Answers
Answers
The questions, with the exception of those from past question papers, and all example answers that appear in this book were
written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.
Non-exact numerical answers should be given correct to three significant figures (or one decimal place for angles in degrees) unless
a different level of accuracy is specified in the question.You should avoid rounding figures until reaching your final answer.
Chapter 1
 4

(vi)  12
 20

4, 1, 0
? (Page 4)
When subtracting numbers, the order in which the
numbers appear is important – changing the order
changes the answer, for example 3 – 6 ≠ 6 – 3.
So subtraction of numbers is not commutative.
The grouping of the numbers is also important,
for example (13 – 5) – 2 ≠ 13 – (5 – 2). Therefore
subtraction of numbers is not associative.
3
You can use more formal methods to prove these
properties. For example, to show that matrix addition
is commutative:
) (
) ( ) ( )
a b
e f
a+e b+f
e f
e+a f+b
a b
+
=
=
=
+
c d
g h
c+g d+h
g h
g+c h+d
c d
(i)
2
(i) 3 × 2
(iv) 5 × 1


(i)  5 −8   2 −3 
(ii)
1
2
0
1
0
1
2
0






 −15 8 


 −4 3 
(ii)






0
1
2
0
0
0
0
0
2
0
0
2
2
0
1
0
1
2
R
2
S
2
1
4
w = 2, x = −6, y = −2, z = 2
5
p = −1 or 6, q = ± 5
(i)






1
0
1
0
0
0
1
1
1
1
0
0






3
0
3
1
1
0
1
2
1 10 7 

4 2 10 
1 11 8 
1 8 6 
4
0
7
3
4
2
5
3






Cambridge International AS & A Level Further Mathematics Further Pure Mathematics 1
© Roger Porkess, Sophie Goldie and Rose Jewell 2018
9781510421783_Answers.indd 1






Q
2
(iii) 1 × 2
(vi) 3 × 2
 3 1 −4 


 4 2 12 
2
0
2
0
P
6
(ii) 3 × 3
(v) 2 × 4
0
1
0
1
 9 7 −17 


 10 5 28 
(ix)
(iii)
Addition of numbers is commutative.
Exercise 1A (Page 4)






 −3 −9 14 


 0 0 4 


 (vii)

(viii) Non-conformable
Matrices follow the same rules for commutativity
and associativity as numbers. Matrix addition is both
commutative and associative, but matrix subtraction
is not commutative or associative. This is true because
addition and subtraction of each of the individual
elements will determine whether the matrices are
commutative or associative overall.
1
(v)
(iv) Non-conformable
? (Page 1)
( ) ( ) (
 −8 5 


 −3 7 
(iii)
1
09/03/18 8:24 PM


(ii) 



1
1
0
0
0
1
0
1
0
0
1
1
2
3
1
2
1
2
2
3
 6 6 
CD = DC = 
.
 −2 −2 






Activity 1.2 (Page 10)
Answers
Stars 2 vs United 1
Cougars 2 vs Town 1
Cougars 1 vs United 1
7
(i)
 15 3 7 15 


 5 9 15 −3 
 19 10 9 3 
The matrix represents the number of
jackets left in stock after all the orders
have been dispatched. The negative
element indicates there was not
enough of that type of jacket in stock
to fulfil the order.
(ii)
 20 13 17 20 


 15 19 20 12 
 19 10 14 8 
(iii)
 12 30 18 0 
 6 18 24 36 


 30 0 12 18 
(i)
 −6 −1 
AB =  −20 4 


(ii)
BC = 

 −8 −15 

(iii) (AB)C = 
 −12 −28 
 −8 −15 
(iv) A(BC) =  −12 −28 


(AB)C = A(BC) so matrix multiplication is
associative in this case
o produce a general proof, use general matrices such
T
as
e f 
a b


 and C =  i j  .
A= 
 , B = 



c d
k l 
 g h 
The assumption is probably not very
realistic, as a week is quite a short time.
? (Page 9)
 a b   e f   ae + bg af + bh

=
AB =  c d   g h   ce + dg cf + dh

 

,


 e f   i j   ei + fk ej + fl

 
BC = 
  k l  =  gi + hk gj + hl
g
h

 






and so
The dimensions of the matrices are A (3 × 3), B
(3 × 2) and C (2 × 2). The conformable products
(AB)C =
are AB and BC. Both of these products would have
dimension (3 × 2), even though the original matrices
are not the same sizes.
 ae + bg af + bh   i j 
Activity 1.1 (Page 10)
 ae + bg af + bh   i j   aei + bgi + afk + bhk aej +
 ce + dg cf + dh  
 =

  k l   cei + dgi + cfk + dhk cej +
 aei + bgi + afk + bhk aej + bgj + afl + bhl 
= 
 ce + dg cf + dh  



  k l   cei + dgi + cfk + dhk cej + dgj + cfl + dhl 
 2 − 1   −4 0   − 6 − 1 
AB = 


=
 3 4   −2 1   −20 4 
 −4 0   2 −1   −8 4 
BA = 


=
 −2 1   3 4   −1 6 
 −4 −8 
0 −1 
and
 a b   ei + fk ej + fl
A(BC) = 

 c d   gi + hk gj + hl
  aei + afk + bgi + bhk aej + afl
=
  cei + dgi + cfk + dhk cej + dgj
 
 a b   ei + fk ej + fl   aei + afk + bgi + bhk aej + afl + bgj + bhl
These two matrices are not equal and so
matrix
=
 
 c dThere
  gi + hk gj + hl   cei + dgi + cfk + dhk cej + dgj + cfl + dhl
multiplication is not usually commutative.
are some exceptions, for example if
Since (AB)C = A(BC) matrix multiplication is
2 0
 3 3 
associative and the product can be written without
 and D = 
 then
C= 
 −1 −1 
0 2
brackets as ABC.
2




Cambridge International AS & A Level Further Mathematics Further Pure Mathematics 1
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Exercise 1B (Page 11)
2
(i)
(iii)
3
4
 21 6 


 31 13 
(ii)
(c) 2 × 3
(f) 3 × 5
non-conformable
non-conformable
 −54 


 −1 
(iv)
(v) non-conformable (vi)
 28 −18

 26 2
 16 25




 −38 −136 −135

 133 133 100
 273 404 369
6
 2x 2 + 12 −9 
 (ii) x = 2 or 3
(i) 
3 
−4

 8 12 
 18 18 
 or 

(iii) BA = 
 8 15 
 12 15 
7
(i) (a)
(c)
 16 15 


 0 1 
(b)
1
0
1
0
(ii)






4
2
2
1
3
2
1
1
 8 7 


 0 1 




2
1
0
1
0
0
2
0
3
2
5
0
4
2
0
2











2 represents the number of two-stage
M
routes between each pair of resorts.
(iii)M3 would represent the number of threestage routes between each pair of resorts.
9
 26 37 16 


 14 21 28 
 −8 −11 2 
5
 4 3 


 0 1 
1
1
1
0
AB ≠ BA so matrix multiplication is noncommutative.
 5 25 
 −7 26 

(i) 
(ii) 

 16 22 
 2 34 
 31 0 


 65 18 
(i)





(−30 −15)
 3 −56 
 −25 8 
 , BA = 

AB = 
 20 −73 
 28 −45 
(iii)
8
 1024 1023   2n 2n − 1 

 =

0   0
1 
 0
Answers
1 (i) (a) 3 × 3 (b) 1 × 3
(d) 2 × 4 (e) 2 × 1
(ii) (a) non-conformable
(b) 3 × 5
(c)
(d) 2 × 3
(e)
(iii)
(i)
 8 + 4 x −20 + x 2

 − 8 + x −3 − 3 x



x = −3 or 4
 −4 −11 
 24 −4 
(iii) 
 or 

−
11
6


 −4 −15 
(ii)
10




(ii)
 0 1 0 


(iii)  0 0 1  ,
 1 0 0 


b

 c
a




 0 0 1  c
(iv)  1 0 0  ,  a

 
 0 1 0  b




(i)
(v)
b

 a
c
 1 0 0

 0 1 0
 0 0 1
 1 0 0

 0 0 1
 0 1 0








he strands are back in the original order
T
at the end of Stage 6.
 2n 2n − 1 

(ii) 
1 
 0
Cambridge International AS & A Level Further Mathematics Further Pure Mathematics 1
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?
(Page 18)
Answers
   
The image of the unit vector  1  is  a  and
0 c 
0  
the image of the unit vector   is  b  .
1 d 
The origin maps to itself.
Rotations of 45° clockwise about the origin
and 135° anticlockwise about the origin are also
1
represented by matrices involving ± 2 .
This is due to the symmetry about the origin.
(i)
he diagram for a 45° clockwise rotation about
T
the origin is shown below.
y
Activity 1.3 (Page 18)
B
The diagram below shows the unit square with
two of its sides along the unit vectors i and j . It is
rotated by 45° about the origin.
B′
45°
y
B
A′
A′
45°
O
x
1
B′
A
x
You can use trigonometry to find the images of the
unit vectors i and j.
For A′, the x-coordinate satisfies cos 45 = x so
1
x = cos 45 = 1 .
2
1
In a similar way, the y-coordinate of A′ is
.
2
For B′, the symmetry of the diagram shows that the
1
1
.
x-coordinate is − 2 and the y-coordinate is
2
 1 
1  2 
Hence, the image of   is 
 and the image
0  1 
 2 
0




of 1 is
 
 1 
−

2  and so the matrix representing an

 1 
 2 


anticlockwise rotation of 45° about the origin is
 1

− 1 

2 .
 2
1 
 1
 2
2 

4
A
O

1  1


The image of 0 is  2
 
− 1

2



 and the image



 1 
 0  

is  2  and so the matrix
of 

1

  1 
 2 


representing an anticlockwise rotation of 45°
about the origin is
 1

 2
− 1

2

1
2
1
2


.



he diagram for a 135° anticlockwise rotation
(ii) T
about the origin is shown below.
y
B
A′
135°
O
A
x
B′
Cambridge International AS & A Level Further Mathematics Further Pure Mathematics 1
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y


1 − 1 
2  and the image of
The image of  0  is 
   1 
 2 


B′
Answers


0 − 1 
  is 
2  and so the matrix representing
1
− 1 

2 

an anticlockwise rotation of 45° about the origin is
 1

− 1 
−
2
2 

 1 − 1 .
 2
2 

?
B
j=
(Page 20)
Activity 1.4 (Page 20)
The diagram below shows the effect of the
2 0
matrix 
 on the unit vectors i and j.
0 1
y
A′
A
x
0
the vector j has image   . Therefore this
5
matrix represents a stretch of scale factor
5 parallel to the y-axis.


