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DIFFERENTIATION
DIFFERENTIATION..............................................................
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....... 4
IMPLICIT DIFFERENTIATION
DIFFERENTIATION .............................................................
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............. 5
PARAMETRIC
PARAMETRIC DIFFERENTIATION
DIFFERENTIATION .........................................................
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....... 7
TRIGONOMETRIC
TRIGONOMETRIC DIFFERENTIATION
DIFFERENTIATION ............................................................
..........................................................................................
..........................................................
............................ 9
DIFFERENTIATION
DIFFERENTIATION OF INVERSE TRIGONOMETRIC
TRIGONOMETRIC FUNCTIONS
FUNCTIONS .......................................
................................................................
......................... 10
DIFFERENTIATION
DIFFERENTIATION OF EXPONENTIAL
EXPONENTIAL FUNCTIONS
FUNCTIONS ..........................................................
.........................................................................................
...............................11
DIFFERENTIATION
DIFFERENTIATION OF NATURAL LOGARITHMS
LOGARITHMS ...........................................................
..........................................................................................
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..... 13
PARTIAL DERIVATIVES
DERIVATIVES ..........................................................
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..........................15
First Partial Derivative ............................................................
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..................... 15
Second PARTIAL DERIVATIVE
DERIVATIVE ............................................................
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.....16
INTEGRATION
INTEGRATION RESULTS....................................................................
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............... 17
Even powers of
Odd powers on
Even powers of
and
............................................................
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..... 19
and
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..... 20
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..........................21
INTEGRATION
INTEGRATION BY PARTS.......................................................................
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..........22
Reduction Formulae ............................................................
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..........................25
PARTIAL FRACTIONS
FRACTIONS..........................................................
.........................................................................................
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...............................28
Denominator with Linear Factors .......................................
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..............................................................
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.................... 28
Denominator with unfactorizable quadratic factor.............................................................
......................................................................................
..........................29
Denominator with a repeated factor ...........................................................
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........................................................
..........................30
Improper Fractions
Fractions (degree of numerator
≥
degree of denominator).................................
denominator)................................................
............... 31
TRAPEZIUM RULE (NUMERICAL INTEGRATION).................
INTEGRATION)................................................
..............................................................
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............... 33
COMPLEX NUMBERS...........................................................
..........................................................................................
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...............................34
SQUARE ROOT OF NEGATIVE NUMBERS ................................................................................
..........................................................................................................
..........................34
Operations on Complex Numbers ......................................................
.....................................................................................
..............................................................
....................................
..... 34
Adding and Subtracting Complex Numbers.............................. .......................................................................................... 34
Multiplying Complex Numbers ................................................................................................................................................. 34
34
Dividing C omplex Numb ers ....................................................................................................................................................... 34
Square Roots of Complex Numbers .....................................
...................................................................
.............................................................
...................................................
.................... 35
Quadratic Equations ............................................................
...........................................................................................
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..........................35
Equations with Real Coefficients .................................................................................................................................. ...........35
Equations with Complex Coeffic ients .................................................................................................................................... 35
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Roots of Equations......................................
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....................36
Argand Diagram......................................
Diagram.....................................................................
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.........................37
–
–
De Moivre’s Theorem
Representing Sums and Differences on Argand Diagrams .......................... ................................................................ 37
Modulus Argument Form ....................................................................
..................................................................................................
.............................................................
....................................
..... 37
The Modulus of a Complex Number
N umber ....................................................................................................................................... 37
The Argument
Ar gument of Complex Number
N umber ........................................................................................................................................ 38
Modulus Argument Form ...................................................................................................................... .................................. 39
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..........................41
Multiples of Sine and
a nd Cosine ...................................................................................................................................................... 43
The Exponential
Exponential Form of a Complex Number ...............................................
.............................................................................
...................................................
..................... 43
Locus on the Argand diagram ................................
..............................................................
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...... 44
SEQUENCES.............................................................
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..................... 49
SEQUENCES..............................................................
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................ 50
Types of Sequences .........................................................................
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............... 51
Convergent Sequences
S equences ...................................................................................................................................................... ...........51
Divergent Sequences ..................................................................................................................................................................... 51
Convergence of a Sequence .......................................................................................
.....................................................................................................................
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................ 52
Recurrence Relations ..........................................................
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..........................52
SERIES ............................................................
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...............................54
SERIES ............................................................
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..........................55
Using Sigma Notation..........................................................................
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..........55
Sum of a Series ..................................................................
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...............................55
Mathematical Induction ..........................................................
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.....................56
Method of Differences ....................................................
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............................... 57
ARITHMETIC
ARITHMETIC PROGRESSIONS
PROGRESSIONS.............................................................
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..... 58
GEOMETRIC PROGRESSIONS
PROGRESSIONS....................................................................................
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................ 60
MACLAURIN’S SERIES
PASCAL’S TRIANGLE
.............................................................
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..........................63
TAYLOR SERIES ...........................................................
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..... 66
BINOMIAL THEOREM..............................................................
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...............................69
FACTORIALS ............................................................
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................ 69
THE BINOMIAL THEOREM .........................................................................
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..... 71
Extension of the Binomial Expansion .................................
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..................... 72
ROOTS OF EQUATIONS
EQUATIONS ...........................................................
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THE INTERMEDIATE
INTERMEDIATE VALUE THEOREM .............................................................
...........................................................................................
...................................................
..................... 77
DETERMINING
DETERMINING THE ROOTS OF AN EQUATION
EQUATION ....................................................................
...................................................................................................
...............................78
BISECTION METHOD................................................................................
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..... 78
LINEAR INTERPOLATION
INTERPOLATION ...........................................................
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................78
NEWTON RAPHSON...................................................................
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..................... 79
DERIVING AN ITERATIVE
ITERATIVE FORMULA ................................................
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..... 80
MATRICES...........................................................
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..........................81
MATRICES...........................................................
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..................... 82
Matrix Multiplication.................................................................
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.....................82
THE DETERMINANT OF A
×
MATRIX ...........................................................
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................ 83
The Transpose of a Matrix...........................................................
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................ 84
Finding the inverse of A Matrix (Cofactor Method) ............................................................................
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.......... 85
SYSTEMS OF EQUATIONS
EQUATIONS............................................................
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................86
ROW REDUCTION
REDUCTION ............................................................
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...............................87
Row Reduction and Systems of Equations .......................................................................................................................... 87
DIFFERENTIAL EQUATIONS
EQUATIONS...........................................................
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.............................................................
...................................................
..................... 90
DIFFERENTIAL
DIFFERENTIAL EQUATIONS
EQUATIONS ...........................................................
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................ 91
Separable Differential Equations....................................
Equations...................................................................
..............................................................
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.........................91
The Integrating
Integrating Factor..................................................................................
................................................................................................................
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...............................92
Linear Differential Equations with Constant Coefficients...........................................................
...........................................................................
................ 93
–
Homogenous Differential Equations ...................................................................................................................................... 93
Non Homogeneous Differential Equations ...................................................................................................................... 94
Differential Equations Requiring a Substitution.........................................................
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..... 98
Mathematical Modelling..........................................................
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.................. 101
DIFFERENTIATION
Page |4
At the end of this section students should be able to:
ln
–
,,
1. find the derivative of
, where
is a differentiable function of ;
2. find the derivative of
(to include functions of
polynomials or trigonometric);
trigonometric);
3. apply the chain rule to obtain gradients and equations of tangents and normals to curves
given by their parametric equations;
4. use the concept of implicit differentiation, with the assumption that one of the variables is a
function of the other;
5. differentiate
differentiate any combinations of polynomials, trigonometric,
trigonometric, exponential and logarithmic
functions;
6. differentiate
differentiate inverse trigonometric functions
7. obtain second derivatives,
, of the functions in 3, 4, 5 above;
.
8. find the first and second partial derivatives of
Page |5
IMPLICIT DIFFERENTIATION
0 5
A function which is written in the form
is
called an explicit function: is stated explicitly in
terms of . However, functions such as
or
are implicit
LESSON 4
functions.
LESSON 1
Differentiate
2
SOLUTION
LESSON 2
determine
SOLUTION
SOLUTION
Determine
3 3 5
22 3 3 0
333322
SOLUTION
3
3 5
6
6 6 2 3
66 2 3
2 2 62 0
2 2 2 6 2
2 262
26 2 6 2
2 62
6 2 2 663602 2 3
31,13, 3 1 1 0
1
2611262 32
1
LESSON 5
Find the equations of the
tangents at the points where
Use implicit differentiation to
1
for
Determine
for
with respect to .
.
SOLUTION
.
1
1
2 2 0
2 22
2
5 5 0
5 5 0
4 1110 10 0
1 4
LESSON 3
410 1 1
When
for
.
Gradient at (6,
Equation of line:
Using (6,
)
)
on the curve
.
Page |6
1 32 6
103
2 10
66232262 32
3 32 6
12 3
2 12
22 6
22 6
2 2 2
2 0
2 1212 2 2
2 1 1
0
2221 0
2 2 0
6
2
22 60 6 6 0
33,33, 2 2 2 0
3,3,3,333,2 2,2, 2
22 2 2221 1122 222
Gradient at (6, 3)
Equation of line
Using (6, 3)
LESSON 6
Find and classify the stationary
points on the curve
.
SOLUTION
Stationary points occur when
Sub.
into
and
.
We have already stated that
2 222220 1 2
3,3,2331 1 2 1 1
2332 1
25 → Maximum
2,2, 2 2
222 1 1
5 → Minimum
For
For
0
and
Page |7
×
5 3 2
5
12 2
2 3
×
22 3 31
22 3 3
2
22 3 3 1
44 1 1 √2 7 7 1
4 11
4
122 7 27 2
√
221 7
2 2 7 7−−2
22 7 7
PARAMETRIC DIFFERENTIATION
Given that
a parameter, then
and
where is called
LESSON 1
Find the gradient of the stated
curve at the point defined.
;
where
SOLUTION
when
LESSON 2
Find the gradient of the stated
curve at the point defined.
;
SOLUTION
when
÷
122 272−
2
266 1√2 7
1
26261 112
1 7 7
2
301
, 2211
8 8−− 24
24
24 1 1
4××
24
6
1
88
18
2
2221211 1
7
when
LESSON 3
to the curve
Find the equation of the normal
at the point where
the curve crosses the line
SOLUTION
when
when
(1, 7)
.
Page |8
6
2
At (1, 7):
6
16
3
7 31
10916 16
163 10916
44 1 1 5 5
444 1 1
355
3×
4
32
332 × 14
8
3
1
1
6 6 3
36 1 1
6 6 3
Gradient of normal is
LESSON 4
Find
and
for the
parametric equations
and
SOLUTION
LESSON 5
Find
and
for the
parametric equations
and
.
Hence find and classify the stationary point(s).
