PURE MATHEMATICS 1 - Imago Education...PURE MATHEMATICS 1 WORKED SOLUTIONS FOR CHAPTERS 1 TO 3 OF...
Transcript of PURE MATHEMATICS 1 - Imago Education...PURE MATHEMATICS 1 WORKED SOLUTIONS FOR CHAPTERS 1 TO 3 OF...
PURE MATHEMATICS 1
WORKED SOLUTIONS FOR
CHAPTERS 1 TO 3 OF
Pure Mathematics 1: Coursebook
by Hugh Neill, Douglas Quadling and Julian Gilbey
revised edition
Cambridge University Press, 2016,
ISBN 9781316600207
Authors: Charissa Button (BSc);
Dr Bruce Button (BSc (Hons), Hons BA, MA, PhD)
Β© Imago Education (Pty) Ltd, January 2018
This is a free sample comprising chapters 1 to 3 of the complete worked
solutions. Solutions for all the chapters may be purchased from https://imago-
education.com/products-and-services/as-maths-code-9709-p1.html
This document may be freely distributed provided that the copyright notices are
retained and the document is not changed in any way.
NOTATION AND TERMINOLOGY
strict inequality one which doesnβt include βequal toβ
< and > are strict inequalities
weak inequality one which includes βequal toβ
β€ and β₯ are weak inequalities
β΄ therefore
β΅ because, or since
eqn equation
β substitute into
E.g (1) β (2) means βsubstitute equation (1) into
equation (2)β.
β tends to or approaches
β implies
E.g. A β B means βstatement A implies statement Bβ.
In other words, if statement A is true, statement B
must be true (but not necessarily the other way
around).
β is implied by
E.g. A βΈ B means βstatement A is implied by
statement Bβ. In other words, if statement B is true,
statement A must be true (but not necessarily the
other way around).
βΊ implies and is implied by
E.g. A β B means βstatement A implies and is
implied by statement Bβ. In other words, if statement
A is true, statement B must be true AND if statement
B is true, statement A must be true.
TBP to be proved (used at beginning of proof)
QED which was to be shown (used at end of proof)
(from Latin quod erat demonstrandum)
π₯ change in
π₯ The discriminant (= π2 β 4ππ) of a quadratic
expression in the form ππ₯2 + ππ₯ + π.
Note: the context will indicate which use of Ξ is
applicable in any particular situation.
Ξ΄ a small change in
β₯ perpendicular to
β‘ identical to
β for all
β there exists
Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1A
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CHAPTER 1
EXERCISE 1A
Question 1
b)
lππππ‘β = β(π₯2 β π₯1)2 + (π¦2 β π¦1)2
= β(1 β (β3))2
+ (β1 β 2)2
= β42 + 32
= β16 + 9
= β25
= 5
d)
πππππ‘β = β(π₯2 β π₯1)2 + (π¦2 β π¦1)2
= β(β7 β (β3))2
+ (3 β (β3))2
= β(β4)2 + 62
= β16 + 36
= β52
= β4 Γ 13 = β4β13 = 2β13
f)
πππππ‘β = β(π₯2 β π₯1)2 + (π¦2 β π¦1)2
= β((π β 1) β (π + 1))2
+ ((2π β 1) β (2π + 3))2
= β(β2)2 + (β4)2
= β4 + 16 = β20
h)
πππππ‘β = β(π₯2 β π₯1)2 + (π¦2 β π¦1)2
= β(3π β 12π)2 + (5π β 5π)2
= β(β9π)2 + 02
= β81π2
= 9π
j)
πππππ‘β = β(π₯2 β π₯1)2 + (π¦2 β π¦1)2
= β((π β 3π) β (π + 4π))2
+ (π β (π β π))2
= β(β7π)2 + π2
= β50π2
= πβ50
Page 1
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1A
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Question 2
Plot the points on the co-ordinate plane:
Now show that a pair of opposite sides is parallel and equal in length.
Take AB, formed by (1,-2) and (4,2):
ππππππππ‘ = π¦2 β π¦1
π₯2 β π₯1
=2 β (β2)
4 β 1
=4
3
πππππ‘β = β(π₯2 β π₯1)2 + (π¦2 β π¦1)2
= β(4 β 1)2 + (2 β (β2))2
= β32 + 42 = 5
Now take DC, formed by (6,-1) and (9,3):
ππππππππ‘ = π¦2 β π¦1
π₯2 β π₯1
=3 β (β1)
9 β 6
=4
3
πππππ‘β = β(π₯2 β π₯1)2 + (π¦2 β π¦1)2
= β(9 β 6)2 + (3 β (β1))2
= β32 + 42 = 5
Therefore AB and DC are parallel and equal in length, so the four points
form a parallelogram.
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1A
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Question 3
Plot the three points:
From the diagram it appears most likely that sides AC and AB will be
equal. Therefore calculate their lengths:
πππππ‘β π΄π΅ = β(2 β (β3))2
+ (β7 β (β2))2
= β52 + 52 = β50
πππππ‘β π΄πΆ = β(β2 β (β3))2
+ (5 β (β2))2
= β12 + 72 = β50
Therefore AB = AC, which means that triangle ABC is isosceles.
Question 4
Plot the points and the centre of the circle:
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1A
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To show that A, B & C lie on a circle with centre D, we must prove that
AD = BD = CD:
π΄π· = β(7 β 2)2 + (12 β 0)2
= β52 + 122 = 13
π΅π· = β(2 β (β3))2
+ (0 β (β12))2
= β52 + 122 = 13
πΆπ· = β(14 β 2)2 + (β5 β 0)2
= β122 + 52 = 13
Therefore AD = BD = CD, which means that A, B, C all lie on a circle
with centre D.
(Notice that, when calculating the length of a line segment, one can use
either point as the first one. Thus, in calculating CD above, D was used
as the first point and C as the second, giving (14 β 2) and (β5 β 0)
rather than (2 β 14) and (0 β (β5)). The result is the same either way
because 122 = (β12)2.)
Question 5
a) Midpoint is (π₯, π¦) = (1
2(π₯1 + π₯2),
1
2(π¦1 + π¦2)):
π₯ =1
2(2 + 6) = 4
π¦ =1
2(11 + 15) = 13
Therefore the midpoint is (4,13).
c)
π₯ =1
2(β2 + 1) = β
1
2
π¦ =1
2(β3 + (β6)) = β
9
2
Therefore midpoint is (β1
2, β
9
2).
e)
π₯ =1
2(π + 2 + 3π + 4) =
1
2(4π + 6) = 2π + 3
π¦ =1
2(3π β 1 + π β 5) =
1
2(4π β 6) = 2π β 3
Midpoint is (2π + 3, 2π β 3).
g)
π₯ =1
2(π + 2π + 5π β 2π) =
1
2(6π) = 3π
π¦ =1
2(2π + 13π + (β2π β 7π)) =
1
2(6π) = 3π
Midpoint is (3π, 3π).
Question 6
The centre of the circle (letβs call it C) is at the midpoint of AB.
π₯πΆ =1
2(β2 + 6) = 2
π¦πΆ =1
2(1 + 5) = 3
πΆ = (2, 3)
Page 4
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1A
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Question 7
We use the midpoint formula to set up two equations, which we solve
for π₯π΅ and π¦π΅:
π₯π =1
2(π₯π΄ + π₯π΅)
5 =1
2(3 + π₯π΅)
10 = 3 + π₯π΅
π₯π΅ = 7
π¦π =1
2(π¦π΄ + π¦π΅)
7 =1
2(4 + π¦π΅)
14 = 4 + π¦π΅
π¦π΅ = 10
Therefore π΅ = (7,10).
Question 10
a)
ππππππππ‘ = π¦2 β π¦1
π₯2 β π₯1
= 12 β 8
5 β 3=
4
2= 2
c)
ππππππππ‘ = β1 β (β3)
0 β (β4)=
2
4=
1
2
e)
ππππππππ‘ =(βπ β 5) β (π β 3)
(2π + 4) β (π + 3)=
β2π β 2
π + 1
=β2(π + 1)
π + 1= β2
g)
ππππππππ‘ = (π β π + 3) β (π + π β 3)
(π β π + 1) β (π + π β 1)=
β2π + 6
β2π + 2
=β2(π β 3)
β2(π β 1)=
π β 3
π β 1
Question 11
ππππππππ‘π΄π΅ = π¦π΅ β π¦π΄
π₯π΅ β π₯π΄=
6 β 4
7 β 3=
2
4=
1
2
ππππππππ‘π΅πΆ = π¦πΆ β π¦π΅
π₯πΆ β π₯π΅=
1 β 6
β3 β 7=
β5
β10=
1
2
The gradients of AB and BC are equal, which means that the lines AB
and BC are parallel. Since they have point B in common, A, B and C
must be on the same straight line (i.e. they are collinear).
Question 12
ππππππππ‘π΄π =π¦π β π¦π΄
π₯π β π₯π΄=
π¦ β 0
π₯ β 3=
π¦
π₯ β 3
ππππππππ‘ππ΅ = π¦π΅ β π¦π
π₯π΅ β π₯π=
6 β π¦
5 β π₯
Since A, P & B are all on the same straight line,
ππππππππ‘π΄π = ππππππππ‘ππ΅. Therefore
π¦
π₯ β 3=
6 β π¦
5 β π₯
π¦(5 β π₯) = (6 β π¦)(π₯ β 3)
5π¦ β π₯π¦ = 6π₯ β 18 β π₯π¦ + 3π¦
2π¦ = 6π₯ β 18
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1A
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π¦ = 3π₯ β 9
Question 13
Draw the points:
(Note: it looks like the points are almost on the same straight line, rather
than forming a triangle. Geometrically, this can be a little confusing, but
the algebraic calculations are exactly the same.)
The median AM will join A and M, with M being the midpoint of BC (since
BC is the side opposite to A). We must first find the coordinates of M
using the midpoint formula:
π₯π =1
2(π₯π΅ + π₯πΆ) =
1
2(0 + 4) = 2
π¦π =1
2(π¦π΅ + π¦πΆ) =
1
2(3 + 7) = 5
β΄ π = (2,5)
Now we can find the length of AM using the distance formula:
πππππ‘βπ΄π = β(π₯π β π₯π΄)2 + (π¦π β π¦π΄)2
= β(2 β (β1))2
+ (5 β 1)2
= β32 + 42 = β9 + 16 = 5
Question 16
We first find the coordinates of P, Q, R & S using the midpoint formula:
π₯π =1
2(1 + 7) = 4
π¦π =1
2(1 + 3) = 2
β΄ π = (4,2)
π₯π =1
2(7 + 9) = 8
π¦π =1
2(3 + (β7)) = β2
β΄ π = (8, β2)
π₯π =1
2(9 + (β3)) = 3
π¦π =1
2(β7 + (β3)) = β5
β΄ π = (3, β5)
π₯π =1
2(β3 + 1) = β1
π¦π =1
2(β3 + 1) = β1
β΄ π = (β1, β1)
Now we calculate the gradients of the four sides using the gradient
formula with the coordinates of the points just calculated:
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1A
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ππππππππ‘ππ =π¦π β π¦π
π₯π β π₯π=
β2 β 2
8 β 4=
β4
4= β1
ππππππππ‘ππ = β5 β (β2)
3 β 8=
β3
β5=
3
5
ππππππππ‘π π = β1 β (β5)
β1 β 3=
4
β4= β1
ππππππππ‘ππ = 2 β (β1)
4 β (β1)=
3
5
Thus, the opposite sides PQ and RS are parallel, and the opposite sides
QR and SP are parallel; therefore the quadrilateral PQRS is a
parallelogram.
Question 18
a) We use the distance formula:
ππ = β(6 β 0)2 + (4 β 0)2 = β62 + 42 = β60
πΏπ = β(4 β (β2))2
+ (7 β 3)2 = β62 + 42 = β60
Thus ππ = πΏπ (qed).
b) We calculate gradients using the gradient formula:
ππππππππ‘ππ = 4 β 0
6 β 0=
2
3
ππππππππ‘πΏπ =7 β 3
4 β (β2)=
4
6=
2
3
Therefore ON is parallel to LM (since their gradients are equal).
c) We use the distance formula:
ππ = β(4 β 0)2 + (7 β 0)2 = β42 + 72 = β16 + 49 = β65
πΏπ = β(6 β (β2))2
+ (4 β 3)2 = β82 + 12 = β65
d) OLMN is a quadrilateral with two opposite sides (ON and LM)
equal and parallel. This means that OLMN is a parallelogram. In
addition, the diagonals OM and LN are equal. Therefore OLMN is
a rectangle.
Question 20
We start by plotting the points:
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1A
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We calculate the coordinates of M and N using the midpoint formula:
π₯π =1
2(π₯π + π₯π) =
1
2(2 + 8) = 5
π¦π =1
2(π¦π + π¦π) =
1
2(5 + 7) = 6
β΄ π = (5,6)
π₯π =1
2(π₯π + π₯π) =
1
2(6 + 8) = 7
π¦π =1
2(π¦π + π¦π) =
1
2(1 + 7) = 4
β΄ π = (7,4)
To help visualize the points and the triangle TMN, it helps to add the
points M and N, and to draw in the triangle:
From the diagram we can see that TM and TN are likely to be the equal
sides, therefore we calculate the lengths of these sides using the
distance formula:
ππ = β(5 β 3)2 + (6 β 2)2 = β22 + 42 = β20
ππ = β(7 β 3)2 + (4 β 2)2 = β42 + 22 = β20
Therefore TM=TN and triangle TMN is isosceles.
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1A
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Question 22
a) We start off using the midpoint formula:
π₯π =1
2(2 + 10) = 6
π¦π =1
2(1 + 1) = 1
β΄ π = (6,1)
b) First, we use the distance formula:
π΅πΊ = β(6 β 6)2 + (4 β 10)2 = β36 = 6
πΊπ = β(6 β 6)2 + (1 β 4)2 = β9 = 3
β΄ π΅πΊ = 2πΊπ
To prove that BGM is a straight line, we notice that B, G & M all
have the same π₯-coordinate. Therefore they all lie on the line π₯ =
6. Note that we cannot prove this using gradients because a
vertical line has an infinite gradient. (Try calculating the gradient
of BG, and you will see that you get a division-by-zero error.)
c) We use the midpoint formula:
π₯π =1
2(6 + 10) = 8
π¦π =1
2(10 + 1) =
11
2
β΄ π = (8,11
2)
d)
ππππππππ‘π΄πΊ = 4 β 1
6 β 2=
3
4
ππππππππ‘πΊπ =
112
β 4
8 β 6=
3
2Γ
1
2=
3
4
Therefore A, G, N are all on the same straight line.
Now we use the distance formula:
π΄πΊ = β(6 β 2)2 + (4 β 1)2 = β16 + 9 = 5
πΊπ = β(8 β 6)2 + (11
2β 4)
2
= β4 +9
4= β
25
4=
5
2
Therefore π΄πΊ = 2πΊπ.
Page 9
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1B
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EXERCISE 1B
Question 1
To check whether a point lies on a line, we must check whether the left
and right hand sides of the equation are equal when the π₯ and π¦ values
of the point are substituted into the equation of the line.
a) (1,2) on π¦ = 5π₯ β 3:
πΏπ»π = 2
π π»π = 5(1) β 3 = 2
Therefore the point (1,2) does lie on the line π¦ = 5π₯ β 3.
b) (3, β2) on π¦ = 3π₯ β 7:
πΏπ»π = β2
π π»π = 3(3) β 7 = 9 β 7 = 2
Therefore the point (3, β2) does not lie on the line π¦ = 3π₯ β 7.
d) (2,2) on 3π₯2 + π¦2 = 40:
πΏπ»π = 3(2)2 + (2)2 = 3(4) + 4 = 16
π π»π = 40
Therefore the point (2,2) does not lie on the line 3π₯2 + π¦2 = 40.
f) (5π,5
π) on π¦ =
5
π₯:
πΏπ»π =5
π π π»π =
5
5π=
1
π
Therefore the point (5π,5
π) does not lie on the line π¦ =
5
π₯.
h) (π‘2, 2π‘) on π¦2 = 4π₯:
πΏπ»π = (2π‘)2 = 4π‘2
π π»π = 4(π‘2) = 4π‘2
Therefore the point (π‘2, 2π‘) does lie on the line π¦2 = 4π₯.
Question 2
Use either π¦ = ππ₯ + π, where π is the gradient of the line and π is the π¦
intercept, or π¦ β π¦1 = π(π₯ β π₯1). The second is usually easier.
b)
π = β3
π¦ = β3π₯ + π
π¦ β π¦1 = π(π₯ β π₯1)
π¦ β (β2) = β3(π₯ β 1)
π¦ + 2 = β3π₯ + 3
π¦ = β3π₯ + 1
Therefore the equation is π¦ = β3π₯ + 1.
d)
π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 1 = β3
8(π₯ β (β2))
π¦ β 1 = β3
8π₯ β
3
4
8π¦ β 8 = β3π₯ β 6
3π₯ + 8π¦ = 2
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1B
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f) If the gradient is zero (i.e. the line is horizontal), the equation of
the line has the form π¦ = π. We can take the π¦-value of any point
on the line to get the value of π. Therefore the equation of the line
is
π¦ = 8
h) Use π¦ = ππ₯ + π:
π¦ =1
2π₯ + π
Then substitute (β3,0) into eqn.
