
2
= 70x
4 Equate the two expressions for distance.
Note: When the second train catches
up with the frst train, they are the same
distance from station A — that is, d
1
= d
2
.
When the second train catches up with the frst
train, d
1
= d
2
.
5 Solve the equation. On a Calculator page,
press:
• MENU b
• 3:Algebra 3
• 1:Solve 1
Then complete the entry line as:
solve(60(x + 0.5) = 70x,x).
Then press ENTER ·.
6 Substitute 3 in place of x into either of the
two expressions for distance, say into d
2
.
Substitute x = 3 into d
2
= 70x
d
2
= 70 × 3
7 Evaluate. = 210
8 Answer the questions. a The second train will catch up with the frst
train 3 hours after leaving station A.
b Both trains will be 210 km from station A.
20 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 23
Two hamburgers and a packet of chips cost $8.20, while one hamburger and two packets of chips
cost $5.90. Find the cost of a packet of chips and a hamburger.
THINK WRITE
1 Defne the two variables. Let x = the cost of one hamburger.
Let y = the cost of a packet of chips.
2 Formulate an equation from the frst
sentence and call it [1].
Note: One hamburger costs $x, two
hamburgers cost $2x. Thus, the total cost of
cost of two hamburgers and one packet of
chips is 2x + y and it is equal to $8.20.
2x + y = 8.20 [1]
3 Formulate an equation from the second
sentence and call it [2].
Note: One packet of chips costs $y, two
packets cost $2y. Thus, the total cost of
two packets of chips and one hamburger
is x + 2y and it is equal to $5.90.
x + 2y = 5.90 [2]
4 Solve for the simultaneous equations. On
a Calculator page, press:
• MENU b
• 3:Algebra 3
• 1:Solve 1
Then complete the entry line as:
solve(x + 2y = 5.90 and 2x + y = 8.20,x).
Then press ENTER ·.
5 Answer the question and include
appropriate units.
A hamburger costs $3.50 and a packet of chips
costs $1.20.
CHAPTER 4 • Algebra 21
WORKED EXAMPLE 25
Simplify
a
−
− x x
3 2
1
b
+
−
−
a
x x
2
3
2
3
.
THINK WRITE
a
&
b
1 On a Calculator page, press:
• MENU b
• 2:Number 2
• 7:Fraction Tools 7
• 4:Common Denominator 4
Complete the entry lines as:
comDenom −
−






x x
3 2
1
comDenom
+
−
−






a
x x
2
3
2
3
Press ENTER · after each entry.
a
&
b
2 Write the answer. a
b
−
−
=
−
− x x
x
x x
3 2
1
3
2
+
−
−
=
− − −
−
a
x x
ax x a
x
2
3
2
3
2 2 6 6
9
2
WORKED EXAMPLE 26
Simplify
a
×
x
x
3
4
20
9
2
b
+
×
+
x
y
y
x
4
6
2
5 20
2
2 2
.
THINK WRITE
a
&
b
1 On a Calculator page, complete the
entry lines as:
×
x
x
3
4
20
9
2
+
×
+
x
y
y
x
4
6
2
5 20
2
2 2
Press ENTER · after each entry.
a
&
b
2 Write the answers. a
b
× =
x
x
x 3
4
20
9
5
3
2
+
×
+
=
x
y
y
x y
4
6
2
5 20
1
15
2
2 2
CHAPTER 5 • Trigonometric ratios and their applications 23
CHAPTER 5
Trigonometric ratios and
their applications
WORKED EXAMPLE 1
Determine the value of the pronumerals, correct to 2 decimal places.
a
x
4
50°
b
7
h
24° 25′
THINK WRITE
a 1 Label the sides, relative to the
marked angles.
a
x
O
4
H
50°
2 Write what is given. Have: angle and hypotenuse (H)
3 Write what is needed. Need: opposite (O) side
4 Determine which of the
trigonometric ratios is required,
using SOH–CAH–TOA.
sin (θ ) =
O
H
5 Substitute the given values into the
appropriate ratio.
sin (50°) =
x
4
6 Transpose the equation and solve
for x.
4 × sin (50°) = x
x = 4 × sin (50°)
7 Round the answer to 2 decimal places. = 3.06
b 1 Label the sides, relative to the
marked angle.
b
7
A
H
24° 25′
h
2 Write what is given. Have: angle and adjacent (A) side
3 Write what is needed. Need: hypotenuse (H)
4 Determine which of the
trigonometric ratios is required,
using SOH–CAH–TOA.
cos (θ ) =
A
H
5 Substitute the given values into the
appropriate ratio.
cos (24°25′ ) =
h
7
Contents
Trigonometric ratios and
their applications 23
24 Maths Quest 11 Advanced General Mathematics
6 Solve for h. On a Calculator page,
complete the entry line as:
solve

′ =




h
h cos (24 25 )
7
,
Then press ENTER ·.
7 Round the answer to 2 decimal
places.
h ≈ 7.69
WORKED EXAMPLE 2
Find the angle θ, giving the answer in degrees and minutes.
THINK WRITE
1 Label the sides, relative to the marked
angles.
12
A
O 18
θ
2 Write what is given. Have: opposite (O) and adjacent (A) sides
Need: angle
3 Write what is needed.
4 Determine which of the trigonometric
ratios is required, using SOH–CAH–TOA.
tan (θ ) =
O
A
5 Substitute the given values into the
appropriate ratio.
tan (θ ° ) =
18
12
6
Transpose the equation and solve for θ,
using the inverse tan function.
To calculate tan
−
1
, on a Calculator page,
complete the entry line as:










−
tan
18
12
1
¢
DMS
Then press ENTER ·.
Note:
¢
DMS is located in the catalogue.
7 Write the answer to the nearest minute.
θ ° =






−
tan  
18
12
1
= 56° 19′
12
18
θ
CHAPTER 5 • Trigonometric ratios and their applications 25
WORKED EXAMPLE 8
In the triangle ABC, a = 10 m, c = 6 m and C = 30°.
a Show that the ambiguous case exists.
b Find two possible values of A, and hence two possible values of B and b.
THINK WRITE
a
&
b
1 Draw a labelled diagram of the triangle
ABC and fll in the given information.
a = 10
c = 6
A
30°
C
B
2 In part a it was shown that the
ambiguous case of the sine rule exists.
Therefore, on a Calculator page,
complete the entry line as:
solve
=






| ≤ ≤
a
a a
10
sin ( )
6
sin (30)
, 0 180
Then press ENTER ·.
3 Convert the angles to degrees and
minutes.
A = 56°27′ or A = 123°33′
4 Calculate the size of the angle B given
each angle A.
If A = 56°27′, B = 180 − (30 + 56°27′)
= 93°33′
If A = 123°33′, B = 180 − (30 + 123°33′)
= 26°27′
5 To fnd the side length b, on a Calculator
page, complete the entry lines as:
solve