SOLUTION
2 6
×
23636
11333 3 33
3 33399× 1
1 6 9 6
0
33 3 0
3 3 0
333366131 3 32626 1212
26,26,1212
613 1621
Stationary points occur when
Minimum point
Page |9
TRIGONOMETRIC DIFFERENTIATION
sec 22 33 sec tan
sec
s
ec
22 3 tan
se
c
′cos
cos
sec
s
ec
os
2
3
t
a
n
sec
si
n
sseinc
t
a
n
≡
2
1
s
e
c
csccot
cscsect
cotacotn 11 tan sec
secccsc tann
sec 1 sec 2sec
2secsect
sectann
2se
2
sec
c
2sec
2
sec
tan
tan
csc
1tan
2se
2
sec
c
1t
s
e
c
1
t
a
n
cotcosec4t 5
21tan
1tan
1
tan
1tan
3cs+
3+csc−−12
12
2
tan 1 tan
n
22 tan
tan1 tann
22 1
sec4
4sec4t
4sec4 tan4
an4
cot5csc5 5
3cs33csc6612
12 csc12
cot12
1
2
1
2
18 csc12
1 2 cot12
1 2
−−
4
+112csc
2csc4
−2 csc
2csc4
−
−
4
4csc
4
csc
4
4
cot
4
4
2csc4 4
4
cot44
4
4csc2csc
(e)
LESSON 2
SOLUTION
LESSON 1
(a)
(b)
(c)
(d)
(e)
SOLUTION
(a)
(b)
(c)
(d)
Differentiate the following w.r.t
Recall:
Given that
, show that
P a g e | 10
DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS
LESSON 1
PROOF
Let
then
Differentiate
sin−
.
sisinsin−, – < <
sisin
1
cos
cos1
cos
sisi
n
1
coscos11si sinn 1
√1 1
sin− √1 1
−
cos
coscoscos,− , 0 ≤ ≤
cos
cos
si n 1 1
cossinsisi n 1
ssiinn11cos
cos 1
coscos− √1 1
−
t
a
n
tantan−, < <
tann
sec 1 1
sescec 1 tan
11
tan 1
1
But
LESSON 2
PROOF
Let
then
Differentiate
But
LESSON 3
PROOF
Let
then
But
Differentiate
.
−
cos− 3 3 cos
LESSON 4
Show that if
cos− 3
, then
SOLUTION
iff
cos 33
sincos
3
si3n
sin 111cos
3
3
coscos− 33 1
3
1
3
3
√193
sin− √1 1 . ′′
cos− √1 1 . ′′
tan− 1 1 . ′′
In general, given that
is a function of , we have
P a g e | 11
DIFFERENTIATION OF EXPONENTIAL FUNCTIONS
FUNCTION
DERIVATIVE
LESSON 1
(a)
(b)
(c)
(d)
(e)
Differentiate the following
2−
+
11
2−−
223cos3
663cos3
cos3
1 [2]
SOLUTION
(a)
(b)
(c)
(d)
2 2
12
1 2
+2− 1 1 −1
2−2 121− −
1
1
2 1
−−2221121
1 13
1131 55
15
151
5
5
1 3 2 15
3 3 2 2 15
33 2 2 15
15153 22
sin
2 2 2 0
sisin coscos
sisin c cos
sisin co coss coscos s sin
2 cos
(e)
LESSON 2
Determine
for
SOLUTION
LESSON 3
that
Given that
, prove
SOLUTION
2 2 2 2
20 coscos 2 2coscos sisi n 2 sin
P a g e | 12
LESSON 4
Determine
defined parametrically by
for the equation
1 1 sin− 2
1 1
2
sin− 2
1
2
1 22
×
√142 × 21
1
√14
and
SOLUTION
.
P a g e | 13
DIFFERENTIATION OF NATURAL LOGARITHMS
2×2 ×+1 1 2l n 1
1
222212122llnn n
2211lnn
1
l
n
2
1
1
3
3
2
llnn22 11 11 2l lnn33332222
221 1 2 332 2
222321 1223662222 1
66 24 1 11233 6 2
221911331 2
22 13 2 2
+−
l
n
1 ≡ 2
ln11 1ln1
111 1 1
1
1 2 1
1 2 . 1
2
(vii)
LESSON 1
following.
(ii)
(iii)
(iv)
(v)
(vi)
1
Differentiate each of the
l3lnn377112
llnnsisin4411
ln
+
ln22 133 2
ln33 1 1
3 1
3l33ln7772 2
7217 2 2
(vii)
(viii)
SOLUTION
(ii)
(iii)
(iv)
(v)
(vi)
ln2 1 11
1
ln4co4siscosins444
4co4sicotnt44
22lnlnn 1
22 lnn
2l2ln 1
(viii)
LESSON 2
that
SOLUTION
Given that
, show
P a g e | 14
LESSON 3
Differentiate
SOLUTION
3
l1n ln 3
ln3
3l nln33
3
w.r.t
ln
−
ln ln √3 3
l1n ln1 l1ln2 3 3
1 2 3 3
3 3
3
33 3
3 3
3 3 3
Sub 3 3
3 33 3 3
33 3
In general, if
LESSON 4
SOLUTION
then
If
, find
P a g e | 15
PARTIAL DERIVATIVES
For partial derivatives we differentiate with
respect to one variable and treat the other
variable(s) as constants.
,, 4ln
,, 44 ln 2
4ln 4
4ln 8
,, 442ln
48
LESSON 3
Given that
, determine
(i)
(ii)
SOLUTION
LESSON 1
Given that
,, 2 2
,,
2
,, 2 2 5
3 2 2 52
,, 22 2 −
4 25 −
4 5
33 2 2
33 22
8843333 2233 2 266 22
33 22
88433 2232222 22
, evaluate
(i)
(ii)
SOLUTION
(i) For
we differentiate
, treating
as a constant.
(ii)
LESSON 2
with respect to
(i)
(ii)
LESSON 4
determine
(i)
(ii)
(iii)
Given that
, determine
SOLUTION
(i)
(i)
(ii)
(ii)
SOLUTION
(i)
(ii)
Given that
(iii)
2 4 4
3
,
P a g e | 16
2
,,
2,2, 1
,, 3 2 2
2,2, 1 321 12
LESSON 5
For
determine
.
,
SOLUTION
LESSON 1
Given that
,, 4 4
,, 3 4442
2 −−
6 6
,, 4 42
3 4 2
3 8
8
,, 8844−
3 8
8 2
, determine
(i)
(ii)
(iii)
SOLUTION
(i)
(ii)
(iii)
NB:
LESSON 2
Given that
l ln s sinn
(i)
, determine
(ii)
(iii)
(iv)
(v)
(vi)
SOLUTION
l ln1 s sin
48
l ln12s sin 2
2223−coscos3−
44 322 cos 23 32 322 sin 23
4 32 cos23 94 sin23
l ln2 s sin 2−
2 2 3cos2 3
2
cos3
3
22 − cos −
2 34 cos23 32 32sin23
22 34 cos23 49 sin 23
(i)
(ii)
(iii)
(iv)
(v)
(vi)
P a g e | 17
INTEGRATION RESULTS
SOLUTION
+
′
√′
FUNCTION
LESSON 1
(i)
(ii)
(v)
(vi)
ln||
ln||
1 1 ++
1 +
llnn||cos|
csec|
ossec||
1 lnsecsec
ln|sesectanttan|
lntan2
ln|−sisin||
sin
1 tan−
(vii)
(i)
(ii)
(iii)
or
Determine
∫
∫ −
∫ −+
∫∫ tanan+−
∫ 12∫ 332l2l 5n5
(iii)
(iv)
INTEGRAL
(iv)
(v)
(vi)
∫ 1
22
l2ln n||
∫−ln|44 1|
∫ −5 2
2 12
+52 ln|12|
1 2|
∫ 1+−46
2 2 6 6 9
12 ln|22 6 6 9|
∫ tansin
cos
sicosn
∫ 12ln|cos|
c33os 5| 5
3335 5
2 ln
∫ 12ln
33 2 ln
2 ln4
332ln
(vii)
P a g e | 18
LESSON 2
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Evaluate each of the following.
∫∫ 2−
∫ 2
∫ ++
∫∫ 3 √ 6 6
+
∫∫ csc
(v)
(vii)
(viii)
(vi)
SOLUTION
(i)
(ii)
(iii)
(iv)
∫ 1
2 −
∫ 2 −
2
2
22 11−−
2
∫ 222
4
∫ ++
5−1 −−−
25 2−1 −−−−−
52 1 1
52 1 5−−−
21 −
∫ √ 6 6
6 6
14 46 6
1 4 636+
2
16 6 6
∫3
∫ +
∫csc
∫ −
∫∫ −− +
−
∫ − 1
√2
2 sin− 2
∫ −− 1 + 3 2
22 1 23 3 222
2 1 1−− 2 3 22
(vii)
(viii)
LESSON 3
(a)
(b)
(c)
SOLUTION
(a)
(b)
Determine
P a g e | 19
(c)
12 1222 1−− 32113 tan−223
4 22 1−− 2 tan− 3
∫−
Even powers of
and
sin
sin
csions211cos2
2 sisin
2
2
sin 1 cos2
2sin22
2 4
cos
cos
cos
co
s
2
2co
2
cos
s
1
1
1
cos2
cos 2 2
cos 1 cos2
2sin22
2 4
LESSON 1
Determine
SOLUTION
Using the identity
LESSON 2
Determine
SOLUTION
Using the identity
P a g e | 20
sin
sinsin 1si ncossisi n
sinsin
sin 1cos
1cos
s sinn sincn cos
sinn cos sisin cos
co coss 3
cos
cos
co coss cos
cos
cos 1si
1sin
c cos sin cos sin
sin 3
sin 2
sin 2
sin2sisi2 n 2
sisi n 22 1cos
1cos 22
1sisi n 2 cos sin22cos
2cos 2
2 cos2 6
LESSON 1
Determine
SOLUTION
Since
LESSON 2
Determine
SOLUTION
LESSON 3
SOLUTION
Determine
cos sin
cos sin
co coss coscos sin
cos
cos 1si
1si n sin
scoincos s sisisnin cos sin
5 7
LESSON 4
SOLUTION
Determine
P a g e | 21
tan
tansec 1sec
se tcan 1 1
t
a
n
1si
n
cos1 sin
cos cos
lsen|seseseccttttaann| ln|sec|
|
s
ec
|
|
ln sec
sec sec t tan
tan 2
tan 2
1se secc 2 1
2 tan 2
LESSON 1
Determine
SOLUTION
Since
LESSON 2
SOLUTION
Determine
P a g e | 22
––
––
–
INTEGRATION BY PARTS
When choosing
we use the following acronym.