The matrix  m 0  represents a stretch of scale
0
1


factor m parallel to the x-axis.


The matrix  1 0  represents a stretch of scale
0 n
B′
B
( )
factor n parallel to the y-axis.
0
1
O
( )
1
i=
0
You can see that the vector i is unchanged and


origin is  cos θ sin θ  .
 − sin θ cos θ 
j=
0
1
O
The matrix for a rotation of θ ° clockwise about the
(i)
( )
i=
( )
1
0
A
A′
x
 
You can see that the vector i has image  2 
0
and the vector j is unchanged. Therefore this
matrix represents a stretch of scale factor
2 parallel to the x-axis.
Activity 1.5 (Page 22)
Point A: 6 ÷ 2 = 3
Point B: 6 ÷ 2 = 3
Point C: 3 ÷ 1 = 3
Point D: 3 ÷ 1 = 3
The ratio is equal to 3 for each point.
(ii)The diagram below shows the effect of the


matrix  1 0  on the unit vectors i and j.
0 5
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Exercise 1C (Page 24)
1
(i) (a)
y
6
(d)
B′
A′
 0 −1 


 −1 0 
(iv) (a)
5
Answers
(c) x′ = –y, y′ = –x
y
4
3
3
2
2
1
1
–3
O
1
3
2
4
5
–2
x
6
3
(v)
2
(a)
2
1
B′
x
3
2
1
–3
A′
–2
2
–2
A′
2
1
–1 O
–1
1
2
–2
–3
(b)
6
2
3
x
A′ = (3, 1), B′ = (0, 1)
1
(c) x′ = 3x, y′ = 2 y
(d)
3
–3
1
(b)
y
B′
A′
–3
1 0 


 0 −1 
(iii) (a)
–1 O
–1
B′
–2
(b) A′ = (1, –2), B′ = (0, –2)
(c) x′ = x, y′ = –y
(d)
A′
y
–3
x
3
–1 O
–1
–2
3
 0 1


 −1 0 
1
–2
2
(b) A′ = (2, –1), B′ = (2, 0)
(c) x′ = y, y ′ = –x
(d)
y
–3
1
–3
3 0


0 3
(ii) (a)
–1 O
–1
–2
(b) A′ = (3, 6), B′ = (0, 6)
(c) x′ = 3x, y′ = 3y
(d)
B′
3
x
3 0 


0 1
2 

(i)
Reflection in the x-axis
(ii) Reflection in the line y = –x
(iii) S tretch of factor 2 parallel to the x-axis
and stretch factor 3 parallel to the y-axis
(iv) Enlargement, scale factor 4, centre the
origin
(v) Rotation of 90° clockwise (or 270°
anticlockwise) about the origin
A′ = (–2, –1), B′ = (–2, 0)
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(i)
4
(i)  1 4   1 1 −1 −1  =  5 −3 −5 3 



 
 0 1   1 −1 −1 1 
y
D′
(ii)
7
A′(4, 5), B′(7, 9), C′(3, 4). The original square and
the image both have an area of one square unit.
8
(i)
B′
–3
The transformation is a shear with the
x-axis fixed and the point A(1, 1) has
image A′(5, 1).
(ii)
9
6
4
10
1
2
(i)
0
0.5 1 1.5
x
–1
2
x
B
1
The gradient of A′C′ is , which is the
2
reciprocal of the top right-hand entry of
the matrix M.
5 0
Any matrix of the form 
 or
0 k
k 0

.
0 5
If k = 5 the rectangle would be a square.
(ii)
1
B′ C
1
0
1 0
.
The matrix for the transformation is 
1 0
y
2
–1
A′
1
( x, y ) → ( x, x )
x
(b) The image of the unit square has
vertices (0, 0), (1, 0), (0.5, 1), (1.5, 1) as
shown in the diagram below.
–2
C′
(i) (a) The image of the unit square has
vertices (0, 0), (1, 5), (0, 1), (1, 6) as
shown in the diagram below.
0
(ii)
y
x
2

3 
1 
(i) A′ (2 3 − 1, 2)
A
y

 1
 0
6
A′
O
C′
Matrix A represents a shear with the
y-axis fixed; the point (1, 1) has image
(1, 6). A has shear factor 5.
Matrix B represents a shear with the
x-axis fixed; the point (1, 1) has image
(1.5, 1).
B has shear factor 0.5.
 1 −1 −1 1 
so the transformed square would look like this:
5
(ii)
otation of 60° anticlockwise about the
R
origin
(ii) Rotation of 55° anticlockwise about the
origin
(iii) Rotation of 135° clockwise about the
origin
(iv) Rotation of 150° anticlockwise about the
origin
Answers
3
 1 0 

 2 1  , 
 ,

 0 1 1 2 

 1 2
1 0
 0 1


 or 
 2 1

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
7 3 3
2
(iii) 
 0 3
2



,



3
0 2

 7 3 3
2



,


Answers


 3

0
 3 3 7
2


 2
 or 

 3 0 
 3 3 7
 2

 2

?
(Page 27)
A represents a reflection in the line y = x.
B
The transformation A is represented by the
1 0 
 and the transformation
matrix A = 
 0 −1 
?
(Page 28)
AB represents ‘carry out transformation B followed
by transformation A.
(AB)C represents ‘carry out transformation C
followed by transformation AB, i.e. ‘carry out C
followed by B followed by A’.
BC represents ‘carry out transformation C followed
by transformation B’.
A(BC) represents ‘carry out transformation BC
followed by transformation A, i.e. carry out C
followed by B followed by A’.
Activity 1.7 (Page 29)
B is represented by the matrix


A =  cos θ − sin θ  ,
 sin θ cos θ 
 0 −1 
. The matrix product
B = 
 1 0 
 cos φ − sin φ
B = 
 sin φ cos φ
 0 −1   1 0   0 1 
=
.
BA = 
 1 0   0 −1   1 0 
This is the matrix which represents a reflection in the
line y = x.
Activity 1.6 (Page 28)
(i)




(ii)
cos θ cos φ − sin θ sin φ − sin θ cos φ − cos θ sin φ
BA = 

sin θ cos φ + cos θ sin φ − sin θ sin φ + cos θ cos φ
 cos (θ + φ) −sin (θ + φ)

(iii) C = 
 sin (θ + φ) cos (θ + φ)




sin (θ + φ) = sin θ cos φ + cos θ sin φ
(i)
 a b   x   ax + by 
P′ = 

  =
 c d   y   cx + dy 
(ii)
(v)  A rotation through angle θ followed by
 p q   ax + by   pax + pby + qcx + qdy
P′′ = 
=
 rotation through angle φ has the same effect as
 

 r s   cx + dy   rax + rby + scx + sdy  a rotation through angle φ followed by angle θ .
 p q   ax + by   pax + pby + qcx + qdy 
P′′ = 

 =

 r s   cx + dy   rax + rby + scx + sdy 
 p q   a b   pa + qc pb + qd 
(iii) U = 


 =
 r s   c d   ra + sc rb + sd 
(iv)
cos (θ + φ) = cos θ cos φ − sin θ sin φ
Exercise 1D (Page 30)
1 (i) A: enlargement centre (0,0), scale factor 3
B: rotation 90° anticlockwise about (0,0)
C: reflection in the x-axis
D: reflection in the line y = x
and so
0 1
(a)
 BC =  1 0  , reflection in the line
 pa + qc pb + qd   x   pax + qcx + pby (ii)
+ qdy
UP = 
   = 
sdy  y = x
 ra + sc rb + sd   y   rax + scx + rby +
 pa + qc pb + qd   x   pax + qcx + pby + qdy 
= 

   = 
 ra + sc rb + sd   y   rax + scx + rby + sdy  .
Therefore UP=P″
8
 0 −1 
(b)
CB =  −1 0  , reflection in the line


y = −x
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 0 −1 
(c) DC = 
 , rotation 90°
1 0 
anticlockwise about (0, 0)
9 0
 , enlargement centre (0, 0),
(d)
A2 = 
0 9
The matrix Q interchanges the coordinates,
i.e. (x, y) → (y, x)
It does not matter what order these two
transformations occur as the result will be
the same
4
x-axis
1 0
 returns the object to
(f)
DC2D = 
0 1
its original position
5
(i)
1 0 
X =  0 −1 


(ii)
 −1 0 
XY =  0 −1  , rotation of 180°


 0 1 
,
Possible transformations are B = 
 −1 0 
3 0
 , which is a stretch of scale factor
A= 
0 1
3 parallel to the x-axis. The order of these is
important as performing A followed by B leads
 0 1
 . Could also have
to the matrix 
 −3 0 
 −1 0 

Y= 
 0 1
1 0
 . which represents a stretch of
B= 
0 3
factor 3 parallel to the y-axis, followed by
about the origin
 −1 0 
(iii) YX =  0 −1 


(iv) W
hen considering the effect on the unit
vectors i and j, as each transformation
only affects one of the unit vectors the
order of the transformations is not important in this case.
3
(ii) (32, −33)
which is a rotation of 90° clockwise about the
origin, followed by
(iii) For example, B4, C2 or D2
2
 8 −4 
(i) 

 −3 12 
0 1

Q= 
1 0
(i)
 −1 0 

P= 
 0 −1 
(ii)
 0 −1 
PQ = 
 , reflection in the line
 −1 0 
6
 0 −1 
(iii) QP =  −1 0 