0 =1
2(β3) + π
0 = π β3
2
π =3
2
Therefore the equation of the line is
π¦ =1
2π₯ +
3
2
or
2π¦ = π₯ + 3
j)
π¦ β π¦1 = π(π₯ β π₯1)
π¦ β 4 = β1
2(π₯ β 3)
2π¦ β 8 = βπ₯ + 3
π₯ + 2π¦ = 11
l)
π¦ β (β5) = 3(π₯ β (β2))
π¦ + 5 = 3π₯ + 6
π¦ = 3π₯ + 1
n)
π¦ = βπ₯ + π
2 = β(0) + π
π = 2
Equation is:
π¦ = βπ₯ + 2
p)
π¦ β 0 = β3
5(π₯ β 3)
5π¦ = β3π₯ + 9
3π₯ + 5π¦ = 9
r)
π¦ = ππ₯ + π
4 = π(0) + π
π = 4
Equation is:
π¦ = ππ₯ + 4
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1B
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t)
π¦ = ππ₯ + π
(Use π instead of π here to avoid confusion with the value of π₯,
which is given as π.)
0 = ππ + π
π = βππ
Equation is:
π¦ = ππ₯ β ππ
Question 3
b)
ππππππππ‘ =π¦2 β π¦1
π₯2 β π₯1=
β7 β 5
β2 β 4=
β12
β6= 2
Therefore the equation of the line is π¦ β π¦1 = π(π₯ β π₯1).
Substitute the π₯ and π¦ values of one of the points into this eqn.:
π¦ β 5 = 2(π₯ β 4)
π¦ = 2π₯ β 3
f)
ππππππππ‘ =20 β (β1)
β4 β 3=
21
β7= β3
Equation of line: π¦ β π¦1 = π(π₯ β π₯1). At (3, β1):
π¦ + 1 = β3(π₯ β 3)
β΄ π¦ = β3π₯ + 8
j)
ππππππππ‘ = β3 β (β1)
5 β (β2)=
β2
7= β
2
7
Equation of line is π¦ = β2
7π₯ + π. At (β2, β1):
β1 = β2
7(β2) + π
β7 = 4 + 7π
π = β11
7
β΄ π¦ = β2
7π₯ β
11
7
7π¦ = β2π₯ β 11
2π₯ + 7π¦ + 11 = 0
n)
ππππππππ‘ = β1 β 4
β2 β (β5)=
β5
3= β
5
3
Equation of line is π¦ = β5
3π₯ + π.
At (β5, 4):
4 = β5
3(β5) + π
12 = 25 + 3π
3π = β13
π = β13
3
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 1B
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π¦ = β5
3π₯ β
13
3
3π¦ = β5π₯ β 13
5π₯ + 3π¦ + 13 = 0
p)
ππππππππ‘ = π β 0
π β 0=
π
π
Equation of line is π¦ =π
ππ₯ + π. At (0,0):
0 = π
β΄ π¦ =π
ππ₯
ππ ππ¦ = ππ₯
ππ ππ₯ β ππ¦ = 0
q)
ππππππππ‘ = π β 1 β π
π + 3 β π= β
1
3
Equation of line is π¦ = β1
3π₯ + π. At (π, π):
π = β1
3π + π
3π = βπ + 3π
3π = 3π + π
π =3π + π
3
π¦ = β1
3π₯ +
3π + π
3
3π¦ = βπ₯ + 3π + π
π₯ + 3π¦ β π β 3π = 0
r) Notice that both points have the same π₯-coordinate, π. This means
that both points lie on the vertical line, π₯ = π. So, the equation of
the line joining these points is π₯ = π. This question cannot be
answered in the same way as above, since the gradient of a
vertical line is infinite.
s)
ππππππππ‘ =(π + 2) β π
(π + 2) β π
= 1
Using (π, π):
π¦ β π = π₯ β π
β΄ π¦ = π₯ + π β π
ππ π₯ β π¦ + π β π = 0
t)
ππππππππ‘ =π β 0
0 β π
= βπ
π
Equation of line is π¦ = βπ
ππ₯ + π. At (0, π):
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π = βπ
π(0) + π
π = π
β΄ π¦ = βπ
ππ₯ + π
ππ ππ₯ + ππ¦ β ππ = 0
Question 4
In each case write the given equation in the form π¦ = ππ₯ + π and then
read off the gradient (it is the coefficient of π₯).
a)
2π₯ + π¦ = 7
π¦ = β2π₯ + 7
From the form of the equation it is now easy to see that the
gradient is -2.
c)
5π₯ + 2π¦ = β3
2π¦ = β5π₯ β 3
π¦ = β5
2π₯ β
3
2
So, the gradient is β5
2.
f) The line 5π₯ = 7 is a vertical line at π₯ =7
5, so the gradient is infinite
or undefined.
h)
π¦ = 3(π₯ + 4)
π¦ = 3π₯ + 12
Gradient is 3.
k)
π¦ = π(π₯ β π)
π¦ = ππ₯ β ππ
Gradient is π.
l)
ππ₯ + ππ¦ = ππ
ππ¦ = βππ₯ + ππ
π¦ = βπ
ππ₯ + π
Gradient is βπ
π.
Question 5
Since the line must be parallel to the line π¦ =1
2π₯ β 3, the gradient is
1
2.
So the equation of the line is π¦ =1
2π₯ + π, and we find the value of π by
substituting for π₯ and π¦ at the given point (β2,1):
1 =1
2(β2) + π
π = 1 + 1
= 2
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So the equation of the line is π¦ =1
2π₯ + 2.
Question 6
First, we find the gradient. The line is parallel to π¦ + 2π₯ = 7, which can
be written as π¦ = β2π₯ + 7. So the gradient is β2.The equation of the line
is π¦ = β2π₯ + π. At (4, β3):
β3 = β2(4) + π
π = β3 + 8
= 5
The equation of the line is π¦ = β2π₯ + 5.
Question 7
Here the gradient is the same as for the line joining the points (3, β1)
and (β5,2):
ππππππππ‘ =2 β (β1)
(β5) β 3
= β3
8
The equation of the line is π¦ = β3
8π₯ + π. At (1,2):
2 = β3
8(1) + π
π = 2 +3
8
=19
8
The equation of the line is π¦ = β3
8π₯ +
19
8.
Question 8
Here the gradient is the same as for the line joining the points (β3,2)
and (2, β3):
ππππππππ‘ =2 β (β3)
(β3) β 2
= β1
The equation of the line is π¦ = βπ₯ + π. At (3,9):
9 = β1(3) + π
π = 9 + 3 = 12
The equation of the line is π¦ = βπ₯ + 12.
Question 9
A line parallel to the π₯-axis is a horizontal line with a gradient of zero.
Thus, the equation of the line that passes through (1,7) and is parallel
to the π₯-axis is π¦ = 7.
Question 10
Since the line must be parallel to π¦ = ππ₯ + π, the gradient of the line is
π. The equation of the line is π¦ = ππ₯ + π (using π to avoid confusion).
At (π, 0):
0 = ππ + π
π = βππ
The equation of the line is π¦ = ππ₯ β ππ.
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Question 11
The point of intersection of two lines is the point (π₯, π¦) that satisfies both
equations.
b) The simplest approach here is to equate the right hand sides of
each equation and solve for π₯:
3π₯ + 1 = 4π₯ β 1
π₯ = 2
Now substitute this value of π₯ into either of the given equations:
π¦ = 3(2) + 1
π¦ = 7
Check this answer by substituting these values for π₯ and π¦ back
into the second equation (the one that was not used to solve for π¦)
and check that both sides are equal:
π π»π = 4(2) β 1 = 7
πΏπ»π = 7
β΄ πΏπ»π = π π»π
So the point of intersection is (2,7).
d) Use the same approach as above.
3π₯ + 8 = β2π₯ β 7
5π₯ = β15
π₯ = β3
Substitute for π₯ in the first equation:
π¦ = 3(β3) + 8
π¦ = β1
Check:
π π»π = β2(β3) β 7 = β1
πΏπ»π = β1
β΄ πΏπ»π = π π»π
So the point of intersection is (β3, β1).
f) Here it is simpler to multiply the first equation by 5 and the second
equation by β2 and then to add the equations to eliminate π₯:
10π₯ + 35π¦ = 235
β10π₯ β 8π¦ = β100
Adding these equations and solving for π¦:
27π¦ = 135
π¦ = 5
Substituting for π¦ in the first equation and solving for π₯:
2π₯ + 7(5) = 47
2π₯ = 12
π₯ = 6
Check by substituting these values for π₯ and π¦ into the second
equation:
πΏπ»π = 5(6) + 4(5) = 30 + 20 = 50 = π π»π
So the point of intersection is (6,5).
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g) Multiply the first equation by β3 and add the two equations to
eliminate π₯:
β6π₯ β 9π¦ = β21 (1)
6π₯ + 9π¦ = 11 (2)
(1) + (2):
0 = β10
This is obviously not true. These equations cannot be solved
simultaneously, so we conclude that these two lines do not
intersect. This can be understood geometrically by writing the
equations in the form π¦ = ππ₯ + π:
π¦ = β2
3π₯ +
7
3
π¦ = β2
3π₯ +
11
9
These lines have the same gradient so they are parallel lines and
never intersect. (The π¦-intercepts are different so they are not
collinear. In the case of collinear lines, there are infinitely many
points which will simultaneously satisfy both the equations.)
h) Solve the first equation for π¦:
π¦ = β3π₯ + 5
Then substitute this expression for π¦ into the second equation and
solve for π₯:
π₯ + 3(β3π₯ + 5) = β1
π₯ β 9π₯ + 15 = β1
β8π₯ = β16
π₯ = 2
Then solve for π¦ by substituting this value of π₯ into the first
expression for y:
π¦ = β3(2) + 5
π¦ = β1
Check by substituting these values for π₯ and π¦ into the second
equation:
πΏπ»π = 2 + 3(β1) = β1 = π π»π
The lines intersect at (2, β1).
i) The first equation is already written as an expression for π¦, so we
just substitute this expression into the second equation:
4π₯ β 2(2π₯ + 3) = β6
4π₯ β 4π₯ β 6 = β6
β6 = β6
This is always true. Since the π₯βs cancel out in the second line, any
value of π₯ will simultaneously satisfy these equations. Thus, any
value of π¦ will also simultaneously satisfy both these equations.
There are thus infinitely many solutions.
This can be seen geometrically as well. Rewriting the equations in
the form π¦ = ππ₯ + π gives:
π¦ = 2π₯ + 3
π¦ = 2π₯ + 3
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So the given equations actually represent the same line.
j) The second equation here is already written as an expression for
π¦, so we just substitute that expression into the first equation:
ππ₯ + π(2ππ₯) = π
(π + 2ππ)π₯ = π
π₯ =π
π + 2ππ
To find π¦ we substitute this expression back into the second
equation:
π¦ = 2π(π
π + 2ππ)
π¦ =2ππ
π + 2ππ
π¦ =2π
1 + 2π
To check we substitute these expressions for π₯ and π¦ into the first
equation:
πΏπ»π = π (π
π + 2ππ) + π (
2π
1 + 2π)
=ππ + 2πππ
π + 2ππ= (
π + 2ππ
π + 2ππ) π
= π = π π»π
The lines intersect at (π
π+2ππ,
2ππ
π+2ππ).
k) Add the two equations to eliminate π₯:
2π¦ = π + π
π¦ =π + π
2
Substitute this expression for π¦ into the first equation:
π + π
2= ππ₯ + π
ππ₯ =π + π
2β π
π₯ =π + π β 2π
2π
π₯ =π β π
2π
Check by substituting for π₯ and π¦ into the second equation:
π π»π = βπ (π β π
2π) + π
=π β π
2+
2π
2
=π + π
2= πΏπ»π
So the lines intersect at (πβπ
2π,
π+π
2).
l) Substitute the expression for π¦ from the second equation into the
first equation:
ππ₯ β ππ₯ = 1
(π β π)π₯ = 1
π₯ =1
π β π
Substitute this back into the second equation and to obtain π¦:
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π¦ =1
π β π
Check, using the first equation:
πΏπ»π = π (1
π β π) β π (
1
π β π) =
π β π
π β π= 1 = π π»π
So the lines intersect at (1
πβπ,
1
πβπ).
Question 12
Since π (π. π) satisfies the equation π¦ = ππ₯ + π, we have
π = ππ + π. (1)
The point π (π, π ) is any other point on the line π¦ = ππ₯ + π so it also
satisfies the equation π¦ = ππ₯ + π and we can write
π = ππ + π. (2)
Now the gradient of the line joining the points π and π is given as usual
by:
ππππππππ‘ =π β π
π β π.
Substituting for π and π from eqns (1) and (2) gives:
ππππππππ‘ = ππ + π β (ππ + π)
π β π
ππππππππ‘ =ππ β ππ
π β π
ππππππππ‘ = π β π
π β ππ
ππππππππ‘ = π
Thus, the gradient of the line segment joining the points π and π is π.
(Since π and π are arbitrary points on the line, this shows that if you
have an equation of a line in the form π¦ = ππ₯ + π, then the coefficient of
π₯ is the gradient of that line, a fact that we used in exercise 4.)
Question 13
If π = π = π = 0, then any value of π₯ and π¦ will satisfy the equation ππ₯ +
ππ¦ + π = 0. This equation then represents the entire π₯π¦-plane rather
than a straight line.
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EXERCISE 1C
Question 1
The gradient of a line that is perpendicular to a given line, is the negative
of the reciprocal of the gradient of the given line.
c) β4
3
d) 6
5
g) π
h) β1
π
k) 1
π
l) β1π
πβπ
= βπβπ
π=
πβπ
π
Question 2
b) The gradient of the given line is β1
2, so the gradient of the line is
2. So the equation of the line is π¦ = 2π₯ + π. At (β3,1):
1 = 2(β3) + π
π = 7
β΄ π¦ = 2π₯ + 7
d) The line π¦ = 21
2=
5
2 is a horizontal line. So, any vertical line is
perpendicular to this line. So the equation of the line is π₯ = 7.
f) Rewrite the given line in the form π¦ = ππ₯ + π:
3π₯ β 5π¦ = 8
5π¦ = 3π₯ β 8
π¦ =3
5π₯ β
8
5
So the gradient of the perpendicular line is β5
3. Using (4,3):
π¦ β 3 = β5
3(π₯ β 4)
3π¦ β 9 = β5π₯ + 20
5π₯ + 3π¦ = 29
h) The gradient is β1
2. The equation of the line is π¦ = β
1
2π₯ + π. At
(0,3):
3 = β1
2(0) + π
π = 3
β΄ π¦ = β1
2π₯ + 3
ππ 2π¦ + π₯ = 6
j) The gradient is β1
π. The equation of the line is π¦ = β
1
ππ₯ + π. At
(π, π):
π = β1
ππ + π
π = π +π
π
π =ππ + π
π
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l)
ππ₯ + ππ¦ = π
ππ¦ = βππ₯ + π
π¦ = βπ
ππ₯ +
π
π
So the gradient of the line is π
π. The equation of the line is
π¦ =π
ππ₯ + π. At (β1, β2):
β2 =π
π(β1) + π
π =π
πβ 2 =
π β 2π
π
β΄ π¦ =π
ππ₯ +
π β 2π
π
ππ ππ₯ β ππ¦ = 2π β π
Question 3
The gradient of the line is β1
3. The equation of the line is
π¦ = 3π₯ + π. At (β2,5):
5 = β1
3(β2) + π
π = 5 β2
3
π =13
3
So the equation of the line is π¦ = β1
3π₯ +
13
3.
To find the point of intersection we equate right hand sides of the
two equations:
3π₯ + 1 = β1
3π₯ +
13
3
9π₯ + 3 = βπ₯ + 13
10π₯ = 10
π₯ = 1
Using this, we can solve for π¦:
π¦ = 3(1) + 1
π¦ = 4
So the two lines intersect at the point (1,4).
Question 4
Rewrite the given equation in the form π¦ = ππ₯ + π:
2π₯ β 3π¦ = 12
3π¦ = 2π₯ β 12
π¦ =2
3π₯ β 4 (1)
So the gradient of the line will be β3
2. The equation is
π¦ = β3
2π₯ + π. At (1,1):
1 = β3
2(1) + π
π = 1 +3
2=
5
2
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β΄ π¦ = β3
2π₯ +
5
2(2)
To find the point of intersection, substitute eqn (1) into eqn (2):
2
3π₯ β 4 = β
3
2π₯ +
5
2
4π₯ β 24 = β9π₯ + 15
13π₯ = 39
π₯ = 3
Substitute this result into eqn (1) to solve for π¦:
π¦ =2
3(3) β 4
π¦ = β2
So the lines intersect at (3, β2).
Question 5
First plot the points of the triangle in a coordinate plane:
Now we are looking for the line that is perpendicular to the line segment
π΅πΆ and passes through the point π΄ (2,3). So the first step is to find the
gradient of the line segment π΅πΆ:
ππππππππ‘ ππ π΅πΆ =β1 β (β7)
4 β 1=
6
3= 2
So the gradient of the line we are looking for is β1
2. The equation of the
line is π¦ = β1
2π₯ + π. At (2,3):
3 = β1
2(2) + π
π = 4
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So the equation of the altitude through the vertex A is
π¦ = β1
2π₯ + 4.
Question 6
Plot the triangle:
a) The line segment ππ is opposite to the vertex π . Since the line
segment ππ is a horizontal line, the altitude through π is a vertical
line passing through (8, β7). The equation of this line is π₯ = 8.
The altitude through the vertex π is perpendicular to the line
segment ππ .
ππππππππ‘ ππ ππ = 5 β (β7)
2 β 8=
12
β6= β2
So the gradient of the altitude through π is 1
2. The equation of the
line is π¦ =1
2π₯ + π. At π (12,5):
5 =1
2(12) + π
π = β1
So the equation of the altitude through π is π¦ =1
2π₯ β 1.
b) Find the point of intersection:
π₯ = 8 (1)
π¦ =1
2π₯ β 1 (2)
(1) β (2):
π¦ =1
2(8) β 1 = 3
So these two altitudes intersect at the point (8,3).
c) To show that the altitude through π also passes through the point
(8,3), we first find the equation of this altitude and then show that
the point (8,3) satisfies that equation.