Logarithms
Inverses
Algebra
Trigonometric Ratios
Exponentials
LESSON 1
SOLUTION
Determine
1
LESSON 2
SOLUTION
Determine
121 2 3 3 1681 2 3 3
1681 2 3 31414 2 2 3
1681 2 3 31212 3 3
56 2 3 34 1 1
√312 1
√312 1 33 11 22−
3 1 − 3
1 2
12
312
2
√121
3 1 11 2
2
3312
1 2
3 1 11 2
2 1 2
2
11 2
23 1 1 2
12
1 2 2 2
c
cos
cos
1
coscos sisin
c
cos sin sin
s sin cos
cos
LESSON 3
SOLUTION
LESSON 4
∫ 22 3
2 3 3
1 1
22 3 3 12 2 3 3
2 3 3
121 2 3 3 121 22 3
Evaluate
SOLUTION
Evaluate
P a g e | 23
LESSON 5
Determine
ln
SOLUTION
lnn 1.ln
ln 1
ln1
lnn 1
1
lnn
LESSON 6
SOLUTION
SOLUTION
Evaluate
sisin
Determine
lnn
ln 1
3
lnn
3 ln 3
13 ln 19
9 3l3ln 1
LESSON 7
lnn
sin
2
sisin coscos
sin
coscos 2 cos
22 2
22cosco s cos sisin
2222 ssisiin—
nn — 22sicosconns
sisin
2cos s0sisin 2 coscos
2cos
2cos—
— cocoss 022 2co
244 2 2
sisin
sisisisinn
coscos 1 12 1
coscosisisn 2 sisin 2 cos
sisin 12
LESSON 8
SOLUTION
Evaluate
P a g e | 24
1sin 1 1 1
2 sisin 2 12 coscos 12 sisin 1
5 sisin 2 1 sisin 4 1 coscos 4 sisin
4 sisinsisin 2 2 sisinsisin 1 4 cocoss
5 5
P a g e | 25
LESSON 9
Establish a reduction formula
that could be used to find
∫
and use it
4
−
−
−
4 −−
44 44 3 3
44 44 1212
1212 2 2
24
44 44 12
12
12
24
12
24
24
4 4 12
12 24 24 2 24
4 4 12
12 24 2 24
44
4 412
12
12 12 24
24
24
24
24 2 24
≡
cos
∫
sincos∫ cos− − −−
when
.
SOLUTION
LESSON 10
If
show that
.
Hence, find
.
SOLUTION
STEP 1: Write the integral in the product form
co coss cos−
cos− 1 1 cos− sisin
STEP 2: Integrate by parts or an appropriate
method
coscos sisin
sin coscos−− 1 1 cos− sin
STEP 3: Simplify
sincos− 1 1 cos− 1 co coss
sin coscos−− 1 1 cos− 1cos
1cos
sincos−− 1 1 cos− 1 1 cos
sincos sincos 1−1−−111−1−
1sin coscos−− 1111−−
sincos −−
STEP 4: Apply the derived formula to the rest of
the question
51 4
5 sincos 5
31 2
3 sincos 3
STEP 5: When you have reduced your integral to
its lowest form, go back to the original integral &
1
c cos sisin
15 sincos 45 13 sincos 23 sin
∫ 1 1
2 1 1 −−
1 1
plug in the final value of .
LESSON 11
that
SOLUTION
If
, show
P a g e | 26
1 1 1 11−−
1 2 −− 1
2 1 1 11 . 2
12 1 1 2 11−−
12 1 1 2 −−
2 1 1 −−
∫lnln , ≥ 0.0.
≥ 1,1, 12 12 −−
ln 22ln ln −−. 1
ln 22ln ln −−
22ln ln ln −−
.
.
2 ln ln
−−
ln
1 1
ln 2 ln 1 2 −−
112 32 −−
12 32 1 1 3 3
2 1 2 32 1 1 2 4 2
4 2 2 2
LESSON 12
It is given that
By considering
or otherwise, show that for
Hence, find
SOLUTION
in terms of
METHOD 1
Rearrange the equation
Take integrals of both sides
14 34 34
12 34 ln
121 34312 3 12
12 83 8
8 8
ln
lln 1ln −−. 1
1 2
2 ln 1 2 ln −−
112 ln 1 2 −−
2 2 −−
4 sisin 1
√16
METHOD 2
is found the same way
LESSON 13
By using the substitution
, find the exact value of
−16 16√16
1− √16
Show that
∫1616 − 11−,− 22≥√23
44cos4cosisin s
2;2;4cos
4cos
6
Deduce, or prove otherwise, that if
for
SOLUTION
, then
. Hence find
.
P a g e | 27
0; 0
√161
1
16
4 cos
16
4si
4
si
n
4cos
√1616sin
4cos
16
16
1
sisi
n
cos
√1sin
cos
cos
1 1
0 6
−16
−. 12 −16
1 6. 22 16 1 1−
2√216 1 116 −−
1616
√16 1 1√16
√16 1 1√161616−−
1 1 √16
1 1
√16 − 1616√16
16 √16
16
1 1 √16
−16 1616√16
1− √16
√16
−
1616 1 √16 −16
2
1616 1 1−− −−16 0
1616 1 1−− 2 2 1√12
1616 1 1−− 2 2. 2 . √4.4. √3
2 1616162121√−3− 2 2 √2
2 16 4 4√3
√161 sin− 4 20 6
2 483 4√3
3 2√3
P a g e | 28
PARTIAL FRACTIONS
LESSON 2
Denominator with Linear Factors
LESSON 1
fractions.
Express
−−−
in partial
2 27 2227 1 1 ≡ 2 2 1 1
2 7 1 1 1 2 2 2
3 3 9 3 13
1 2 13 1
2 27 2
272 21 2 13 1
ln|112|2|3l n| 1 1|
ln 2 2
233 1 111333 3 1 3 3
2 11 113 33 33 1 3 3 3 33 1
24 112
2 0
3 13
33 333 2 2
2
1
2
233 1 111333 1 2311 1 213 3
SOLUTION
When
(a) Express
(b) Hence determine
SOLUTION
(a)
Comparing coefficients of
in partial fractions
∫ +−−
++
−−
13 2195195 6 1113132219 3 3
13
2
After factorizing the denominator we get
When
13
19
1313 1 19
119 223232 1 331 22 1133
3 121 2 2
453 15
3
202 10
1
616
13 2195195 6 11 1 23 2 32 3
13
2
+−−
∫ ++
ln|11111|3l n|223 22| 2l n|33233|
When
When
When
When
When
++
−− +
(b)
P a g e | 29
LESSON 3
++
+++++
+
Express
++
+
+++
++
in partial
5
2 2662 22 5
6
5
5
6
2
5
4
5353413 3 333313333 1 5 2626 22 25 2 2225 22 5 2 2
2 2 5 2
3
3
3
3
3
2 5 6
3 3 1 5
254 6
3 321 5
5
2
2
5
8
10
36 4 4
2 5
2
3
3 1
5
5
4
3
2
2
5
6
6
2
3 3 1 3 3 3 3 1 3 3
222 232 4 5
2
2
2
2
5
−+
+
3
1
1
7
l
n
2
2
−+
2
2
5
∫ +
3
1
1
7
l
n
2
2
2
2
5
2
2
5
25 5 2 25511 2
211 5 2 27 5
32 22
l
n
2
2
25511 2 11
ln 2 2 32 ln| 2 2 5| 7 1 11 4 4
2 5 5 2 11
ln 2 2 32 ln| 2 2 5| 7 1 11 2 2
3
7
−
|
|
1 1
l
n
2
2
l
n
2
2
5|
5
t
a
n
25
2
2
2 22
0
2 5 5 2 2 51 1−
ln 5t5 tan
LESSON 1
Express
fractions.
fractions and hence determine
in partial
SOLUTION
SOLUTION
LESSON 2
Express
and hence determine
SOLUTION
in partial fractions
.
P a g e | 30
LESSON 1
Express
fractions
−−−+
in partial
6363 9 99 3636 999 3 3339
3 3339 3 3 33
3 393 3 3
330
93 3
91 9
11 1
26363 9 99 1 32 3 333
SOLUTION
When
When
Equating coefficients of
LESSON 2
Express
+++
−−
fractions and hence determine
in partial
5
2 2662 22 5
9
2
9
24
2
9
9
24
24
2 4 94933218 2 2 3 3
2 2929324323233322333 3 2 2
3
2
1535
502 25
22 2
02 9 9 24 2 3
4 4 3 3 18 2 2 3 3
2
4 499332184
SOLUTION
When
Equating coefficients of
When
22 2 3
3 3 3−−
2ln| 2 2| 33 3
P a g e | 31
≥
+−−
+
+
2 3311 4
2 3311 4 1 1
2 3 3 4 1 1 1 1 1
24
4 4 1111
3 3
23
3 2 3
2 3311 4 22 3 4 12 1
2 3311 4
2 2 3 4− 12 1
22 l ln 4 2l 2ln| 1 1|
+
+
+−
−+
∫ ++−−
If for
,
has degree
and
has degree
, then quotient has degree
LESSON 1
Express
in partial
fractions and hence determine
SOLUTION
LESSON 2
2
3
2
32 21 122222 12221 1 2222 2 1 1
2
1414 5
11
5
4
2
5
311102 →
14
2
3
2
3
11
2 1 312222 2 101022 1 5 2 2
2 1 1 2 2
32 1120 221 1 145 21 2
32 1120 ln|22 1| 145 ln| 2 2|
SOLUTION
Express
fractions and hence determine
in partial
When
When
Equating coefficients of
−−−
++
++
++
+
19
19 4 44 21
LESSON 3
Express
in
partial fractions and hence determine
3 3 2 2 11
33119224441 121
3 3 2 2 1 1
1 19 4 44 21
3 3 2 2 1 1
1193 344242211 1 2 2 1 1
3 33 1 1 3 3222
63 2
11
1
42 2
SOLUTION
When
When
When
P a g e | 32
1
11 666
5
3319
19 2244
44 1 1221
5 33 3 21 2 12 1
19 224441 121
3319
5 5 3
3 1 1 22 1
Equating coefficients of
Equating coefficients of
3 3 2 2
1 1
2 5 5 3ln| 3 3| ln| 2 2| 2l n| 1 1|
P a g e | 33
TRAPEZIUM RULE (NUMERICAL INTEGRATION)
Introduction
1 1
The area under the curve
equal width (as shown above)
can be estimated by finding the sum of the areas of trapeziums of
−
ℎ2 ℎ2 ⋯ ℎ2 −
ℎℎ2 ⋯−
2 2 ⋯
⋯ −
The area of a trapezium with parallel sides
The width of each trapezium is
where
and
and width
≈ 2 2 ⋯−
1
1
− 0.4
Using 5 trapeziums, estimate
0.00.4 00.141111 1.1616
0.1.01..82 0.1.82 11 11 1.2.64446444
1.12.6 21.16 1115 3.5656
1 1 ≈ 222501 5 5 2 21.1.161.642.443.566
≈ 4.72
SOLUTION
Width of each trapezium is
is given by the formula
is the number of trapeziums.
Thus we have area under curve is
LESSON 1
ℎ
P a g e | 34
COMPLEX NUMBERS
INTRODUCTION
̅
If
then
is its conjugate and vice
versa. It is also important to note that the product of a
complex number and its conjugate is
which is
always a Real number. The conjugate is also denoted
.
∗
√1 1
+
:
2 → ̅ 22
,,
3
2
26 32
62131
3 5 6
5 5
√16 111616 √1√1√16 4
+
−
√18 9211 3√2
37
52
552
2
3151355 7272414152
5
2
11 4512 214
± ±± ±±
1 2929 4129
62i
6 2i 54i
5 4i 116 5 25 2 4
1
18
1 82
8 2 18
12
12
9
22334 368861 6166 6 8
32
3 254
5 4 15151515 81212 12121010 10 810 8
23
23
2
2
Complex numbers are written in the form
where and are real numbers and is the
imaginary unit such that
or
Sometimes the letter is used to denote a complex
number,
. A complex number can also
be written as an ordered pair of its real numbers,
.
is also known as the real part i.e.