′
=






b
b
sin (93 33 )
6
sin (30)
,
solve

′
=






b
b
sin (26 27 )
6
sin (30)
,
Press ENTER · after each entry.
6 Write the answers. If B = 93°33′, b = 11.98 m
If B = 26°27′, b = 5.35 m
26 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 9
Find the third side of triangle ABC given a = 6, c = 10 and B = 76°, correct to 2 decimal places.
THINK WRITE
1 Draw a labelled diagram of the triangle
ABC and fll in the given information.
b
a = 6 c = 10
A C
B
76°
2 Write the appropriate cosine rule to
fnd side b.
b
2
= a
2
+ c
2
− 2ac cos (B)
3 On a Calculator page, complete the entry
line as:
solve(b
2
= 6
2
+ 10
2
− 2 × 6 × 10 × cos (76),b)
Then press ENTER ·.
4 Since b represents the side length of a
triangle, then b > 0.
b = 10.34, correct to 2 decimal places.
WORKED EXAMPLE 10
Find the smallest angle in the triangle with sides 4 cm, 7 cm and 9 cm.
THINK WRITE
1 Draw a labelled diagram of the triangle
ABC and fll in the given information.
a = 4 c = 7
b = 9
A C
B
2 Write the appropriate cosine rule to fnd
the angle A.
a
2
= b
2
+ c
2
− 2bc cos (A)
3 On a Calculator page, complete the entry
line as:
solve(4
2
= 9
2
+ 7
2
− 4 × 9 × 7 × cos (a),a)
|0 ≤ a ≤ 180
Then press ENTER ·.
4 Round the answer to degrees and
minutes.
A = 25.2088°
= 25°13′
CHAPTER 6 • Sequences and series 27
CHAPTER 6
Sequences and series
WORKED EXAMPLE 1
a Find the next three terms in the sequence, b: {14, 7,
7
2
, . . .}.
b Find the 4th, 8th and 12th terms in the following sequence: e
n
= n
2
− 3n, n ∈ {1, 2, 3, . . .}.
c Find the 2nd, 3rd and 5th terms for the following sequence: k
n + 1
= 2k
n
+ 1, k
1
=
−
0.50.
THINK WRITE
a 1 In this example the sequence is listed
and a simple pattern is evident. From
inspection, the next term is half the
previous term and so the sequence
would be 14, 7,
7
2
,
7
4
,
7
8
,
7
16
.
a
The next three terms are
7
4
,
7
8
,
7
16
.
2 On a Calculator page, complete the
entry lines as:
14
Ans × 0.5
Press ENTER · repeatedly to
generate the sequence.
b 1 This is an example of a functional
defnition. The nth term of the
sequence is found simply by
substitution into the expression
e
n
= n
2
− 3n.
b e
n
= n
2
− 3n
2 Find the 4th term by substituting
n = 4.
e
4
= 4
2
− 3 × 4
= 4
3 Find the 8th term by substituting
n = 8.
e
8
= 8
2
− 3 × 8
= 40
4 Find the 12th term by substituting
n = 12.
e
12
= 12
2
− 3 × 12
= 108
Contents
Sequences and series 27
28 Maths Quest 11 Advanced General Mathematics
5 On a Calculator page, press
• MENU b
• 6: Statistics 6
• 4: List Operations 4
• 5: Sequence 5
Complete the entry line as:
seq(e(n) = n
2
− 3n,n,4,12,4)
Then press ENTER ·.
Note: The frst 4 and the 12 are the
highest and lowest values required,
and the last 4 is the step value.
An alternative method to the one
above for generating a sequence is
shown below.
Insert a new Lists and Spreadsheets
page.
Go to the title cell of column A and
press:
• MENU b
• 3: Data 3
• 1: Generate Sequence 1
Complete the sequence felds as
shown.
Then press OK.
6 Scroll down to fnd the required
terms.
c 1 This is an example of an iterative
defnition. We can fnd the 2nd, 3rd
and 5th terms for the sequence
k
n + 1
= 2k
n
+1, k
1
=
−
0.50 by iteration.
c k
n + 1
= 2k
n
+ 1,
k
1
=
−
0.50
2
Substitute k
1
=
−
0.50 into the formula
to fnd k
2
.
k
2
= 2 × −0.50 + 1
= 0
3 Continue the process until the value
of k
5
is found.
k
3
= 2 × 0 + 1
= 1
k
4
= 2 × 1 + 1
= 3
k
5
= 2 × 3 + 1
= 7
CHAPTER 6 • Sequences and series 29
4 Write the answer. Thus k
2
= 0, k
3
= 1 and k
5
= 7.
5 Follow the same steps as for
the alternative method in part b;
however, because the formula is
defned as u(n), the rule needs to be
entered as:
u(n) = 2 × u(n − 1) + 1
with the initial term as:
−
0.5
Press OK to view the sequence.
30 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 2
Given that a = 2 and t
0
= 0.7, use the logistic equation to generate a sequence of 6 terms, and
state whether the sequence is convergent, divergent or oscillating. If the sequence is
convergent, state its limit.
THINK WRITE
1 Insert a new Lists & Spreadsheet page.
Go to the title cell of column A and press:
• MENU b
• 3: Data 3
• 1: Generate Sequence 1
Complete the sequence felds as shown.
Then press OK.
2 Scroll down to fnd the required terms.
Note: Here the initial term corresponds to
cell A1; however, we need to remember
that this is actually t
0
. Hence it is the
sixth term when the sequence frst
converges to 0.5, not the ffth term.
CHAPTER 6 • Sequences and series 31
WORKED EXAMPLE 6
Find the 16th and nth terms in an arithmetic sequence with the 4th term 15 and 8th term 37.
THINK WRITE
1 Write the two equations that represent
t
4
and t
8
.
t
4
: a + 3d = 15 [1]
t
8
: a + 7d = 37 [2]
2 To solve equations [1] and [2]
simultaneously, open a new Calculator
page and press:
• MENU b
• 3: Algebra 3
• 1: Solve 1
Complete the entry line as:
solve(a + 3d = 15 and a + 7d = 37,a)
Then press ENTER ·.
3 Write the answer. If =
−
a
3
2
and = d
11
2
,
=
−
t
n 11 14
2
n
t
16
= 81
32 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 7
Find the sum of the frst 20 terms in the sequence t
n
: {12, 25, 38, . . .}.
THINK WRITE
1 On a Lists & Spreadsheet page, place the
cursor in the grey header cell of column A.
Then press:
• MENU b
• 3: Data 3
• 1: Generate sequence 1
Complete the sequence felds as shown.
2 Scroll down to the 20th term.
3 Write the answer. S
20
= 2710
CHAPTER 6 • Sequences and series 33
WORKED EXAMPLE 10
The ffth term in a geometric sequence is 14 and the seventh term is 0.56.
Find the common ratio, r, the frst term, a, and the nth term for the sequence.
THINK WRITE
1 Write the general rule for the nth term of
the geometric sequence.
t
n
= ar
n − 1
2 Use the information about the 5th term to
form an equation. Label it [1].
When n = 5, t
n
= 14
14 = a × r
5 − 1
14 = a × r
4
[1]
3 Similarly, use information about the 7th
term to form an equation. Label it [2].
When n = 7, t
n
= 0.56
0.56 = a × r
7 − 1
0.56 = a × r
6
[2]
4 Solve equations simultaneously: Divide
equation [2] by equation [1] to eliminate a.
[2]
[1]
gives
=
ar
ar
0.56
14
6
4
5 Solve for r. r
2
= 0.04
r 0.04 = ±
= ± 0.2
6 As there are two solutions, we must perform
two sets of computations. Consider the
positive value of r frst. Substitute the value of
r into either of the two equations, e.g. [1], and
solve for a.
If r = 0.2
Substitute r into [1]:
a × (0.2)
4
= 14
0.0016a = 14
a = 8750
7 Substitute the values of r and a into the
general equation to fnd the expression
for the nth term.
The nth term is:
t
n
= 8750 × (0.2)
n − 1
8 Now consider the negative value of r. If r =
−
0.2
9 Substitute the value of r into either of the
two equations, e.g. [1], and solve for a.
(Note that the value of a is the same for
both values of r.)
Substitute r into [1]
a = (
−
0.2)
4
= 14
0.0016a = 14
a = 8750
10
Substitute the values of r and a into
the general formula to fnd the second
expression for the nth term.
The nth term is:
t
n
= 8750 × (
−
0.2)
n − 1
11 Write the two equations that represent
t
5
and t
7
.
t
5
: 14 = a × r
4
[1]
t
7
: 0.56 = a × r
6
[2]
34 Maths Quest 11 Advanced General Mathematics
12
To solve equations [1] and [2]
simultaneously, open a new Calculator
page and press:
• MENU b
• 3: Algebra 3
• 1: Solve 1
Complete the entry line as:
solve(14 = a × r
4
and 0.56 = a × r
6
,r)
Then press ENTER ·.
13 Write the answer. When r =
−
0.2 and a = 8750,
t
n
= 8750 × (
−
0.2)
n − 1
When r = 0.2 and a = 8750,
t
n
= 8750 × (0.2)
n − 1
CHAPTER 6 • Sequences and series 35
WORKED EXAMPLE 11
Find the sum of the frst 5 terms (S
5
) of these geometric sequences.
a t
n
: {1, 4, 16, . . .} b t
n
= 2(2)
n − 1
, n ∈ {1, 2, 3, . . .} c t
n + 1
=
1
4
t
n
, t
1
=
1
2
−
THINK WRITE
a 1 On a Lists & Spreadsheet page, place
the cursor in the grey header cell of
column A.
Then press:
• MENU b
• 3: Data 3
• 1: Generate sequence 1
Complete the sequence felds as
shown.
a
2 To sum the terms of any sequence,
place the cursor in cell B1 and type =
3 Then press Catalogue and scroll
down to sum(.
Then press ENTER ·.
36 Maths Quest 11 Advanced General Mathematics
4 To sum the frst 5 terms, complete
the entry line as:
sum(a1:a5)
Then press ENTER ·.
5 The answer will appear in cell B1.
6 Write the answer. If t
n
: {1, 4, 16, . . .} then, S
5
= 341.
b 1 Write the sequence. b t
n
= 2(2)
n − 1
, n ∈ {1, 2, 3, . . .}
2 Compare the given rule with the
general formula for the nth term of
the geometric sequence t
n
= ar
n − 1
and
identify values of a and r; the value of n
is known from the question.
a = 2; r = 2; n = 5
3 Substitute values of a, r and n into
the general formula for the sum and
evaluate.
=
−
−
=
−
=
S
2(2 1)
2 1
2(32 1)
1
62
5
5
c 1 Write the sequence. c t
n + 1
=
1
4
t
n
, t
1
=
1
2
−
2 This is an iterative formula, so the
coeffcient of t
n
is our r; a = t
1
; n is
known from the question.
r =
1
4
; a =
− 1
2
; n = 5
3 Substitute values of a, r and n into
the general formula for the sum and
evaluate.