ππππππππ‘ ππ ππ =5 β (β7)
12 β 8= 3
So the gradient of the altitude is β1
3 and the equation of the line is
π¦ = β1
3π₯ + π. At (2,5):
5 = β1
3(2) + π
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π = 5 +2
3=
17
3
So the equation of the altitude through π is π¦ = β1
3π₯ +
17
3.
Now we show that the point (8,3) satisfies this equation by finding
the value of π¦ if we substitute π₯ = 8 into the equation:
π¦ = β1
3(8) +
17
3
π¦ =9
3= 3
So the altitude through π also passes through (8,3).
MISCELLANEOUS EXERCISE 1
Question 1
In order to show that a triangle is right-angled, we have to show that two
of the line segments joining the vertices are perpendicular. First we plot
and label the vertices of the triangle:
From the graph it looks most likely that the line segments π΄π΅ and π΅πΆ will
be perpendicular. To show this we calculate the gradients of each line
segment:
ππππππππ‘ ππ π΄π΅ =3 β 5
1 β (β2)= β
2
3
ππππππππ‘ ππ π΅πΆ =9 β 3
5 β 1=
6
4=
3
2
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We see that the gradient of π΄π΅ is the negative of the reciprocal of the
gradient of π΅πΆ. So these two line segments are perpendicular.
Question 2
Find the point of intersection:
2π₯ + π¦ = 3 (1)
3π₯ + 5π¦ β 1 = 0 (2)
Rearrange eqn (1):
π¦ = β2π₯ + 3 (3)
(3) β (2):
3π₯ + 5(β2π₯ + 3) β 1 = 0
3π₯ β 10π₯ + 15 β 1 = 0
7π₯ = 14
π₯ = 2 (4)
(4) β (3):
π¦ = β2(2) + 3
π¦ = β1 (5)
So the lines intersect at (2, β1).
Question 3
a) To show that the angle π΄πΆπ΅ is a right angle, we have to show that
the line segments π΄πΆ and πΆπ΅ are perpendicular:
ππππππππ‘ ππ π΄πΆ =8 β 3
0 β (β1)= 5
ππππππππ‘ ππ πΆπ΅ = 8 β 7
0 β 5= β
1
5
So the gradient of π΄πΆ is the negative of the reciprocal of the
gradient of πΆπ΅. So these line segments are perpendicular to each
other and it follows that the angle between them (angle π΄πΆπ΅) is a
right angle.
b) First we need to find the equation of the line that is parallel to π΄πΆ
and passes through π΅. Since it is parallel to π΄πΆ it has the same
gradient as π΄πΆ which is 5 (from part (a)). So the equation of the
line is π¦ = 5π₯ + π. At π΅ (5,7):
7 = 5(5) + π
π = 7 β 25 = β18
So the equation of the line is π¦ = 5π₯ β 18.
To find the point where this line crosses the π₯-axis, we set π¦ to
zero and solve for π₯:
0 = 5π₯ β 18
5π₯ = 18
π₯ = 3.6
So this line crosses the π₯-axis at the point (3.6,0).
Question 4
a) The diagonals of a square have the same length and bisect each
other at 90Β°. So the diagonal π΅π· is perpendicular to π΄πΆ and passes
through the midpoint of π΄πΆ.
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ππππππππ‘ ππ π΄πΆ = (1
2(7 + 1),
1
2(2 + 4))
= (4,3)
ππππππππ‘ ππ π΄πΆ =4 β 2
1 β 7=
2
β6= β
1
3
So the gradient of the diagonal π΅π· is 3. The equation of the
diagonal is π¦ = 3π₯ + π. At (4,3):
3 = 3(4) + π
π = β9
So the equation of the diagonal π΅π· is π¦ = 3π₯ β 9.
b) Since the diagonals bisect each other and have the same length,
the distance from the midpoint of π΄πΆ to either point π΅ or π· is half
the distance from point π΄ to point πΆ. Both points π΅ and π· also lie
on the diagonal π΅π·. First we plot all the information we know:
πππ π‘ππππ ππππ π΄ π‘π πΆ = β(7 β 1)2 + (2 β 4)2
= β40
= 2β10
So the distance from the midpoint (4,3) to either point π΅ or π· is
β10. Suppose we have a point (π₯, π¦) that lies on the diagonal π΅π·
and is a distance β10 from the midpoint. Then it must satisfy the
following equations:
β(π₯ β 4)2 + (π¦ β 3)2 = β10 (1)
π¦ = 3π₯ β 9 (2)
Starting from eqn (1):
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(π₯ β 4)2 + (π¦ β 3)2 = 10 (3)
(2) β (3):
(π₯ β 4)2 + (3π₯ β 9 β 3)2 = 10
(π₯ β 4)2 + (3π₯ β 12)2 = 10
π₯2 β 8π₯ + 16 + 9π₯2 β 72π₯ + 144 = 10
10π₯2 β 80π₯ + 150 = 0
π₯2 β 8π₯ + 15 = 0
(π₯ β 5)(π₯ β 3) = 0
π₯ = 5 ππ π₯ = 3
This gives two values of π₯ that satisfy equations (1) and (2) Thus
there are two points that will satisfy the conditions listed above.
This is what we were expecting since π΅ and π· satisfy those
conditions. So we can find the corresponding π¦ values:
At π₯ = 5:
π¦ = 3(5) β 9 = 6
At π₯ = 3:
π¦ = 3(3) β 9 = 0
So the coordinates of π΅ are (3,0) and the coordinates of π· are
(5,6).
[Note: The answer at the back of the text book is incorrect.]
Question 6
a) Rearrange the equation of π1:
3π₯ + 4π¦ = 16
π¦ = β3
4π₯ + 4
So the gradient of π2 is 4
3 and its equation is π¦ =
4
3π₯ + π. At (7,5):
5 =4
3(7) + π
π = 5 β28
3
π = β13
3
So the equation for π2 is π¦ =4
3π₯ β
13
3.
b)
π¦ = β3
4π₯ + 4 (1)
π¦ =4
3π₯ β
13
3(2)
(2) β (1):
4
3π₯ β
13
3= β
3
4π₯ + 4
16π₯ β 52 = β9π₯ + 48
25π₯ = 100
π₯ = 4 (3)
(3) β (1):
π¦ = β3
4(4) + 4
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π¦ = 1 (4)
So the lines intersect at (4,1).
c) The perpendicular distance of π from the line π1 is the distance
from π to the point of intersection:
π = β(7 β 4)2 + (5 β 1)2
= β9 + 16
= 5
Question 9
Rearrange the given equation:
2π₯ + 7π¦ = 5
π¦ = β2
7π₯ +
5
7
So the gradient of the line is β2
7. The equation is π¦ = β
2
7π₯ + π. At (1,3):
3 = β2
7(1) + π
π = 3 +2
7=
23
7
So the equation of the line is
π¦ = β2
7π₯ +
23
7
ππ 2π₯ + 7π¦ = 23
Question 10
First we find the gradient of the line joining (2, β5) and (β4,3):
ππππππππ‘ =3 β (β5)
β4 β 2=
8
β6= β
4
3
So the gradient of the line we are looking for is 3
4. The line passes through
the midpoint of the line joining (2, β5) and (β4,3):
ππππππππ‘ = (1
2(2 + (β4)),
1
2(β5 + 3))
= (β1, β1)
The equation of the line is π¦ =3
4π₯ + π. At (β1, β1):
β1 =3
4(β1) + π
π = β1 +3
4= β
1
4
The equation of the line is π¦ =3
4π₯ β
1
4.
Question 11
ππππππππ‘ ππ π΄πΆ = (1
2(1 + 6),
1
2(2 + 6))
= (31
2, 4)
To find the coordinates of π·, first plot the information on a graph:
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The point π· lies on the diagonal π΅π· and is the same distance from π as
π΅ is from π. These conditions give two equations which can be solved
to find the coordinates of π·.
πππ π‘ππππ ππππ π΅ π‘π π = β(3 β 3.5)2 + (5 β 4)2
= β0.25 + 1
= β1.25
Equation of the diagonal π΅π·:
ππππππππ‘ =4 β 5
3.5 β 3= β
1
0.5= β2
The equation of the diagonal π΅π· is π¦ = β2π₯ + π. At (3,5):
5 = β2(3) + π
π = 11
Equation of π΅π·:
π¦ = β2π₯ + 11
So the point π· (π₯, π¦) satisfies the following equations:
β(π₯ β 3.5)2 + (π¦ β 4)3 = β1.25 (1)
π¦ = β2π₯ + 11 (2)
Starting with eqn (1):
(π₯ β 3.5)2 + (π¦ β 4)2 = 1.25 (3)
(2) β (3):
(π₯ β 3.5)2 + (β2π₯ + 7)2 = 1.25
π₯2 β 7π₯ + 12.25 + 4π₯2 β 28π₯ + 49 = 1.25
5π₯2 β 35π₯ + 60 = 0
π₯2 β 7π₯ + 12 = 0
(π₯ β 3)(π₯ β 4) = 0
π₯ = 3 ππ π₯ = 4
The π₯ coordinate of π· is 4 since the π₯ coordinate of π΅ is 3. So
π¦ = β2(4) + 11 = 3
The coordinates of π· are (4,3).
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Question 12
a) The line π΄π is perpendicular to the line π¦ = 3π₯. So the gradient of
π΄π is β1
3. The equation of π΄π is π¦ = β
1
3π₯ + π. At (0,3):
3 = β1
3(0) + π
π = 3
So the equation of π΄π is π¦ = β1
3π₯ + 3.
b) The lines π΄π and π¦ = 3π₯ intersect at the point π:
3π₯ = β1
3π₯ + 3
9π₯ + π₯ = 9
π₯ = 0.9
Solve for π¦:
π¦ = 3(0.9)
π¦ = 2.7
The coordinates of π are (0.9,2.7).
c) The perpendicular distance of π΄ from the line π¦ = 3π₯ is the
distance between the point π΄ and the point π:
πππ π‘ππππ = β(0.9 β 0)2 + (2.7 β 3)2
= β0.92 + 0.32
= β0.9
Question 13
To show that the given points are collinear, we must show that one of
the points lies on the line passing through the other two points.
Equation of the line passing through (β11, β5) and (4,7):
ππππππππ‘ =7 β (β5)
4 β (β11)=
12
15=
4
5
Equation of line: π¦ =4
5π₯ + π. At (4,7):
7 =4
5(4) + π
π = 7 β16
5=
19
5
So the equation of the line passing through (β11, β5) and (4,7) is π¦ =4
5π₯ +
19
5.
To see if this line passes through the point (β1,3) find the value of π¦
when π₯ = β1:
π¦ =4
5(β1) +
19
5
=15
5= 3
So this line does indeed pass through the point (β1,3). So the given
points are collinear.
Question 14
i) To find the value of π substitute the coordinates of the point π΄ into
the equation of the line:
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4π₯ + ππ¦ = 20
4(8) + π(β4) = 20
32 β 4π = 20
4π = 12
π = 3
So the equation of the line is 4π₯ + 3π¦ = 20.
Now we can find the value of π, by substituting (π, 2π) into the
equation of the line:
4(π) + 3(2π) = 20
10π = 20
π = 2
ii)
ππππππππ‘ = (1
2(8 + 2),
1
2(β4 + 4)) = (5,0)
[Note: The answer at the back of the text book is incorrect.]
Question 15
The point π· is the point of intersection of the lines π΄π· and π·πΆ. Since π΄π·
is perpendicular to π΄π΅ and the coordinates of both π΄ and π΅ are given,
we can find the equation of π΄π·:
ππππππππ‘ ππ π΄π΅ =8 β 2
3 β 6=
6
β3= β2
So the gradient of π΄π· is 1
2. The equation of π΄π· is π¦ =
1
2π₯ + π. At (3,8):
8 =1
2(3) + π
π = 8 β3
2=
13
2
So the equation of π΄π· is π¦ =1
2π₯ +
13
2.
The line π·πΆ is parallel to π΄π΅, so its gradient is β2. The equation of π·πΆ
is π¦ = β2π₯ + π. At (10,2):
2 = β2(10) + π
π = 2 + 20 = 22
So the equation of π·πΆ is π¦ = β2π₯ + 22.
Now the point π· is the intersection of these two lines. So we can equate
the right hand sides of the two equations and solve for π₯:
1
2π₯ +
13
2= β2π₯ + 22
π₯ + 4π₯ = 44 β 13
5π₯ = 31
π₯ =31
5= 6
1
5
β΄ π¦ = β2 (31
5) + 22
π¦ =48
5= 9
3
5
So the coordinates of π· are (61
5, 9
3
5).
[Note: The answer at the back of the text book is incorrect.]
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Question 17
The point π΅ is the intersection of the line segments π΄πΆ and π΅π·. The
equation of the line segment π΅π· can be obtained since it is
perpendicular to π΄πΆ and the coordinates of the point π· are given.
Rearrange the equation of π΄πΆ:
2π¦ = π₯ + 4 (1)
π¦ =1
2π₯ + 2
So the gradient of π΅π· is β2. The equation of π΅π· is π¦ = β2π₯ + π. At
(10, β3):
β3 = β2(10) + π
π = β3 + 20 = 17
So the equation of π΅π· is
π¦ = β2π₯ + 17 (2)
Now find the point at which π΄πΆ intersects π΅π·:
(2) β (1):
2(β2π₯ + 17) = π₯ + 4
β4π₯ β π₯ = 4 β 34
β5π₯ = β30
π₯ = 6
β΄ π¦ = β2(6) + 17
π¦ = 5
So the coordinates of π΅ are (6,5).
The point πΆ lies on the line 2π¦ = π₯ + 4 and is the same distance from the
point π΅ as the point π΄ is from π΅. Thus, to find the coordinates of πΆ we
first need to find the coordinates of π΄ and the distance from π΄ to π΅.
The point π΄ lies on the π¦-axis, so its π₯ coordinate is zero. We can find
the π¦ coordinate by substituting π₯ = 0 into eqn (1):
2π¦ = 4
π¦ = 2
So the coordinates of π΄ are (0,2).
πππ π‘ππππ ππππ π΅ π‘π π΄ = β(6 β 0)2 + (5 β 2)2
= β36 + 9 = β45
So the point πΆ (π₯, π¦) satisfies the following equations:
(π₯ β 6)2 + (π¦ β 5)2 = 45 (1)
π¦ =1
2π₯ + 2 (2)
(2) β (1):
(π₯ β 6)2 + (1
2π₯ β 3)
2
= 45
π₯2 β 12π₯ + 36 +1
4π₯2 β 3π₯ + 9 = 45
5
4π₯2 β 15π₯ = 0
π₯2 β 12π₯ = 0
π₯(π₯ β 12) = 0
π₯ = 0 ππ π₯ = 12
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The π₯ coordinate of πΆ is 12, since the π₯ coordinate of π΄ is zero. So we
can substitute this value of π₯ into eqn (2) to solve for π¦:
π¦ =1
2(12) + 2
π¦ = 6 + 2 = 8
So the coordinates of the point πΆ are (12,8).
[Note: The answer at the back of the text book is incorrect.]
Question 18
Plot the given points on a coordinate plane:
A rectangle has two sets of parallel lines, each set having the same
length and all its corners are right angles. To show that the given points
satisfy these conditions we must show that the distance between π and
π is the same as the distance between π and π and that the distance
between π and π is the same the distance between π and π. We must
also show that π π is parallel to ππ and that ππ is parallel to π π and that
ππ and π π are perpendicular to π π and ππ. We start by finding the
distances between the points:
πππ π‘ππππ ππππ π π‘π π = β(β1 β 0)2 + (4 β 7)2 = β10
πππ π‘ππππ ππππ π π‘π π = β(5 β 6)2 + (2 β 5)2 = β10
πππ π‘ππππ ππππ π π‘π π = β(β1 β 5)2 + (4 β 2)2 = β40
πππ π‘ππππ ππππ π π‘π π = β(0 β 6)2 + (7 β 5)2 = β40
So we see that
πππ π‘ππππ ππππ π π‘π π = πππ π‘ππππ ππππ π π‘π π
πππ π‘ππππ ππππ π π‘π π = πππ π‘ππππ ππππ π π‘π π
Next we find the gradients of the line segments joining the points:
ππππππππ‘ ππ ππ =4 β 7
β1 β 0= 3
ππππππππ‘ ππ π π =5 β 2
6 β 5= 3
ππππππππ‘ ππ ππ =4 β 2
β1 β 5= β
1
3
ππππππππ‘ ππ ππ =7 β 5
0 β 6= β
1
3
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So it is clear that ππ and π π are parallel and ππ and ππ are
parallel. It is also clear that ππ and π π are perpendicular to ππ and
ππ. So the parallel line segments have the same length and all the
corners are right angled. Thus these points do form a rectangle.
Question 20
a) The diagonals of a rhombus bisect each other at 90Β°, but unlike a
square they are not the same length. Start by plotting the points π΄
and πΆ and completing a rhombus using arbitrary points for π΅ and
π·. Note: at this stage we are not certain whether or not π· lies on
the π¦-axis.
Since we are given the points π΄ and πΆ we can find the midpoint of
π΄πΆ and then the equation of the diagonal π΅π·:
ππππππππ‘ ππ π΄πΆ = (1
2(β3 + 5),
1
2(β4 + 4)) = (1,0)
ππππππππ‘ ππ π΄πΆ =β4 β 4
β3 β 5= 1
π΅π· is perpendicular to π΄πΆ so the gradient of π΅π· is β1. The
equation of π΅π· is π¦ = βπ₯ + π. At (1,0):
0 = β1(1) + π
π = 1
So the equation of the diagonal π΅π· is π¦ = βπ₯ + 1.
b) The point π΅ is the intersection of the lines π΅π· and π΅πΆ. We are
given the gradient of π΅πΆ and the coordinates of πΆ so we can find
the equation of π΅πΆ. The equation is π¦ =5
3π₯ + π. At (5,4):
4 =5
3(5) + π
π = 4 β25
3= β
13
3
So the equation of π΅πΆ is π¦ =5
3π₯ β
13
3.