(iv) The matrix P has the effect of making the
coordinates of any point the negative of
their original values,
 0 1
 , which represents a rotation of
A = 
 −1 0 
90° clockwise about the origin; again the order
is important.
 1 0 

(i) PQ = 
 −3 −1 
1 0 
(ii) P =  0 −1  represents a reflection in


the x-axis.
1 0
Q= 
 represents a shear with the
3 1
y-axis fixed; point B(1,1) has image (1,–4).
y = −x
7
 1
1 


2 
X=  2
 1 − 1 
 2
2 

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Answers
scale factor 9
1 0 
(e)
BCB = 
 , reflection in the
 0 −1 
i.e. (x, y) → (−x, − y )
9
09/03/18 8:24 PM
Answers
A matrix representing a rotation about the
 cos θ − sin θ 
 and so
origin has the form 
 sin θ cos θ 


the entries on the leading diagonal would be
equal. That is not true for matrix X and so this
cannot represent a rotation.
8
(i)
(ii)
10
9
2(
1
2 + 2)
By comparison to the matrix
 cos θ − sin θ 

 for an anticlockwise
 sin θ cos θ 


A reflection in the x-axis and a stretch of
scale factor 5 parallel to the x-axis
11
Reflection in the x-axis; stretch of scale
factor 5 parallel to the x-axis; stretch of scale
factor 2 parallel to the y-axis.The outcome
of these three transformations would be the
same regardless of the order in which they
are applied.There are six different possible
orders.
(iv)
2 + 2 and b =
4
D represents an anticlockwise rotation of 22.5°
about the origin.
1 0


0 2
5 0 

(iii) 
 0 −2 
a=
rotation of θ about the origin, a and b are
the exact values of cos 22.5° and sin 22.5°
respectively.




3 
3 
 1
 −1
2
2 
2 2 
(i) P = 
Q= 
 3


1
3 1 
 2 −2 
 2
2 



(ii)
1

5 0 


 0 −1 
2 



 1 − 3 
2
2
QP = 
 , which represents
 3
1 
 2
2 

a rotation of 60° anticlockwise about the
origin.
(i)
 1 −R
1

0 1



(ii)
(iii)
 1


1
 − R2

−R1
R1
+1
R2





(iv)

R
 1+ 1
R
2


1
 − R2



3 
 1
2  , which represents a
(iii) PQ =  2

3 1 
− 2
2 

rotation of 60° clockwise about the origin.
 1 0


 − 1 1 
 R2


−R1 


1 

12
?
A reflection in a line followed by a second
reflection in the same line returns a point to its
original position.
(Page 33)
In a reflection, all points on the mirror line map to
themselves.
In a rotation, only the centre of rotation maps to itself.
The effect of Type B followed by Type A
is different to that of Type A followed by
Type B.
10
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Exercise 1E (Page 35)
1
(ii)
(i)
Points of the form (λ , − 2λ)
(ii)
(0, 0)
10
x-axis, y-axis, lines of the form y = mx
(ii)
x-axis, y-axis, lines of the form y = mx
y = − –94 x
6
(i)
(iii)
(iv)
y = x , lines of the form y = −x + cw
(v)
y = −x, lines of the form y = x + c
B
A
2
y =x
1
10 12
x
y =x
(ii)
y=x
C′
C
B′
B
–1 O
1A 2
–1 A′
(i)
8
y
2
9
6
4
C′
3
(vi) x-axis, lines of the form y = c
Any points on the line y = 1 x , for
2
example (0, 0), (2, 1) and (3, 1.5)
4
2
C
–4 –2 O
–2
(iii) no invariant lines
(ii)
B′
Answers
(i)
(i)
y=x
6
(iv) Points of the form (2λ , 3λ)
3
A′
8
(iii) Points of the form (λ , − 3λ)
2
y
3
x
x ′ = x + a, y ′ = y + b
(iii) (c) a = −2b
y = 1x
2
(iii) Any line of the form y = −2x + c
(iv) Using the method of Example 1.12 leads
to the equations
2m 2 + 3m − 2 = 0 ⇒ m = 0.5 or − 2
(4 + 2m) c = 0 ⇒ m = −2 or c = 0
If m = 0.5 then c = 0 so y = 1 x is
2
invariant.
If m = −2 then c can take any value and
so y = −2x + c is an invariant line.
Chapter 2
?
(Page 40)
A decreasing sequence of fractions starting at 1 or a
decreasing sequence of percentages starting at 100%
e.g. 100%, 95%, 80%, 55%, 50%, 20%, 5%
?
(Page 41)
Start at 2 and add 3 each time.
4
(i)
5

 
Solving  4 11   x  =
 
 11 4   y 
to the equations y = − 3x
11
x
  leads
y
and y = − 11x .
3
The only point that satisfies both of these
is (0, 0).
(ii)
y = x and y = −x
(i)
y = x , y = −9x
4
Exercise 2A (Page 45)
1(i)6, 11, 16, 21, 26
Increasing by 5 for each term
(ii) −3, −9, −15, −21, −27
Decreasing by 6 for each term
(iii) 8, 16, 32, 64, 128
Doubling for each term
(iv) 8, 12, 8, 12, 8
Oscillating
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(v)
Answers
2, 5, 11, 23, 47
Increasing
(vi) 5, 5 , 5 , 5 , 1
2 3 4
Decreasing, converging to zero
2 (i) 21, 25, 29, 33
(ii) u1 = 1, ur + 1 = ur + 4
(iii) ur = 4r – 3
3 (i)
(a)
(b)
(c)
(d)
(ii) (a)
(b)
(c)
(d)
(iii) (a)
(b)
(c)
(d)
0, –2, –4, −6
u1 = 10, ur + 1 = ur − 2
ur = 12 − 2r
–28
32, 64, 128, 256
u1 = 1, ur + 1 = 2ur
ur = 2 r − 1
524 288
31 250, 156 250, 781 250, 3 906 250
u1 = 50, ur + 1 = 5ur
ur = 10 × 5r
9.54 × 1014
4 (i) 25
(iii) 363
5 (i)
62500
7
∑ (56 − 6r )
(ii)
(iv)
–150
–7.5
(ii)
224
r =1
7 (i) –5, 5, –5, 5, –5, 5
Oscillating
(ii) (a) 0
(b) –5
5
5
n
(iii) − + ( −1)
2 2
8 (i) 0, 100, 2, 102, 4, 104
Even terms start from 100 and increase
by 2, odd terms start from 0 and increase
by 2.
(ii) 201
(iii) 102
9749 cm
10 1 n(n 3 + 1)
2
1110, 5, 16, 8, 4 (This will reach 1 at c7 and then
repeat the cycle 4, 2, 1)
(ii)
2, 12, 36
1 n(n + 1)(3n + 1)(n + 2)
12
4 n4
5 1 n(n + 1)(n + 2)
3
1 n(n + 1)(n + 2)(n + 3)
6
4
7 1 n(3n + 1)
2
8
n 2 (4n + 1)(5n + 2)
9 (ii)
7 layers, 125 left over
10 (i)
$ 227.50
1 n(35(n + 1) + 30I )
24
(ii)
11
S1 = 3
S3 = 21
S2 = 10
S4 = 36
ur = 4 r − 1
2n
∑u
r = n +1
?
r
= 6n 2 + n
(Page 51)
As n becomes very large, the top and bottom of
n
n + 1 are very close, so the sum becomes very close to
1 (it converges to 1).
?
(Page 53)
As n becomes very large, the expression
n(3n + 7)
3n 2
2(n + 1)(n + 2) becomes close to 2n 2 (since
terms in n² are much bigger than terms in n). So the
3
sum becomes very close to 3 it converges to 2 .
2
)
)
Exercise 2C (Page 54)
1 (ii)
(1 – 0) + (4 – 1) + (9 – 4) + … +
[(n − 2)2 − (n − 3)2 ] +
[(n − 1)2 − (n − 2)2 ] + [n 2 − (n − 1)2 ]
(iii) n2
Exercise 2B (Page 49)
12
3 (i)
2 (i)
1 (i)
1, 3, 5
(ii)
n²
2 (i)
4, 14, 30
(ii)
n (n + 1)2
First term: r = 1, last term: r = 10
(iii) 20
21
3 (ii)
n (n 2 + 4n + 5) (iv)
99
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Activity 2.2 (Page 60)
n (n + 2)
(n + 1)2
4 (ii)
Assume true for n = k, so
(i)
2+4+6+ … +
( )
2k + 2 (k + 1) = k + 1
2
16n 4 + 32n 3 + 24n 2 + 8n
1
1
10 (ii) A = 2 , B = − 2
(3n + 2)(n − 1)
(iii)
4n (n + 1)
(iv) As n → ∞, the sum → 3
4
n
1
1
11 ∑ uk = −
5
2n + 1
k = 13
∞
= 1
5
∑u
k = 13
12 (i)
(ii)
k
4.820
55
13 1 n 2 (n + 1)
2
?
(Page 56)
Activity 2.1 (Page 56)
1
1× 2
1
1× 2
1
1× 2
1
1× 2
Exercise 2E (Page 65)
5 (ii) M is a shear, x-axis fixed, (0, 1) maps to (1, 1).
Mn is a shear, x-axis fixed, (0, 1) maps to (n, 1).
1
1
1
6 (i) u2 = , u3 = , u4 =
2
3
4
1
(ii) un = n
7 (i)
 −5 −12 
 −3 −8 
, A3 = 
A2 = 