(i)
First we find the conjugate of the denominator
Multiply the numerator and the denominator by
the conjugate
is also known as the imaginary part i.e.
SQUARE ROOT OF NEGATIVE
NUMBERS
Thus complex numbers can be used to find the
square roots of negative numbers. Examples
(ii) Express
With this extension of the number system we can
now solve equations which we once unsolvable.
For example,
(i)
(ii)
(i)
(ii)
(iii)
(iv)
in the form
P a g e | 35
SOLUTION
1 11 0
±√±±√1
3333±33 40 413
3±
2
1
3 ±2√3
3±
4 232±2√311
4 22221 ± 022 4 441
2±2 ± 12 24
√8
2±
48311
2±22 ±4
1 ±8√2√33
4± 4
22
2 2 1515 10 10 0
22
222
2
1510
1
510
0
222
2
1
5
1
0
0
1510
2 2 ± 2221 44111510
2 2 ± 22
2 222 2 2 6 60 40
22
2 2 ± √28 6 60 40
22
2 2 ±2 √6032
22
√606060323232
(a)
LESSON 1
Find
√1515 88
SOLUTION
We assume that the square root of a
complex number is a complex number
√1515 8
15 8
15 8 2 2
15
2 4 8
4 15
16 15
15
15 16
16 0
16 16 1 1 0
16
1
±1
±4
√15 8 4 or 4
±±4
Invalid since
is real
Thus we see that a complex number has 2 square
roots, which are complex numbers.
LESSON 1
(a)
(b)
(c)
Solve the following equations
31 1330 0
4 2 2 1
(b)
(c)
LESSON 2
SOLUTION
Determine
such that
P a g e | 36
6060 3232 60 2 2
16 → 16
25616 60
6025
60 2566 0
60
64464 4 4 0
±2
2
82
162 8
2
82 216 8
2 2 ±2 82
82
22
2 2 2 82
82 5
22
2 2 2 82
82 3 2
22
0
±√± √2 44
4 4,, > 0 −±−±√−
2 ± 2
When
When
For the equation
Letting
, we have
As a result we can conclude that if a quadratic
equation has complex roots they occur in conjugates.
2
2
In general, if a polynomial has complex roots they
occur in conjugate pairs. For example, if
is the
root of a polynomial equation then
is also a root
of the same equation.
Recall:
0
sum of root s prproduct of roots 0
0
,, andand
543
5
5
5555555525 0
25 25 0
4
3
4
3
3
43
4344 34334 3 25 8
8 8 25 25 0
12
15 0
If
, then
where
and
and
are the roots of the equation
i.e.
Also, if
then
,
where
and
are the roots of the
equation.
LESSON 1
Given one root find the equation
(i)
(ii)
SOLUTION
(i) Let
then
Equation is
(ii) Let
then
Equation is
LESSON 2
equation
roots.
Given that
is a root of the
, find the other 2
SOLUTION
Since complex roots occur in
conjugate pairs and a cubic polynomial has 3 roots
one root must be real.
21, 1 2, ∈ ℝ
12
1 2 112
2 3 1 1 2 1
Let
P a g e | 37
INTRODUCTION
,,
A complex number
can be represented
on a diagram called an Argand diagram as
(i) a point with coordinates
(ii) a vector
5 8 2
58
58
58
2
7
9
58 2 3 7
LESSON 1
Find
and
and
and
for
. Hence, represent
on Argand diagrams.
SOLUTION
(i)
(ii)
| |
| | √
13 4
511 √3
|| 11 1 1 √2
The modulus of a complex number,
, is a
measure of the magnitude of , and is written as
.
Thus modulus
LESSON 1
(a)
(b)
(c)
(d)
SOLUTION
(a)
.
Determine the modulus of
P a g e | 38
|| |334
344 | 33 4 4 5
11 √3
|| 51
1 √3 11
√√3 2
|| 5
| | 3 4 22
3,3,4 2,2,11
(b)
SOLUTION
(a)
(c)
(d)
LESSON 2
what is
If
?
and
,
(b)
SOLUTION
We are trying to find the distance between
and
. In other words, what is the distance between
the points
and
on the Argand
Diagram?
(c)
(d)
| | |34 2 |
|32
32 41
41|
32
32 41
41
√50
arg
t– <an≤
113 4
511 √3
The angle is called the argument of (
)
where is the angle the vector representing the
complex number on the Argand diagram makes
with the positive real axis. Thus
. To
avoid complications we use
and this
is known as the principal argument of .
LESSON 1
(a)
(b)
(c)
(d)
Determine the argument of
arg 1
tan− 11
4
arg 34
2. 2t1 an− 43
arg 1 1√3
2 tan− √13
3
arg 5
P a g e | 39
5c
5
c
os
o
s
s
s
i
n
n
2
3
2
3
cosco s sisin
6
5c5 cosos6si
n
∴ cosco s sisin
|| 5c55cos
os 6 sisin 6
cos sinn
ar
g
6
1 –
co
c
o
s
s
i
n
√2cos4 sinn4
c
o
s
s
i
n
– coscoscoscossinsin
113 4
511 √3
coscoscossinncosscions sinn
sisinncos
si
n
coscocscossisincoscos sisisisin ncossisin
√112cos sin
4 4
coscos sin
53coscos42.2.211 sin2.2.211
+
+
2cos
11 √3 2 sinn 2
c+
o+
s
s
s
i
n
c
o
s
s
i
n
c
o
s
s
i
n
n
c
o
s
s
i
n
n
55cos s3inn 3
scocoss co
cosssisicoscosnnsisi
sisi nnco
ncos s sinsin
cosccoosco
os coss sisi n sisinn sisincos cossin
coscos sin
c
c
os
o
s
sisi
n
cos5cosisisnsi
5cos
sinn
̅
c
o
s
s
i
n
n
cos
c
os
s
i
n
|| cosco1s si
sinn
|
|
|
|
|
3
|
|
ar g 5sisin55cos
arg || argarg a argg
arg arg arg
5sin 3cos3
If
Therefore,
is
has modulus
and
and argument
in modulus
then
argument form
LESSON 3
and
LESSON 1
Write the following in modulus
argument form
(a)
(a)
(b)
SOLUTION
(c)
(d)
(a)
Prove that for
(b)
SOLUTION
(i)
(ii)
(iii)
SOLUTION
(b)
(iv)
LESSON 2
Find the modulus and argument
of the following
NB:
If
(i)
(ii)
SOLUTION
(i)
Furthermore, we can conclude
(a)
(b)
(ii)
(c)
(d)
then
P a g e | 40
LESSON 4
11
1 √3
Given that
, determine (i)
argument form.
and (ii)
and
in modulus
SOLUTION
|| 1 √3 2
arg tan− √13 3
|| 1 1 1 1√2
arg t tan− 1 4
2√2cos7 si7n
2√
2
2
c
os
s
i
n
√
12
12
√ cos si n
√2cos 12 sin 12
(i)
(ii)
–
P a g e | 41
INTRODUCTION
ccosos sisin
coscos sisi n 2 2 sincn coss
coscos 2 s sin22
then
This can be extended to give
cos
cos sin
Use De Moivre’s Theorem to prove the fol owingidentities
coscos 4 ≡ 8cos 8 coscos 1 1
LESSON 1
SOLUTION
4 c
cos4 sin4 cos sinn
cos 4 coscos sinn 6co
6coss sin 4c4 coss sin sin
cos 4 4cosos sin 6 6 cos sin 4cosos sisin sisi n
cos4 cos 6 cossin sisi n
cos 6 6 cos 1 co coss 1cos
1cos 1 cosos
cos 6 6 cos 6co6 coss co coss 2 cos 1
8co8 coss 8 coscos 1 1
Use de Moivre’s theorem to show that
sisin 5 co coss sinn cos sin s sin
,,
5
c
cos5
s
i
n
5
c
o
s
s
i
n
n
1co1 coss 5co
5 coss sin 10cos sin 10cos sin 5c 5cos
os sin 1 1sin
cos 5cos sin 10cos sin 10cos sin 5cos sin sin
sisin 5 5co5 coss sisin 10co
10coss sin sisi n
5, 10,
10, 1
When
Equating real parts
LESSON 2
where
and
are integers determined.
SOLUTION
When
Equating Imaginary Parts
P a g e | 42
Use de Moivre’s theorem to show
3t
3
t
a
n
t
a
n
tan3 13tan
c
cos3
s
i
n
3
c
o
s
s
i
n
n
cos 3co
3 coss sin 3cos
3cos sin sin
LESSON 3
that
(ii)
−√
1 √3 –
1t√3tan − √13
√√3 2
By DeMoi3 vre’s Theorem
3−
cos 3 3cosos sisinn 3co
3cosssisin sin
11 √3− −2 cos
os 3si
n
3 sin3 3
2
cos
cos
3
cos3 cos 3 3 cos sin
118 cos sinn
1
1
8
sisin 3 3co3 coss sisin sisi n
18
sin3
–
√3√3
tan3 cos3
3 coscoss 3s3incossss ssiinn
3co
3 coscoss si n cossin
3co
1
3
3
1
2
√
√
coscos 3cossi
n
cos
1
−
t
a
n
an t tan
√3 6
3t3 t13tan
√3 2 cos6 s sinn6
By De Moivre’s Theorem
cos si n
√3 2cos6si n 6
−√
6
cos106 sisinn10
2
1
0
si n
5
5
cos
1024cos
s
i
n
n
3
3
BycosDeMoisivren’sTheor
e
m
12 √23
1024
4 4 cos12× sin12×
1coscos 34sin3 4
51512 512√12√3
Rewriting
form
in Modulus
Argument
SOLUTION
Equating Real Parts:
Equating Imaginary Parts
LESSON 5
Express
argument form. Hence, find
.
SOLUTION
LESSON 4
(i)
(ii)
SOLUTION
(i)
Find the value of
in the modulus
in the form
P a g e | 43
INTRODUCTION
sin cos
1 cos 1 sisin coscos sisin
cos sinnco. coscos s sinsisin
∴ 1 ccos sin ccos sinn
2c
2
c
o
s
1 ccos sinn cos sinn
22 sisin
By De Moivre’s Theorem 1
coscos sisin ,so
, so that
c1os
os sisin 1
2co2 coss and 2sin
cos
1 4 4. 1 6 6. 1 4 4..1 1
4 4 6 6 4 1
14 1 6
cos sin,n , 2co2 coss
2cos 2cos2cos 4 42cos2
∴ 2cos
2cos2 6
16cos 1 2cos2cos 4 8 coscos 2 6
cos 8 coscos 4 4 cos 2 3 3
Expressions for powers of
and
in terms
of sines and cosines of multiples of can be
derived using the following results
If
then
LESSON 1
Express
cosines of multiples of .
SOLUTION
If
in terms of
s
i
n
1 3 3. 1 3 3..1 1
3 3 3 1
1 3 1
cos sin,n , 2sin
∴ 22sin 2sin 3 3 322sin
8sin 1 2sin336sin
sin 4 sin3 4 sin
LESSON 2
multiples of .