−
−

× −

−
¸
¸

_
,


¸

1
]
1
−
¸
¸

_
,

−
−
S
1
1
1
5
1
2
1
4
1
4
1
2
1
1024
3
4
341
512
5
CHAPTER 7 • Variation 37
CHAPTER 7
Variation
WORKED EXAMPLE 1
For the given data, establish whether direct variation exists between x and y using:
a a numerical approach (clearly specify k, the constant of variation, if applicable) and
b a graphical approach.
c Confrm your result using a CAS calculator.
x 4 7 8 10
y 5.2 9.1 10.4 13
y
x
THINK WRITE/DRAW
a 1 Find the ratio
y
x
for each of the
four pairs of values.
One variable varies directly as the
other if the ratio between any two
corresponding values is constant.
a Ratio =
y
x
First pair:
=
5.2
4
1.3
Second pair:
=
9.1
7
1.3
Third pair:
=
10.4
8
1.3
Fourth pair:
=
13
10
1.3
2 Compare each of the four ratios and
answer the question.
Since all four ratios are the same (that is, 1.3),
y varies directly as x.
3 Copy and complete the table.
x 4 7 8 10
y 5.2 9.1 10.4 13
y
x
1.3 1.3 1.3 1.3
b 1
2
Plot the information from the table
onto a set of axes.
Join the given points and see if a
straight line is obtained.
Note: For a direct variation to exist
between two variables x and y, a
straight line passing through the
origin (0, 0) must be obtained.
b
13
10.4
0 10
9.1
5.2
8 7 4
x
y
The given points ft perfectly on a straight line. If
the straight line is extended it will pass through the
origin. Therefore y varies directly as x.
Contents
Variation 37
38 Maths Quest 11 Advanced General Mathematics
3 Calculate the gradient using any
two points on the straight line.
Answer the question.
Note: The gradient of the straight
line will equal k, the constant of
variation, if direct variation exists
between the variables x and y.
Let (x
1
, y
1
) = (4, 5.2) and let
(x
2
, y
2
) = (10, 13)
=
−
−
m
y y
x x
2 1
2 1
=
−
−
13 5.2
10 4
=
7.8
6
= 1.3
The gradient of the straight line is equal to k, the
constant of variation.
c 1 On a Lists & Spreadsheet page enter
the data into the spreadsheet. Label
the columns x and y.
c
2 To draw the scatterplot of the data,
on a Data & Statistics page:
Tab to each axis to select ‘Click
to add variable’. Place x on the
horizontal axis and y on the
vertical axis.
3 To check there is a linear relationship,
press:
• MENU b
• 4: Analyze 4
• 6: Regression 6
• 1: Show linear (mx + b) 1
In X List select x; in Y List select y.
Interpret the graph; if all points are
on the line, then it confrms that the
relationship is one of direct variation.
The relationship is one of direct variation.
CHAPTER 7 • Variation 39
WORKED EXAMPLE 5
For the given data, establish the rule relating the variables x and y then graph the relationship
using a CAS calculator.
x
3 6 7 10
y
28.8 115.2 156.8 320
THINK WRITE/DISPLAY
1 On a Calculator page, complete the entry
lines as:
{3,6,7,10} → x
{28.8,115.2,156.8,320} → y
Press ENTER · after each entry.
2 Find the rule that fts the data. To do this
press:
MENU b
• 6: Statistics 6
• 1: Stat Calculations 1
• 9: Power Regression 9
Enter x in X List and y in Y List
Then press ENTER ·.
3 Write down the rule for x and y. y = 3.2x
2
4 Graph the rule. On a Graphs page, use
the NavPad and the up arrow ` to display
f 1(x), which has already been defned in
this document.
Then press ENTER · to display the
graph.

40 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 8
For the data represented in the table below, establish whether an inverse variation exists between x
and y using:
a a numerical approach (clearly specify k, the constant of variation, if applicable)
b a graphical approach.
x 1 2 3 4 5
y 20 10
6
2
3
5 4
xy
THINK WRITE/DISPLAY
a 1 Find the product of xy for each of the
5 pairs of values.
Note: One variable varies inversely
as the other if the product between
any 2 corresponding values is
constant.
a Product = xy
First pair: 1 × 20 = 20
Second pair: 2 × 10 = 20
Third pair: 3 × 6
2
3
= 20
Fourth pair: 4 × 5 = 20
Fifth pair: 5 × 4 = 20
2 Compare each of the fve products
and answer the question.
Since the product of the corresponding values
is the same in each case (that is 20), y varies
inversely as x.
3 Copy and complete the table.
x 1 2 3 4 5
y 20 10
6
2
3
5 4
xy 20 20 20 20 20
b 1 Calculate the values of
x
1
.
Place these values into a table.
b
x 1 2 3 4 5
x
1
1
1
2
1
3
1
4
1
5
y 20 10
6
2
3
5 4
2 On a Calculator page, enter the
x
1