 3 7 
 2 5 
8 (i)
 7 0
,
M2 = 
 0 7 
 −7 14 
 49 0 
M3 = 
M4 = 

21
7
,


 0 49 
 1 0
 −1 2 
(ii) M 2m = 7m 
, M 2 m + 1 = 7m 
 0 1 
 3 1 
9 (i)
10
3, 5, 17, 257, 65 537
2!− S1 = 1, 3!− S2 = 1, 4!− S3 = 1, 5!− S4 = 1.
Sn = (n + 1)!− 1
= 1
2
1 = 2
2×3 3
+ 1 + 1 = 3
2×3 3×4
4
1
1
+
+
+ 1 = 4
2×3 3×4 4×5 5
+
2
It is not true for n = 1.
(ii) It breaks down at the inductive step.
If she was 121 last year then it would be fine, but we
don’t know if this is true. If she were able to provide
any evidence of her age at a particular point then we
could work from there, but we need a starting point.
+ 2 (k + 1)
( )
= (k + 1 + 1)
2
8n 3 + 12n 2 + 6n
9 (ii)
2
= k 2 + k + 1 + 2k + 2
4
2
= k + 3k + 9
4
2
= k+ 3
2
n (n + 3)
4 (n + 1)(n + 2)
(iii) 0.249 95..., 0.249 9995… The sum looks
as if it is approaching 0.25 as n becomes
large.
7 (ii)
8 (ii)
( )
2
2 + 4 + 6 + … + 2k = k + 1 .
2
For n = k + 1,
11 2
1+ x
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Answers
n (3n + 5)
5 (ii)
4 (n + 1)(n + 2)
(iii) 0.7401, 0.7490, 0.7499. The sum looks as if
it is approaching 0.75 as n becomes large.
6 (ii) 13
120
13
09/03/18 8:24 PM
Chapter 3
Answers
?
Activity 3.3 (Page 75)
(Page 69)
4x 3 + x 2 − 4x − 1 = 0
Looking at the graph you may suspect that x = 1
is a root. Setting x = 1 verifies this. The factor
theorem tells you that (x – 1) must be a factor, so
factorise the cubic ( x − 1)(4 x 2 + 5x + 1) = 0 .
Now factorise the remaining quadratic factor:
( x − 1)(4 x + 1)( x + 1) = 0 , so the roots are
1
x = 1, – 4 , –1.
4x 3 + x 2 + 4x + 1 = 0
This does not have such an obvious starting
point, but the graph suggests only one real root.
Comparing with previous example, you may spot
1
that x = – 4 might work, so you can factorise giving
(4 x + 1)( x 2 + 1) = 0 . From this you can see that the
other roots must be complex. x 2 = −1, so the three
1
roots are x = – 4 , ±i.
Activity 3.1 (Page 71)
Equation
−3 ± i 31 , −3 ± i 31
4
2
(ii)
2 ± 7,5± 7
3
3
Exercise 3A (Page 75)
1 (i)
(ii)
(iii)
α + β = − 7,
2
1
α +β = ,
5
α + β = 0,
24
(iv) α + β = − 5 ,
αβ = 3
1
5
2
αβ =
7
αβ = −
αβ = 0
α + β = −11,
αβ = −4
8
(vi) α + β = − 3 ,
αβ = −2
(v)
2 (i)
(ii)
z 2 − 10z + 21 = 0
z 2 − 3z − 4 = 0
(iii) 2z 2 + 19z + 45 = 0
(iv) z 2 − 5z = 0
(v)
z 2 − 6z + 9 = 0
(vi) z 2 − 6z + 13 = 0
Two
roots
Sum of Product
roots
of roots
2
(i) z − 3z + 2 = 0
1, 2
3
2
(ii)
2
(ii) z + z − 6 = 0
2, –3
–1
–6
(iii) 2z 2 + 13z + 9 = 0
2
(iii) z − 6z + 8 = 0
2, 4
6
8
3
–10
2
(iv) z − 3z − 10 = 0 –2, 5
2
(v) 2z − 3z + 1 = 0
1, 1
2
3
2
1
2
2
(vi) z − 4 z + 5 = 0
2±i
4
5
?
(Page 71)
If the equation is ax2 + bx + c = 0, the sum appears
b
c
to be – a and the product appears to be a .
?
(Page 72)
You get back to the original quadratic equation.
14
(i)
3 (i)
2z 2 + 15z − 81 = 0
2 z 2 − 5z − 9 = 0
(iv) z 2 − 7z − 12 = 0
4 z 2 − 20z + 4 = 0
5 (i)
(ii)
(iii)
(α
+ β ) − 2αβ
2
α +β
αβ
(α
+ β ) − 2αβ
(αβ)2
2
(iv) αβ (α + β )
(v)
(α
+ β ) − 3αβ (α + β )
3
6 (i)Roots are real, distinct and negative (since
αβ > 0 ⇒ same signs and α + β < 0 ⇒
both <0)
(ii)
α = −β
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(iii) One of the roots is zeros and the other
is − b .
a
12Roots are – p and ± −q (note ± −q is not
­necessarily imaginary, since q is not necessarily >0)
(iv) The roots are of opposite signs.
az + bkz + ck = 0
az 2 + (b − 2ka )z + (k 2a − kb + c ) = 0
11 (ii)
z 2 − (5 + 2i)z + (9 + 7i) = 0
2
13 (i)
(
β
1
q = 8 + αβ +
2
2α
r = −4 β
Exercise 3B (Page 80)
3
3
15 ac = b d
z 3 − 7z 2 + 14 z − 8 = 0
(ii)
z 3 − 3z 2 − 4 z + 12 = 0
(iii)
2z 3 + 7z 2 + 6z = 0
(iv)
2z 3 − 13z 2 + 28z − 20 = 0
(v)
z 3 − 19z − 30 = 0
(vi)
z − 5z + 9 z − 5 = 0
z = 1, 3, 9
2 2 2
16 p = −15, q = 71 ; α = 3 , r = −105
?
2
(Page 82)
See derivation of formulae on page 83.
∑ α = α + β + γ + δ = − ba
∑ αβ = αβ + αγ + αδ + βγ + βδ + γδ =
∑ αβγ = αβγ + βγδ + γδα + δαβ = − da
3 (i)
z = 2, 5, 8
2
(ii) z = − , 2 , 2
3 3
(iii) z = 2 − 2 3 , 2, 2 + 2 3
(iv) z = 2 , 7 , 5
3 6 3
c
a
αβγδ = e
a
z =w−3
(ii) (w − 3)3 + (w − 3)2 + 2(w − 3) − 3 = 0
(iii) w 3 − 8w 2 + 23w − 27 = 0
(iv) α + 3, β + 3,γ + 3
Exercise 3C (Page 84)
−3
2
(ii) 3
(iii) 5
2
(iv) 2
1 (i)
5 (i)
(ii)
w 3 − 4w 2 + 4w − 24 = 0
z3 - 2z2 - 11z - 9 = 0
6 (i)
2w 3 − 16w 2 + 37w − 27 = 0
2 (i)
2w 3 + 24w 2 + 45w + 37 = 0
(ii)
(ii)
)
14 z = 3 , 7 , − 2
7 3
2 (i)
4 (i)
)
(iii) r = 9; x = 1, 1 , − 9
2
4
1
r = −6; x = −2, − , 3
4 2
−3
1 (i)
2
1
(ii) −
2
7
(iii) −
2
3
(
p = −8 α + 1 + β
2α
Answers
10 (i)
(ii)
2
(iii) z3 - 28z2 + z - 1 = 0
7 The roots are 3 , 2, 5 k = 47
2
2
2
8 z = 1 , 1 , − 3
4 2 4
9 α = –1, p = 7, q = 8 or α = p = q = 0
z 4 − 6z 3 + 7z 2 + 6z − 8 = 0
4 z 4 + 20z 3 + z 2 − 60z = 0
(iii) 4 z 4 + 12z 3 − 27z 2 − 54 z + 81 = 0
(iv) z 4 − 5z 2 + 10z − 6 = 0
3 (i)
(ii)
z 4 + 4 z 3 − 6z 2 + 8z + 48 = 0
2z 4 + 12z 3 + 21z 2 + 13z + 8 = 0
4 (i)Let w = x + 1 then x = w − 1
new quartic: x 4 − 6x 2 + 9
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Answers
(ii)
5 (i)
(ii)
Solutions to new quartic are x = ± 3
(each one repeated), solutions to original
quartic are therefore: α = β = 3 − 1
and γ = δ = − 3 − 1 .
Chapter 4
α = −1, β =
The population of rabbits fluctuates, but eventually
approaches a stable number.
3
p = 4 and q = −9
(iii) Use substitution y = x − 3α (i.e.
y = x + 3 then x = y − 3) and
y 3 − 8y 2 + 18y – 12 = 0
6 (i) α + β + γ + δ + ε = − b
a
αβ + αγ + αδ + αε + βγ + βδ + βε +
c
γδ + γε + δε =
a
αβγ + αβδ + αβε + αγδ + αγε + αδε +
d
βγδ + βγε + βδε + γδε = −
a
e
αβγδ + βγδε + γδεα + δεαβ + εαβγ =
a
f
αβγδε = −
a
∑α = − b
a
∑ αβ = c
a
–4
12
4
12
y 4 − 4 y 2 + 4 = 0;
y = ± 2 (twice)
x = ± 2 − 1 (twice)
8 (i)
(ii)
S2 = 6 , S4 = 26
(Page 90)
?
(Page 90)
(i) y = 0
(ii) y = 1
(iii) y = −2
?
(Page 93)
(i)positive, positive, positive
(ii)negative, negative, negative
(iii)positive
(iv)negative
(Page 93)
( x + 2)
( x + 2)
=
( x − 2)( x + 1) ( x 2 − x − 2)
v du − u dv
dy
u
= dx 2 dx
The quotient rule for y = is
v dx
v
Using this with u = x + 2 and v = x2 −x −2 gives
∑ αβγδ = e
= αβγδε = −
(Page 88)
(i) x = −2
(ii) x = 1 and x = −2
(iii) x = 1
2
y =
∑ αβγ = − d
a
7 (i)
(ii)
(iii)
(iv)
?
?
a
∑ αβγδε
?
f
a
dy
( x 2 − x − 2) × 1 − ( x + 2)(2x − 1)
=
dx
( x 2 − x − 2)2
−x 2 − 4x
( x 2 − x − 2)2
− x( x + 4)
= 2
( x − x − 2)2
=
dy
= 0 ⇒ − x( x + 4) = 0 ⇒ x = 0, x = −4
dx
(0 + 2)
When x = 0 then y = (0 − 2)(0 + 1) = −1
( −4 + 2)
1
When x = −4 then y = ( −4 − 2)( −4 + 1) = − 9