Next we find the point of intersection of π΅π· and π΅πΆ:
βπ₯ + 1 =5
3π₯ β
13
3
5π₯ + 3π₯ = 3 + 13
8π₯ = 16
π₯ = 2
β΄ π¦ = β2 + 1 = β1
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The coordinates of π΅ are (2, β1).
The point π· is the intersection of the line segments π΅π· and π΄π·.
Since π΄π· is parallel to π΅πΆ we can find the equation of π΄π·. It is
given by π¦ =5
3π₯ + π. At π΄ (β3, β4):
β4 =5
3(β3) + π
π = β4 + 5 = 1
So the equation of π΄π· is π¦ =5
3π₯ + 1.
Now we find the point of intersection between π΄π· and π΅πΆ:
βπ₯ + 1 =5
3π₯ + 1
3π₯ + 5π₯ = 0
π₯ = 0
β΄ π¦ = β(0) + 1 = 1
The coordinates of π· are (0,1).
Question 21
Note: a median is the line joining a vertex of a triangle to the midpoint of
the opposite side.
First plot and label the points:
Next we find the midpoints of each side:
π1 = ππππππππ‘ ππ π΄π΅ = (1
2(0 + 4),
1
2(2 + 4)) = (2,3)
π2 = ππππππππ‘ ππ π΅πΆ = (1
2(4 + 6),
1
2(4 + 0)) = (5,2)
π3 = ππππππππ‘ ππ π΄πΆ = (1
2(0 + 6),
1
2(2 + 0)) = (3,1)
Now find the equation of each median:
ππππππππ‘ ππ π΄π2 =2 β 2
0 β 5= 0
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The equation of π΄π2 is π¦ = π. Since the line passes through (0,2), the
value of π is 2. So the equation of π΄π2 is
π¦ = 2 (1)
ππππππππ‘ ππ π΅π3 =4 β 1
4 β 3= 3
The equation of π΅π3 is π¦ = 3π₯ + π. At (4,4):
4 = 3(4) + π
π = 4 β 12 = β8
So the equation of π΅π3 is
π¦ = 3π₯ β 8 (2)
ππππππππ‘ ππ πΆπ1 =0 β 3
6 β 2= β
3
4
The equation of πΆπ1 is π¦ = β3
4π₯ + π. At (6,0):
0 = β3
4(6) + π
π =9
2
So the equation of πΆπ1 is
π¦ = β3
4π₯ +
9
2(3)
To show that all the medians are concurrent, we find the point of
intersection of π΄π2 and π΅π3 and show that πΆπ1 passes through this
point:
(1) β (2):
2 = 3π₯ β 8
3π₯ = 10
π₯ =10
3
So π΄π2 and π΅π3 intersect at (10
3, 2). To show that πΆπ1 passes through
(10
3, 2), we find the π¦ value of πΆπ1 when π₯ =
10
3.
π¦ = β3
4(
10
3) +
9
2=
4
2= 2
So the medians are concurrent and all pass through the point (10
3, 2).
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 2A
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CHAPTER 2
EXERCISE 2A
Question 1
a) β3 Γ β3 = β3 Γ 3 = 3
c) β16 Γ β16 = β16 Γ 16 = 16
e) β32 Γ β2 = β32 Γ 2 = β64 = 8
g) 5β3 Γ β3 = 5β3 Γ 3 = 5(3) = 15
i) 3β6 Γ 4β6 = 12β6 Γ 6 = 12(6) = 72
k) (2β7)2
= 22(β7)2
= 4(7) = 28
m) β53
Γ β53
Γ β53
= β5 Γ 5 Γ 53
= 5
o) (2β23
)6
= 26(β23
)6
= 64(2)2 = 256
Question 2
a) β18 = β9 Γ 2 = β9 Γ β2 = 3β2
c) β24 = β4 Γ 6 = β4 Γ β6 = 2β6
e) β40 = β4 Γ 10 = β4 Γ β10 = 2β10
g) β48 = β16 Γ 3 = β16 Γ β3 = 4β3
i) β54 = β9 Γ 6 = β9 Γ β6 = 3β6
k) β135 = β9 Γ 15 = β9 Γ β15 = 3β15
Question 3
a) β8 + β18 = β4 Γ 2 + β9 Γ 2 = 2β2 + 3β2 = 5β2
c) β20 β β5 = β4 Γ 5 β β5 = 2β5 β β5 = β5
d) β32 β β8 = β16 Γ 2 β β4 Γ 2 = 4β2 β 2β2 = 2β2
f) β27 + β27 = 2β27 = 2β9 Γ 3 = 2(3β3) = 6β3
g)
β99 + β44 + β11 = β9 Γ 11 + β4 Γ 11 + β11
= 3β11 + 2β11 + β11
= 6β11
i) 2β20 + 3β45 = 2β4 Γ 5 + 3β9 Γ 5 = 4β5 Γ 9β5 = 13β5
j) β52 β β13 = β4 Γ 13 β β13 = 2β13 β β13 = β13
l)
β48 + β24 β β75 + β96 = β16 Γ 3 + β4 Γ 6 β β25 Γ 3 + β16 Γ 6
= 4β3 + 2β6 β 5β3 + 4β6
= 6β6 β β3
Question 4
a) β8
β2=
2β2
β2= 2
c) β40
β10=
2β10
β10= 2
e) β125
β5=
5β5
β5= 5
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g) β3
β48=
β3
4β3=
1
4
Question 5
a) 1
β3=
1
β3Γ
β3
β3=
1
3β3
c) 4
β2=
4
β2Γ
β2
β2=
4β2
2= 2β2
e) 11
β11=
11
β11Γ
β11
β11=
11β11
11= β11
g) 12
β3=
12
β3Γ
β3
β3=
12β3
3= 4β3
i) β6
β2=
β6
β2Γ
β2
β2=
β12
2=
2β3
2= β3
k) 3β5
β3=
3β5
β3Γ
β3
β3=
3β15
3= β15
m) 7β2
2β3=
7β2
2β3Γ
β3
β3=
7β6
2(3)=
7
6β6
o) 9β12
2β18=
9(2)β3
2(3)β2=
18β3
6β2Γ
β2
β2=
18β6
6(2)=
3
2β6
Question 6
a) β75 + β12 = 5β3 + 2β3 = 7β3
c) 12
β3β β27 =
12β3
3β 3β3 = 4β3 β 3β3 = β3
d) 2
β3+
β2
β6=
2β3
3+
β12
6=
2β3
3+
2β3
6=
(2+1)β3
3= β3
f)
(3 β β3)(2 β β3) β β3 Γ β27 = (6 β 3β3 β 2β3 + 3) β β3 Γ 3β3
= (9 β 5β3) β 3(3)
= 9 β 9 β 5β3
= β5β3
Question 7
a)
ππππ = 4β5 Γ β10
= 4β5 Γ 10 = 4β50
= 4β25 Γ 2 = 20β2
b) The diagonal π΄πΆ is the hypotenuse of the right angled triangle
π΄π΅πΆ. We use Pythagorasβs theorem to find its length:
π΄πΆ2 = π΄π΅2 + π΅πΆ2
= (4β5)2
+ (β10)2
= 16(5) + 10
= 90
β΄ π΄πΆ = β90 = 3β10
Question 8
a)
π₯β2 = 10
π₯ =10
β2=
10β2
2= 5β2
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c)
π§β32 β 16 = π§β8 β 4
4π§β2 β 2π§β2 = 12
2π§β2 = 12
π§ =6
β2= 3β2
Question 9
a) β243
= β8 Γ 33
= 2β33
d) β30003
β β3753
= β1000 Γ 33
β β125 Γ 33
= 10β33
β 5β33
= 5β33
Question 10
Since each of the triangles is right-angled, we can use Pythagorasβs
theorem. Letβs call the length of the unknown side π₯ in each case.
a)
π₯2 = 122 + 82
= 144 + 64
= 208
β΄ π₯ = β208 = 4β13 ππ
b)
π₯ = (10β2)2
+ (5β3)2
= 100(2) + 25(3)
= 200 + 75
= 275
β΄ π₯ = β275 = 5β11 ππ
d)
(3β7)2
= π₯2 + (3β2)2
9(7) = π₯2 + 9(2)
π₯2 = 45
β΄ π₯ = 3β5 ππ
Question 11
To use the information that we are given, we have to first write each
expression in the form πβ26.
a)
β104 = β4 Γ 26 = 2β26 = 2(5.099019513593)
= 10.198 039 027 2
b)
β650 = β25 Γ 26 = 5β26
= 5(5.099019513593)
= 25.495 097 568 0
c)
13
β26=
13β26
26=
1
2β26 =
1
2(5.099019513893)
= 2.549 509 756 8
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 2A
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Question 12
7π₯ β (3β5)π¦ = 9β5 (1)
(2β5)π₯ + π¦ = 34 (2)
Rearrange eqn (2):
π¦ = 34 β (2β5)π₯ (3)
(3) β (1):
7π₯ β (3β5)(34 β (2β5)π₯) = 9β5
7π₯ β 102β5 + 6(5)π₯ = 9β5
37π₯ = 111β5
π₯ = 3β5 (4)
(4) β (3):
π¦ = 34 β (2β5)(3β5)
= 34 β 6(5)
= 4 (5)
So the solution is π₯ = 3β5 and π¦ = 4.
Question 13
a)
(β2 β 1)(β2 + 1) = 2 β β2 + β2 β 1
= 2 β 1 = 1
c)
(β7 + β3)(β7 β β3) = 7 β β7β3 + β3β7 β 3
= 7 β 3 = 4
e)
(4β3 β β2)(4β3 + β2) = 16(3) + 4β3β2 β 4β2β3 β 2
= 48 β 2 = 46
g)
(4β7 β β5)(4β7 + β5) = 16(7) + 4β7β5 β 4β5β7 β 5
= 112 β 5 = 107
Question 14
All the parts in question 13 show the same pattern:
(π + π)(π β π) = π2 β π2
We use the pattern that we found in question 13 to answer this question.
(This is a very useful pattern and is called the difference of two squares.)
a) (β3 β 1)(β3 + 1) = 3 β 1 = 2
c) (β6 β β2)(β6 + β2) = 6 β 2 = 4
e) (β11 + β10)(β11 β β10) = 11 β 10 = 1
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Question 15
a) For any real number, π, π Γ 1 = π. Since any real number divided
by itself is equal to 1, we have β3+1
β3+1= 1. Therefore
1
β3β1Γ
β3+1
β3+1=
1
β3β1.
πΏπ»π =1
β3 β 1=
1
β3 β 1Γ
β3 + 1
β3 + 1=
β3 + 1
3 β 1=
β3 + 1
2= π π»π
b)
πΏπ»π =1
2β2 + β3Γ
2β2 β β3
2β2 β β3=
2β2 β β3
4(2) β 3=
2β2 β β3
5= π π»π
(Notice that in choosing what to multiply the given expression by,
we used the difference of two squares so that the denominator
would be rational.)
Question 16
a)
1
2 β β3=
1
2 β β3Γ
2 + β3
2 + β3=
2 + β3
4 β 3= 2 + β3
b)
1
3β5 β 5=
1
3β5 β 5Γ
3β5 + 5
3β5 + 5=
3β5 + 5
9(5) β 25=
3β5 + 5
20
c)
4β3
2β6 + 3β2=
4β3
2β6 + 3β2Γ
2β6 β 3β2
2β6 β 3β2
=8β18 β 12β6
4(6) β 9(2)
=24β2 β 12β6
6
= 4β2 β 2β6
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 2B
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EXERCISE 2B
Question 1
a) π2 Γ π3 Γ π7 = π2+3+7 = π12
c) π7 Γ· π3 = π7β3 = π4
d) π5 Γ π4 = π5+4 = π9
f) (π₯3π¦2)2 = (π₯3)2(π¦2)2 = π₯3Γ2π¦2Γ2 = π₯6π¦4
g) 5π5 Γ 3π3 = 15π5+3 = 15π8
i) (2π2)3 Γ (3π)2 = (8π2Γ3)(9π2) = 72π6π2 = 72π8
j) (π2π3)2 Γ (ππ3)3 = π2Γ2π3Γ2π3π3Γ3 = π4π6π3π9 = π7π15
l) (6ππ3)2 Γ· (9π2π5) = 36π2π6 Γ· 9π2π5 = 4π2β2π6β5 = 4π
m) (3π4π2)3 Γ (2ππ2)2 = 27π12π62 Γ 4π2π4 = 108π14π10
o) (2π₯π¦2π§3)2 Γ· (2π₯π¦2π§3) = 4π₯2π¦4π§6 Γ· 2π₯π¦2π§3 = 2π₯π¦2π§3
Question 2
a) 211 Γ (25)3 = 211 Γ 215 = 211+15 = 226
c) 43 = (22)3 = 22Γ3 = 26
e) 27Γ28
213 =215
213 = 22
g) 42 Γ· 24 = (22)2 Γ· 24 = 24 Γ· 24 = 24β4 = 20
Question 3
a) 2β3 =1
23 =1
8
c) 5β1 =1
5
e) 10β4 =1
104 =1
10 000
g) (1
2)
β1=
1
2β1 = 2
i) (21
2)
β1= (
5
2)
β1=
2
5
k) 6β3 =1
63 =1
216
Question 4
a) 4π₯β3 =4
π₯3 =4
23 =4
8=
1
2
c) 1
4π₯β3 =
1
4π₯3 =1
4(23)=
1
4(8)=
1
32
d) (1
4π₯)
β3= (
1
4)
β3π₯β3 =
43
π₯3 =64
23 =64
8= 8
f) (π₯ Γ· 4)β3 = (π₯
4)
β3=
43
π₯3 =64
23 =64
8= 8
Question 5
a) (2π¦)β1 =1
2π¦=
1
2(5)=
1
10
c) (1
2π¦)
β1= (
1
2)
β1π¦β1 =
2
π¦=
2
5
d) 1
2π¦β1 =
1
2π¦=
1
2(5)=
1
10
f) 2
(π¦β1)β1 =2
π¦β1Γβ1 =2
π¦=
2
5
Question 6
a) π4 Γ πβ3 = π4β3 = π
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c) (πβ2)3 = πβ2Γ3 = πβ6 =1
π6
d) πβ1 Γ 2π =2π
π= 2
f) πβ2
π3 =1
π2π3 =1
π5
g) 12π3 Γ (2π2)β2 = 12π3 Γ 2β2πβ4 =12π3
4π4 =3
π
i) (3πβ2)β2 = 3β2π4 =1
9π4
j) (1
2πβ2)
β3= (
1
2)
β3π6 = 23π6 = 8π6
l) (π2π4π3)β4 = (π2Γβ4π4Γβ4π3Γβ4) = πβ8πβ16πβ12
m) (4π2)β1 Γ 8π3 =8π3
4π2 = 2π
o) (2π₯π¦2)β1 Γ (4π₯π¦)2 =16π₯2π¦2
2π₯π¦2 = 8π₯
p) (5π3πβ1)2 Γ· (2πβ1π2) =25π6πβ2
2πβ1π2 =25π6π
2π2π2 =25π7
2π4
r) (3π₯β2π¦)2 Γ· (4π₯π¦)β2 = (9π₯β4π¦2)(42π₯2π¦2) = 9(16)π₯β2π¦4 =144π¦4
π₯2
Question 7
a) We can rewrite the given equation so that the right-hand side is
expressed in powers of 3:
3π₯ = 3β2
So π₯ = β2.
c) Rewriting the given equation:
2π§ Γ 2π§β3 = 25
2π§+π§β3 = 25
So we have:
2π§ β 3 = 5
π§ = 4
d) Rewriting the given equation:
73π₯βπ₯+2 = 7β2
So we have
2π₯ + 2 = β2
π₯ = β2
f) Rewriting the given equation:
3π‘ Γ (32)π‘+3 = (33)2
3π‘+2π‘+6 = 36
So we have:
3π‘ + 6 = 6
π‘ = 0
Question 8
a)
π£πππ’ππ = (3 Γ 10β2 π)3
= 33 Γ 10β6 π3
= 27 Γ 10β6 π3
= 2.7 Γ 101 Γ 10β6 π3
= 2.7 Γ 10β5 π3
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b)
π π’πππππ ππππ = 6 Γ (3 Γ 10β2 π)2
= 6 Γ (9 Γ 10β4 π2)
= 54 Γ 10β4 π2
= 5.4 Γ 10β3 π2
Question 9
π ππππ =2 Γ 10β1 ππ
7.5 Γ 10β3 β
= 0.2667 Γ 10β1 Γ 103 ππ ββ1
= (2.667 Γ 10β1) Γ 102 ππ ββ1
= 2.67 Γ 101 ππ ββ1
= 26.7 ππ ββ1
Question 10
a)
π = π(2 Γ 10β3 π)2(80 π)
= 80 Γ 4 Γ π Γ 10β6 π3
= 1005.3 Γ 10β6 π3
= 1.0 Γ 10β3 π3 π‘π 2 π . π.
b)
8 Γ 10β3 π3 = π(5 Γ 10β3 π)2π
π =8 Γ 10β3 π3
π(25 Γ 10β6 π2)
π = 0.1019 Γ 10β3 Γ 106 π
π = 0.1019 Γ 103 π
π = 101.9 π π‘π 1 π. π.
c)
6 Γ 10β3 π3 = ππ2(61 π)
π2 =6 Γ 10β3 π3
π(61 π)
π2 = 0.03131 Γ 10β3 π2
π2 = 3.131 Γ 10β5
π = β3.131 Γ 10β5 π2
π = 5.6 Γ 103 π π‘π 2 π . π.