Express
in terms of sines of
SOLUTION
If
From Maclaurin’s Theor em
cos 1 2!2! 4!4! 6!6! ⋯
sin 3!3! 5!5! ⋯ ⋯
! ! ! !
c
o
s
si
n
1
1
⋯
coscos sisi n
1
1 ! ! ! ! ⋯ ⋯
1
11 1
Then
This series
the expansion of
appears to be similar to
i.e.
Looking at the powers of
P a g e | 44
111 1
Now let’s try the expans ion
11 2!2! 3!3! 4!4! 5!5! ⋯
11 2!2! 3!3! 4!4! 5!5! ⋯
11 2 4 ⋯ 3!3! 5!5! ⋯ ⋯
cos sisin
cos sinn
11+
√−
|11| 1 1 1 √2
arg tan−1 4
√+2−
√ 1 √3
arg || 1 1 1 √2
4
|| −√31 11 2
arg t tan √3 6
|| √22 5
arg√2 arg a arg 4
6 12
2
Grouping Imaginary and Real terms
NB: If
∗ −
then
We will be using the notation
LESSON 1
If the point in the complex
plane corresponds to the complex number , find
the locus of in each of the following situations.
|||223| 4
| 3
3 | 2
| | 3
0
,0,
0
,,
∴ | 0 0 0| 3
| | 9 3
(a)
(b)
(c)
SOLUTION
(a)
The distance between the point
and the
point
representing the complex
number
is 3
CARTESIAN FORM
LESSON 1
Express the following complex
numbers in the form
.
(a)
i.e. a circle with centre at (0, 0) and radius 3
(b)
SOLUTION
(a)
(b)
Let
and
(b)
| 2 2| 4
2,0 2,2, 0
,,
|| 2|2 4 0| 4
| 2 22 | 4 4
2 2 16
Circle with centre
and radius 4
The distance between the point
and the
point
representing the complex
number
is 4
CARTESIAN FORM
P a g e | 45
7 77 11227 0
2 4
, 00
Circle with centre
(c)
| 3
3 | 2 3,3,1
3,
3,
1
,,
|| 3
33 1 1|| 2 2
3 3 1 1 2
Circle with centre
and radius 2
The distance between the point
and
the point
representing the complex
number
is 2.
CARTESIAN FORM
LESSON 2
Determine the Cartesian
equation of the locus of points satisfying the
following conditions.
(a)
(b)
2−|− 3√|3 ||
+
2| 3| ||
22|| 33 33||||||
4444324
3 24 3636
3 33 8 824
2412 36360 0
4 4 40,0, 4
−+− √3
|| 1 1| √1|3| √232|| 2|
| 1 11 | √33|222|
3 1
2 222 21
1 14
1
2
123
12
3
14 11 11 0
SOLUTION
(a)
CARTESIAN FORM
Circle with centre
(b)
and radius 2
and radius
√
.
,,
LESSON 3
Sketch the locus of the point
representing the complex number
, given that
.
Write down the Cartesian equation of the locus.
| 3| | 2 5|
| 3| | 2 5|
| 3 3| | 25
|
25
,,
0,
0
,
3
,, ⊥ 2,2,55
| 0 3| | 25
25|
| 3 3| | 2 2 55|
3 3 2 2 5 5
6 6 9 4 4 4 10
10 2 25
1616 4 4 20 0
4 5 5 0
arargg 3 332
SOLUTION
Rewriting
The distance between the point
,
representing the complex number
, and
the point
is equal to the distance between
and the point
. Therefore, we are
finding the bisector of
.
LESSON 4
Describe and sketch the locus of
the points satisfying the following conditions.
(a)
(b)
CARTESIAN FORM
SOLUTION
(a)
arg 3 3
P a g e | 46
arg 30
3 0 4
arg− 3 3 4
tan 3 3 4
3 3 t3an>43 1
3,3, 0
;
This is the half line starting at
positive real axis.
(b)
SOLUTION
| 2| 1
| 0 2 2| 1
Circle with centre
, not
including (3, 0), making an angle of
| 2| ≤ 1
LESSON 6
Shade on an Argand diagram the
region in which
.
with the
arargg32 32
32 3
ar 2g2 3 3 2 2 3
3 23 t√a3n333√3√3
√3– 2 3√3 > 33,3,2
;
The half line starting at
, exclusive,
which makes an angle of
with the positive
0,0, 2
and radius 1.
LESSON 7
(a) Sketch on one Argand diagram:
(i) the locus of points satisfying
| | | 2 2|
arg 4
| | ≤ | 2 2| – ≤ arargg ≤
||| |022|| | 20
2 0|
0
,0,
1
ar2g,2, 0
arg– 0 0,0, 14
(ii) the locus of points satisfying
real axis.
(b) Shade on your diagram the region in which
and
SOLUTION
(a) (i)
22 1
21 11
11 1 2,2, 1
LESSON 5
where
Describe and sketch the locus of
SOLUTION
Using vectors
This is the line passing through the point
and parallel to the vector
, i.e
This is the perpendicular bisector of the
line segment joining the points
and
(ii)
Half
line starting at
making an angle of
axis.
, excluding,
with the positive -
P a g e | 47
(b)
(ii)
LESSON 8
arg s sin−
LESSON 9
(a) Sketch on an Argand diagram the locus of
points satisfying the equation
| 6 6| 3
| 6 6| 3
| |
1 < ≤ 1
(b) It is given that satisfies the equation
.
(i) Write down the greatest possible value of
.
(ii) Find the greatest possible value of
,
giving your answer in the form
, where
.
SOLUTION
(a) Circle with centre
arg
0,0, 6
(a) On the same Argand diagram, sketch the loci
of points satisfying:
(i)
(ii)
|arg3
3 33| 5
(b) (i) From your sketch, explain why there is
only one complex number satisfying
both equations.
(ii) Verify that this complex number is
74
–
SOLUTION
and radius 3
3,3,3,3,110
(a) (i) Circle with centre
(ii) Half line, starting at
making an angle of
real axis.
(b) (i) 9 is the largest possible value of
| |
and radius 5
, exclusive,
with the positive
.
(b) (i) There is only one complex number
satisfying both equations since there is
only one point of intersection due to the
P a g e | 48
73
||743
|
4
3
3
|
4434 | 33
5
ar garg744
444 3
tan− 44 34
half-line which starts within the circle.
(ii) If
is the point of intersection it
must satisfy both conditions.
SEQUENCES
P a g e | 49
At the end of this section, students should be able to:
{}
1. define the concept of a sequence
of terms
as a function from the positive integers to
the real numbers;
2. write a specific term from the formula for the th term, or from a recurrence relation;
3. describe the behaviour of convergent and divergent sequences, through simple examples;
4. apply mathematical induction to establish properties
properties of sequences.
P a g e | 50
SEQUENCES
INTRODUCTION
A sequence is a list of numbers which obey a
particular pattern. Each number in the sequence is
called a term of the sequence. These are usually
denoted
where
is the first
term,
is the second term and
is the
term.
In some cases the sequence can be defined by a
formula an expression for the
term.
, , , … , −,
–
+4+ 1
11 ++
+
441 11 1 3
4423 11 11 71111
4445 11 11 15191159
3, 7,++11, 15, 19, ….
1122 1 11 23
33 2 1 42
44 3 1 53
55 4 1 64
3 54 55 6
2,2, 2, 3, 4, 5, …
1 1
21 12
21 14
21 81
2 16
(d)
LESSON 1
Write down the first 5 terms of
the following sequences:
(a)
(b)
(c)
(d)
SOLUTION
(a)
(b)
(c)
1 1211 3211 1
2, 41, 18,++16, +32, …
11 1 12 1 12 2
11 2 3 1 3 3
11 3 4 1 4 4
11 4 5 1 5 5
1 112 35 14 56
2 , 3, 4 , 5, 6, …
5, 8, 11, 14, …. 2, …
, , , , , …
1,1,1, , ,, , , , , …, …
× , ×, × , × , × , ….
2,2, , , , …
LESSON 2
For each of the following
sequences determine an expression for the
term, .
(a)
(b) 8,
(c)
(d)
(e)
(f)
(g)
6,
4,
2,
0,
SOLUTION
3 3 2
2
2 10 2
+
(a) Consecutive terms differ by 3 therefore we try
. To create the right formula we add 2 i.e.
(b) Consecutive terms differ by
therefore we
try
. To create the correct expression we
need to add 10 i.e.
(c) The numerators are the natural numbers
and the denominators are two more than the
numerator i.e.
(d) Ignoring the signs, each numerator is 1 and
the denominators are the natural numbers .
Since the signs alternate between positive and
P a g e | 51
11++
11++ +
1
1
++
1 1
+
negative, starting with positive, we use
. Therefore
(e) Each numerator is 1 and the denominators
are powers of 2 i.e.
(f) Each numerator is 1 and the first number of
the denominator is and the second is
.
Therefore
(g) The numerators are the natural numbers but
they begin with 2, i.e.
and the
denominators are the square numbers.
Therefore
The sequence above diverges since it does not
converge to any specific value.
A sequence can be classified as being convergent,
divergent, oscillating or periodic.
Convergent sequences as the name suggests
converge to a definite limit.
→lim
This oscillating sequence above is divergent.
The sequence above is convergent because it is
tending to a value.
This divergent sequence is PERIODIC as it consists
of a set of values which are constantly repeated.
The repeating pattern of the sequence consists of
three values therefore the sequence is said to have
a period of 3.
The sequence above is OSCILLATING and
converges.
Divergent sequences are sequences which are not
convergent.
P a g e | 52
LESSON 1
Determine which of the following
functions is convergent or divergent. If the
sequence is convergent, determine the limit of the
sequence.
(a)
(b)
(c)
(d)
+
−
−
√+
lim 3
→
→lim 31 1
→lim 1
→lim 1 3 1
3
lim
→
→lim 77
→lim 7
1
→lim 1 7
0
lim √−+12
→
1
2
→lim √3 4
→lim 434 44 1
4
4
4
1
→lim 3 4
SOLUTION
(a)
4 1
4
→lim 3 4
4
1
4
→lim 3 4
l→im
→lim 1 By L’Hopital
0→lim 3
DOES NOT EXIST
Not convergent
(d)
Convergent and converges to 0.
LESSON 1
following
A sequence is given by the
4
+ 3
converges and it converges to 3
(b)
is convergent and it converges to 0.
(c)
Write down the first four terms of the sequence.
SOLUTION
4
+ 3 3 7
+ 3 3 1010
+ 3 3 1313
1,1,
+ 33 1, 1 ∈1ℤ+ 1,1,
∈ ℤ+
1
1 1 1 1
1
LESSON 2
defined by
A sequence of positive integers is
Prove by induction that
SOLUTION
When
,
P a g e | 53
1 11 1
+ 1 1 1 1
+ 33 1
11 1
133 1
13
1 3
2 2 1
2 2 1 1
1 1 1 1
+
1 1 1, ∈ ℤ+
5
+
2++11 2 2
2+ 1 1
2 1 1
5
+
2 +1 1
+ 2 1 1
++ 2 2++
2 +12
22+2 1
2 1 1
+
Therefore
is true
Assume true for
Now,
Therefore,
is true when
Mathematical Induction
LESSON 3
and
is true. Hence, by
A sequence is defined by
. Prove by induction that
.