S3 = −15, S5 = −75
−15
16
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Step 4:
Exercise 4A (Page 96)
(
y
)
Step 5: y ≠ 0
(
Step 5: y ∈
x
3
2
−
3
)
( )
2
3
−3
Step 5: y ≠ 1
x
3
x
2
2
5 Step 1: (2, 0), 0, 3
Step 2: x = –3
Step 3: y → –1 from above as x → ∞
y → –1 from below as x → –∞
Step 4:
y
2 Step 1: 0, − 2
9
Step 2: x = 3
Step 3: y → 0 from above as x → ∞
y → 0 from above as x → –∞
Step 4: y
2
9
0
0
−2
Answers
1 Step 1: 0, − 2
3
Step 2: x = 3
Step 3: y → 0 from above as x → ∞
y → 0 from below as x → –∞
Step 4: y
0
−1
x
2
( )
6 Step 1: (5, 0), 0, 65
Step 2: x = –2, x = 3
Step 3: y → 0 from above as x → ∞
y → 0 from below as x → –∞
Step 4:
y
Step 5: y > 0
3 Step 1: (0, 1)
Step 2: none
Step 3: y → 0 from above as x → ∞
y → 0 from above as x → –∞
y
Step 4:
5
6
−2
0
3
5
x
1
Step 5: y 9 − 2 14 , y 9 + 2 14
25
25
x
0
Step 5: 0 < y  0
4 Step 1: (0, 0)
Step 2: x = 2, x = –2
Step 3: y → 0 from above as x → ∞
y → 0 from below as x → –∞
( )
7 Step 1: (3, 0), 0, 83
Step 2: x = 2, x = 4
Step 3: y → 0 from below as x → ∞
y → 0 from above as x → –∞
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Step 4:
Step 4:
y
y
10
Answers
3
8
0
2
3
x
4
3
2
2
5
0
3
2
−1
Step 5: y ∈
8 Step 1: (0, 0)
Step 2: none
Step 3: y → 0 from above as x → ∞
y → 0 from below as x → –∞
Step 4:
y
4
x
Step 5: y  25 − 2 66 , y  25 + 2 66
5
5
11
O
x
Step 1: (3, 0) (repeated), (0, 9)
Step 2: none
Step 3: y → 1 from below as x → ∞
y → 1 from above as x → -∞
Step 4:
y
9
Step 5: − 1 y 1
2 3
2 3
(
)
1
3
9 Step 1: (3, 0), 0, − 16
Step 2: x = 4
Step 3: y → 0 from above as x → ∞
y → 0 from below as x → –∞
Step 4: y
0
Step 5: 0 y 10
( )
0
−
3
3
16
x
4
x
3
3
12 Step 1: (6, 0), 0,
2
Step 2: x = 4
Step 3: y → 1 from below as x → ∞
y → 1 from above as x → -∞
Step 4:
y
Step 5: y − 1
4
(
) (
) (
)
Step 1: 3 , 0 , − 2 , 0 , 0, 3
2
2
5
Step 2: x = –1, x = 4
Step 3: y → 10 from above as x → ∞
y → 10 from below as x → –∞
10
1
−1
0
4
6
x
Step 5: y ≠ 1, y ≠ 7
5
18
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13 (i)
(d)
y
2
y = 2x − 1 + x − 4
y
1
0
x
2
Answers
−2
y = 2x − 1
−1
(ii)
14 (i)
x
O
k −1 or k > 1
y
x=4
1
3
−1
(ii)
0
3
(a)
x
(41 (9 −
)(
)
65), 0 , 1 (9 + 65), 0 ,
4
(0, − 1 )
2
(b) y = 2x − 1; x = 4
(c)
( )
(d)
(ii) x = 1, minimum point = 1, 1
4
(iii) (a) k < 0, k > 41
(b)
x=4
y
y = 2x − 1
2
y = 2x − 1 + 4 − x
k = 41
(c) 0 k < 1
4
15 (i) x  −2 or x >1
(ii) −5  x < 2
(iii) –1 < x  2 or 3 < x  7
1
5
(iv) 2 < x 3
5
(v) −3 < x 3
(vi) x < −6 or − 2 x < 2
3
(
1 (9
−
4
)(
33), 0 , 1 (9 +
4
x
O
2 (i)
(a) (0, −6)
(b)
y = 3 − x; x = 1
(c)
((1 − 3, − 2(1 +
((1 +
Exercise 4B (Page 100)
1 (i) (a)
No turning points
)
(d)
3));
3, 2( 3 − 1))
y
33), 0 ,
y=
x2 − 4x + 6
x−1
(0, − 3 )
2
(b) y = 2x − 1; x = 4
y=x−3
x
O
(c) (3, 3); (5, 11)
x=1
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(ii)
(a)
(0, 6)
(b)
y = 3 − x; x = 1
(c)
((1 −
4 Vertical asymptote: x = 1; oblique asymptote:
y = 2x + 3
5 Vertical asymptote: x = 2; oblique asymptote:
y=x+p+2
p > −5
2
3, 2(1 + 3));
Answers
((1 + 3, 2(1 − 3))
(d)
y
y=
x2 − 4x +
1−x
x=1
y=
6
(i)
(a)
y=3−x x
(d)
− 0.5
x
(21 , 0) , (4, 0), (0, − 43)
(b) y = 2x − 3; x = 3
(c)
x2 − x + 1
x−2
y=x+1
O
3
x=2
y
6 No stationary points when k > 2
vertical asymptote: x = −1; oblique asymptote:
y = 2x + 2; crosses axes at (0, 0) and (–2, 0)
No turning points
y
y = 2x − 3
y=
x = −1
y
y = 2x + 2
(2x − 1)(x − 4)
(x − 3)
2x2 + 4x
y= x+ 1
x
O
x
−2
x=3
(ii) (a)
(b)
(c)
(d)
(21 , 0) , (4, 0), (0, 43)
7 (i) Vertical asymptote: x = −1; oblique asymptote:
y = px + 4 − p
(ii) p = 4
y = 3 − 2x; x = 3
No turning points
y
y=
(2x − 1)(x − 4)
(3 − x)
y
x = −1
4x2 + 4x + 1
x+1
y = 4x
x
O
−
x=1
20
y=
1
2
x
y = 3 − 2x
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(iii) Intersections at ( −2 ±
?
3, 0)
x = −1
y
y=x+3
Using the chain rule:
1 and u = f ( x )
Let y =
f (x )
dy
1
Then y = 1 ⇒
= − 2
u
du
u
and du = f '( x )
dx
dy
dy du
=
×
So
dx
du dx
1
= − 2 × f '( x )
u
f '( x )
= −
[f ( x )]2
x2 + 4x + 1
x+ 1
x
8 Vertical asymptote: x = 2; oblique asymptote:
y = λx +1
x=2
y
y=1−x
?
Answers
y=
(Page 104)
(Page 107)
1 is
f (x )
undefined when x = 1 since f(x) is undefined at that
point.
y = g(x) has a root at x = 1.
The graphs are the same shape, but y =
O
x
3
Exercise 4C (Page 110)
1
y
(i)
0
−2
y = f(x)
2
x
−4
y
(ii)
y = |f(x)|
4
−2
0
2
x
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y
Answers
(iii)
−2
y = f(|x|)
0
x
2
−4
(iv)
y
y=
1
4
x = −2
(v)
1
f(x)
x
x=2
y
y2 = f(x)
−2
22
0
2
x
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y
1
2 (i)
y = f(x)
−2π
−π
−π
2
π
2
3π
2
π
2π x
Answers
−3π
2
−1
(ii)
y
1
y = |f(x)|
x
−2π
−3π
2
−π
−π
2
π
2
3π
2
π
2π
−1
(iii)
y
1
y = f(|x|)
x
−2π
−3π
2
−π
−π
2
π
2
3π
2
π
2π
−1
(iv)
y
y=
1
f(x)
1
−2π
−3π
2
−π
−π
2
(v)
x
π
2
−1
1
3π
2
π
2π
y
y 2 = f(x)
x
−2π
−3π
2
−π
−π
2
π
2
π
3π
2
2π
−1
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3
(i), (ii)
y
Answers
y = f(x)
1
x
0
y
(iii)
y = f(|x|)
1
0
(iv)
x
y
1
y=
f(x)
1
0
24
x
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(v)
(iii)
y
y
y2 = f(x)
−
1 + √3, 3√3
2
2
1 + √3 , 3√3
2
2
x
0
−2
−
−1
1 0
2
x
2
1
2
Answers
1
−2
y = f(|x|)
4 (i)
(iv)
y
x = −1
y
x=
1
2
x=2
1 + √3 , 3√3
2
2
1 − √3 , 2
−
2
3√3
y = f(x)
−1
1
2
0
y=
x
0
x
2
1
f(x)
1 + √3 , 2
2
3√3
−2
1 − √3 , 3√3
−
2
2
y
(v)
(ii)
y
y = |f(x)|
3
1 − √3 , 3√3
2
2
1 + √3 , 3 4
2
√2
1 + √3 , 3√3
2
2
2
−1
−1
0
1
2
2
x
0
1
2
2
x
3
y2 = f(x)
1 + √3 , 3 4
−
2
√2
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x = −1 y
5 (i)
y
(v)
3
√3
y=1
Answers
y=1
−3
x
0
−3
x
0
y = −1
−√3
y2 = f(x)
y = f(x)
y
6 (i)
(ii)
x = −1
y
(−2, 9)
(6, 9)
6
3
y = |f(x)|
y=1
−3
−4
x
0
0
(ii)
2
x
8
y
6
y
(iii)
−2 0
−8
3
8
2
x
y = f(|x|)
y=1
(iv)
x = −3
(−6, −9)
y
1
3
−1 0
y=
x
0
−1
(6, −9)
y
(iii)
y=1
_−2, 1 +
9
x
x
1
_6, − 9 +
1
f(x)
x = −4
26
1
6
0
x=2
x=8
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9 (i)
y
(iv)
(−2, 3)
x = −3
y
√6
−4
2
0
(ii)
x = −1 y
−3
x = −3 and x = 4.5
x=2
−3
x
0
1
3
x = −2
y
10 (i)
x=2
(ii)
y 7 − 2 6 or y 7 + 2 6
y
x=1
3
2
−2
(c)
1
2
x
0
(b)
8 (i) (a)
(b)
(ii) (a)
(b)
y=−
2
y = 2 x + 3x − 2
x −1
(iii)
−
x
0
x  −3 or –1 < x < 2
(ii) (a)
x = 4.5
y
3
−
2
(b)
x
4.5
−√6
(−2, −3)
7 (i) (a)
0
y = −2
x
8
Answers
3
0 1
2
x
–2 < x < 2
y=0
–2 < y < 2
y=0
maximum 2 and minimum − 2
x0
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Chapter 5
Answers
?
π + π = 5π so the diagonals do not meet at
6 4 12
