1
FUNCTIONS
The idea of a function was developed in the 17th century. During that
time, Rene Descartes (1596 – 1650), in his book Geometry (1637), used
the idea of a function to describe many mathematical relationships.
The term ‘function’ was introduced by Gottfried Wilhelm Leibniz
(1646 – 1716) almost fifty years later after the publication of Geometry.
The concept of a function was further formalised by Leonhard Euler
(1707 – 1783) who introduced the notation of a function, that is,
y = f (x).
The idea of a function is very important in Mathematics because it
describes any situation in which one quantity depends on another. For
example, the distance of an object travels (at a constant speed) depends
on the time it travels. When such relationship exists, one variable is said
to be a function of the other. In this case, distance is a function of time.
Concept Map
Functions
Functions
• Relations and functions
• Graphs of algebraic functions
• Domain and codomain of functions
• Onto, one-to-one, odd and even functions
• Graphs of piecewise-defined functions
• Composite functions and their existence
• Inverse functions and their existence
Polynomials and Rational Functions
• Remainder theorem
• Factor theorem
• Partial fractions
• Solving inequalities
Exponential and Logarithmic Functions
Trigonometric Functions
• Graphs of exponential and logarithmic functions
• Laws of exponents and equations involving
exponential functions
• Laws of logarithms and equations and
inequalities involving logarithmic functions
• Graphs of trigonometric functions and inverse trigonometric
functions
• Trigonometric formulae
(d) Half-angle formulae
(a) Basic identities
(b) Compound-angle formulae (e) Multiple-angle formulae
(f) Factor formulae
(c) Double-angle formulae
• Trigonometric equations
• The t = tan q– substitution
2
• Equations in the form a sin q + b cos q
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2
ACE AHEAD Mathematics (T) First Term
1.1 Functions
Relations
A relation from a set (let’s say set A) to another set (let’s say set B) is the
linking (or pairing) of the elements of set A to the elements of set B.
A square root of B
Object
Image
4
2
9
25
3
Domain
5
Range
6
Nonimage
Codomain
Figure 1.1
•
•
•
•
•
The domain is set A = {4, 9, 25}.
The codomain is set B = {2, 3, 5, 6}.
The object of 2 is 4, the object of 3 is 9 and the object of 5 is 25.
The image of 4 is 2, the image of 9 is 3 and the image of 25 is 5.
The range is the set of images, that is, {2, 3, 5}.
A relation between two sets can also be represented by using ordered
pairs and graphs. The graph representing the relation from set A to
set B is as shown in Figure 1.2.
6
Set B
Figure 1.1 shows an arrow diagram representing a relation from set
A = {4, 9, 25} to set B = {2, 3, 5, 6}. The relation is defined by ‘square
root of ’.
5
The ordered pairs representing the relation from set A to set B is
{(4, 2), (9, 3), (25, 5)}.
3
2
There are four types of relations.
4
9
25
Set A
Figure 1.2
Types of Relations
1. One-to-one relation
Each object in the domain has only
one image in the codomain.
Figure 1.3(a)
b
e
c
f
3. One-to-many relation
There are objects in the domain that
have more than one image in the
codomain.
Figure 1.3(c)
a
d
e
c
Codomain
c
a
d
b
e
Domain
4. Many-to-many relation
There are objects in the domain that
have more than one image in the
codomain and there are elements in
the codomain that are linked to more
than one object in the domain.
Codomain
b
Domain
Figure 1.3(b)
CH001_A.indd 2
d
Domain
2. Many-to-one relation
There are more than one object in the
domain having the same image in the
codomain.
Figure 1.3(d)
a
Codomain
a
d
b
e
c
f
Domain
Codomain
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1 Functions
97
By using the half angle formula,
2 tan
tan u =
u
2
1 – tan2
u
2
u
2 tan
12
2
=
5 1 – tan2 u
2
u
tan
6
2
=
5 1 – tan2 u
2
u
2 u
6 – 6 tan = 5 tan
2
2
u
u
6 tan2 + 5 tan – 6 = 0
2
2
u
u
2 tan + 3 3 tan – 2 = 0
2
2
3 2
u
tan = – or
2 3
2
u
Since is an acute angle,
u
3
2
tan = – is not accepted
u
2
2
tan has to be positive.
2
u 2
tan =
2 3
(
)(
)
EXAMPLE 1.75
Prove the identity
sin 4x + sin 2x
= tan 2x.
1 + cos 4x + cos 2x
Solution
LHS =
sin 4x + sin 2x
1 + cos 4x + cos 2x
}
sin 4x = 2 sin 2x cos 2x
=
2 sin 2x cos 2x + sin 2x
1 + (2 cos2 2x – 1) + cos 2x
}
cos 4x = 2 cos2 2x – 1
sin 2x (2 cos 2x + 1)
2 cos2 2x + cos 2x
sin 2x (2 cos 2x + 1)
=
cos 2x (2 cos 2x + 1)
sin 2x
=
cos 2x
= tan 2x
= RHS
sin 4x + sin 2x
tan 2x
1 + cos 4x + cos 2x
=
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1 Functions
99
Checkpoint 1.12
For questions 1–14, prove the given trigonometric identities.