SOLUTION
Therefore
Assume
is true
is true for
Now,
Therefore
is ture whenever
is true.
Hence by mathematical oinduction
2+ 1 1
.
P a g e | 54
SERIES
At the end of this section, students should be able to:
1.
2.
3.
4.
5.
6.
7.
8.
Ʃ
use the summation
notation;
define a series, as the sum of the terms of a sequence;
identify the th term of a series, in the summation notation;
define the th partial sum
as the sum of the first terms of the sequence, that is,
= ;
apply mathematical induction to establish properties of series;
find the sum to infinity of a convergent series;
apply the method of differences to appropriate series,
series, and find their sums;
use the Maclaurin theorem for the expansion of series;
9. use the Taylor theorem for the expansion of series .
P a g e | 55
SERIES
2
= 2
INTRODUCTION
, , , , … ,
⋯
Given the sequence
corresponding series is
, the
⋯
is the
11++
partial sum where:
the first partial sum
the second partial sum
are powers of 2 i.e.
partial
sum
LESSON 1
Write each of the following series
using sigma notation.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
581114⋯
86420
86420
2
2
⋯
⋯
⋯
1
⋯
1 ⋯ ⋯ ⋯
×2 × × ⋯× ×
33 2
3 2
= 2
2 102
102
=
1
++ 1
1
1
=
1
= 2−
(v) Each numerator is 1 and the denominators
the third partial sum
the
(iv) Ignoring the signs, each numerator is 1 and
the denominators are the natural numbers .
Since the signs alternate between positive and
negative, starting with positive, we use
. Therefore
1 1
1
= 11
1 1
1
1
=
(vi) Each numerator is 1 and the first number of
the denominator is and the second is
.
Therefore
(vii)
The numerators are the natural numbers
but they begin with 2, i.e.
and the
denominators are the square numbers.
Therefore
SOLUTION
(i) Consecutive terms differ by 3 therefore we try
. To create the right formula we add 2 i.e.
(ii) Consecutive terms differ by
therefore we
try
. To create the correct expression we
need to add 10 i.e.
(iii) The numerators are the natural numbers
and the denominators are two more than the
numerator i.e.
+
The following standard results can be used to find
the sum of various series.
1
1
,
1
1
2
2
1
,
2
6
=
=
= 4 1
∑∑== 1 1
LESSON 1
Find each of the following sums
(a)
(b)
SOLUTION
(a)
∑== 1 1
P a g e | 56
=
= =
4
(b)
406 4 1 124 1 1 42 4 1
∑=
= =
16164 471161
1 61 94 9 1 1
∑∑====11221 1
∑== 1 1 1 1
= 1 1
= = 1
611112222116
1 1226 1 6
22 3 3616
22 3 36 156
22 65 1 1
∑= 26 2
= 2 2
= 2=
4 1 1 2 2 6 1122 1
LESSON 2
Express each of the following in a
factorized form.
(a)
(b)
SOLUTION
(a)
(b)
31 111334 41121 141222111
1 133 3123 8 4
12
LESSON 1
that
Prove by mathematical induction
1
= 1 1 12 1 1 33 2
1
: = 1 1 12 1 1 33 2
: 11 1 1 121 11 1 131 2 2
0 0
1
: = 1 1 12 1 1 3 2 2
+ 1
+: = 1 12 1 1 1 1 1 3 1 2
121 1 1 2 233 5
121 1 1 2 23 5 5
1 1
for all positive integers .
SOLUTION
Therefore,
Assume
Now,
is true.
is true for
+
term
+ 121 1 13 2 2 1 1 11
1 1
121 1 1 1 133 2 1212121 1
121 1 1 1 13 2 2 12 1
P a g e | 57
121 1 133 2 12 12
121 1 133 11
11 10 10
121 1 1 2 233 5
+
1
1
1
1
1
3
3
2
12
=
1 1
= {={ 1 }
Therefore
is true whenever
is true.
Hence by mathematical induction
:1: 1: 111112 2111 21211
1
= 1 1 3 1 1 22
∑== 1 1 ∑== ∑==
(iii)
for all positive integers .
If
, then
LESSON 1
LESSON 2
(i) Express
in partial fractions.
(ii) Use the method of differences to show that
(i) Show that
1 1 2 2 1 1 1 ≡ 33 1 1
(iii)Write
(iii) Write down the limit to which
(ii) Hence use the method of differences to find an
expression for
1 1
=
∑∑=== 1 ∑1= .
1111222 211111 1
33 1 111 2 2
∑= 1: 1 1 121311120211 1
2:3: 233445 23 231234
4: 456 4 435
converges as
(iv) Find
(iii) Show that you can obtain the same expression
for
using the standard results
for
and
(ii)
tends to infinity.
giving your answer to 3 significant figures.
SOLUTION
SOLUTION
(i)
16 1 12 1 1 2 3 1 1
16 1 122 1 6 1 1
16 1 12 1 3
16 1 122 4
3 1 1 2 2
4
1 1 2 2
4 3 2 1
= 1 1 2 2 2 1 1 2 2
4
= 1 1 2 2
4
= 1 1 2 2
(i)
+ +
4+
+++
+ 1
1 2 2 2 2
1
1
0
44 02 12
2 1
4 11 1112
12
When
When
P a g e | 58
3 3
2
42 222 222 1
1 4 2 3 1
++1+1+2 2 1 1 2 2
∑= +
4
2
3
1
11:1 2 22 3 111 2 2
1
2
3
2
3
2: 22 33 141
3: 32 43 51
4: 4 5 6
1: 2 12 1 3 3 111 1
: 4 1 1 2 2
= 1 1 2 2
21 32 22 11 1 13 1 21 2
When
(ii)
32 12 1 21 2
→ ∞4 3 2 1
= 1 1 2 2 2 1 2
3 2 1 2 1 1 1 2
3 2
∑= +++++ + +
= 141 2 2 = 141 2 2
0.3210410012 10021 32 4912 4921
(iii) As
(iv)
ARITHMETIC
ARITHME
TIC PROGRESSIONS
PROGRESSIONS
, , , … −, , …
−
2 2
1 1 for every > 1
INTRODUCTION
A sequence
is called an
or
there exists a constant , called the
such that
if
That is
Therefore,
LESSON 1
Find the common difference for
each of the following arithmetic progressions.
3, 5, 7,2 9,7,11, … …
2, 5, 8, 11, …
3,3, 5,5, 53
5 37,7, …2
8,8, 3,3,222,23 , 55 5
2,2 , 5,511, 88,8, 3 11,…
11, 15, … is an arithmetic
−
4 1
− 4 1 1 1
44 5 5
− 44 1 44 55
4
4
(a)
(b) 8,
(c)
3,
,
SOLUTION
(a)
(b)
(c)
NB: Any pair of consecutive terms can be used.
LESSON 2
Prove that the sequence 3, 7,
progression.
SOLUTION
We need to show that
Therefore,
is a constant.
P a g e | 59
5 2 2
LESSON 3
The sum, , of the first terms
of a sequence is given by
. Show
that the sequence is an arithmetic progression
with common difference 10.
−
5 2 2 1 15 1 1 2 2
5 2
2 1 15 5 2
5 2
2 1 155 7
5 2
2 55 7
7 55 7 7
5 2
2 55 1 12 7
5 2 2 5 12 12 7
1010 7 7
− 1010 1 1 7
1010 10 7
10 17
− 1010 7 7 1010 17 17
10
SOLUTION
502
1111 120120
502
21 11 13535 70
3636 25050 3636 12
30602
LESSON 2
The last term of an arithmetic
progression of 20 terms is 295 and the common
difference is 4. Calculate the sum of the
progression.
SOLUTION
20,20, 295, 4
201
201 141295
219
202 2219219 2020 14
5140
We need to determine
LESSON 3
The sum of the first 6 terms of an
arithmetic progression is 54.75 and the sum of the
next 6 terms is 63.75. Find the common difference
and the first term.
SOLUTION
622254.756 1 1 54.75
6
15
15
54.
54
.
7
5
5, 9, 13
54.
75
7
5
63.
75
7
5
118.
11
8.
5
12
55101
1 011 14
1222 66
121
121118.118.5 118.5
66
2
41
6
15
15
54.
54
.
7
5
12
66
66
118.
11
8.
, , , … , ⋯ ⋯ ×2: 5
3066 109.118.10119.8.55
30
12
12
66
36
9
2 22 1 1
1
8.4 5
250
120
LESSON 4
If the first three terms of an
arithmetic progression are 5, 9, and 13, what is
the value of the 10 th term?
SOLUTION
(1)
The common difference, , is 4
(2)
Solving (1) and (2) simultaneously
Sum Formulae for Finite Arithmetic Sequence
If
is a finite arithmetic sequence,
then the corresponding series
is called a finite arithmetic series. The sum of
the first terms of the series, which we denote ,
would be stated as
LESSON 1
Find the sum of the even
numbers from 50 to 120 inclusive.
SOLUTION
(1)
P a g e | 60
GEOMETRIC PROGRESSIONS
, , , … , , …
INTRODUCTION
A sequence
is called a
or
there exists a nonzero constant , called the
such that
if
−
, > 1
27, … is a Geometric Progression.
33−, −− 3−, 3, 3
− 33− 3
− 33−−−
−−−
−
3
Therefore,
or
LESSON 1
Prove that the sequence 1, 3, 9,
SOLUTION
We need to show that
is a
constant.
LESSON 2
The first and fourth terms of a
geometric progression are 6 and 20.25
respectively. Determine the 8 th term of the
progression.
SOLUTION
6 20.25
√20.3.6327553.1.3575
63 6561
2 64
Since length cannot be negative
Sum of a Geometric Progression
The sum of the first terms of a G.P is given by
LESSON 1
The fourth term of a geometric
progression is 6 and the seventh term is
.
Calculate
(i)
the common ratio,
(ii)
the first term,
(iii)
the sum of the first eleven terms.
SOLUTION
(i)
486
2 486 8
8 8 6 6
34−−
−3
4 1 22221 1
512.
25
1
51
−1 5 11−
4 5 5 13 5 45 1−3
4 5 413 5 41−4 3
5 44 313− 114 3 13
(ii)
(iii)
LESSON 2
LESSON 3
The lengths of the sides of a
triangle are in geometric progression and the
longest side has a length of 36 cm. Given that the
perimeter is 76 cm, find the length of the shortest
side.
SOLUTION
Let longest side be
and shortest side be .
36 76
3636 363636367676
369 93693610 40 0
3 52 53 2 2 0
3
3636 23 16
1 ,
1
><1 1 or 111 1 ,
48
Given that
, find
and prove that this sequence is a Geometric
Progression.
SOLUTION
P a g e | 61
5 6 13−−
− 556 131−
− 65 31−
1 36 3
(ii)
1 12 14 18 161 ⋯
If we think about it we should realise that the sum
appears to be 2. Since the sum appears to tend
towards a specific number as it goes on
indefinitely we refer to this series as a
CONVERGENT series. The sum of this series can be
given using the formula
1 , 1<1< 1
1 1 12 2
therefore
and
,
Thus we see that our intuitive answer is indeed
correct.