right angles and therefore the shape is not a
rhombus.
(Page 115)
So ABCD is a parallelogram.
See text that follows on page 115.
4
?
y
(Page 116)
There are an infinite number of pairs of polar
co-ordinates (r , θ ) to define a given point P.
B
3π
—
10
–π5
Activity 5.1 (Page 116)
3
2
–π5
3π
—
10
3
All the points are in the position shown below:
A
x
4
C
D
0
By symmetry AB and BC are the same length;
similarly AD and CD are the same length.
1 2 3 4 5
The diagonals of the shape meet at right angles.
ABCD is a kite.
5
(
)(
)(
5π
11π
13π
For example, 6, − 4 , 6, 4 , 6, − 4
Exercise 5A (Page 120)
1
2
)
(i) (0, −8)
(ii) (−4 2, −4 2)
(iii) (4, 4 3)
(iv) (−4 3, 4)
(i) (13, 1.18)
(ii) (5, π)
(iii)
(iv) (5, 0.927)
(2, − 5π6)
3
6
(i) 4
(ii) 16 < r < 170 θ = −27
(iii) (a) 99 < r < 107 153 < θ < 171
(b) 16 < r < 99 −81 < θ < −63 and
107 < r < 162 −81 < θ < −63
(c) 162 < r < 170 45 < θ < 63
5π
C 4, 11π
(i) B 4, − 12
12
π
7π
(ii) B(0, 0) and C either 4, − 12 or 4, 12
(iii) The points could be arranged as shown:
(
)
(
(
A
4
A
B
) ( )
y
C2
y
5
π
–6
–π4
3
)
D
3
O
π
–
4
–π6
C1
C
x
B
x
5
π
–π4 –3
If O is the midpoint of AB then B has
(
)
coordinates 4, − 3π and C has coordinates
4
By symmetry the quadrilateral has two pairs of
parallel sides.
28
(
) (
)
either 4 3, − π or 4 3, 3π .
4
4
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Alternatively the points could be arranged
as shown below:
y
4
1
−4
−2
O
−1
−2
7
4
π
–
4
2
C
π
–
3
4
x
–7
Answers
2
r = 7 is a circle centre (0, 0), radius 7:
A
3
B
Activity 5.2 (Page 123)
7
−3
–7
−4
If O is the midpoint of BC then B has
4 3 3π 
coordinates  3 , 4  and C has
θ = π is a half-line starting at the origin making an
3
π
angle 3 with the initial line:
4 3
π
coordinates  3 , − 4  .
7
(ii) A(5.39, 0.38)
C(8.71, 1.64)
(iii) A(5.00, 2.00)
C(−0.58, 8.69)
8
(i)
r = 2 sec θ
(ii)
tan θ =
(iii) r =
–π3
B(8.71, 1.01)
D(5.39, 2.27)
B(4.63, 7.38)
D(−3.45, 4.14)
For example
3 or θ = ± π
3
3
Activity 5.3 (Page 128)
− 5π θ 7 π using convention 3
2
2
(iv) r 2 = 18cosec2θ
100
r = cot θ cosecθ
5
(vi) r 2 =
1 − sin 2θ
(v)
9
(i)
x =5
(ii)
x 2 + y 2 = 25
50
−50
(iii) y = 4
0
50
100 x
−50
(iv) x + y = 4 x
2
(v)
(x
2
2
+ y 2 ) = 2xy
2
(vi) x 2 + y 2 = 2 ( x 2 + y 2 ) – 2x
a
10 r = a sec θ ⇒ r = cos θ ⇒ r cos θ = a ⇒ x = a
r = b cosecθ ⇒ r =
b ⇒ r sin θ = b ⇒ y = b
sin θ
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Activity 5.4 (Page 129)
Exercise 5B (Page 129)
k = 1, n = 2
1
(i)
(i) Circle centre O, radius 5
3
5
Answers
2
1
–5
−3
−2
−1
O
1
2
5
3
−1
(ii)
−2
−3
‘Radius’ of rhodonea is 1; two petals
–5
Half-line from the origin making an angle
− 3π with the initial line
4
k = 5, n = 2
(ii)
5
4
3
2
1
−5 −4 −3 −2 −10
−1
−2
−3
−4
−5
3π
4
1 2 3 4 5 x
(iii) Circle symmetrical about the x-axis,
passing through the origin and the
point (3, 0)
‘Radius’ of rhodonea is 5; two petals
k = 1, n = 4
3
2
(iii)
1
−2
−1
(iv) Circle symmetrical about y-axis, passing
through the origin and the point (0, 2)
1
2
2
−1
−2
‘Radius’ of rhodonea is 1; four petals
Generally, r = k sin (nθ) has ‘radius’ of k (for k > 1).
For n > 1 and r > 0, the curve has n petals.
If r < 0 included, then the curve has n petals if n is odd
and 2n petals when n is even. For k = 1, n = 1 the curve
is a circle of diameter 1 passing through the origin and
the point (0, 1), symmetrical about the y-axis.
30
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(v)
Extension: When negative values of r are
included, the loops for r = a cos nθ are drawn
twice for odd values of n. There are additional
loops for r = a cos nθ when n is even giving 2n
loops for 0 θ < 2 π.
Spiral starting from the origin
20
10
10
10
Answers
4
20
20
10
20
0
8
16
2
π
12
π
4
π
6
π
3
5π
12
7π 2π 3π 5π 11π π
12 3 4 6 12
π
2
θ
0
r
0 2.07 4 5.66 6.93 7.73 8 7.73 6.93 5.66 4 2.07 0
5
r would be negative
(i) T
he curve has a heart shape, hence the
name cardiod.
The circle has centre (0, 4), radius 4 so the
2
Cartesian equation is x 2 + (y − 4) = 16.
16
O
3
(ii) The curve is a reflection of the curve in
part (i) in the ‘vertical axis’
0
1
2
20
x
3
5π
6
20
r = 3 cos 2θ
10
7π
6
4
6
(i)
10
10
20
10
20
2
4
2
2
4
2
0
4
1 2 3 4 5
r = 3 cos 3θ
r = 3cos 2 θ has two loops and r = 3cos 3 θ has
3 loops.
r = a cos nθ has n loops
r = 2 + 3 cos θ
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10
5
Answers
5
5
5
10
5
5
10
5
5
10
r = 3 + 3 cos θ
(ii)
5
5
(b)
2b > a > b Limacon has a ‘dimple’
(c)
a = b Cardioid
( )
( )
−arc cos − a θ arc cos − a
b
b
5
r = 4 + 3 cos θ
7
5
5
5
5
r = 5 + 3 cos θ
(a) a 2b Convex limacon
(iii) r 0
b cos θ −a
5
r = 7 + 3 cos θ
10
Extension: When r < 0 included, limacon
has a ‘loop’ inside for these values of θ.
(iv) The shape is the same but the curves are
now symmetrical about the y-axis rather
than the x-axis.
Values of r are only defined in the intervals
0 θ π , 3π θ 5π and 7π θ 2π
4
4 4
4
(the value of r2 is negative in the other
intervals). Where the curve is defined, one half
π
of each loop is created for each interval of 4 .
Taking negative square roots instead of positive
square roots produces the same curve. For example,
when θ = 0, r = −a when taking the negative
square root; this is equivalent to the point (π, a)
obtained when taking the positive square root.
5
10
5
5
5
10
–a
a
10
r = 6 + 3 cos θ
r2 = a2 cos 2θ
32
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Exercise 5C (Page 132)
1
2
rea is 25π
A
(i)
5
0
r2 = a2 sin 2θ
8
5
(i)
25π
(ii) 16
P(r, θ)
O
θ
α
5
Answers
a
–a
5
A( p, α)
–α
Initial line
3
(i)
2π
3
L
6
4
π
2
π
3
5π
Let (r, θ ) be a point on the line L that is
2
6
perpendicualr to OA. In triangle OPA,
p
2
6 4 2
cos (θ − α) = ⇒ r cos (θ − α) = p
2
r
(ii) r cos (θ − α) = p ⇒ r cos θ cos α + r sin θ sin α = p ⇒ x cos α + y sin α4 = p
r cos (θ − α) = p ⇒ r cos θ cos α + r sin θ sin α = p ⇒ x cos α + y sin α = p
6
s θ cos α + r sin θ sin α = p ⇒ x cos α + y sin α = p
27π
(ii) 2
9
Intersection at  1 , π ;
 2 3
64π
1
4
C1: Circle centre at the pole and radius
3
2
C2: Curve approx. correct orientation from
20
5
(0, 0) to (1, π)
π
6
4
6
10
20
−2
−1
0
1
2
10
10
20
10
20
Larger part = 24π + 64
Smaller part = 24π − 64
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10
Answers
6
a
–a
1 2
One loop has area 2 a .
7
4 kπ
Common ratio e
9
2
3
4
5
3π ;
4.712, 4.713
Chapter 6
?
1.15
1
Sketch should show upper half of cardioid
through (4, 0) and (0, π);
2
2