Question 11
a)
π¦ =ππ
π
π¦ =(7 Γ 10β7)(5 Γ 10β1)
8 Γ 10β4
π¦ =35 Γ 10β8
8 Γ 10β4
π¦ = 4.375 Γ 10β4
b)
π =ππ¦
π
π =(2.7 Γ 10β4)(10β3)
0.6
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 2C
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π =2.7 Γ 10β7
0.6
π = 4.5 Γ 10β7
Question 12
35π₯+2
91βπ₯=
274+3π₯
729
35π₯+2
(32)1βπ₯=
(33)4+3π₯
729
35π₯+2
32β2π₯=
312+9π₯
36
35π₯+2β2+2π₯ = 312+9π₯β6
37π₯ = 39π₯+6
So we have:
7π₯ = 9π₯ + 6
2π₯ = β6
π₯ = β3
EXERCISE 2C
Question 1
a) 251
2 = β25 = 5
c) 361
2 = β36 = 6
e) 811
4 = β814
= 3
g) 16β1
4 =1
1614
=1
β164 =
1
2
k) 642
3 = (β643
)2
= 42 = 16
l) (β125)β4
3 = (ββ1253
)β4
= (β5)β4 =1
(β5)4 =1
625
Question 2
a) 41
2 = β4 = 2
c) (1
4)
β2= 42 = 16
e) (1
4)
1
2= 4
1
2 = 2
h) ((1
4)
1
4)
2
= (1
4)
1
2=
1
β4=
1
2
Question 3
a) 82
3 = (β83
)2
= 22 = 4
c) 9β3
2 =1
(β9)3 =
1
33 =1
27
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 2C
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e) 322
5 = (β325
)2
= 22 = 4
g) 64β5
6 =1
( β646
)5 =
1
25 =1
32
i) 10 000β3
4 = (β10 0004
)β3
= 10β3 =1
1000
k) (33
8)
2
3= (
27
8)
2
3= (
β273
β83 )
2
= (3
2)
2=
9
4= 2
1
4
Question 4
a) π1
3 Γ π5
3 = π6
3 = π2
c) (6π1
4) Γ (4π)1
2 = 6(2)π3
4 = 12π3
4
d) (π2)1
3 Γ· (π1
3)2
= π2
3 Γ· π2
3 = 1
f) (24π)1
3 Γ· (3π)1
3 =β243
π13
β33
π13
= β(24
3)
3= β8
3= 2
g) (5π2π4)
13
(25ππ2)β13
= 51
3251
3π2
3π1
3π4
3π2
3 = (125)1
3ππ2 = 5ππ2
i) (2π₯2π¦β1)
β14
(8π₯β1π¦2)β12
= 81
22β1
4π₯β1
2π¦β1π₯β1
2π¦1
4 = (23)1
22β1
4π₯β1π¦β3
4 = 25
4π₯β1π¦β3
4
Question 5
a)
π₯12 = 8
π₯ = 82 = 64
c)
π₯23 = 4
π₯ = 432 = 8
e)
π₯β32 = 8
π₯ = 8β23 =
1
4
g)
π₯32 = π₯β2
π₯12 = 2
12
π₯ = (212)
2
= 2
Question 6
a)
π = 2ππ12πβ
12
π = 2π(0.9 π)12(9.81 π π β2)β
12
π = 1.9 π
b)
π12 =
π
2ππβ12
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 2C
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π =π2π
4π2
π =(3 π )2(9.81 π π β2)
4π2
π = 2.2 π
Question 7
π = (3π
4π)
13
π = (3(1150 ππ3)
4π)
13
π = (274.5 ππ3)13
π = 6.5 ππ
Question 8
In each case we rewrite the given equation in the form ππ₯ = ππ where π
is known and then solve for π₯.
a)
4π₯ = 32
22π₯ = 25
β΄ 2π₯ = 5
π₯ =5
2
c)
16π§ = 2
24π§ = 2
β΄ 4π§ = 1
π§ =1
4
e)
8π¦ = 16
23π¦ = 24
β΄ 3π¦ = 4
π¦ =4
3
g)
(2π‘)3 Γ 4π‘β1 = 16
23π‘22π‘β2 = 24
25π‘β2 = 24
β΄ 5π‘ β 2 = 4
5π‘ = 6
π‘ =6
5
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Miscellaneous exercise 2
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MISCELLANEOUS EXERCISE 2
Question 1
c)
(β5 β 2)2
+ (β5 β 2)(β5 + 2) = (5 β 2β5 β 2β5 + 4) + 5 β 4
= 10 β 4β5
d)
(2β2)5
= 25252
= 252212
= 25+2212
= 27β2
= 128β2
Question 2
b) β63 β β28 = β9 Γ 7 β β4 Γ 7 = 3β7 β 2β7 = β7
d) β23
+ β163
= 21
3 + (24)1
3 = 21
3 + 24
3 = 21
3 + 21
323
3 = (1 + 2)21
3 = 3β23
Question 3
a) 9
2β3=
9
2β3Γ
β3
β3=
9β3
2(3)=
3
2β3
c) 2β5
3β10=
2β5
3β10Γ
β10
β10=
2β5β10
3(10)=
2β5β5β2
30=
2(5)β2
30=
10β2
30=
1
3β2
Question 4
c)
1
β2(2β2 β 1) + β2(1 β β8) = 2 β
1
β2+ β2 β β16
= 2 β 4 ββ2
2+ β2
= β2 +β2
2
d)
β6
β2+
3
β3+
β15
β5+
β18
β6=
β12
2+
3β3
3+
β75
5+
β108
6
=2β3
2+ β3 +
5β3
5+
6β3
6
= β3 + β3 + β3 + β3
= 4β3
Question 6
a) β12 Γ β75 = 2β3 Γ 5β3 = 10(3) = 30
b) β12 Γ β75 = (22 Γ 3)1
2 Γ (52 Γ 3)1
2 = 2 Γ 31
2 Γ 5 Γ 31
2 = 2 Γ 5 Γ
3 = 30
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Miscellaneous exercise 2
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Question 8
a)
ππππ =1
2Γ πππ π Γ βπππβπ‘
ππππ =1
2(6 β 2β2)(6 + 2β2) ππ2
ππππ =1
2(36 β 4(2)) ππ2
ππππ =1
2(36 β 8) ππ2
ππππ = 14 ππ2
b) From the figure it is clear that the side ππ is the hypotenuse of a
right-triangle. Thus the length of ππ can be found using
Pythagorasβs theorem:
ππ 2 = (6 β 2β2)2
+ (6 + 2β2)2
= 36 β 24β2 + 4(2) + 36 + 24β2 + 4(2)
= 88
β΄ ππ = β88 = β4 Γ 22 = 2β22
Question 10
From the figure above it is clear that the side π΄πΆ is opposite to the angle
π΅. So the cosine rule gives us:
π΄πΆ2 = π΄π΅2 + π΅πΆ2 β 2(π΄π΅)(π΅πΆ) cos(π΅)
π΄πΆ2 = (4β3)2
+ (5β3)2
β 2(4β3)(5β3) cos(60Β°)
π΄πΆ2 = 16(3) + 25(3) β 2(20)(3) (1
2)
π΄πΆ2 = 63
β΄ π΄πΆ = β63 = β9 Γ 7 = 3β7
Question 11
5π₯ β 3π¦ = 41 (1)
(7β2)π₯ + (4β2)π¦ = 82 (2)
Rearrange eqn (1) to make π¦ the subject:
3π¦ = 5π₯ β 41
π¦ =5
3π₯ β
41
3(3)
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Miscellaneous exercise 2
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(3) β (2):
(7β2)π₯ + (4β2) (5
3π₯ β
41
3) = 82
(7β2)π₯ +20β2
3π₯ β
164β2
3= 82
41β2
3π₯ = 82 +
164β2
3
π₯ =3(246 + 164β2)
41(3)β2
π₯ =6
β2+ 4
π₯ = 3β2 + 4 (4)
(4) β (3):
π¦ =5
3(3β2 + 4) β
41
3
π¦ = 5β2 +20
3β
41
3
π¦ = 5β2 β 7 (5)
So the solution is π₯ = 3β2 + 4 and π¦ = 5β3 β 7.
Question 14
a) The equation of the line is π¦ = β1
2π₯ + π. At the point π΄ (2,3):
3 = β1
2(2) + π
π = 3 + 1 = 4
So the equation of the line π is π¦ = β1
2π₯ + 4.
b) The point π (2 + 2π‘, 3 β π‘) lies on the line π if, for any value of π‘, it
satisfies the equation found in part a. Starting from the right-hand
side gives:
π π»π = β1
2(2 + 2π‘) + 4
= β1 β π‘ + 4
= 3 β π‘
= πΏπ»π
c) The length of π΄π can be found using the distance formula:
π΄π = β(2 + 2π‘ β 2)2 + (3 β π‘ β 3)2
π΄π = β(2π‘)2 + (βπ‘)2
π΄π = β4π‘2 + π‘2
π΄π = β5π‘2
Now we want to find the values of π‘ such that π΄π = 5:
β5π‘2 = 5
5π‘2 = 25
π‘2 = 5
π‘ = Β±β5
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Miscellaneous exercise 2
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So the length π΄π is 5 units if π‘ is either β5 or ββ5.
d) First we need to find an expression for the gradient of the line ππ:
ππππππππ‘ ππ =3 β π‘ β 0
2 + 2π‘ β 0=
3 β π‘
2 + 2π‘
Now we are looking for the value of π‘ such that ππ is perpendicular
to π, i.e. where the gradient of ππ is 2.
3 β π‘
2 + 2π‘= 2
3 β π‘ = 4 + 4π‘
5π‘ = β1
π‘ = β1
5
So the length of the perpendicular from π to π is the length of the
lines segment ππ, when π‘ = β1
5:
πππππ‘β = β(2 + 2 (β1
5))
2
+ (3 β (β1
5))
2
= β(8
5)
2
+ (16
5)
2
= β(64
25) + (
256
25)
= β320
25=
8
5β5
Question 16
We start by plotting the parallelogram:
To find the length of the side π΄π·, we need to find the coordinates of the
points π΄ and π·. We do this by finding the point of intersection of the sides
π΄π· and πΆπ· and the point of intersection of the sides π΄π΅ and π΄π·:
π₯ + π¦ = β4 (1)
π¦ = 2π₯ β 4 (2)
(2) β (1):
π₯ + 2π₯ β 4 = β4
3π₯ = 0
π₯ = 0 (3)
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(3) β (2):
π¦ = 0 β 4 = β4 (4)
So the coordinates of π΄ are (0, β4).
π₯ + π¦ = β4 (5)
π¦ = 2π₯ β 13 (6)
(6) β (5):
π₯ + 2π₯ β 13 = β4
3π₯ = 9
π₯ = 3 (7)
(7) β (6):
π¦ = 2(3) β 13 = β7 (8)
So the coordinates of π· are (3, β7).
Now we can find the length of the side π΄π·:
πππππ‘β = β(3 β 0)2 + (β7 + 4)2
= β9 + 9
= β18 = 3β2
To find the perpendicular distance between the side π΄π· and π΅πΆ we find
the equation of the line that is perpendicular to π΄π· and passes through
the point π΄. Let us call this line π. The equation of the line π΄π· can be
rewritten as
π¦ = βπ₯ β 4.
So the gradient of π is 1. Equation of π: π¦ = π₯ + π. At (0, β4):
β4 = 0 + π
π = β4
So the equation of the line π is π¦ = π₯ β 4.
Now we need to find the point at which the line π intersects the side π΅πΆ:
π₯ + π¦ = 5 (9)
π¦ = π₯ β 4 (10)
(10) β (9):
π₯ + π₯ β 4 = 5
2π₯ = 9
π₯ =9
2(11)
(11) β (10):
π¦ =9
2β 4 =
1
2(12)
So the line π intersects the side π΅πΆ at (9
2,
1
2). The perpendicular distance
between sides π΄π· and π΅πΆ is the distance between the points (0, β4) and
(9
2,
1
2):
πππ π‘ππππ = β(9
2β 0)
2
+ (1
2+ 4)
2
= β(9
2)
2
+ (9
2)
2
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= β81
4+
81
4
= β2 (81
4)
= β81
4β2
=9
2β2
So the area of the parallelogram is
ππππ = 3β2 Γ9
2β2 =
27
2(2) = 27
Question 19
π₯23 β 6π₯
13 + 8 = 0
We cannot add the two terms involving π₯ since they have different
powers of π₯. However, if we make a substitution we can transform this
equation into a quadratic equation which we can solve.
Let π¦ = π₯1
3. Then the equation becomes
π¦2 β 6π¦ + 8 = 0
(π¦ β 4)(π¦ β 2) = 0
π¦ = 4 ππ π¦ = 2
Now we must substitute back into our expression for π¦ to solve for π₯:
π₯ = π¦3
π₯ = 43 ππ π₯ = 23
π₯ = 64 ππ π₯ = 8
[Note: the answer at the back of the text book is partially incorrect.]
Question 20
42π₯ Γ 8π₯β1 = 32
Rewrite this equation in the form 2ππ₯+π = 2π:
(22)2π₯ Γ (23)π₯β1 = 25
24π₯23π₯β3 = 25
27π₯β3 = 25
So we have
7π₯ β 3 = 5
7π₯ = 8
π₯ =8
7
Question 22
b)
(5π)β1
(8π6)13
=1
5π(2π2)=
1
10π3
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d)
(π13π
12)
2
Γ (π16π
13)
4
Γ (ππ)β2 = π23π
22π
64π
43πβ2πβ2
= π43πβ
63π
43πβ1
= πβ23π
13
Question 26
c) 21
3 + 21
3 + 21
3 + 21
3 = 4 (21
3) = 2221
3 = 27
3
e)
80.1 + 80.1 + 80.1 + 80.1 + 80.1 + 80.1 + 80.1 + 80.1 = 8(80.1)
= 81+1.1 = 81.1 = (23)1.1 = 23.3
[Note: the answer to part (e) at the back of the text book is
incorrect.]
Question 27
1253π₯
5π₯+4=
25π₯β2
3125
(53)3π₯(5βπ₯β4) =(52)π₯β2
55
59π₯5βπ₯β4 = 52π₯β45β5
58π₯β4 = 52π₯β9
So we have
8π₯ β 4 = 2π₯ β 9
6π₯ = β5
π₯ = β5
6
Question 28
a) Solve for π in the expression for π:
π3 =3
4
π
π
π = (3
4
π
π)
13
Now substitute this into the expression for π:
π = 4ππ2
= 4π (3
4
π
π)
23
= 22(22)β23(3)
23ππβ
23π
23
= 2233
23π
13π
23
[Note: the answer at the back of the text book is partially incorrect.]
b) Solve for π in the expression for π:
π2 =π
4π
π = (π
4π)
12
Now substitute this into the expression for π:
π =4
3ππ3
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=4
3π (
π
4π)
32
= 4 (4β32) 3β1ππβ
32π
32
= 4β123β1πβ
12π
32
= (22)β123β1πβ
12π
32
= 2β13β1πβ12π
32
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CHAPTER 3
EXERCISE 3A
Question 1
π(π₯) = 2π₯ + 5
a) π(3) = 2(3) + 5 = 11
b) π(0) = 2(0) + 5 = 5
c) π(β4) = 2(β4) + 5 = 3
d) π (β21
2) = 2 (β2
1
2) + 5 = 2 (β
5
2) + 5 = 0
Question 2
π(π₯) = 3π₯2 + 2
a) π(4) = 3(4)2 + 2 = 3(16) + 2 = 50
b) π(1) = 3(1)2 + 2 = 3 + 2 = 5
π(β1) = 3(β1)2 + 2 = 3 + 2 = 5
(Note that (βπ₯)2 = π₯2 so the function has the same value at βπ₯ or
π₯.)
c) π(3) = 3(3)2 + 2 = 3(9) + 2 = 29
π(β3) = 3(β3)2 + 2 = 3(9) + 2 = 29
d) π(3) = 3(3)2 + 2 = 3(9) + 2 = 29
Question 3
π(π₯) = π₯2 + 4π₯ + 3
a) π(2) = (2)2 + 4(2) + 3 = 4 + 8 + 3 = 15
b) π (1
2) = (
1
2)
2+ 4 (
1
2) + 3 =
1
4+ 2 + 3 = 5
1
4
c) π(1) = (1)1 + 4(1) + 3 = 1 + 4 + 3 = 8
π(β1) = (β1)2 + 4(β1) + 3 = 1 β 4 + 3 = 0
(Note: this function has the term 4π₯ and 4(π₯) β 4(βπ₯) so the
function has different values at π₯ and βπ₯.)
d) π(3) = (3)2 + 4(3) + 3 = 9 + 12 + 3 = 24
π(β3) = (β3)2 + 4(β3) + 3 = 9 β 12 + 3 = 0
Question 4
π(π₯) = π₯3 and β(π₯) = 4π₯ + 1
a) π(2) + β(2) = (2)3 + 4(2) + 1 = 8 + 8 + 1 = 17
b)
3π(β1) β 4β(β1) = 3(β1)3 β 4(4(β1) + 1)
= β3 β 4(β4) β 4
= 9
c)
π(5) = 53 = 125
β(31) = 4(31) + 1 = 124 + 1 = 125
β΄ π(5) = β(31)
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d) β(π(2)) = 4(π(2)) + 1 = 4(23) + 1 = 4(8) + 1 = 33
Question 5
We are given π(π₯) = π₯π and π(3) = 81, so we substitute 3 into the
function to determine the value of π:
π(3) = 3π = 81 = 34
So π = 4.