and y values by completing the
entry lines as:
{1,
1
2
,
1
3
,
1
4
,
1
5
} → x
{20, 10, 6
2
3
, 5, 4} → y
Press ENTER · after each entry.
CHAPTER 7 • Variation 41
3 To fnd the rule that fts the data, press:
• MENU b
• 6: Statistics 6
• 1: Stat Calculations 1
• 3: Linear Regression (mx + b) 3
Enter x in X List and y in Y List
Then press ENTER ·.
4 To graph the rule, open a Graphs page.
Complete the entry line as
f 2(x) = f 1(x)
Then press ENTER · to view the
graph.
Adjust the window if necessary.
5 Answer the question. The graph of y against
x
1
is a straight line
directed from, but not passing through, the
origin, hence an open circle at the point (0, 0).
Therefore y ∝
x
1
.
42 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 16
My telephone bill consists of 2 parts: a fxed charge of $32 (paid whether any calls are made or
not) and a charge proportional to the number of calls made. Last quarter I made 296 calls and my
bill was $106.
a Find the equation of variation.
b Find the amount to be paid when 300 calls are made.
THINK WRITE/DISPLAY
a 1 Defne each variable to be used. a Let A = the total amount to be paid, in dollars
Let n = the number of calls
2 Write the equation of variation. A = k n + 32
3 Substitute the values for A and n into
the equation.
When n = 296 and A = 106,
106 = 296k + 32
4 To solve the equation for k, on a
Calculator page, press:
• MENU b
• 3: Algebra 3
• 1: Solve 1
Complete the entry line as:
solve(106 = 296 × k + 32,k)
Then press ENTER ·.
5 Rewrite the equation substituting
1
4
in
place of k.
So A =
1
4
n + 32
b 1
Substitute n = 300 into the given
equation.
b When n = 300,
A =
1
4
(300) + 32
2 Evaluate. =
1
4
× 300 + 32
= 75 + 32
= 107
3 Answer the question and include the
appropriate unit.
The amount to be paid when 300 calls are made
is $107.
CHAPTER 7 • Variation 43
WORKED EXAMPLE 18
The following table shows the values of the total surface area, TSA, of spheres and their
corresponding radii, r.
Radius (r) (cm) 1 2 3 4 5
TSA (cm
2
) 12.57 50.27 113.1 201.06 314.16
Graph the values given in the table and comment on the shape of the graph. Using the graph, or
otherwise, fnd the equation which relates total surface area of the sphere, TSA, and its radius r.
THINK WRITE/DISPLAY
1 On a Lists & Spreadsheet page, enter
values and label in column A:
radius: 1, 2, 3, 4, 5
and in column B:
tsa 12.57, 50.27, 113.1, 201.06, 314.16
2 Draw the scatterplot of the data on a
Data & Statistics page.
Tab to each axis to select ‘Click to add
variable’. Place radius on the horizontal
axis and tsa on the vertical axis.
To check the power relationship, press:
• MENU b
• 4: Analyse 4
• 6: Regression 6
• 7: Show power 7
In X List select x; in Y List select y
3 Comment on the graph obtained. The graph is not a straight line, passing through
the origin, so direct variation does not exist
between the two variables. Hence, there is no
direct variation between the radius and the total
surface area of the sphere.
4 Make assumptions about the graph
obtained.
The graph resembles a parabola, so it is
reasonable to assume that area is directly
proportional to the square of the radius.
5 Write the variation statement for the
assumption made.
TSA ∝ r
2
6 Write the variation equation. TSA = kr
2
7 Transpose the equation to make k the
subject.
k =
r
TSA
2

44 Maths Quest 11 Advanced General Mathematics
8 Test the assumption by fnding the values
of r
2
and check that the ratio
r
TSA
2
is
constant. On the Lists & Spreadsheet page,
label column C: ratio
In the header (grey) cell, complete the entry
line as:
=tsa/(radius)^2
Then press ENTER ·.
9 Comment on the result obtained. The ratio is constant for each corresponding pair
(when rounded to 2 decimal places).
Hence, TSA ∝ r
2
TSA = kr
2
TSA = 12.57r
2
10 Alternatively, once the values of r
2
have
been calculated, rule up a table of values
and plot TSA versus r
2
.
r
2
1 4 9 16 25
TSA 12.57 50.27 113.1 201.06 314.16
11 Comment on the graph obtained.
300
200
100
014 9 16 25 r
2
TSA
The graph is a straight line, passing through the
origin.
12 Establish the value of k by substituting
any pair of values from the table into
the equation of variation and write the
equation relating the two variables.
TSA = kr
2
When r = 1, TSA = 12.57, k = ?
12.57 = k × 1
12.57 = k
k = 12.57
TSA = 12.57r
2
CHAPTER 8 • Further algebra 45
CHAPTER 8
Further algebra
WORKED EXAMPLE 2
If 5x
3
+ 2x
2
− 7x + 1 = (2 a + b)x
3
− ax
2
− (b − c) x + 1, then fnd the values of a, b and c.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 1: Solve 1
Complete the entry line as:
solve
a b
a a
b c
5
= ,
= 7
2 +
−
2
−
¹
,
¹
¹
¹
¹
,
¹
¹
¹
¸
¸



_
,



Then press ENTER ·.
Note: The template used here can be
found in the Maths expression template.
2 Write the answer. a =
−
2, b = 9 and c = 2
WORKED EXAMPLE 4
If x − 4 is a factor of x
3
− 6x
2
+ 2x + 24, fnd the other factor.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 2: Factor 2
Complete the entry line as:
factor(x
3
− 6x
2
+ 2x + 24)
Then press ENTER ·.
2 Write the answer. The quadratic factor of x
3
− 6x
2
+ 2x + 24 is
x
2
− 2x − 6.
Contents
Further algebra 45
46 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 5
Express
+
−
x
x
4 5
3
in the form +
−
A
b
x 3
.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 9: Fraction Tools 9
• 1: Proper Fraction 1
Complete the entry line as:
propFrac
+
−






x
x
4 5
3
Then press ENTER ·.
2 Write the answer in the form
+
−
A
b
x 3
.
+
−
= +
−
x
x x
4 5
3
4
17
3
WORKED EXAMPLE 6
Express
+
− −
x
x x
3
3 40
2
in partial fraction form.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 3: Expand 3
Complete the entry line as:
expand
+
− −






x
x x
3
3 40
2
Then press ENTER ·.
2 Write the answer in the form
−
+
+
A
x
B
x ( 8) ( 5)
.
+
− −
=
−
+
+
x
x x x x
3
3 40
11
13( 8)
2
13( 5)
,
2
x ∈R\{
−
5, 8}
CHAPTER 8 • Further algebra 47
WORKED EXAMPLE 7
Express
−
− +
x
x x
2 1
( 2)( 1)
2
in partial fractions.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 3: Expand 3
Complete the entry line as:
expand
−
− +






x
x x
2 1
( 2)( 1)
2
Then press ENTER ·.
2 Write the answer in the form
−
+
+
+
+
A
x
B
x
C
x ( 2) ( 1) ( 1)
2
.
x
x x x x x
2 1
( 2)( 1)
1
3( 2)
1
3( 1)
1
( 1)
,
2 2
−
− +
=
−
−
+
+
+
x ∈R\{
−
1, 2}
WORKED EXAMPLE 8
Express
+ +
−
x x
x
5 9 10
8
2
3
in partial fractions.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 3: Expand 3
Complete the entry line as:
expand
+ +
−






x x
x
5 9 10
8
2
3
Then press ENTER ·.
2 Write the answer in the form
−
+
+
+ +
A
x
Bx C
x x 2 2 4
.
2
+ +
−
=
−
+
+
+ +
∈
x x
x x
x
x x
x R
5 9 10
8
4
2
3
2 4
, \{2}
2
3 2
48 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 9
Express
+ −
−
x x
x
5 2
1
2
as a partial fraction.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 3: Expand 3
Complete the entry line as:
expand
+ −
−






x x
x
5 2
1
2
Then press ENTER ·.
2 Write the answer.
+ −
−
=
−
+ +
x x
x x
x
5 2
1
4
1
6,
2
x ∈R\{1}
WORKED EXAMPLE 10
Solve simultaneously: y = x and y = x
2
+ 3x + 1.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 1: Solve 1
Complete the entry line as:
solve(y = x and y = x
2
+ 3x + 1, x)
Then press ENTER ·.
2 Write the answer. Solving y = x and y = x
2
+ 3x + 1 for x
and y gives x =
−
1 and y =
−
1.
That is, (
−
1,
−
1).
CHAPTER 8 • Further algebra 49
WORKED EXAMPLE 11
Solve simultaneously: y = x + 1 and x
2
+ y
2
= 4.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 1: Solve 1
Complete the entry line as:
solve( y = x + 1 and x
2
+ y
2
= 4, x)
Then press ENTER ·.
2 Write the answer. Solving y = x + 1 and x
2
+ y
2
= 4 for x and y gives
=
+
−
x
( 7 1)
2
and =
−
−
y
( 7 1)
2
or
=
−
x
7 1
2
and
=
+
y
7 1
2