1 cos4 x – sin4 x cos 2x
2 cot x – cot 2x cosec 2x
sin 2u cos 2u
–
3
sec u
cos u
sin u
4 cot u – tan u 2 cot 2u
tan 2u
5
tan u
1 + sec 2u
6
sin u + sin 2u
tan u
1 + cos u + cos 2u
cos2 u – cos 2u
tan2 u
sin2 u + cos 2u
8 cot u – cot 2u– cosec 2u
7
9 cosec u – 2 cos u cot 2u– 2 sin u
2 tan 1 y
2
10
sin y
1 + tan2 1 y
2
11 cos 3x 4 cos3 x – 3 cos x
12 sin 3x 3 sin x – 4 sin3 x
13 cos 4z 8 cos4 z – 8 cos2 z + 1 8 sin4 z – 8 sin2 z + 1
1 – tan2 3y
14
cos 6y
1 + tan2 3y
15 Given that cos 40° = m and sin 10° = n, express each of the following in terms of m and n.
(a) sin 50°
(b) cos 30°
(c) sin 80°
(d) cos 20°
12
4
16 It is given that cos A = – and sin B = – , where A is in the second quadrant and B is in the third
13
5
quadrant. Without using a calculator, find the value of tan (A – B).
5
a
17 Given that tan a = –
, where a is an obtuse angle, find the value of sin without using a
2
2
calculator.
5
. Without using a calculator, find the value
18 It is given that A is an obtuse angle and cos A = –
3
of each of the following.
(a) cos 3A
(b) sin 3A
(c) tan 3A
(d) cos 4A
19 Express
(a) sin2 x in terms of cos 2x,
(b) cos2 x in terms of cos 2x,
(c) cos2 2x in terms of cos 4x.
1
Hence, prove that sin4 x + cos4 x = (3 + cos 4x).
4
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5 Analytic Geometry
283
2b2 6 ,
=
a
5
5
(iv) length of major axis = 2 and length
of minor axis = 2 3.
(iii) length of latus rectum =
If the centre of the ellipse is at (h, k), the standard equations of the
ellipses are as given below.
Ellipse
Major axis Centre
(x – h)2 ( y – h)2
2
2
2
+
= 1 where c = a – b , a > b Horizontal (h, k)
a2
b2
( y – k)2 (x – h)2 1 where c2 a2 – b2, a > b
Vertical
+
=
=
b2
a2
(h, k)
Foci
Vertices
(h ± c, k)
(h ± a, k)
(h, k ± c)
(h, k ± a)
Table 5.4
EXAMPLE 5.16
Find the equations of the ellipses with the following foci and vertices.
(a) Foci at (4, 2 ± 5) and vertices at (4, 5), (4, –1).
(b) Foci at (–3 ± 7, 2) and vertices at (–7, 2), (1, 2).
Solution
(a) By inspection, the ellipse is vertical with its centre at (4, 2)
where a = 3 and c = 5.
y
(4, 5) (4, 2 + 5)
Using
c2 = a2 – b2
5 = 9 – b2
(y − 2)2 (x − 4)2
b=2
+
=1
b2 = 4 or b = 2
9
4
(4, 2)
x
( y – 2)2 (x – 4)2
O
The equation is
= 1.
+
(4, −1) (4, 2 − 5)
4
9
(b) By inspection, the ellipse is horizontal with its centre at (–3, 2)
y
where a = 4 and c = 7.
2
2
2
Using
c =a –b
b=3
a=4
7 = 16 – b2
(−7, 2)
(1, 2)
(−3, 2)
b2 = 9 or b = 3
x
O
(x 3)2 ( y – 2)2
The equation is +
2
2
+
= 1.
(x + 3)
(y − 2)
9
16
—–—— + —–—— = 1
16
9
The general equation of an ellipse is of the form
Ax2 + By2 + Cx + Dy + E = 0 (A ≠ B)
The method of completing the square is used to convert this
form to the standard form.
EXAMPLE 5.17
Show that the following equations represent ellipses and find their
centres, foci and vertices.
(a) 16x2 + 4y2 – 64x – 40y + 100 = 0
(b) 4x2 + 9y2 – 24x + 36y + 36 = 0
MATHS FILE
The major axis of an ellipse is
vertical if the denominator of
the term containing y 2 is greater
than the denominator of the term
with x 2.
Solution
(a) 16x2 + 4y2 – 64x – 40y + 100 = 0
16(x2 – 4x) + 4(y2 – 10y) = –100
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