Question 6
We are given that π(π₯) = ππ₯ + π and that π(2) = 7and π(3) = 12, so we
substitute 2 and 3 into the function and solve the two resulting equations
simultaneously to obtain the values of π and π:
π(2) = 2π + π = 7 (1)
π(3) = 3π + π = 12 (2)
(1) β (2):
2π + π β 3π β π = 7 β 12
βπ = β5
π = 5 (3)
(3) β (1):
2(5) + π = 7
π = 7 β 10 = β3 (4)
So the function is π(π₯) = 5π₯ β 3.
Question 7
The largest possible domain of a function is the largest set of numbers
for which the function is defined.
a) The function βπ₯ is defined for all positive real numbers π₯ and zero.
To put it the other way around, the square root of a negative
number is undefined. So the largest possible domain of the
function is the set of all real numbers π₯ such that π₯ β₯ 0.
e) Since the square root of a negative number is undefined, the
largest possible domain of the function is the set of all real
numbers π₯ such that
π₯(π₯ β 4) β₯ 0
Now the product of two real numbers is positive if either both
numbers are positive or both numbers are negative. So we have
π₯ β₯ 0 πππ π₯ β 4 β₯ 0 ππ π₯ β€ 0 πππ π₯ β 4 β€ 0
π₯ β₯ 0 πππ π₯ β₯ 4 ππ π₯ β€ 0 πππ π₯ β€ 4
Note that if π₯ β₯ 0 πππ π₯ β₯ 4, then π₯ must be greater than or equal
to 4. Similar reasoning applies to the statement after βORβ above.
Thus the solution of the above inequalities is
π₯ β₯ 4 ππ π₯ β€ 0
So the largest domain of this function is the set of all real numbers
π₯ such that π₯ β€ 0 ππ π₯ β₯ 4.
h) Again, we are looking for the set of values for π₯ that always give a
positive number under the square root:
π₯3 β 8 β₯ 0
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π₯3 β₯ 8
π₯ β₯ 2
So the domain of the function is the set of all real numbers π₯ such
that π₯ β₯ 2.
i) Here the operation is defined unless the denominator is zero β
division by zero is undefined. So we find the value of π₯ where the
denominator is zero:
π₯ β 2 = 0
π₯ = 2
So the domain is the set of all real numbers except 2.
j) Since we are taking the square root π₯ β 2 cannot be less than
zero. However since we are also dividing by βπ₯ β 2, (π₯ β 2)
cannot be zero. So we have:
π₯ β 2 > 0
π₯ > 2
So the domain of this function is the set of real numbers such that
π₯ > 2.
k) Since the square root of a negative number is undefined, π₯ has to
be positive or zero. But, since division by zero is undefined the
denominator cannot be zero. However, since the square root of
any positive number is positive, the denominator will always be
positive. Thus the domain of this function is the set of all real
numbers π₯ such that π₯ β₯ 0.
l) Since division by zero is undefined we need to find those values
of π₯ that would make the denominator zero.
(π₯ β 1)(π₯ β 2) = 0
π₯ = 1 ππ π₯ = 2
So the domain is the set of all real numbers except 1 and 2.
Question 8
a) We are given that the domain is the set of all positive real numbers.
Now clearly the function reaches its minimum value on the given
domain where π₯ goes to zero. Also it is clear that there is no
restriction on the maximum value of the function on this domain.
Since π(0) = 7, the range of this function is π(π₯) > 7.
(Note: zero is excluded in the domain, so the range of π(π₯) is a
strict inequality.)
c) This is the same as question (a): the function reaches its minimum
value on the given domain as π₯ goes to zero and there is no
restriction on its maximum value on this domain. So the range of
this function is π(π₯) > β1.
d) Since π₯2 is always positive, the minimum of this function is β1
where π₯ = 0. There is no restriction on the maximum value on this
domain. So the range is π(π₯) > β1.
f) The term (π₯ β 1)2 is the square of a real number, so it is always
positive. So the minimum of the function is 2 which occurs where
(π₯ β 1)2 = 0 (i.e. where π₯ = 1). So the range is π(π₯) β₯ 2.
(Note: since π₯ = 1 is included in the domain, π(π₯) = 2 is included
in the range.)
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Question 9
a) Since π₯2 is always positive, the lowest value of this function is 4,
which occurs when π₯ = 0. As π₯ increases in the positive direction
and in the negative direction the value of the function also
increases. So the range is π(π₯) β₯ 4.
c) Similarly to question (a), the minimum value of (π₯ β 1)2 is zero
since (π₯ β 1)2 is the square of a real number. (This minimum value
occurs when π₯ = 1.) Therefore, π(π₯) has a minimum of 6, and the
range is π(π₯) β₯ 6.
d) In the same way, the term (1 β π₯)2 has a minimum of zero when
π₯ = 1; for other values of π₯, (1 β π₯)2 > 0. Therefore β(1 β π₯)2 has
a maximum of zero where π₯ = 1, and is negative elsewhere. Thus,
π(π₯) has a maximum of 7 and the range is π(π₯) β€ 7.
f) The term 2(π₯ + 2)4 has a minimum of zero where π₯ = β2 and for
all other values of π₯ it is greater than zero. So the minimum of the
function is -1 and the range is π(π₯) β₯ β1.
Question 10
b) This function is a straight line with a negative gradient. So to find
the range of the function, we evaluate the function at the end
points of the given domain:
π(β2) = 3 β 2(β2) = 3 + 4 = 7
π(2) = 3 β 2(2) = 3 β 4 = β1
So the range of π on the given domain is β1 β€ π(π₯) β€ 7.
c) π₯2 is always greater than or equal to zero, so this function reaches
its minimum value of zero where π₯ = 0 which is included in the
given domain. The greatest value of the function on this domain is
π(4) = 42 = 16. So the range of the function on this domain is
0 β€ π(π₯) β€ 16.
(Note: in this case we cannot simply evaluate the function at the
end points of the domain since π(β1) = (β1)2 = (1)2 = 1 and
π(0) = 0 < π(β1). So here the minimum of the function does not
occur at the left end point of the domain.)
Question 11
a) Since the power of π₯ is even, π(π₯) is greater than or equal to zero
for all real numbers π₯. So its minimum value is 0 where π₯ = 0. So
the range is π(π₯) β₯ 0.
c) Here the largest possible domain is the set of all real numbers, π₯
excluding zero. In the expression π₯3, the power of π₯ is odd, so the
value π₯3 ranges over all real numbers. So the function π(π₯) =1
π₯3
can be either positive or negative but it excludes zero. So the
range is π(π₯) < 0 ππ π(π₯) > 0.
e) The term π₯4 is greater than or equal to zero for all values of π₯. So
the minimum of this function is 5 and there is no restriction on its
maximum. So the range is π(π₯) β₯ 5.
g) The largest possible domain for this function is the set of all real
numbers such that
4 β π₯2 β₯ 0
π₯2 β€ 4
π₯ β₯ β2 πππ π₯ β€ 2
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Now π(Β±2) = β4 β 4 = 0 and π(0) = β4 β 0 = 2. So the range of
the function is 0 β€ π(π₯) β€ 2.
Question 12
Start by drawing a diagram of the rectangle. Here we have labelled the
width π€ and the height β.
Since the length of the wire is 24 ππ, we have a condition on the
perimeter of the rectangle:
2π€ + 2β = 24
2β = 24 β 2π€
β = 12 β π€ (1)
So we can find an expression for the area:
π΄ = π€β = π€(12 β π€)
= 12π€ β π€2 (2)
We can prove that this is equal to 36 β (6 β π€)2 by rearranging the latter
expression:
π΄ = 36 β (6 β π€)2
= 36 β (36 β 12π€ + π€2)
= 12π€ β π€2
Therefore:
π΄ = 36 β (π€ β 6)2 (3)
In order to form a rectangle, both π€ and β have to be greater than zero.
From eqn (1) we see that this requires that π€ < 12. So the domain of the
function is 0 < π€ < 12. The term (π€ β 6)2 is zero where π€ = 6. So the
maximum of π΄ is 36 where π€ = 6 and the minimum is zero where π€ =
0 ππ 12. So the range is 0 < π΄ β€ 36.
(Note the strict inequality on the left-hand side of the range since the
domain does not include 0 or 12 and the weak inequality on the right-
hand side since the domain does include 6.)
Question 13
If you expand the expression for π¦, you will find that the highest power
of π₯ is 3. So the graph has the shape of a cubic function.
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The height, length and width of a box have to be greater than zero. This
leads to three conditions on π₯:
π₯ > 0
22 β 2π₯ > 0 β π₯ < 11
8 β 2π₯ > 0 β π₯ < 4
So an appropriate domain for this function is the set of all real numbers
π₯ such that 0 < π₯ < 4.
EXERCISE 3B
Question 1
The four graphs have been drawn on the same axes to show the relative
characteristics of these functions.
From the graph above we can note the following characteristics of
functions of the form π(π₯) = π₯π where π is a positive integer:
1. Even powers of π₯ are always positive while odd powers of π₯ have
the same sign as π₯.
2. All the graphs pass through the point (1,1).
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3. In the region β1 < π₯ < 1, for the higher the powers of π₯, π(π₯) is
closer to the π₯-axis. (I.e. the absolute value of π(π₯) is smaller.)
4. In the regions π₯ < β1 and π₯ > 1, the higher the power of π₯ the
more quickly the function goes to infinity or negative infinity. (I.e.
the graph is steeper.)
Question 2
To find the order in which the vertical line, π₯ = π, meets the three graphs,
letβs substitute the value of π into the three functions:
a) π = 2:
π: π¦ = (2)β2 =1
22=
1
4
π: π¦ = (2)β3 =1
23=
1
8
π: π¦ = (2)β4 =1
24=
1
16
So the points at which the vertical line π₯ = 2 meets the graphs are
in the following order: π , π, π.
b) π =1
2:
π: π¦ = (1
2)
β2
= 22 = 4
π: π¦ = (1
2)
β3
= 23 = 8
π: π¦ = (1
2)
β4
= 24 = 16
So the points at which the vertical line π₯ =1
2 meets the graphs are
in the following order: π, π, π .
c) π = β1
2:
π: π¦ = (β1
2)
β2
= (β2)2 = 4
π: π¦ = (β1
2)
β3
= (β2)3 = β8
π: π¦ = (β1
2)
β4
= (β2)4 = 16
So the points at which the vertical line π₯ = β1
2 meets the graphs
are in the following order: π, π, π .
d) π = β2:
π: π¦ = (β2)β2 =1
(β2)2=
1
4
π: π¦ = (β2)β3 =1
(β2)3= β
1
8
π: π¦ = (β2)β4 =1
(β2)4=
1
16
So the points at which the vertical line π₯ = β2 meets the graphs
are in the following order: π, π , π.
From the characteristics of functions of the form π₯π where π is a
positive integer listed in question 1 and the observations above,
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we can work out the characteristics of functions of the form
π(π₯) = π₯βπ where π is a positive integer:
1. If π is even then π(π₯) is symmetric about the π¦-axis; if π is
odd then π(π₯) is symmetric about the origin. (That is when π₯
is negative, if π is even the graph is above the π₯-axis and if
π is odd the graph is below the π₯-axis.
2. The function π(π₯) always passes through the point (1,1).
3. The function is always undefined at π₯ = 0. For any π the
function goes to positive infinity as π₯ goes to zero from the
right-hand side. If π is even, then the function goes to
positive infinity as π₯ goes to zero from the left-hand side and
if π is odd, then the function goes to negative infinity as π₯
goes to zero from the left.
4. As π₯ goes to either positive or negative infinity, the function
goes to zero.
5. In the region 0 < π₯ < 1, if π and π are positive integers and
π > π, then π₯π < π₯π. So 1
π₯π >1
π₯π. That is the graph of π₯βπ
lies below the graph of π₯βπ in this region.
6. In the region π₯ > 1, if π and π are positive integers and π >
π, then π₯π > π₯π. So 1
π₯π <1
π₯π. That is the graph of π₯βπ lies
above the graph of π₯βπ in this region.
7. In the region β1 < π₯ < 0, if π and π are positive integers
and π > π, then |1
π₯π| > |1
π₯π|. That is the graph π₯βπ is closer
to the π₯-axis than the graph of π₯βπ.
8. In the region π₯ < β1, if π and π are positive integers and
π > π, then |1
π₯π| < |1
π₯π|. That is the graph π₯βπ is further from
the π₯-axis than the graph of π₯βπ.
These points are illustrated in the graph below:
Question 3
a)
0 < π₯β3 πππ π₯β3 < 0.001
(0)β13 < (π₯β3)β
13 πππ (π₯β3)β
13 < (10β3)β
13
π₯ > 0 πππ π₯ > 10
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β΄ π₯ > 10
(Notice that the inequality sign is reversed because each side of
the inequality is raised to a negative power. This is because, if
π₯ > π¦ and they both have the same sign, then 1
π₯<
1
π¦. Think about
it!)
d)
8π₯β4 < 0.000 05
π₯β4 <0.000 05
8
(π₯β4)β14 > (
0.00005
8)
β14
= 20
β΄ π₯ > 20
Question 4
The properties described in questions 1 and 2 apply here. Some other
graphs with integer powers of π₯ have been plotted for reference.
a)
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c)
d)
f)
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Question 5
a) This function can be rewritten as π¦ = βπ₯23. Since π₯2 is always
positive, this function is defined for π₯ < 0 and it is symmetric about
the π¦-axis.
c) This function can be rewritten as π¦ = βπ₯45. Again, since π₯4 is
always positive, this function is defined for π₯ < 0 and is symmetric
about the π¦-axis. It has a similar shape to the graph in question a
above, except that it goes to infinity more slowly.
d) Here the function can be written as π¦ =1
βπ₯3 . Since it involves the
cube root, it is defined for π₯ < 0. However, it is not symmetric
about the π¦-axis. It is an odd function, and symmetric about the
origin.
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f) This function can be written as π¦ =1
βπ₯3. This involves the square
root of a number that can be either positive or negative. So it does
not exist for π₯ < 0.
Question 6
a) To sketch this function, note that at π₯ = 1 both terms are equal to
1. Between π₯ = 0 and π₯ = 1, the term π₯β1 dominates the function
and the graph goes to plus infinity as π₯ goes to zero. To the right
of π₯ = 1, the term π₯2 dominates the function and the graph looks
like that of the function π₯2. At π₯ = β1, π₯2 = 1 and π₯β1 = β1. So at
π₯ = β1 the graph crosses the π¦-axis. Between π₯ = β1 and π₯ = 0,
the term π₯β1 again dominates the function and the graph to minus
infinity as π₯ goes to zero from the left. To the left of π₯ = β1, the
term π₯2 dominates the function and it again looks like the graph of
π₯2.
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b) At π₯ = 1 both terms are equal to 1 so π¦ = 2. At π₯ = β1 the first
term is equal to β1 while the second term is equal to 1 so π¦ = 0.
In the interval β1 < π₯ < 1, the term π₯β2 dominates the function
and the graph goes to plus infinity as π₯ goes to zero from the left
and the right. In the two regions π₯ < β1 and π₯ > 1, then the term
π₯ dominates the function and the graphs looks like that of a straight
line.
e) In the region β1 < π₯ < 1, the term π₯β3 is greater than the term
π₯β2. so in this region the graph of the function is similar to the
graph of βπ₯β3. At π₯ = 1, π¦ = 0, so the graph crosses the π₯-axis,
and at π₯ = β1, π¦ = 2. In the regions π₯ < β1 and π₯ > 1, the term
π₯β2 dominates the function so the graph is similar to that of π₯β2.
Question 7
A function π(π₯) is an even function if for all values of π₯, π(π₯) = π(βπ₯)
and a function is an odd function if for all values of π₯, π(π₯) = βπ(π₯).
a) This function is odd since π₯ is raised to an odd power. We can also
see this from the definition of an odd function:
π¦(βπ₯) = (βπ₯)7 = β(π₯)7 = βπ¦(π₯)
b) This function is even since the terms involve π₯ raised to even
powers:
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π¦(βπ₯) = (βπ₯)4 + 3(βπ₯)2 = π₯4 + 3π₯2 = π¦(π₯)
c) This function is odd. It can be rewritten as π¦ = π₯3 β π₯ which shows
that the two terms involve π₯ raised to odd powers:
π¦(βπ₯) = (βπ₯)3 β (βπ₯) = β(π₯)3 β (βπ₯) = β(π₯3 β π₯) = π¦(βπ₯)
Question 8
a) |β7| = 7
b) |β1
200| =
1
200= 0.005
c) |9 β 4| = |5| = 5
d) |4 β 9| = |β5| = 5
e) |π β 3| = (π β 3) β΅ π > 3
f) |π β 4| = (4 β π) β΅ π < 4
Note: parts (c) and (d) demonstrate that the modulus (π₯ β π) gives the
distance between these two numbers on the real number line. That is
why |π₯ β π| = |π β π₯|. This is also the reason why if you have actual
numbers, you can rewrite the expression without taking the modulus by
examining which number is greater as in the parts (e) and (f).
Question 9
a) |2 β 22| = |2 β 4| = |β2| = 2
b) |1
2β (
1
2)
2| = |
1
2β
1
4| = |
1
4| =
1
4
c) |1 β 12| = |1 β 1| = 0
d) |β1 β (β1)2| = |β1 β 1| = |β2| = 2
e) |0 β 02| = |0| = 0
Question 10
a) Here we are given that |π₯| > 100, so π₯2 > (100)2. Therefore
0 < π¦ <1
(100)2
Since π¦ = π₯β2, π₯ is raised to a negative power, the inequality sign
is reversed. Since it is an even power of π₯, π¦ will be positive (i.e.
greater than zero). Simplifying the inequality gives
0 < π¦ < 0.0001
b) Here we are given that |π₯| < 0.01. Again the inequality sign is
reversed:
π¦ >1
0.012
π¦ > 10 000
We do not need to state that π¦ > 0 because π¦ > 10000 implies that
π¦ > 0.