 kπ
 kπ
A = a e 2 − 1 , B = a e 4 kπ e 2 − 1 ,
4k 
4k




a 2 8kπ  k2π
C = 4k e e − 1
8
−1
(Page 134)
The triangles are all congruent to each other.
256 yellow triangles make up the purple triangle.
Activity 6.1 (Page 135)
The diagram shows the image of the unit square
 a b
OIPJ under the transformation with matrix 
 c d

 .
y
P′
3
Sketch should show a cardioid through (a, 0) ,
3π
2a, π , (a, π) and 0, 2 ;
2
a2 π + 1 + 1 3
4
8
( )
(
)
( )
6.5 units2
2
1J
O
P
1I
4
J′
units2
8 units2
I′
2
2.5 units2
3
4
5
6
7
8
x
9
The point I(1, 0) is transformed to the point I' (a, c); the
point J(0, 1) is transformed to the point J' (b, d).
P' has coordinates (a + b, c + d).
The area of the parallelogram is given by the area of
the whole rectangle minus the area of the rectangles
and triangles.
Area of first rectangle = b × c
1
Area of first triangle = 2 × b × d
34
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(ii)
Area of second triangle = 1 × a × c
2
Area of whole rectangle = (a + b ) × (c + d )
(a)
y
6
Therefore the area of the parallelogram is
)
?
4
2
(i)A rotation does not reverse the order of the
 0 −1 
,
vertices, e.g. for A = 
 1 0 
det A = 1, which is positive.
(ii)A reflection reverses the order of the vertices,
 1 0 
,
e.g. for B = 
 0 −1 
det B = −1, which is negative.
(iii)An enlargement does not reverse the order of
 2 0
,
the vertices, e.g. for C = 
 0 2 
det C = 4, which is positive
Exercise 6A (Page 139)
(a)
y
B′
5
3
1
–4 –3 –2 –1 O
–1
–1 O
–1
B
1
A
2
3
4
5
x
6
A′
–2
(b) Area of parallelogram = 16
(c) 16
(iii) (a)
y
3
2
1C
–8 –7 –6 –5 –4 –3 –2 –1 O
–1
B′
–2
C′
–3
B
1
A
A′
2
3
4
5
6
x
4
5
6
x
(b) area of parallelogram = 0
(c) 0
(iv) (a)
A′
4
2
1 C
–4
6
C′
B′
3
(Page 137)
1(i)
C′
Answers
(
5
(a + b ) × (c + d ) − 2 bc + 1 bd + 1 ac = ad − bc .
2
2
y
C
B
1
A
2
3
(b) Area of parallelogram = 11
(c) 11
4
x
3
C′
2
1C
–8 –7 –6 –5 –4 –3 –2 –1 O
–1
B′
–2
B
A
1 2
3
–3
A′
–4
(b) area of parallelogram = 11
(c) –11
2x = 2, x = 6
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Answers
3 (i)
8(i)
 1 0 
A =
 0 −1 
 0 1
C=
 1 0 
 −1 0 
B=
 0 1 
 0 −1 
D=
 −1 0 
(iii) A
y
3
2
1C
–7 –6 –5 –4 –3 –2 –1 0
–1
B′
–2
J
P
O
I
I′
J′
P′
x
J′ J
P′
I′
I
y
x
J
O
P′
P
I J′
y
D
J′
P′
O
12 (i)
P
(ii)
I′
5 ad = 1
(ii)
 1 3
 0 1 
det M = –2,
det N = 7
 9 13 
MN = 
, det MN = –14
 8 10 
and –14 = –2 × 7
36
5
x
6
–4
 1 2
T=
 3 6

 ,
y = 3x − 3 s + t
(
)
9
3 3
1
P' 8 s − 8 t, 8 s − 8 t
(
) (
)
Activity 6.2 (Page 141)
(i)
 1 0 
P= 
 0 −1 
(ii)
 1 0
2
P = 
 0 1 
(ii) determinant = 1 so area is preserved
7 (i)
4
 9

−3 

8
8
(iii) 
, which has determinant
3 −1 
 8 8


9
1
3
3
×− − ×− =0
8
8
8
8
x
4 66 cm2
6 (i)
3
det T = (1 × 6) − (3 × 2) = 0
(iii) (3, 9)
x
I
2
 5 p − 10q 
 − p + 2q 


1
(ii) y = − x
5
(iii) det N = 0 and so the image has zero area
10 (i)
y=–x
J
A
9 (i)
y=x
I′
1
(ii)The image of all points lie on the line
y = 1 x . The determinant of the matrix is
2
zero, which shows that the image will have
zero area.
P
O
A′
B
–3
C′
y
B
C
y
(iii)Reflecting an object in the x-axis twice takes
it back to the starting position and so the final
image is the same as the original object. Hence
the matrix for the combined transformation is I.
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Activity 6.3 (Page 142)
?
 a b   d −b   ad − bc
0 
 c d   −c a  = 
0
ad − bc 
Therefore M
1  d −b  .
=
ad − bc  −c a 
−1
Activity 6.4 (Page 143)
(i)
 11 3   2 −3 
AA −1 = 1 
4  6 2   −6 11 
Exercise 6B (Page 144)
(ii)
2 (i)
 4 0
= 1
= I
4  0 4 
(ii)
 0 1
− 1
2  2 5 
(iv)
(v)
(vi)
(vii)
1  d −b 
ad − bc  −c a 
1  a b   d −b 
=
ad − bc  c d   −c a 
=
1  ad − bc −ab + ab 
ad − bc  cd − dc − cb + ad 
=
0 
1  ad − bc

ad − bc 
0
ad − bc 
 1 0
= 
= I
 0 1 
non-singular,
singular
singular
singular
non-singular,
1  11 −3 
112  −3 11 
 −8 −4 a 
1
16 (1 − ab)  −4b −2 
provided ab ≠ 1
1  3 −6 
3  −2 5 
(ii)
1  −1 −5 
2  2 8 
(iii)
 28 19 
 10 7 
(iv)
 50 63 
 −12 −15 
(v)
1  7 −19 
6  −10 28 
(vi)
1  −15 −63 
6  12 50 
3 (i)


(vii) 1 −15 −63
6  12 50 


(viii) 1 7 −19
6  −10 28 
1 2 2 
8  −3 1 
1  d −b   a b 
ad − bc  −c a   c d 
4 (i)
=
1  da − bc db − bd 
ad − bc  −ca + ac −cb + ad 
5 k = 2 or k = 3
=
0 
1  ad − bc
ad − bc 
0
ad − bc 
M −1M =
1  2 −3 
24  4 6 
singular
(iii) non-singular,
 a b  1  d −b 
MM −1 = 
 c d  ad − bc  −c a 
(10, −6)
(iii) (1, 2)
1  2 −3   11 3 
−1
A A = 4  −6 11   6 2 
M −1 =
(MN)−1 = N −1M −1
1 (i)
 4 0
= 1
= I
4  0 4 
(ii)
First reverse the reflection by using the
transformation with the inverse matrix of the
reflection. Secondly, reverse the rotation by using
the transformation with the inverse matrix of the
rotation.
Answers
To turn this into the identity matrix it would need to
be divided by ad − bc, which is the value |M|.
(Page 143)
6  2 1 0 −1 
 1 0 −3 4 
7 (i)
(3, 1), (1, 1) and (–6, –2)
 1 0
=
= I
 0 1 
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Exercise 6C (Page 151)
y
1 (i)
(ii)
3
2
1
B B′
Answers
C
–6 –5 –4 –3 –2 –1 0
–1
1A 2
3
4
5
x
–2
C′
(ii)
A′
ratio of area T’ to T is 3 : 1.5 or 2 : 1
The determinant of the matrix M = 2 so
the area is doubled.
(iii)
 1 −1 
M −1 = 1 
2  −1 3 
8 (ii)
Mn = (a + d )n −1 M
9 (iii)

1
 0 5
 1 11
 6 − 10






(iv)
 33 6 
 5 0 
 1
2
5
 − 17 − 17 17

1
6
=  − 8
17 17 17

8 − 3
 4
17
 17 17
B is singular.
C −1
D−1
 1

−
−2 1
 15 3 5 


= 2
0 −1 
5
5


 − 1 1 1 
 15 3 5 


−1 −3 4

3 
=
2
 1 2 −3 


 1 3 −1 
 3 0 6
1

3  −4 2 3 
 2 −1 0 
2 (i)
(ii)
Matrix is singular
 −0.06 −0.1 −0.1 
(iii)  0.92 0.2 0.7 


 0.66 0.1 0.6 
(iv)
 34 11 32 
1
6
6 
21  9

 −38 −16 −37 


−10
4
12
1

3 M = 28 − 10k − (4k − 8) 8 − (4 + 2k)  ;




−k
7
−3k
k = 2.8
4 (i)The columns of the matrix have been
moved one place to the right, with the
final column moving to replace the first.
This is called cyclical interchange of the
columns.
(ii) det A = det B = det C = −26
Cyclical interchange of the columns leaves
the determinant unchanged.
−1
Activity 6.5 (Page 146)
det A = −17 A −1
(a) 5
(b) 5
(a) −5
(b) −5
Interchanging the rows and columns has
not changed the determinant.
(iii) (a) 0
(b) 0
If a matrix has a repeated row or column
the determinant will be zero.