That is,
+ −






− −
( 7 1)
2
,
( 7 1)
2
or
− + 





7 1
2
,
7 1
2
.
50 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 12
Solve simultaneously: y = 2x − 1 and =
−
y
x
2
3
.
THINK WRITE
1 On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 1: Solve 1
Complete the entry line as:
solve y x y
x
x 2 1and
2
3
, = − =
−






Then press ENTER ·.
2 Write the answer.
Solving y x y
x
2 1and
2
3
= − =
−
for x and y gives
x y
x y
(
41 7
)
4
and
(
41
5)
4
or
41 7
4
and
41 5
4
=
−
=
−
−
=
+
=
+
−
That is,
(
41 7
)
4
,
(
41
5)
4
or
41 7
4
,
41 5
4
.
−
− 





+ + 





−
−
CHAPTER 10 • Linear and non-linear graphs 51
CHAPTER 10
Linear and non-linear graphs
WORKED EXAMPLE 11
Convert [2,
2
3
π
] to Cartesian coordinates.
THINK WRITE
1 Find the x-coordinate. x = r cos (θ )
= 2 cos
2
3
π






= 2 × −
1
2
x =
−
1
2 Find the y-coordinate. y = r sin (θ )
= 2 sin
2
3
π






= 2 ×
3
2
y =
3
3 State the Cartesian coordinates. Hence, the Cartesian coordinates are
(
−
1, 3).
4 Alternatively, open a Calculator page.
Make sure the setting is on AUTO.
Complete the entry lines as:
P
¢
Rx 2,
2
3
π






P ¢ Ry 2,
2
3
π






Press ENTER · after each entry.
Note: P
¢
Rx and P
¢
Ry can be found in
the catalogue.
[2, ]
x
y
2
3
2
—
3
π
2
—
3
π
52 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 16
Sketch the graph of r = θ for 0 < θ < 4π using a CAS calculator.
THINK WRITE
Open a new Graphs page.
Ensure the angle setting is on radians.
Then press
• MENU b
• 3: Graph Entry/Edit 3
• 4: Polar 4
Complete the entry line as:
θ θ
θ π θ
=
≤ ≤ =





r
step
1( )
0 4 0.13
Then press ENTER ·.
The increment step is
6
π
(0.13) because this was
used in the introductory table.
Compare the graphs of worked examples 15 and 16.
CHAPTER 10 • Linear and non-linear graphs 53
WORKED EXAMPLE 17
Sketch the graph of r = 8 for 0 ≤ θ ≤ 2π.
THINK WRITE/DISPLAY
1
Construct a table of values for 0 ≤ θ ≤ 2π
and fnd the corresponding r values to
2 decimal places.
θ 0
6
π
3
π
2
π 2
3
π 5
6
π
r 8 8 8 8 8 8
θ π
7
6
π 4
3
π 3
2
π 5
3
π 11
6
π
2π
r 8 8 8 8 8 8 8
2 Sketch the graph using a protractor and
ruler to plot each of the points from the
table. Remember r is the distance from
the centre (the origin).
−8
8
8 −8
3
—
2
π
π

—
2
π
3 On a Graphs page, press
• MENU b
• 3: Graph Entry/Edit 3
• 4: Polar 4
Complete the entry line as:
θ
θ π θ
=
≤ ≤ =





r
step
1( ) 8
0 2 0.13
Then press ENTER ·.
Zoom out to see the whole circle.
54 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 18
Sketch the graph of r = 2 sin (θ ) for 0 ≤ θ ≤ 2π using a calculator.
THINK WRITE
On a Graphs page, press
• MENU b
• 3: Graph Entry/Edit 3
• 4: Polar 4
Complete the entry line as:
θ θ
θ π θ
π
=
≤ ≤ =





r
step
1( ) 2sin ( )
0 2
12
Then press ENTER ·.
Zoom out to see the whole circle.
WORKED EXAMPLE 21
Express the following complex numbers in polar form.
a z = 1 + i b z i 3 = +
−
THINK WRITE
a
&
b
1 Open a Calculator page.
Enter the real and imaginary parts
using the [ ] parentheses. Complete
the entry lines as:
[1,1] ¢ Polar
3,1
−





¢
Polar
Press ENTER · after each entry.
Note: ¢ Polar can be found in the
catalogue.
a
&
b
2 Write the answers. a
b
z = 1 + i in polar form is z 2cis
4
π
=




.
z i 3 = +
−
in polar form is
z 2cis
5
6
π
=





 .
CHAPTER 10 • Linear and non-linear graphs 55
WORKED EXAMPLE 22
Express 3cis
5
6
π





 in Cartesian form.
THINK WRITE
1 On a Calculator page, complete the entry
lines as:
P ¢ Rx
π






3,
5
6
P ¢ Ry
π






3,
5
6
Press ENTER · after each entry.
2 Write the answer. i 3cis
5
6
3
2
3
2
π






= +
−
56 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 24
Sketch the graph that results from the addition
of ordinates of the functions y = x and y =
x
1
,
whose graphs are shown at right.
THINK WRITE
1 Add the ordinates at the LHS end points
of the graph: A small negative value and a
large negative value should give a slightly
larger negative value. Mark this point on
the set of axes.
x
y
3
2
1
−1
−2
−3
1 −1 −2 −3 2 3
0
y = x
y =
1
–
x
2 Add the ordinates at the frst point of
intersection. Because they are the same
we simply double the y-value of this
point. Mark the point on the set of axes.
x
y
3
2
1
−1
−2
−3
1 −1 −2 −3 2 3
0
y = x
y =
1
–
x
3 At its x-intercept, the straight line graph
y = x has an ordinate of zero while the
non-linear graph has an undefned ordinate.
This indicates that our graph will also be
undefned at the point where x = 0, giving a
vertical asymptote.
Vertical asymptote at x = 0
4 Add the ordinates at the second point of
intersection. Mark the point on the set
of axes.
x
y
3
2
1
−1
−2
−3
1 −1 −2 −3 2 3
0
y = x
y =
1
–
x
x
y
y = x
y =
3
2
1
−1
−2
−3
1 −1 −2 −3 2 3
0
1
–
x
CHAPTER 10 • Linear and non-linear graphs 57
5 Add the ordinates at the RHS end points
of the graph: A small positive value and a
large positive value should give a slightly
larger positive value. Mark this point on
the set of axes.
x
y
3
2
1
−1
−2
−3
1 −1 −2 −3 2 3
0
y = x
y =
1
–
x
6 Join the points with a smooth curve
noting that the graph will asymptote
toward the y-axis as x approaches zero,
and toward the line y = x as
x approaches positive or negative infnity.
0
–
x
y = x +
1
x
y
3
2
1
−1
−2
−3
1 −1 −2 −3 2 3
0
7 To perform the addition of ordinates on
the CAS calculator, open a new Graphs
page.
Complete the function entry lines as:
f x x
f x
x
f x f x f x
1( )
2( )
1
3( ) 1( ) 2( )
=
=
= +
Press ENTER · after each entry.
58 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 25
Sketch the reciprocal of the graph of the function y = x
2