Question 11
a) Here we are given that |π₯| < 1000 and that π¦ = π₯β3. We are
dealing with an odd power of π₯, so we have to be more careful in
our working.
Note that |π₯| < 1000 can be written as
β1000 < π₯ < 1000
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(Think about this carefully: the distance of π₯ from zero has to be
less than 1000 but π₯ can be either to the right or to the left of zero.)
Now we must consider two cases: Case I, where π₯ < 0 (i.e π₯ is
negative), and Case II, where π₯ > 0, i.e. π₯ is positive.
Case I (π > βππππ and π is negative):
β1000 < π₯ πππ π₯ < 0
β1
1000>
1
π₯ πππ
1
π₯< 0
Note that the first inequality is reversed because have inverted
both sides and we are dealing with non-zero numbers (test this out
with a few actual numbers). We cannot invert the second
inequality because that would involve dividing by zero. But we can
say that if π₯ is negative, then 1
π₯ is also negative, i.e.
1
π₯< 0. Now this
gives that
1
π₯3< β
1
10003 πππ
1
π₯3< 0
β΄ π¦ < β1
109 πππ π¦ < 0
which can be written simply as
π¦ < β1
109
Case II (π > ππππ and π is positive):
π₯ < 1000 πππ π₯ > 0
1
π₯>
1
1000 πππ
1
π₯> 0
1
π₯3>
1
109 πππ
1
π₯3> 0
β΄ π¦ >1
109
These inequalities mean that π¦ is further away from zero than the
points β10β9 and 10β9. In other words, the first inequality states
that π¦ is to the left of β10β9 and the second inequality states that
π¦ is to the right of 10β9. (Draw this on a number line, if you are
struggling to visualize this.) Now these two inequalities can be
stated in a single statement using the modulus:
|π¦| > 10β9
b) Now we are given that |π¦| > 1000. We follow the same method as
part (a):
|π¦| > 1000 can be stated as
π¦ < β1000 ππ π¦ > 1000
Once again, we must consider two cases, π¦ < β1000 and
π¦ > 1000.
Case I (π > ππππ):
π¦ > 1000
βΉ 0 <1
π¦<
1
1000(1)
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It is essential to include the proviso that 1
π¦> 0 because we are
given that π¦ > 1000, meaning that π¦ is positive. If we did not
include 0 <1
π¦ the statement
1
π¦<
1
1000 would allow for negative
values of 1
π¦, and these would not satisfy the original condition that
π¦ > 1000.
Now we have that π¦ =1
π₯3, so π₯ =1
π¦13
. We have from (1):
0 <1
π¦13
<1
100013
0 < π₯ <1
10(2)
Case II (π < βππππ):
π¦ < β1000
0 >1
π¦> β
1
1000(3)
(Note again that we must add 1
π¦< 0, i.e. negative.)
From (3) we get
β1
100013
<1
π¦13
< 0
β1
10< π₯ < 0 (4)
Now we can combine (2) and (4) to give
β1
10< π₯ <
1
10
|π₯| <1
10
Question 12
If π is reported as 37 000 to the nearest thousand, it means that π lies
between 36 500 and 37 500, which can be stated as
36 500 < π < 37 500
This can be understood as a statement that the distance from π to
37 000 cannot be more than 500. Using the modulus function, this can
be written as
|π β 37 000| < 500
Question 13
The difference between the marks is less than or equal to 5 regardless
of which twin receives the higher mark. This is easily stated using a
modulus sign as
|π β π| β€ 5
Question 14
This question the same as question 12 except that it is asked the other
way around.
|π₯ β 5.231| < 0.005
means that the length π₯ of the line is within 0.005 cm of 5.231.
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EXERCISE 3C
Question 1
Notice that these functions only differ in the value of π.
Question 2
These graphs again show how varying the value of π changes the shape
of the graph.
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Question 3
The given graph has π = 0 so it crosses the π¦-axis at the origin. Graph
a has π = 4 so it will lie above the given graph. Graph b has π = β6 so
it lies below the given graph.
Question 4
The questions above demonstrate that the effect of changing the value
of π is to shift the graph along the π¦-axis. The value of π also determines
the π¦-intercept of the graph.
Question 5
The following graphs differ only in the value of for π.
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Question 6
The figure above shows the graph of π¦ = 2π₯2 + ππ₯ + 4 for four different
values of π. This and question 5 show that the effect of changing the
value of π is to shift the axis of symmetry of the graph along the π₯-axis.
If π and π have the same sign, then the axis of symmetry is to the left of
the origin and if π and π have the opposite sign axis of symmetry is to
the right of the origin.
Question 7
Here the graphs all have different values for π:
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Question 8
Again, these graphs only differ in the value of π:
Question 9
The value of π determines how elongated the graph is. The greater the
value of |π| the more the graph is elongated or lengthened in the π¦-
direction. Furthermore if π is positive the vertex is at the minimum point
of the graph and if π is negative the vertex is at the maximum point of
the graph.
Question 10
In this graph the vertex is at the minimum point and the axis of symmetry
is to the left of the π¦-axis. This requires that π is positive and that π and
π have the same sign. So only the curve in (c) could be the equation of
the curve shown in the diagram.
Question 11
Here the vertex is at the maximum point of the curve and the axis of
symmetry is to the right of the π¦-axis. So this requires that π is negative
and that π and π have the opposite sign. Only the curve in (a) satisfies
these requirements.
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Worked solutions to Pure Mathematics 1: Coursebook, by Neill, Quadling & Gilbey Exercise 3D
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EXERCISE 3D
Question 1
a)
π₯ = 3 (1)
π¦ = π₯2 + 4π₯ β 7 (2)
(1) β (2):
π¦ = (3)2 + 4(3) β 7
π¦ = 9 + 12 β 7
π¦ = 14 (3)
So the curve π¦ = π₯2 + 4π₯ β 7 intersects the vertical line π₯ = 3 at
the point (3, 14).
d)
π¦ + 3 = 0 (1)
π¦ = 2π₯2 + 5π₯ β 6 (2)
Rearrange eqn (1):
π¦ = β3 (3)
(3) β (2):
β3 = 2π₯2 + 5π₯ β 6
2π₯2 + 5π₯ β 3 = 0
(2π₯ β 1)(π₯ + 3) = 0
π₯ =1
2 ππ π₯ = β3 (4)
The horizontal line π¦ = β3 intersects the curve π¦ = 2π₯2 + 5π₯ β 6
at the points (1
2, β3) and (β3, β3).
Question 2
a)
π¦ = π₯ + 1 (1)
π¦ = π₯2 β 3π₯ + 4 (2)
(1) β (2):
π₯ + 1 = π₯2 β 3π₯ + 4
π₯2 β 4π₯ + 3 = 0
(π₯ β 3)(π₯ β 1) = 0
π₯ = 3 ππ π₯ = 1 (3)
At π₯ = 3,
π¦ = 3 + 1 = 4
At π₯ = 1,
π¦ = 1 + 1 = 2
So the line π¦ = π₯ + 1 intersects the curve π¦ = π₯2 β 3π₯ + 4 at the
points (3, 4) and (1, 2).
c)
π¦ = 3π₯ + 11 (1)
π¦ = 2π₯2 + 2π₯ + 5 (2)
(1) β (2):
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3π₯ + 11 = 2π₯2 + 2π₯ + 5
2π₯2 β π₯ β 6 = 0
(2π₯ + 3)(π₯ β 2) = 0
π₯ = β3
2 ππ π₯ = 2
At π₯ = β3
2,
π¦ = 3 (β3
2) + 11 = β
9
2+ 11 =
13
2
At π₯ = 2
π¦ = 3(2) + 11 = 6 + 11 = 17
So the curve π¦ = 2π₯2 + 2π₯ + 5 intersects the line π¦ = 3π₯ + 11 at
the points (β3
2,
13
2) and (2, 17).
e)
3π₯ + π¦ β 1 = 0 (1)
π¦ = 6 + 10π₯ β 6π₯2 (2)
Rearrange eqn (1):
π¦ = β3π₯ + 1 (3)
(3) β (2):
β3π₯ + 1 = 6 + 10π₯ β 6π₯2
6π₯2 β 13π₯ β 5 = 0
(3π₯ + 1)(2π₯ β 5) = 0
π₯ = β1
3 ππ π₯ =
5
2
At π₯ = β1
3,
π¦ = β3 (β1
3) + 1 = 1 + 1 = 2
At π₯ =5
2,
π¦ = β3 (5
2) + 1 = β
15
2+ 1 = β
13
2
So the line π¦ = β3π₯ + 1 intersects the curve π¦ = 6 + 10π₯ β 6π₯2 at
the points (β1
3, 2) and (
5
2, β
13
2).
Question 3
a)
π¦ = 2π₯ + 2 (1)
π¦ = π₯2 β 2π₯ + 6 (2)
(1) β (2):
2π₯ + 2 = π₯2 β 2π₯ + 6
π₯2 β 4π₯ + 4 = 0
(π₯ β 2)(π₯ β 2) = 0
π₯ = 2
At π₯ = 2,
π¦ = 2(2) + 2 = 6
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So there is only one point that simultaneously satisfies equations
1 and 2. So the line π¦ = 2π₯ + 2 and the curve π¦ = π₯2 β 2π₯ + 6
meet only at the point (2, 6).
b)
π¦ = β2π₯ β 7 (1)
π¦ = π₯2 + 4π₯ + 2 (2)
(1) β (2):
β2π₯ β 7 = π₯2 + 4π₯ + 2
π₯2 + 6π₯ + 9 = 0
(π₯ + 3)(π₯ + 3) = 0
π₯ = β3
At π₯ = β3,
π¦ = β2(β3) β 7 = 6 β 7 = β1
Again, there is only one point that satisfies both equations
simultaneously. Thus, the line and the curve meet at only one
point: (β3, 1).
Question 4
a)
π¦ = π₯ (1)
π¦ = π₯2 β π₯ (2)
(1) β (2):
π₯ = π₯2 β π₯
π₯2 β 2π₯ = 0
π₯(π₯ β 2) = 0
π₯ = 0 ππ π₯ = 2
At π₯ = 0:
π¦ = 0
At π₯ = 2:
π¦ = 2
The two points of intersection are (0, 0) and (2, 2).
b)
π¦ = π₯ β 1 (1)
π¦ = π₯2 β π₯ (2)
(1) β (2):
π₯ β 1 = π₯2 β π₯
π₯2 β 2π₯ + 1 = 0
(π₯ β 1)(π₯ β 1) = 0
π₯ = 1
At π₯ = 1:
π¦ = 1 β 1 = 0
So the only point of intersection is (1, 0).
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The curve π¦ = π₯2 β π₯ along with the two lines, π¦ = π₯ and
π¦ = π₯ β 1, are plotted below:
Question 5
a)
π¦ = β3π₯ + 2 (1)
π¦ = π₯2 + 5π₯ + 18 (2)
(1) β (2):
β3π₯ + 2 = π₯2 + 5π₯ + 18
π₯2 + 8π₯ + 16 = 0
(π₯ + 4)(π₯ + 4) = 0
π₯ = β4
At π₯ = β4:
π¦ = β3(β4) + 2
π¦ = 12 + 2 = 14
So the line and curve intersect at (β4, 14).
b)
π¦ = β3π₯ + 6 (1)
π¦ = π₯2 + 5π₯ + 18 (2)
(1) β (2):
β3π₯ + 6 = π₯2 + 5π₯ + 18
π₯2 + 8π₯ + 12 = 0
(π₯ + 2)(π₯ + 6) = 0
π₯ = β2 ππ π₯ = β6
At π₯ = β2:
π¦ = β3(β2) + 6 = 12
At π₯ = β6:
π¦ = β3(β6) + 6 = 24
So the line and the curve meet at the points (β2, 12) and (β6, 24).
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The curve and the two lines are plotted below:
Question 6
a)
π¦ = π₯ + 5 (1)
π¦ = 2π₯2 β 3π₯ β 1 (2)
(1) β (2):
π₯ + 5 = 2π₯2 β 3π₯ β 1
2π₯2 β 4π₯ β 6 = 0
(2π₯ β 6)(π₯ + 1) = 0
π₯ = 3 ππ π₯ = β1
At π₯ = 3:
π¦ = 3 + 5 = 8
At π₯ = β1:
π¦ = β1 + 5 = 4
So the curve and the line intersect at the points (3, 8) and (β1, 4).
b)
π¦ = π₯ + 5 (1)
π¦ = 2π₯2 β 3π₯ + 7 (2)
(1) β (2):
π₯ + 5 = 2π₯2 β 3π₯ + 7
2π₯2 β 4π₯ + 2 = 0
2(π₯2 β 2π₯ + 1) = 0
2(π₯ β 1)(π₯ β 1) = 0
π₯ = 1
At π₯ = 1:
π¦ = 1 + 5 = 6
The curve and the line at the point (1, 6).
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The curve and the two lines are plotted below:
Question 7
a)
π¦ = π₯2 + 5π₯ + 1 (1)
π¦ = π₯2 + 3π₯ + 11 (2)
(1) β (2):
π₯2 + 5π₯ + 1 = π₯2 + 3π₯ + 11
2π₯ β 10 = 0
π₯ = 5
At π₯ = 5
π¦ = (5)2 + 3(5) + 11
π¦ = 25 + 15 + 11
π¦ = 51
So the two curves intersect at the point (5, 51).
c)
π¦ = 7π₯2 + 4π₯ + 1 (1)
π¦ = 7π₯2 β 4π₯ + 1 (2)
(1) β (2):
7π₯2 + 4π₯ + 1 = 7π₯2 β 4π₯ + 1
8π₯ = 0
π₯ = 0
At π₯ = 0:
π¦ = 7(0)2 + 4(π₯) + 1 = 1
So these two curves intersect at (0, 1).
Question 8
a)
π¦ =1
2π₯2 (1)
π¦ = 1 β1
2π₯2 (2)
(1) β (2):
1
2π₯2 = 1 β
1
2π₯2
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π₯2 = 1
π₯ = 1 ππ π₯ = β1
At π₯ = 1:
π¦ =1
2(1)2 =
1
2
At π₯ = β1
π¦ =1
2(β1)2 =
1
2
So the curves intersect at (1,1
2) and (β1,
1
2).
c)
π¦ = π₯2 + 7π₯ + 13 (1)
π¦ = 1 β 3π₯ β π₯2 (2)
(1) β (2)
π₯2 + 7π₯ + 13 = 1 β 3π₯ β π₯2
2π₯2 + 10π₯ + 12 = 0
(2π₯ + 6)(π₯ + 2) = 0
π₯ = β3 ππ π₯ = β2
At π₯ = β3
π¦ = 1 β 3(β3) β (β3)2 = 1 + 9 β 9 = 1
At π₯ = β2
π¦ = 1 β 3(β2) β (β2)2 = 1 + 6 β 4 = 3
So the curves intersect at (β3, 1) and (β2, 3).
e)
π¦ = (π₯ β 2)(6π₯ + 5) (1)
π¦ = (π₯ β 5)2 + 1 (2)
Multiply out the factors in each expression so that it is in the form
ππ₯2 + ππ₯ + π:
π¦ = 6π₯2 β 12π₯ + 5π₯ β 10
π¦ = 6π₯2 β 7π₯ β 10 (3)
π¦ = π₯2 β 10π₯ + 25 + 1
π¦ = π₯2 β 10π₯ + 26 (4)
(4) β (3):
π₯2 β 10π₯ + 26 = 6π₯2 β 7π₯ β 10
5π₯2 + 3π₯ β 36 = 0
(5π₯ β 12π₯)(π₯ + 3) = 0
π₯ =12
5 ππ π₯ = β3
At π₯ =12
5:
π¦ = (12
5)
2
β 10 (12
5) + 26
π¦ =144
25β 24 + 26
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π¦ =144
25+ 2 =
194
25
At π₯ = β3,
π¦ = (β3)2 β 10(β3) + 26
π¦ = 9 + 30 + 26
π¦ = 65
So the curves intersect at the points (12
5,
194
25) and (β3, 65).
Question 9
a)
π¦ = 8π₯2 (1)
π¦ = 8π₯β1 (2)
(2) β (1):
8
π₯= 8π₯2
8π₯3 = 8
π₯3 = 1
π₯ = 1 (3)
(3) β (1):
π¦ = 8(1)2 = 8
So the curves intersect at the point (1, 8) as illustrated in the graph
below:
c)
π¦ = π₯ (1)
π¦ = 4π₯β3 (2)
(1) β (2):
π₯ = 4π₯β3
π₯π₯3 = 4
π₯4 = 4
π₯ = β2 ππ π₯ = ββ2 (3)
At π₯ = β2
π¦ = β2
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At π₯ = ββ2
π¦ = ββ2
So the line π¦ = π₯ intersects the curve π¦ = 4π₯β3 at the points
(β2, β2) and (ββ2, ββ2).
e)
π¦ = 9π₯β3 (1)
π¦ = π₯β5 (2)
(2) β (1):
π₯β5 = 9π₯β3
π₯3π₯β5 = 9
π₯β2 = 9
π₯ = 9β12
π₯ = Β±1
3
At π₯ =1
3
π¦ = (1
3)
β5
= 35 = 243
At π₯ = β1
3
π¦ = (β1
3)
β5
= (β3)5 = β243
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So the curves intersect at the points (1
3, 243) and (β
1
3, β243), as
illustrated in the graph below:
EXERCISE 3E
Question 1
a) First, we need to find the values of π₯ when π¦ = 0:
(π₯ β 2)(π₯ β 4) = 0
π₯ = 2 ππ π₯ = 4
So the graph passes through the points (2, 0) and (4, 0). By
inspection it is easy to see that in the expanded expression the
coefficient π₯2 will be +1. So the graph faces upwards and is not
elongated. Furthermore, the graph intersects the π¦-axis at (0, 8).