LESSON 1
The first and fourth terms of a
geometric progression are 500 and 32
respectively. Find
(i)
the values of second and third terms
(ii)
the sum to infinity of the progression
500
3232
500
32
250032 1258
5
500500 2
200 5
SOLUTION
(i)
LESSON 2
The first term of a geometric
progression is and the common ratio is . Given
that
and that the sum to infinity is 4,
calculate the third term.
SOLUTION
Sum to Infinity
What would be the sum of the infinite series
For our series above we have
2
8085005000 5
−500
1 2500 25
3
12
121
44 41 1212
4 116
4
121212 1
3 4
31
34
16
LESSON 3
The first term of a geometric
series is 120. The sum to infinity of the series is
480. Given that the sum of the first terms is
greater than 300, determine the smallest possible
value of .
SOLUTION
1 120
4804804801 1 120
1
1 3 14
4
1
> 3001
P a g e | 62
1201334> 300
143
12011 4 > 300
4
3
4801
>
300
4
3
1343 > 58
43 < 8 3
ln 43 < ln 83
ln43< ln 8
> llnn 834 ln 34 is negative
> 43.4
1
= 2
1 < < 1
1 1 1 1 1 ⋯
= 12 2 4 8 16
412 12
11< < 1
1 2 12 1
LESSON 4
Determine whether the
geometric series
is convergent. If it converges, determine its sum.
SOLUTION
We need to show that
.
Since
, the series converges.
P a g e | 63
MACLAURIN’S
MACLAURIN’S SERIES
INTRODUCTION
0 0 2!2! 0 3!3! 0 ⋯ ! 0 ⋯ ⋯
must be differentiable
0
0
Us– e Maclaurin’s Theorcosem to find
coscos0.0.2
coscos 0 coscos0 1
s sin 0 sisi n0 0
cos
cos 0 co coss 0 1
sin 0 sin0 0
coscos 0 coscos 0 1
sin 0 sisi n0 0
c cos 0 co coss0 1 1
0 0 2!2! 0 3!3! 0 ⋯ ⋯
! 0 ⋯
1 0 112!2! 0 3!3! 1 4!4!
0 5!5!
11 6!6!
11 2!2! 4!4! 6!6! ⋯ ⋯
0.
0
.
2
0.
0
.
2
0.
0
.
2
0.0.2 1 2 24 720
0.98
must exist at
The derivatives of
must exist at
Only within specific values of
is the series valid.
LESSON 1
the first four non zero terms for
determine an approximation for
SOLUTION
, hence
.
1 cos
LESSON 2
Find the Maclaurin expansion for
up to and including the term in .
SOLUTION
11 cos
11 2 cos
11 2 11 2!2! ..
using result Ques tion 1
11 2!2! 2 2 ⋯ ⋯
11 2 12
Fi
n
d
t
h
e
Macl
a
ur
i
n
’
s
s
e
r
i
e
s
f
o
r
tan
LESSON 3
up to
.
SOLUTIONhttp://sirhunte.teachable.com/courses
/93027/lectures/2211764
0 0 2!2! 0 3!3! 0 ⋯ ⋯
! 0 ⋯
tan
sec
2tanse
n sec
2se2secc
4sec tan
0 tan0 0
0 sec 0 1
0 2tann0 sec 0
0
2se2sec0c0 4sec 0tan 0
2
0 0 2!2! 2 2 3!3! ⋯
P a g e | 64
23!3!
+
Obtain the Maclaurin’s series expansion for
0.0.1
LESSON 4
A function is defined as
.
(a)
up to and including the term in .
(ii) Hence, estimate
to four decimal places.
SOLUTION
++ 0
2 + 0 2
4 + 0 4
8 + 0 8
16 0 16
0 0 2!2! 0 3!3! 0
⋯ ! 0 ⋯ ⋯
2 4 4 2!2! 8 84 3!3! 16
1624!4!
0.0.12 2 22 20.02.1 220.03.1 0.30.1 0.0.1
3.3201
−−
,′
13−−
13
0 1 −− −−
22 13
1 3 33 6 13
1 3
(i)
Valid for
LESSON 6
l
n
1
ln1
–
ln +−
ln 3+− 3
(i) Use the Maclaurin series for
and
to obtain the first three non zero
terms in the Maclaurin series for
You are given that
Find
and
Maclaurin series for
(ii) Find the value of
for which
. Hence
find an approximation to
, giving your
answer to three decimal places.
SOLUTION
(i)
.
. Hence obtain the
as far as the term in .
By considering the equivalent binomial expansion,
give the set of values of for which the Maclaurin
series is valid.
SOLUTION
.
State the range of validity for this series.
(ii)
LESSON 5
0 6 −−
18
1813
1 3 33 545413
1 3−−
0 54 −−
96
9612
1 2 33 28828813
1 3−−
0 288
0 0 2!2! 0 3!3! 0 ⋯ ⋯
! 0 ⋯
1 6 36 4 288 6
16
1 6 9 48
48
1 < 3 < 1 → < <
ln11
1
1 1
0 ln11 0
0 1 0 1
01 1 10
1 2 20 1 020
1 6 0 661 60
P a g e | 65
1 24 0 241 240
0 0 2!2! 0 3!3! 0
⋯ ! 0 ⋯ ⋯
1
0 2!2! 2 3!3! 6 4!4!
2424 5!5! ⋯ ⋯
2 3 4 5 ⋯
ln1 0 ln1 0
1 1 10 1 1 0
1 1 10 1 10
1 2 20 1 20
6 0 6 6
1 1 0
24 024 24
1 1 0
0 0 2!2! 0 3!3! 0
⋯ ! 0 ⋯ ⋯
0 2!2!⋯2 3!3! 6 4!4! 2424 5!5!
2 3 4 5
ln 11 ln1 ln1
2 3 4 5 ⋯
2 3 4 5
2 23 25 1 < < 1
+− 3
14 2 3 33
12 1
ln3 ln 11 212
1
2
1
2
1
221.0296 3 2 5 2
(ii)
P a g e | 66
INTRODUCTION
Maclaurin’s Series is:
0 0 2!2! 0 3!3! 0 ⋯ ⋯ ! 0 ⋯ ⋯
For Taylor’s Series, we let
2!2! 3!3! ⋯ ⋯ ! ⋯
1.
2.
The function
The function
has to be infinitely differentiable
has to be defined in a region near the value
Furthermore, replacing
with
.
we get
2!2! 3!3! ⋯ ⋯ ! ⋯ ⋯
= !!
–
ln 3 3
LESSON 1
Find the first three non
terms of the Taylor expansion of
SOLUTION
zero
.
LESSON 2
+
Find the first four non
–
zero
1
+ 3 3−−
31 3 1 14
313 1 161
323 1 321
363 1 1283
terms for the Taylor expansion of
.
with centre
ln
ln 3 3 → 3
ln1 3 ln1 3
3 3
1 3 19
2!2! 3!3!
⋯ ! ⋯
ln 3 3 ln3
n3 13 19 2!2!1
2!2!
ln3
n3 13 181
3!3! ⋯ ⋯ ! ⋯ ⋯
SOLUTION
Let
1 3 3 14 161 1 1 321 2!2!1 1 1 1283 3!3!1 1 1
P a g e | 67
2
2
2
3
2
3
√
√
√
sin16 2 2√2 163 3 4 16
12
16
0.1920
14 161 1 1 641 1 1 2563 1 1
– sin
LESSON 3
(i) Obtain the first four non zero terms of the
Taylor Series expansion of
in ascending
powers of
.
(ii) Hence, calculate an approximation to
SOLUTION
sin
3!3! ⋯ ⋯ 2!2!!
4 sin √2
4 sin4 2
coscos 4 cos4 √22
s sin 4 sisin 4
√22
cos
cos
co
s
cos √2 4
2
sin √22 √22 √241 √22 2!2!1 4
2
3!3!
4
√22 √22 4√2√42 4
12
4
(i) let
(ii)
sin
sin → 3
4 16 4 16
.
P a g e | 68
BINOMIAL THEOREM
At the end of this section, students should be able to:
∈ ℚ
1. explain the meaning and use simple properties of
2. recognise that
that is,
! , ∈ ℤ
and
, that is,
, is the number of ways in which
where
objects may be chosen
from distinct objects;
3. expand
for
;
4. apply the Binomial Theorem to real-world problems, for example, in mathematics of
finance, science.
P a g e | 69
PASCAL’S TRIANGLE
INTRODUCTION
For any positive integer :
1
111 1
11 2 2 1 1
11 3 3 3 3 1 1
11 4 4 6 6 3 3 1 1
! 1 1 2 2 3 3 … 321
8! 8×7×
8× 7×6×6×5×5×4×4×3×3× 2×1
2× 1
0!0!
!!
FACTORIALS
For example,
Is defined as 1.
LESSON 1
Simplify
SOLUTION
!
9!9!6!6! 9×8×7×6×5×4×3×2×1
6×5×4×3×2
6×5×4 9 ×3×2
× 8 × 7 ×1 504
!! 2 2!
! 2 2!
1 1 2 2 3 3 … 321 2 2!
11 2 2! 2 2!
2 2! 1 1 1
2 2! 11 −−!
−
−! 72
13 13 24 24 35 35 …… 332211 72
1 1 2 2 72
LESSON 2
Simplify
SOLUTION
LESSON 3
SOLUTION
Solve the equation
.
PASCAL’S TRIANGLE
1
1
1
1
1
1
2
1
3
4
3
6
1
4
1
3
3 2 7272
3
3 70 0
10
10 7 7 0
10 7
since
LESSON 4
in invalid
Show that
1 2 2! 31 31! 32 3!
1 2 2! 3131!
3 3332 2! 31 31!
33 33! 3131!
3331! 1
3
32 3!
SOLUTION
P a g e | 70
LESSON 5
+
+!!
(a) Express
and
in the form
are integers.
+
+! +
+!
, where
(b) Hence find
1
= 2 2!!
+
+ !! +
+ ! +
+ !
SOLUTION
(a)
11 11 212!! 2 2
2 2 1 1! 2 2!
1 1 2 2
1
2 1
2 1
1
1
1
1
2 2!! 1 1! 2 2!
1 1 1
= 2 2!! = 11! = 2 2!
1: 2!2!1 3!3!1
2: 3!3!1 4!4!1
3: 4!4!1 5!5!1
Equating coefficients of :
Equating constants:
(b)
1: !1 11 1!
: 11 1! 21 2!
1 1 1
= 2 2!! 2 2 2!
P a g e | 71
THE BINOMIAL THEOREM
INTRODUCTION
We now look at an alternative to Pascal’s Triangle using!Factorials.
! !!
For any positive integer :
= −
0 1 − 2 − 3 − ⋯ 1 1 −
32
3 2
LESSON 1
Determine the expansion
.
32
3 25 5 5 5 5 5
0 3 1 3 22 2 322 3 3 22 43 22 5 22
243810
24243 581811080
22 101072
2727720440 24
240101009883232 531616 13232
2
3
3
23
2
3
5
0
105 2 23
3
1959552
2522523232243
13
1 32
13
1 3 16
1 6 9
2 64 2 65 2 66
60151541 16262 1
6
960
1 6 9…60
…60 1 12 172
6
12
12
6
0
54
72
540
0
469469
SOLUTION
LESSON 2
Find the 6th term of the expansion
SOLUTION
The 6th term of the expansion
begins with
LESSON 3
.
occurs when
since the summation index
.