shown at right.
THINK WRITE
1 Sketch the given function and draw the
horizontal line y = 1 on the same axes.
y
3
−2
x
2
1
−1
1 −1 −2 2
0
3 3
2 2
1 1
–
2
1
–
3
–
3
–
2
1
2 Mark any points where the ordinates of
the graph are equal to 1.
3 Select several points on the original graph
with ordinates greater than 1 (above the
line). Estimate their value and mark their
reciprocals below the line.
4 Select several points on the original graph
with ordinates between 0 and 1 (between
the line and the x-axis). Estimate their
value and mark their reciprocals above
the line.
x
y
3 3 3 3 3
2 2 2 2 2
1
−1
1 −1 −2 2
0
–
2
1
–
3
1
–
3
–
3
1
–
3
1 1
–
2
1
–
2
1
–
2
1
5 Join the points with a smooth curve
noting that the graph will asymptote
toward the y-axis as x approaches zero,
and toward the x-axis as x approaches
positive or negative infnity.
x
y
3 3 3 3 3
2 2 2 2 2
1
−1
1 −1 −2 2
0
–
2
1
–
3
1
–
3
–
3
1
–
3
1 1
–
2
1
–
2
1
–
2
1
–
x
2
1
y =
6 Alternatively, on a Calculator page,
complete the entry line as:
Defne f(x) = x
2

Then press ENTER ·.
On a Graphs page, complete the function
entry lines as:
f x f x
f x
f x
1( ) ( )
2( )
1
( )
=
=
Press ENTER · after each entry.
x
y
3
2
1
−1
1 −1 −2 2
0
y = x
2
CHAPTER 10 • Linear and non-linear graphs 59
WORKED EXAMPLE 26
Sketch the square of the graph of the function y =
x
2
shown below.
THINK WRITE
1 Sketch the given function and draw the
horizontal lines y = 1 and y =
−
1 on the
same axes.
x
y
3
2
1
−1
−2
−3
1
1 1
−1 −2 −3 2 3
2 Mark any points where the ordinates of
the graph are equal to 1 and change any
ordinates of
−
1 to 1. (Squaring 1 results
in 1.)
3 Select, estimate, and square several
ordinates between the horizontal lines
y =
−
1 and y = 1 marking these positive
values closer to the x-axis.
x
y
3
4
2
1
−1
−2
−3
−4
1
1 1
−1 −2 −3 −4 2 3 4
1
– —
2
1
–
2
1
–
4
1
–
4

4 Select, estimate, and square several
ordinates outside the horizontal lines
y =
−
1 and y = 1 marking these positive
values further away from the x-axis.
5
6
7
8
9 9 9
4 4 4
3
3
2
2
1
1 1
−1
−2 −2
−3 −3
−4
1 −2
–1
−3 −4 2 3 4
x
y
0
1
–
2
1
–
4
1
–
4
1
– —
2
3
2
1
−1
−2
−3
1 −2
−1
−3 2 3
x
y
0
y =
2
–
x
60 Maths Quest 11 Advanced General Mathematics
5 Join the points with a smooth curve
noting that the graph will asymptote
toward the y-axis as x approaches zero.
5
6
7
8
9 9 9
4 4 4
3
3
2
2
1
1 1
–1
–2
–2
–3 –3
–4
1 –2
–1
–3 –4 2 3 4
x
y
0
1
–
2
1
–
4
—
x
2
y =
4
1
–
4
1
– —
2
6 Alternatively, on a Calculator page,
complete the entry line as:
Defne f x
x
( )
2
=
Then press ENTER ·.
On a Graphs page, complete the function
entry lines as:
f1(x) = f(x)
f 2(x) = (f(x))
2
Press ENTER · after each entry.
CHAPTER 11 • Linear programming 61
CHAPTER 11
Linear programming
WORKED EXAMPLE 3
Sketch the graph of the inequation y − 4x ≤ 8 and indicate the required region.
THINK WRITE/DRAW
1 Make y the subject of the inequation by
adding 4x to both sides.
y − 4x ≤ 8
y ≤ 8 + 4x
2
On a Graphs page, delete = and
complete the entry line as:
y ≤ 8 + 4x
Then press ENTER ·.
3 Indicate the required region.
Note: The calculator shades the required
region.
The required region.
Contents
Linear programming 61
62 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 5
Sketch the following pair of simultaneous linear inequations, determine the point of intersection
and indicate the required region.
2x + 3y ≤ 6
x − y ≥ 3
THINK WRITE/DRAW
1 Make y the subject of the inequations. 2x + 3y ≤ 6
3y ≤ 6 − 2x
≤
−
y
x 6 2
3
[1]
x − y ≥ 3
−
y ≥ 3 − x
y ≤ x − 3 [2]
2
On a Graphs page, delete = and
complete the entry lines as:
≤
−
y
x 6 2
3
y ≤ x − 3
Press ENTER · after each entry.
To change the colour of the inequations,
highlight the line and press:
• Ctrl /
• MENU b
• 4: Colour 4
• 2: Fill colour 2
Select a colour and press ENTER ·.
3 Indicate the required region. The required region.
4 To determine the point of intersection
between the lines, press:
• MENU b
• 8: Geometry 8
• 1: Points & Lines 1
• 3: Intersection Point(s) 3
Move the fashing cursor to each line,
press ENTER · to select each line
and press ENTER · again to lock in
the point of intersection.
State the point of intersection. The point of intersection between ≤
−
y
x 6 2
3
and
y ≤ x − 3 is (3, 0).
CHAPTER 11 • Linear programming 63
WORKED EXAMPLE 8
a Sketch the following system of linear inequations and indicate the required region.
x + y ≤ 10, y ≥ x − 4, y ≤ 2x + 1, x ≥ 0, y ≥ 0
b Determine the coordinates of the vertices of the feasible region.
c Determine the maximum and minimum values of z = 3x − y subject to the above constraints,
using the corner-point method.
THINK WRITE
a 1 Make y the subject of the frst
inequation and write the others.
a x + y ≤ 10
y ≤ 10 − x [1] x ≥ 0 [4]
y ≥ x − 4 [2] y ≥ 0 [5]
y ≤ 2x + 1 [3]
2 On a Graphs page, complete the
entry lines as:
y ≤ 10 − x
y ≥ x − 4
y ≤ 2x + 1
y ≥ 0
Press ENTER · after each entry.
Note: x ≥ 0 cannot be drawn on the
calculator.
Change the colours as required.
3 Indicate the required region. The required region.
b 1 To determine the points of
intersection between the lines,
press:
• MENU b
• 8: Geometry 8
• 1: Points & Lines 1
• 3: Intersection Point(s) 3
Move the fashing cursor to each
line, press ENTER · to select each
line and press ENTER · again to
lock in the point of intersection.
Repeat for each pair of lines and
state the points of intersection.
Zoom out to see all the coordinates.
b
The points of intersection are: (3, 7), (7, 3), (4, 0).
2 To determine the points of
intersection between the line x = 0
and y = 2x + 1, substitute the value
x = 0 into the equation y = 2x + 1.
y = 2x + 1
x = 0 ⇒ y = 2(0) + 1
⇒ y = 1
That is the point (0, 1).
From the graph, (0, 0) is also a corner point.
64 Maths Quest 11 Advanced General Mathematics
c 1 To determine the maximum and
minimum values, on a Calculator
page, complete the entry line as:
Defne z = 3x − y
z | x = 0 and y = 0
z | x = 0 and y = 1
z | x = 3 and y = 7
z | x = 7 and y = 3
z | x = 4 and y = 0
Press ENTER · after each entry.
c
2 Write the maximum and minimum
values of z.
z
min
=
−
1 at (0, 1)
z
max
= 18 at (7, 3)
CHAPTER 12 • Coordinate geometry 65
CHAPTER 12
Coordinate geometry
WORKED EXAMPLE 20
ABCD is a parallelogram. The coordinates of A, B and C are (1, 5), (4, 2) and (2,
−
2) respectively.
Find:
a the equation of AD b the equation of DC c the coordinates of D.
THINK WRITE/DRAW
a 1 Draw the parallelogram ABCD.
Note: The order of the lettering of
the geometric shape determines the
links in the diagram. For example:
ABCD means join A to B to C to D
to A. This avoids any ambiguity.
a
x
y
A
B
C
D
2 1
−41
4
5
2
−42
2 Find the gradient of BC. m
2 2
2 4
BC
=
−
−
−
4
2
=
−
−
= 2
3 State the gradient of AD. Since m
BC
= 2
and AD
||
BC
then m
AD
= 2
4 Using the given coordinates of A
and the gradient of AD fnd the
equation of AD.
y = 2x + c
Let (x, y) = (1, 5):
5 = 2(1) + c
c = 3
Hence, the equation of AD is y = 2x + 3.
b 1 Find the gradient of AB. b m
2 5
4 1
AB
=
−
−