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c) This graph passes through the points (0,0) and (2,0). It also faces
upwards and crosses the π¦- axis at 0:
e) This graph passes through the points (0,0) and (β3,0) and also
faces upwards:
Question 2
a) This graph passes through the points (β1, 0) and (5, 0) and the π¦-
intercept is (0, β15). It faces upwards and is elongated since the
coefficient of π₯2 is 3:
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c) This graph passes through the points(β3,0) and (β5,0) and it
crosses the π¦-axis at (0, β15). Since the coefficient of π₯2 is β1, it
faces downwards but is not elongated:
e) This graph only touches the π₯-axis at one point (4,0) which has to
be the vertex of the parabola. Because of the coefficient of π₯2 is
β3, the graph faces downwards and is elongated. It crosses the
π¦-axis at (0, β48):
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Question 3
a) We first factorise the function:
π¦ = (π₯ β 4)(π₯ + 2)
So the graph passes through (β2,0) and (4,0) and the π¦-intercept
is at (0, β8). Since the coefficient of π₯2 is 1 the graph faces
upwards and is not elongated:
d)
π¦ = 2π₯2 β 7π₯ + 3
π¦ = 2 (π₯2 β7
2π₯ +
3
2)
π¦ = 2(π₯ β 3) (π₯ β1
2)
So the graph passes through the points (1
2, 0) and (3,0) and
crosses the π¦-axis at (0,3). It faces upwards an is somewhat
elongated:
g)
π¦ = βπ₯2 β 4π₯ β 4
π¦ = β(π₯2 + 4π₯ + 4)
π¦ = β(π₯ + 2)(π₯ + 2)
Here the graph touches the π₯-axis at the point (β2,0) and crosses
the π¦-axis at (0, β4). It faces downwards and is not elongated:
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Question 4
a) Since the we know the values of π₯ at which π¦ is zero we can work
out the factored form of the equation:
π¦ = (π₯ β 2)(π₯ β 5)
Now we can multiply it out to get the equation in the form
π¦ = ππ₯2 + ππ₯ + π.
π¦ = π₯2 β 5π₯ β 2π₯ + 10
π¦ = π₯2 β 7π₯ + 10
c)
π¦ = (π₯ + 5)(π₯ β 3)
π¦ = π₯2 β 3π₯ + 5π₯ β 15
π¦ = π₯2 + 2π₯ β 15
Question 5
a) This function has three factors so it is going to be a cubic. It
crosses the π₯-axis at the three points (β3,0), (2,0), and (3,0). The
constant 1 in front of the factors is implied and since it is positive
the graph goes to positive infinity as π₯ becomes very large (i.e. the
graph lies in the first quadrant as π₯ becomes very large). Since the
graph goes to positive infinity for very large π₯, it comes from
negative infinity for very small π₯:
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c) Here the first factor π₯ is squared, so at the point π₯ = 0, the graph
touches the π₯-axis so that the π₯-axis is tangent to the graph at that
point. From the other factor (π₯ β 4) we know that the graph
crosses the π₯-axis at the point (4,0). Again the coefficient before
the factors is +1 so the graph goes to positive infinity as π₯ goes to
infinity and the graph goes to negative infinity as π₯ goes to
negative infinity:
e) This graph crosses the π₯-axis at the points (β6,0), (β4,0) and
(β2,0). It crosses the π¦-axis at β48. Because the coefficient before
the factors is β1, the graph is opposite to those above: as π₯ goes
to negative infinity, the graph goes to positive infinity and as π₯ goes
to infinity, the graph goes to negative infinity.
Question 6
a) Here we can again find the factored from of the equation of a
parabola that passes through the given points.
π¦ = (π₯ β 1)(π₯ β 5)
π¦ = π₯2 β 6π₯ + 5
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But we were given that the parabola crosses the π¦-axis at the point
(0,15), so we have to multiply our equation by 3 to obtain the
correct result:
π¦ = 3π₯2 β 18π₯ + 15
c) The equation of a parabola that passes through the given points
is
π¦ = (π₯ + 6)(π₯ + 2)
π¦ = π₯2 + 8π₯ + 12
Now we are given that the graph also passes through the point
(0, β6), so we need to multiply the equation we found by β1
2:
π¦ = β1
2π₯2 β 4π₯ β 6
e) The equation of a parabola that passes through (β10,0) and (7,0)
is
π¦ = (π₯ + 10)(π₯ β 7)
π¦ = π₯2 + 3π₯ β 70
But the graph must also pass through the point (8,90), so we need
to multiply the equation by a factor π such that the point (8,90) lies
on the curve:
π¦ = π(π₯2 + 3π₯ β 70)
90 = π(82 + 3(8) β 70)
90 = π(18)
π = 5
So the equation of the parabola is
π¦ = 5π₯2 + 15π₯ β 350
Question 7
c) We first factorise the given equation:
π¦ = β(π₯2 + 3π₯ β 18)
π¦ = β(π₯ + 6)(π₯ β 3)
So the graph passes through the points (β6,0),(3,0) and (0,18)
and it faces downwards:
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d) Again we start by factorising the given function:
π¦ = 2π₯2 β 9π₯ + 10
π¦ = 2 (π₯2 β9
2π₯ + 5)
π¦ = 2(π₯ β 2) (π₯ β5
2)
So the graph passes through the points (2,0), (5
2, 0) and (0.10)
and faces upwards:
Question 8
To answer the following questions, we will use the properties of
parabolas that were found in Exercise 3C.
a) If the value of π is positive then the parabola crosses the π¦-axis at
a positive value of π¦. So the following parabolas cross the π¦-axis
at a positive π¦-value: A (expand the given equation to see the
value of π), B, G, H.
b) The vertex is at the highest point of the graph if π is negative.
B, D, F (expand the given equation)
c) The vertex is to the left of the π¦-axis if π and π have the same sign:
F, G, H
d) A parabola passes through the origin if π is zero:
D
e) A parabola does not cross the π₯-axis at two separate points if its
two factors are the same in tis factored form:
G
f) A parabola has the π¦-axis as its axis of symmetry if π is zero:
I
g) The axis of symmetry of a parabola lies exactly half-way between
the points at which the parabola crosses the π₯-axis. If it only
touches the π₯-axis rather than crosses the π₯-axis then the axis of
symmetry is at the point at which the graph touches the π₯-axis:
B and E
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h) For the vertex of a parabola to lie in the fourth quadrant, the axis
of symmetry has to lie to the right of the π¦-axis and the graph has
to face upwards and it has to cross the π₯-axis:
A, C, E
Question 9
a) Here both points at which the graph crosses the π₯-axis lie to the
left of the π¦-axis. So the roots of the function are negative. The
graph also points upward so π has to be positive and π has to be
positive since the π¦-intercept is positive. Here we can choose π to
be 1 since the graph has no given scale. In factor form a suggested
equation is
π¦ = (π₯ + 4)(π₯ + 2)
So in expanded form the equation is
π¦ = π₯2 + 6π₯ + 8
e) This function is a cubic and it passes through the origin, so one of
it is factors is π₯. The other roots of the function are negative. The
constant in front of the factors has to be positive, since the graph
is in the first quadrant when π₯ is very large. A suggested equation
is
π¦ = π₯(π₯ + 2)(π₯ + 4)
MISCELLANEOUS EXERCISE 3
Question 1
π(π₯) = 7π₯ β 4
a)
π(7) = 7(7) β 4 = 49 β 4 = 45
π (1
2) = 7 (
1
2) β 4 =
7
2β 4 = β
1
2
π(β5) = 7(β5) β 4 = β35 β 4 = β39
b)
π(π₯) = 10
7π₯ β 4 = 10
7π₯ = 14
π₯ = 2
c)
π(π₯) = π₯
7π₯ β 4 = π₯
6π₯ = 4
π₯ =4
6=
2
3
d)
π(π₯) = 37
7π₯ β 4 = 37
7π₯ = 41
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π₯ =41
7
[Note: the answer for part (d) at the back of the text book is
incorrect.]
Question 2
π(π₯) = π(4)
π₯2 β 3π₯ + 5 = 42 β 3(4) + 5
π₯2 β 3π₯ + 5 = 9
π₯2 β 3π₯ β 4 = 0
(π₯ β 4)(π₯ + 1) = 0
π₯ = 4 ππ π₯ = β1
Thus, at π₯ = 4 and at π₯ = β1, π(π₯) = π(4).
Question 3
We are given that the curve passes between the point (2,200) and
(2,2000), that is when π₯ = 2, the value of π¦ is greater 200 and less than
2000. So we can write
200 < π¦(2) < 2000
200 < 2π < 2000
The given graph is not symmetrical about the π¦-axis but it is symmetrical
about the origin. So π has to be odd. Now by inspection we see that
27 = 128, 29 = 512, and 211 = 2048. So clearly π = 9.
Question 4
π¦ = π₯2 β 7π₯ + 5 (1)
π¦ = 1 + 2π₯ β π₯2 (2)
(2) β (1):
βπ₯2 + 2π₯ + 1 = π₯2 β 7π₯ + 5
2π₯2 β 9π₯ + 4 = 0
(2π₯ β 1)(π₯ β 4) = 0
π₯ =1
2 ππ π₯ = 4
At π₯ =1
2:
π¦ = (1
2)
2
β 7 (1
2) + 5
π¦ =1
4β
7
2+ 5
π¦ =7
4
At π₯ = 4:
π¦ = (4)2 β 7(4) + 5
π¦ = 16 β 28 + 5
π¦ = β7
So the two curves intersect at (1
2,
7
4) and (4, β7).
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Question 7
π¦ = (π₯ β 2)(π₯ β 4) (1)
π¦ = π₯(2 β π₯) (2)
Rearrange these equations:
π¦ = π₯2 β 6π₯ + 8 (3)
π¦ = 2π₯ β π₯2 (4)
(4) β (3):
2π₯ β π₯2 = π₯2 β 6π₯ + 8
2π₯2 β 8π₯ + 8 = 0
2(π₯2 β 4π₯ + 4) = 0
2(π₯ β 2)(π₯ β 2) = 0
π₯ = 2
At π₯ = 2:
π¦ = 2(2) β 22
π¦ = 4 β 4 = 0
So the two curves intersect at (2,0). This is illustrated in the graph below.
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Question 8
a) The graph passes through the points (βπ, 0), (2π, 0) and
(0, β2π2). The axis of symmetry lies exactly between the first two
points, where π₯ = βπ +3
2π. Now
3
2π > π, so for any value of π, the
axis of symmetry will lie to the right of the π¦-axis. The graph will
also faces upwards:
d) Since the factor (π₯ β 2π) is squared the graph does not cross the
π₯-axis at (2π, 0) but it only touches the π₯-axis at that point. (The π₯-
axis is tangent to the graph at this point.) It crosses the π₯-axis at
(βπ, 0) and the π¦-axis at (0, 4π3). Since π > 0 the π¦-intercpet is
always positive. The constant before all the factors is 1 so the
graph goes down to negative infinity for very small π₯ and to positive
infinity for very large π₯.
Question 9
We are given a function, π(π₯) = ππ₯2 + ππ₯ + π, and the value of the
function at three points: π(0) = 6, π(β1) = 15, and π(1) = 1. So we can
substitute the value of the function at these points into the expression
for the function and solve the three resulting equations to find the values
of π, π, and π:
π(0) = π(0)2 + π(π₯) + π = π = 6 (1)
π(β1) = π(β1)2 + π(β1) + π = 15 (2)
π(1) = π(1)2 + π(1) + π = 1 (3)
Starting with eqn (2):
π β π + π = 15
But from eqn (1) we have π = 6 so we get
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π β π + 6 = 15
π β π = 9 (4)
Now from eqns (1) and (3) we get
π + π + 6 = 1
π = β5 β π (5)
(5) β (4):
(β5 β π) β π = 9
β2π = 14
π = β7 (6)
(6) β (5):
π = β5 β (β7) = 2
So the function is π(π₯) = 2π₯2 β 7π₯ + 6.
Question 11
a) Notice that the given function is the sum of two parabolas. But, the
sum of two parabolas is still a parabola, so the easiest way to
graph this function is to expand and simplify the given expression
and then to factorise the resulting expression:
π¦ = (π₯ + 4)(π₯ + 2) + (π₯ + 4)(π₯ β 5)
π¦ = π₯2 + 6π₯ + 8 + π₯2 β π₯ β 20
π¦ = 2π₯2 + 5π₯ β 12
π¦ = (2π₯ β 3)(π₯ + 4)
Now it is easy to see that the graph intersects the π₯-axis at the points
(3
2, 0) and (β4, 0). The graph intersects the π¦-axis at (0, β12) and faces
upwards.
Question 13
Again, we can find three equations which we can use to solve for π, π,
and π:
At π₯ = β4 we have that π¦ = 0:
π(β4)2 + π(β4) + π = 0
16π β 4π + π = 0 (1)
At π₯ = 9, π¦ = 0:
π(9)2 + π(9) + π = 0
81π + 9π + π = 0 (2)
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At π₯ = 1, π¦ = 120:
π(1)2 + π(1) + π = 120
π + π + π = 120 (3)
Rearrange eqn (1):
π = 4π β 16π (4)
(4) β (2):
81π + 9π + 4π β 16π = 0
65π + 13π = 0
13(5π + π) = 0
5π + π = 0
π = β5π (5)
(5) β (4):
π = 4(β5π) β 16π
= β20π β 16π
= β36π (6)
(6) & (5) β (3):
π β 5π β 36π = 120
β40π = 120
π = β3 (7)
(7) β (5):
π = β5(β3) = 15
(7) β (6):
π = β36(β3) = 108
So the equation of the curve is π¦ = β3π₯2 + 15π₯ + 108. So the curve
crosses the π₯-axis at (0,108).
Question 14
We can solve for π, π, and π using the same method as in question 13:
π(β1)2 + π(β1) + π = 22
π β π + π = 22 (1)
π(1)2 + π(1) + π = 8
π + π + π = 8 (2)
π(3)2 + π(3) + π = 10
9π + 3π + π = 10 (3)
Rearrange eqn (1):
π = 22 + π β π (4)
(4) β (2):
22 + π β π + π + π = 8
2π = β14
π = β7 (5)
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(5) β (4):
π = 22 β 7 β π
π = 15 β π (6)
(6) & (5) β (3):
9(15 β π) + 3(β7) + π = 10
135 β 9π β 21 + π = 10
β8π = β104
π = 13 (7)
(7) β (6):
π = 15 β 13 = 2
So the equation of the curve is π¦ = 2π₯2 β 7π₯ + 13. The value of π is the
value of the curve at π₯ = β2:
π¦ = 2(β2)2 β 7(β2) + 13
π¦ = 8 + 14 + 13
π¦ = 35
So π = 35.
The possible values of π are those values of π₯ such that π¦ = 17:
17 = 2π₯2 β 7π₯ + 13
2π₯2 β 7π₯ β 4 = 0
(2π₯ + 1)(π₯ β 4) = 0
π₯ = β1
2 ππ π₯ = 4
So the possible values of π are β1
2 and 4.
Question 17
Let us first we find the points of intersection of the first two curves:
π¦ = π₯2 + 3π₯ + 14 (1)
π¦ = π₯2 + 2π₯ + 11 (2)
(1) β (2):
π₯2 + 3π₯ + 14 = π₯2 + 2π₯ + 11
π₯ + 3 = 0
π₯ = β3
At π₯ = β3:
π¦ = (β3)2 + 3(β3) + 14
π¦ = 9 β 9 + 14
π¦ = 14
So the first two curves intersect at only one point, (β3,14). Thus the third
curve also has to pass through this point. This enables us to find the
value of π:
14 = π(β3)2 + π(β3) + π
14 = 9π β 3π + π
14 = 7π
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π = 2
So the equation of the third curve is π¦ = 2π₯2 + 2π₯ + 2.
Question 23
π¦ = 2π₯2 β 7π₯ + 14 (1)
π¦ = 2 + 5π₯ β π₯2 (2)
(1) β (2):
2π₯2 β 7π₯ + 14 = 2 + 5π₯ β π₯2
3π₯2 β 12π₯ + 12 = 0
3(π₯2 β 4π₯ + 4) = 0
3(π₯ β 2)(π₯ β 2) = 0
π₯ = 2
At π₯ = 2:
π¦ = 2 + 5(2) β 22
π¦ = 2 + 10 β 4
π¦ = 8
So the two curves meet at only one point, (2,8).
a) Here the first curve is shifted down by two units while the second
curve remains the same. Now vertex of the second curve is at the
maximum point of the graph. So shifting the first curve down by
two units means that the two curves will now intersect at two
points.
b) Here the first curve remains the same while the second curve is
shifted down by one unit. Since the vertex of the first curve is at
the minimum point of the graph, shifting the second curve down
results in the two curves no longer intersecting.
c) Now both curves are shifted up by twenty units. Since there is no
relative change, the curves will still intersect at only one point.
Question 24
a) If π₯ > 0, then we know that |π₯| > 0 and that 1
π₯> 0 (the inequality
sign is not reversed here since we are dealing with zero and 1
divided by a positive number is still positive). Now π₯ and |π₯| have
the same magnitude so we can say that
|π₯|
π₯= 1
b) If π₯ < 0, we know that |π₯| > 0 and that 1
π₯< 0. Now a negative
number times a positive number is a negative number, so we can
say that
|π₯|
π₯= β1
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