−1 ± 41
5 x =
2
6 x = 1, x = 4
7 1 < k < 5
8 (ii)
P −1



=





Q −1 = 



38
−1
9
2
9
−1
3
1 −4 
6
9
1 −1 
6
9

1 −1 
2
3
3 −4 1 
2
2 
1 −2 0 

−3 5 −1 
2
2
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10 (i) 10 × 43 = 430
(ii) 4 × 5 × −7 × 43 = −6020
(iii) x × 2 × y × 43 = 86xy
(iv) x 4 × 1 × 4 y × 43 = 86x 3 y
2x
Chapter 7
?
(Page 156)
A three-legged stool is the more stable. Three
points, such as the ends of the legs, define a plane
but a fourth will not, in general, be in the same
plane. So the ends of the legs of a three-legged
stool lie in a plane but those of a four-legged
stool need not.The four-legged stool will rest on
three legs but could rock on to a different three.
Three points on the plane are sufficient to
specify a plane. So too would two lines in
the plane, or one point on the plane and the
direction of the normal vector. These ideas are
explored in this chapter.
?
Repeat the work in Example 7.6 replacing
(7, 5, 3) by (α, β, γ), so 7 by α, 5 by β and 3 by γ;
and (3, 2, 1) by (n1, n2, n3) and 6 by d.
Exercise 7A (Page 165)
1 (i)
(ii)
(iii)
(iv)
(v)
(vi)
Parallel, line in plane
Parallel, line not in plane
Not parallel
Parallel, line in plane
Not parallel
Parallel, line not in plane
 2
→   →
2 (i) L M =  2 ; L N =
 –2
(iii) x − 4y − 3z = −2
 5
 2
 
 –1
3 (iii) B
(Page 155)
Grimshaw architects designed the Eden project
using as their starting point the geodesic design of
Buckminster Fuller. Fuller used spherical geometry
to enclose the most space with the least material
and believed that this would put an end to the
dominance of the right angle in architecture.
?
Activity 7.1 (Page 164)
(Page 158)
(i) 90° with all lines.
(ii)No, so long as the pencil remains perpendicular
to the table.
4 (i)Accept any points that satisfy the equations.
(iii) Three points define a plane.
(iv) (1, 0, −1)
5 (i)
(ii)
(iii)
(iv)
(v)
6 (i)
(0, 1, 3)
(1, 1, 1)
(8, 4, 2)
(0, 0, 0)
(11, 19, −10)
2

(a) r = 2 +
3

(b) (1, 3, 1)
(c)
(ii)
 1
 
λ –1
 2
 
6
2
2


(a) r = 3 + λ 5
0
3


(b) (1, 0.5, –1.5)
(c) 3.08
3

(iii) (a) r = 1 +
3

(b) (0, 1, 3)
(c) 3
1

λ 0
0

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Answers
 11 1

−
− − 7
 9
6 18 


(PQ)−1 = Q−1P −1 =  − 95 − 61 − 92 


5 
 13 1
3 18 
 9
9 (ii)Multiplying only the first column by k
equates to a stretch of scale factor k in one
direction, so only multiplies the volume by k.
(iii) Multiplying any column by k multplies
the determinant by k.
39
09/03/18 8:27 PM
Answers
(iv) (a)
(v)
2
 3

 
r = 1 + λ  – 4
0
 1

 
(b) (2, 1, 0): A is in the plane
(c) 0
1
(a) r = λ 1

1

(b)
(c)
14 (ii)
6x + y − 8z = 6
15 (i)
 –1
 3
 
 
r =  3 + λ  −1
 5
−4
 
 
(ii)
12
13
 
r =  5 +
 0
 
 3
 
λ  1
–2
 
4.1°
32.3°
16 (i) 3i + 2j + k
(ii) 72.2°
(iii) r = 3i + 2j + k + λ(6i + 2j − k)
?
(Page 172)
π3 is parallel to π1 and π2
(the common line is at infinity).
Exercise 7B (Page 173)
1 (i)
(iii) 35.6°
 –1
 8
→  

→   
10 (ii) A B = 2 ; AC = –4 ;
 
 
 1
 1
in both cases the scalar
product = 0
(iii) 132.9°
(iv) 8.08
40
 5
 
r =  1
−3
 
(iii) 7x − 11y + 8z = 0
(ii) (4, 2, 6)
(iii) 11.2
9 (i)
(ii)
b = −2, c = 3
(2, 2, 2)
7 (i) x + 2y + 3z = 25
(ii) 206 = 150 + 56
(iii)
W is in the plane;

→ 
→
UW . UV = 0
8 (i)
13 (i)
11 (i)
(ii)
(iii)
(iv)
5, 89
62.2°
20.9
(4, 6, −3)
12 (i)
 2
 –1


PQ: r = 2 + λ  2 ;
 
 
 4
 2
–2
1
 

XY: r = –2 + µ 2
–3
3
 

(iii) Yes
(iv) Yes, (1, 4, 6)
(ii)
 15 
3




r = 1 + λ  27 
 
 7
0


 0
 1


r = −3 + λ  1 


 −4 


 5
 16
 0



(iii) r =  −1  + λ  15
 −1 
 13







2
 11 


(iv) r = 0 + λ  4 
 


 21 
4
2 (i) 56.5° (ii) 80.0° (iii) 24.9° (iv) 63.5°
3 (i)
(ii)
 −2 
 1
r =  3  + λ 1 




 −1 
 5
 4
 3




r =  −3  + λ  2 
 2
 −6 




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4 41x − 19y + 26z = 33
5 x + 3y − z = −8

(a × b).b = 


760x + 11y + 100z = 900;
60x − 11y − 100z = −300;
5
 0

r = 0 + t 100 ; 6.3°
 
6
 –11

j × i = −k
j× j = 0
j×k = i
k×i= j
k×k =0
k × j = −i
Exercise 7C (Page 181)
8 (i) x + 3z = −800
(ii)Normal is approx. 18.4° to the horizontal
(iii)14x − 15y + 3450z = 15 950
(iv)
x = 15λ,
y = −1136λ − 62 396.7,
z = −5λ − 266.7
(v)62 km (assuming seam is sufficiently
extensive)
9 (i)
(ii)
Activity 7.2 (Page 179)
a = −2
3
1 (i)
(iii)
2 (i)
10 (i)
(ii)
4x + 2y + z = 8
77.4°
11 (i)
(ii)
57.7°
r = 2i − k + λ(4i − 7j + 5k)
(iii)
12 (i) 2x − 3y + 6z = 2
(ii) 2
(iii) r = λ(6i + 2j − k)
(Page 178)
Use the scalar product to check that the vector a × b
is perpendicular to both a and b:

a × b).a = 


4
 −23 
 13 


 2 
(ii)
 37 


 41 
 19 
 −8 
 34 


 27 
(iv)
 21 
 −29 


 9 
 5
 19 


 −2 
(ii)
 14 
 −62 


 −9 
 −3 
 −2 


 3
(iv)
 −18 
 57 


 47 
 1 

→
→
A
B ==  −4  ,AC
AC ==
AB


 3
3 (i)
 11 


→  5

→ 
 
−
4
13 OP = 
 , OQ =  −7 
 1
 3




?
Answers
 4
 21 


 4
+
λ
−
2
6 r =




 11 
 −7 
 24   1 
(a × b).b =  − 1  .  −4  = (24 × 1) + (−1 × −4) + (−14 × 2) = 0

 

 −14   2 
24   1 
− 1  .  −4  = (24 × 1) + (−1 × −4) + (−14 × 2) = 0
 

−14   2 
(ii)
1
 3
 12  or  4



 15 
5
(iii)
x + 4 y + 5z = 7
 4
 −1 


 0




 19 
1  15 

635 
 −7 
5 74
 24   3 
× 5)(i)= 0 5x + 4 y − 8z = 5
(a × b).a =  − 1  .  2  = (24 × 3) + (−1 × 2) + (−14 6

  
−14   5 

(ii) 24 x + y − 29z = 1
24   3 
(iii) 19x + 40y + 3z = 188
− 1  .  2  = (24 × 3) + (−1 × 2) + (−14 × 5) = 0
  
(iv) 30x − 29y − 24 z = 86
−14   5 
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Answers
7 (i) −8j
(ii) 6 j − 4 k
(iii) 2i − 12 j
(iv) 6i + 14 j − 2k
1 717
9 (i)
(ii)
717
2
10
i j k
1 0 1 = i + 3j − k
1 −1 −2
Equation of Π1 : x + 3y − z = 12
1 38
2
6 (i) 4, the lines are skew
(ii) 0.4, the lines are skew
(iii) 0, the lines intersect
77 , the lines are parallel
(iv)
6
7 (i) 2 69
(ii) M(−3, 2, 6)
5 (ii)
8 (i)
cos θ = 2 − 3 − 1
11 6
= 2 ⇒ θ = 75.7°or 1.32 rad
66
(ii)
(
9 (i)
Points on both planes is, e. g. (6, 2, 0)
Activity 7.3 (Page 185)
( )
(
or equivalent
)
A = 0, − c , A′ = 0, − c , 0
b
b
(ii)
(ii)
y = − a x − c , d = b, e = −a, f = 0
b
b
10 (ii)
Because it is perpendicular to both l1 and l2.
Exercise 7D (Page 190)
29
)
 2
 −1 
AB: r =  −3  + λ  0  and




 1
 4
 0
 2


CD: r =
3 + λ  0  or


 
 −2 
 7
equivalent; 6
3.08 m – No, the cable is not long enough
1 (10k + 105)
15
−21
(iii) k = 2
Activity 7.4 (Page 189)
42
(i)
1 (i)
 2
 4
r =  0  + λ  −5 




 −5 
 2
(iii) M 82 , 10 , − 229
45 45
45
i j k
1 3 −1 = 2 i + 3 j − 7 k
2 −1 1
r = 6i + 2j + t(2i - 3j - 7k)
2 5
15
(iv)
(ii)
7
(iii)
2 26
2 (i)
13
(ii)
3 5
(iii)
1
3 (i)
5
(ii)
5
(iii)
5 2
4 (i)
 1
 3
r =  1  + λ  −2 


 
 −1 
 0
(ii)
3
 4
 2
r =  12  + α  10  or equivalent




 5
 11 
11 16 ,36.7°
3
12 m = 2 or 2 , but m is an integer so m = 2
19
shortest distance 18
17
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