3
3
=
−
=
−
1
2 State the gradient of DC. Since m
AB
=
−
1
and DC||AB
then m
DC
=
−
1
3 Using the given coordinates of C
and the gradient of DC fnd the
equation of DC.
y =
−
x + c
Let (x, y) = (2,
−
2):
−
2 =
−
(2) + c
c = 0
Hence, the equation of DC is y =
−
x.
c 1 Solve simultaneously to fnd D,
the point of intersection of the
equations AD and DC.
c Equation of AD: y = 2x + 3 [1]
Equation of DC: y =
−
x [2]
66 Maths Quest 11 Advanced General Mathematics
2 On a Calculator page, complete the
entry line as:
solve( y = 2x + 3 and y =
−
x, x)
Then press ENTER ·.
3 Write the solution. Hence, the coordinates of D are (
−
1, 1).
CHAPTER 13 • Vectors 67
CHAPTER 13
Vectors
WORKED EXAMPLE 14
a Draw a vector to represent a i j 3 . = −


b Find the magnitude and direction of the vector a

.
THINK DRAW/WRITE
a 1 Draw axes with i

and j

as unit
vectors in the x- and y-directions
respectively.
a
x
y
3 2 1
1
2
−2
−1
O
−1
j
~
i
~
2 Represent i j 3


− as a vector from
0 that is 3 units in the positive
x-direction and 1 unit in the negative
y-direction, and mark the angle
between a

and the x-axis as θ. x
y
θ
3
1
2
−1
O
−1 −2
j
~
i
~
a
~
b 1 The magnitude of a

(that is,

a) may
be found using Pythagoras’ theorem.
b

a 3 ( 1)
10
2 2
= +
=
−
2 On a Calculator page, complete the
entry line as: 3 ( 1)
2 2
+
−
Then press ENTER ·.
3
Find the value of angle θ using the
tangent ratio.
tan (θ) =
1
3
Contents
Vectors 67
68 Maths Quest 11 Advanced General Mathematics
4 On a Calculator page, complete the
entry line as:






−
tan
1
3
1
Then press ENTER ·.
Note: Ensure your calculator is in
Degree mode.
5 Give the direction of vector a

relative
to the positive x-axis.
Vector a

makes an angle of
−
18.4° from the
positive x-axis.
CHAPTER 14 • Statics of a particle 69
CHAPTER 14
Statics of a particle
WORKED EXAMPLE 8
The forces acting on two bobs suspended from light
inextensible strings are shown in the diagram at right.
If the connecting string is at an angle of 10° to the
horizontal fnd T and m.
THINK WRITE/DRAW
1 Separate the two bobs. Draw a diagram
representing the forces applied to the
right bob.
5g
T
1
sin (30°)
T
1
cos (30°) T

cos (10°)
T

sin (10°)
2 Calculate the force, T. Set up two
equations.
To solve these equations simultaneously,
using a CAS calculator, let T
1
= x and
T = y.
On a Calculator page, complete the entry
line as:
solve(x × sin(30) = 5g + y × sin(10) and
y × cos(10) = x × cos(30), x) | g = 9.8
Then press ENTER ·.
T
1
sin (30°) = 5g + T sin (10°) [1]
T cos (10°) = T
1
cos (30°) [2]
T = 124.1 N
10°
T
2
N
T

N
T

N
T
1
N
5g N
mg N
60°
30°
Contents
Statics of a particle 69
70 Maths Quest 11 Advanced General Mathematics
3 Draw a diagram representing the forces
applied to the left bob.
mg
T

sin (10°) T
2
sin (60°)
T
2
cos (60°) T

cos (10°)
4 Calculate the value of m given
T = 124.1 N.
T cos (10°) = T
2
cos (60°)
Given T = 124.1 N
⇒ = T
124.1cos (10)
cos (60)
2
⇒ T
2
= 244.4 N
T
2
sin (60°) + T sin (10°) = mg
⇒ =
+
m
244.4sin (60) 124.1sin (10)
9.8
⇒ m = 23.80 kg
5 State the answers. T = 124.1 N
m = 23.80 kg
CHAPTER 15 • Kinematics 71
CHAPTER 15
Kinematics
WORKED EXAMPLE 11
A particle is travelling in a straight line with its velocity, v cm/s, at any time, t seconds, given as
=
+
≥ v t
t
t ( )
8
1
,  0.
Find the acceleration of the particle after 1 second.
THINK WRITE/DISPLAY
1 Given the expression, =
+
v t
t
( )
8
1
we
want a(1).
=
+
v t
t
( )
8
1
2 Find the acceleration equation by
differentiating velocity with respect
to time (a(t) = v′(t)). To do this, on
a Calculator page, complete the entry
lines as:
Defne =
+
v t
t
( )
8
1
d
dt
v t ( ( ))
Defne =
+
−
a t
t
( )
8
( 1)
2
Press ENTER · after each entry.
3
Substitute t = 1 seconds into the
formula for a(t).
To do this, complete the entry line as:
a (1)
Then press ENTER ·.
4 State the solution. The acceleration of the particle at t = 1 seconds
is
−
2 cm/s
2
.
Contents
Kinematics 71
72 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 13
A particle is travelling in a straight line with its velocity, v (in m/s), at any
time, t seconds, given as:
v(t) = t
2
+ t, t ≥ 0
Calculate the exact distance travelled during the frst 4 seconds of its motion.
THINK WRITE/DISPLAY
1 On a Calculator page, press:
• MENU b
• 4: Calculus 4
• 3: Integral 3
Complete the entry line as:
∫
+ t t dt ( )
2
0
4
Then press ENTER ·.
2 State the exact distance travelled. The exact distance travelled during the frst
4 seconds of its motion, is 29
1
3
metres.
CHAPTER 15 • Kinematics 73
WORKED EXAMPLE 14
A car accelerates from rest at 2 m/s
2
for 5 seconds.
a Write an equation for the acceleration.
b Write the equation for the velocity.
c Calculate the distance covered in the frst 5 seconds.
THINK WRITE
a 1
The acceleration is 2 m/s
2
. a a(t) = 2
b 1 To determine the equation for
velocity, given a(t) = 2, on a
Calculator page, complete the entry
line as:
∫
(2)dt + c
Then press ENTER ·.
b
2 Write the equation for velocity. v(t) = 2t + c
3
It is given that v = 0 when t = 0.
Calculate the constant, c, using this
information.
0 = 2(0) + c
c = 0
∴ v(t) = 2t, 0 ≤ t ≤ 5
c 1 To calculate the distance, d(t),
covered in the frst 5 seconds, on a
Calculator page, complete the entry
line as:
∫
t dt (2 )
0
5
Then press ENTER ·.
c
2 State the distance covered in the frst
5 seconds.
The distance travelled in the frst 5 seconds is
25 metres.
CHAPTER 16 • Geometry in two and three dimensions 75
CHAPTER 16
Geometry in two and
three dimensions
WORKED EXAMPLE 4
a Use a ruler and a pair of compasses to bisect a line, AB.
b Use a CAS calculator to bisect a line segment.
THINK DRAW/DISPLAY
a 1 (a) Draw a line AB.
(b) Place the compass point at A,
with any radius (more than half
the length of the line).
(c) Draw a circle.
a
A B
2 With the same radius as in step 1,
repeat for point B.
3 The two circles will intersect at
two points. Join these points with a
straight line.
b 1 Open a new Graphs page. To access
the Geometry view, press:
• MENU b
• 2: View 2
• 2: Plane Geometry 2
b
2 To draw a line segment, complete the
following steps.
Press:
• MENU b
• 4: Points and Lines 4
• 5: Segment 5
Move the pencil to the position for
the end of the line segment and press
CLICK a.
Repeat the process for the other end.
Contents
Geometry in two and
three dimensions 75
76 Maths Quest 11 Advanced General Mathematics
3 To bisect the line, press:
• MENU b
• 7: Constructions 7
• 3: Perpendicular bisector 3
Place the cursor over each of the end
points of the line segment and press
CLICK a to view the bisected line.
WORKED EXAMPLE 5
a Use a ruler and compasses to construct a line parallel to a given line.
b Use a CAS calculator to construct a line parallel to a given line.
THINK DRAW/DISPLAY
a 1 Pick any two points, A and B, on the
given line.
a
A B
2 From point A, draw a circle of any radius
(more than half the distance from A to B).
3 With the same radius repeat step 1 at
point B.
4 Join the highest points (or lowest points)
of the circles with a straight line. This
will be parallel to AB.
b 1 Construct a straight line segment
in the same manner as for worked
example 4.
To construct a parallel line, press:
• MENU b
• 7: Constructions 7
• 2: Parallel
Press e until the word ‘segment’
appears.
b
2
Then press CLICK a and use the
NavPad to place the parallel line in
the required position.
CHAPTER 16 • Geometry in two and three dimensions 77
WORKED EXAMPLE 6
a Use a ruler and compasses to bisect any angle. b Use a CAS calculator to bisect any angle.
THINK DRAW/DISPLAY
a 1 With any radius, and the point of
the compass at the vertex V, draw
an arc of a circle which crosses both
arms of the angle. The crossings are
labelled A and B.
a
V
B
A
2 With any radius and the point of the
compass at A, draw an arc inside
the angle. This arc should be long
enough so that the line representing
half the angle would cross it.
V
B
A
3 With the same radius, repeat step 2,
putting the point of the compass at B.
The two arcs will cross at point C.
4 Join the vertex V to C. This line
bisects the angle, namely:
∠
∠
AVC =
AVB
2
V
B
A
C
b 1 Open a new Geometry page in a
Plane Geometry view.
Press:
• MENU b
• 7: Construction 7
• 4: Angle bisector 4
Move the cursor to the end point of
the frst arm of the angle and press
CLICK a.
Using the navigation pad move the
cursor to where the vertex of the
angle will be and press CLICK a.
Move the cursor to the end of the
other arm and again press CLICK a.
The bisector of the angle will
automatically appear.
b
2
Pressing ENTER · will show the
three points and the bisecting line.
78 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 8
Construct the perpendicular bisectors and median bisectors of each side of any triangle and
investigate their properties.
THINK DRAW/DISPLAY
1 Draw any triangle and add circles centred
at each vertex. The radius should be large
enough so that perpendicular bisectors
can be drawn.
2 Use the construction circles to draw
perpendicular bisectors. The black lines
join pairs of intersecting arcs. It should be
clear that the bisectors all meet at a point.
This point is called the circumcentre.
3 Use this point as a centre and draw a
circle which just touches each vertex.
To do this, put the compass point at the
point where the bisectors met and the
pencil point at any vertex. You should
observe that the resultant circumcircle
(or outcircle) just touches each vertex.
4 Alternatively, open a new Geometry
page.
Construct a triangle by pressing:
• MENU b
• 5: Shapes 5
• 2: Triangle 2
Press CLICK a and use the NavPad to
generate the 3 vertices of the triangle.
5 To construct the perpendicular bisectors,
press:
• MENU b
• 7: Construction 7
• 3: Perpendicular Bisector 3
Move the cursor to the end points of each
side of the triangle (vertices) and press
CLICK a.
CHAPTER 16 • Geometry in two and three dimensions 79
6 To draw the circumcircle, press:
• MENU b
• 5: Shapes 5
• 1: Circle 1
Move the cursor over the intersection
point (circumcentre) and press
CLICK a.
Move the cursor until it is over one of
the vertices of the triangle and press
CLICK a.
7 Furthermore, from step 2, we can
determine the midpoint of each side from
the perpendicular bisectors. Note the use
of short bars to indicate the bisection of
the sides, at points P, Q and R.
P
R
Q
8 Join each midpoint to its opposite
vertex. These lines also meet at a single
point, called the centroid, which has
applications in physics and engineering,
as it is effectively the point of symmetry
of the triangle.
P
R
Q
80 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 9
Construct the incentre of the triangle shown at right:
a by hand
b using a CAS calculator.
THINK DRAW/DISPLAY
a 1 (a) Construct the angle bisectors
by drawing arcs centred at each
vertex (A, B and C).
(b) From the intersection of
these arcs and the sides of the
triangles, draw intersecting arcs
between pairs of sides. In the
fgure at right this has been done
to vertex A only, to keep the
drawing uncluttered.
a
C
B
A
2 Complete the construction of the
angle bisectors and observe that they,
too, meet at a point — the incentre.
C
B
A
3 By placing the compass point at the
incentre and carefully drawing a
circle, it is possible to construct the
incircle, which just touches each side
of the triangle.
C
B
A
b 1 Construct a triangle using the method
set out in worked example 8.
Then press:
• MENU b
• 7: Construction 7
• 4: Angle Bisector 4
Move the cursor over each vertex
and press
CLICK a.
This will bisect the angle at the
second vertex in the order in which
they were clicked.
b
2 Repeat this process for each of the
other angles of the triangle.
C
B
A
CHAPTER 16 • Geometry in two and three dimensions 81
3 To draw the incircle, press:
• MENU b
• 5: Shapes 5
• 1: Circle 1
Move the cursor over the intersection
point (incentre) and press CLICK a.
Move the cursor until the circle fts
inside the triangle, touching each
side and press CLICK a.
82 Maths Quest 11 Advanced General Mathematics
WORKED EXAMPLE 18
Construct, with a straight edge and compasses, a tangent to a circle at any point.
This task is relatively easy, relying on the earlier construction of a perpendicular bisector.
THINK DISPLAY
1 To construct a circle, open a new
Geometry page, and then press:
• MENU b
• 5: Shapes 5
• 1: Circle 1
Press CLICK a to locate the centre of
the circle. Use the navigation pad to
increase the size of the circle then press
CLICK a.
2 To draw a tangent to the circle, press:
• MENU b
• 4: Points and Lines 4
• 7: Tangent 7
Press CLICK a to set the position of the
tangent.
3 To add the line OB, press:
• MENU b
• 4: Points and Lines 4
• 4: Line 4
Place the cursor over the centre of the
circle and press CLICK a then move
the cursor to the point of intersection
of the tangent and the circle and press
CLICK a.