Find the coefficient of
in the expansion of
SOLUTION
We only need the terms which will result in a
term after multiplication.
We isolate the multiplications which would create an
term.
.
P a g e | 72
LESSON 4
Find the term independent of
in the expansion
3
.
SOLUTION
86 33 121
28286399 64
16
The term independent of :
1 11 2!2! 1 1 13!3! 2 2 ⋯ ⋯
1 < < 1
1 1
1
1 2!2! 1 1 13!3! 2 2 ⋯ ⋯
1 < < 1 → < <
12
1 2
12
1 23 11 2
23 3 1 3 3 3 1
1 2 22 2 2 121 22 2 2 12 2321 22
1 3 32 12 12 4 32 12 12 312 8
13
1 3 32 12 1 < 2 < 1 12 < < 12
For any real number
provided that
LESSON 1
SOLUTION
Find the binomial expansion of
up to and including the term in
.
P a g e | 73
LESSON 2
Find the first three terms of the expansion of
SOLUTION
−
.
423 423
2 3−− 421
421 32 − 424 2− 11 32 −
−
3
21
21 2
3
1
3
221 11 2 1111
11 21 2
212 1 32 94 ⋯ ⋯
23
2 3 92
+
++
+
+ +
+
++
+
9
1 25
2 5
+
++
+ + +
99 2225
25
5
1
5
2 0
5 3 9
69 3 6
1 2 5 5 1 25
+331 −−
33331
11 11 1111
11 211 1111
1112
12 3121
33
3 3 3 3 3
LESSON 3
(i) Express
in the form
where
(ii) Hence, or otherwise, find the expansion of
including the term in .
(iii)Find
(iii) Find the range of values of
SOLUTION
(i)
Equating constants:
Equationg coefficients of :
(ii)
and
are integers.
as a power series in ascending order up to and
for which the series expansion of
P a g e | 74
625
6625
2 55−− −
621
21 2 −
62− 11 −52
31
31 52 5
1
5
1
5
31
31 511 252 1251111
11 21 2 1111
1112
12 321 2
313 115 2 754 3758
33 2 4 8
31 256
933
3 363 3 33513 3 152 754 3758
2 4 8
1 < < 1 1 < < 1 → < 2< 2
5 < < 5
√1
√82
√1 11
11 14 14 14 1211 14 14 114 23211
11 14 323 1287
82 8 1 1
181
1 81 811
811 811
1
81 11 81
(iii) Valid for
LESSON 4
and
Use the expansion of
SOLUTION
This works for small .
to setimate
to four decimal places.
P a g e | 75
1
31
31 81
1
1
3
1
7
1
331 4 81 32 81 128 81
31.1.003
3.009
16
186
6
8 6
LESSON 5
(a) Find the binomial expansion of
√144
1 6
6 14
11 4664 1 66
86
8 6 8 1
8 11 68
4141 618
4141 4118 418
42
4
2
4
81441446 121122 12
6 24
3
√144≅ 4223 14 23 ≅ 479
(b) Find the binomial expansion of
(a)
(b)
(c)
.
up to and including the term in
(c) Hence, find an estimate for the the value of
SOLUTION
up to and inclusing the term in
in the form where
and
.
are integers.
P a g e | 76
ROOTS OF EQUATIONS
At the end of this section, students should be able to:
0
1. test for the existence of a root of
where f is continuous using the Intermediate
Value Theorem;
2. use interval bisection to find an approximation for a root in a given interval;
3. use linear interpolation to find an approximation for a root in a given interval;
4. explain, in geometrical terms, the working of the Newton-Raphson method;
5. use the Newton-Raphson method to find successive approximations to the roots of
, where is differentiable;
differentiable;
6. use a given iteration to determine a root of an equation to a specified degree of accuracy.
0
P a g e | 77
,, 0
2 2 1
1 1 2 21 1 1 1 11
2 2 2 22 2 2 1 1
1 ×× 2 1
,
THE INTERMEDIATE VALUE THEOREM
If
in
is a continuous function on the closed interval
such that
.
LESSON 1
and the product
< 0
then there exists
Use the Intermediate Value Theorem to show that
has a root between 1 and 2.
SOLUTION
is a polynomial and therefore continuous on the
By the Intermediate Value Theorem there must be some
root between 1 and 2.
1
3 4 4 1 0
3 4 4 1
0 1
1 31 4 41 1 1 6
interval
∈ 1,1, 2
such that
0
1,1, 2
. Therefore there is a
LESSON 2
Use the Intermediate Value Theorem to verify that there is a root of the equation
between 0 and 1.
3 4 4
SOLUTION
Let
0,0, 1
0 ×× 1 6
∈ 0,0, 1 0
is a polynomial and therefore continuous on the interval
By the Intermediate Value Theorem there must be some
root between 0 and 1.
.
such that
.
. Therefore there is a
P a g e | 78
DETERMINING THE ROOTS OF AN EQUATION
3 3 1
LESSON 1
Given that
has
a root between 2 and 3, find this root to 1 decimal
place using the bisection method.
LESSON 1
Use Linear Interpolation, twice
over, to determine the root of the equation
in the interval
.
3 3 1
3 3 1
3 3 1 0
SOLUTION
SOLUTION
3 3 1
1
3 3 1 0
3 3 1
2 2 3 32 2 2 1 33
3 3 3 33 3 3 1 2
–
2.2.5 2.5 3 3 2.2.5 2.2. 5 1 1.1. 625625
–
2.2.755 2.75 3 32.2.755 2.2.7575 1
0.140625
–
2.2.87575 2.875 3 3 2.2.87575 2.875 1
0.8418
2.75
–
2.2.8125125 2.8125 3 3 2.2.8125125 2 2.8125 1
0.33
2,2, 3
Let
Mid
point of interval is 2.5
Since there is a sign change between 2.5 and 3 the
root occurs within this interval. We therefore now
bisect this interval. Mid point is 2.75
Due to sign changes the root must be in the
interval 2.75 and 3.
Mid
point is 2.875
Due to sign changes the root is between
2.875
Mid
and
point is 2.8125
3 3 1
2 2 3 32 2 2 1 3
3 3 3 33 3
311 2
23 33 2 2
2 2 2 33
2 4 9 3
5 1313
5 2.6
Let
Using similar triangles
Therefore root lies between 2.75 and 2.8125 and
since to 1 decimal place both limits are the same
the root is approximately 2.8.
13
13
13
5 5 3 3 5 135 1 1.1. 1
P a g e | 79
21.1 32.2. 6
2 2.2. 6 1.13
2 5.5.2 3.3.3 1.1.11
3.1 8.5
2.74
INTRODUCTION
44 10900900 0
–
LESSON 1
The equation
has exactly one real root,
. Taking
as the first approximation to ,
use the Newton Raphson method to find a
second approximation, , to . Give your answer
to four significant figures.
4 4 900900
3 2 2 4
10
49000
1010 1031010110024241011010090
1010 28440
9.859
2 5 5 3 3 13 0
– 1<<2
2 51.55 3 3 13 0
1,1, 2
5
3
2
5
3
1
3
12 2215 552513 33231311313 13 2929 33
1 ×× 2 < 0
1,1,2 2 5 5 3 3 13
1.56 1 10 3
1.1.5 1921.1.561.15.551.1.51010 31.1.351.1.53 13 13
1.1.5 63
SOLUTION
Let
LESSON 2
The equation
has exactly one real root
.
(i) Show that lies in the interval
.
(ii) Using the Newton Raphson method with
initial estimate
to estimate the root
of the equation
in
the interval
, correct to 2 decimal places.
SOLUTION
(i) Let
Since
, by the Intermediate
Value Theorem there is a root in the interval
(ii)
P a g e | 80
1.1984
1.1984984 21.1.198498461.1.1598498451.1.110984984101.13.198498431.1.19849843 13
1.1.114679840.0516
1.1467 21.1.1467467 5 51.1.1467467 3 31.1.1467467 13
10
6
1.
1
.
1
467
467
10
1.
1
.
1
467
467
3
1.1984984 0.001357
01357
1.1453
1.15
+ +
2 4 4 1 0 1.2
Since
and
correct to 2 decimal places are
both equal to 1.15,
.
DERIVING AN ITERATIVE FORMULA
EXAMPLE 1
Show that
is an
approximate iterative formula for finding the root
of
. Apply the iterative
formula with initial approximation
, to
obtain an approximation of the root to 2 decimal
places.
SOLUTION
2 4 4 1 0
2 4 1
42 1
42 1
+ 421 1
1
41.1.22 1 1 1.2386
41.1.23863862 1 1 1.2551
41.1.25515512 1 1 1.2621
1.26
Therefore, the approximation is
decimal places.
correct to 2
P a g e | 81
MATRICES
At the end of this section, students should be able to:
1.
2.
3.
4.
5.
6.
××
2,2, 3
reduce a system of linear equations to echelon form;
;
row-reduce the augmented matrix of an
system of linear equations,
determine whether the system is consistent, and if so, how many solutions it has;
find all solutions of a consistent system;
invert a non-singular
matrix;
solve a
system of linear equations, having a non-singular coefficient matrix,
matrix, by using
its inverse.
3×3 3×3
P a g e | 82
MATRICES
A matrix is a rectangular array of elements enclosed in brackets. A matrix is defined: number of rows
×
Two matrices are equal if they contain the same corresponding elements.
The addition and subtraction of matrices is only possible if the matrices are of the same size.
number of columns (this is the size/order of a matrix).
A square matrix contains the same number of rows and columns.
o
Matrix addition is commutative and associative.
Two matrices can be multiplied if the number of columns in the first matrix is equal to the number of
rows in the second.
o
Matrix multiplication is not commutative but it is associative.
2 × 2 10 01 3 × 3
11 1 222 311 210 111 101
−
The identity matrix for
LESSON 1
(i) find
If
matrices is
and for
matrices it is
100 010 001
and
(ii) deduce
SOLUTION
(i)
(ii)
11 1 222 131 210 111 101
1
2
2
2
1
1
1
0
1
1
1
2
1
1
1
1
1
1
2
2
0
1
1
1
11122222211113300 111111122211 13 1311 11111222200 13 1311
4
0
0
00 40 04
−4
14 2 1 1
− 10 11 01
Therefore
.
P a g e | 83
×
(i)
ℎ
ℎ ℎ
||| 0
| |||
|
||| 0
||| ×
|
|
∈
ℝ
|
||| ||| |
|
|
|
0
–
det−
311 121 213
–
INTRODUCTION
The determinant of
is
.
o
If the matrix
zeros then
contains a row with ALL
.
o
For square matrices
o
If
o
.
contains two identical rows or
columns then
.
Interchanging two rows or columns of a
matrix
changes the sign of the
determinant.
o
For
,
where
is a
matrix.
o
If a row or a column of a matrix is
multiplied by a constant, , then the
determinant of the matrix is multiplied by
.
o
Adding a multiple of one row to another
does not affect the determinant.
o
o
11 21
| 323| 13121 13111311 11312
2
1
1
1
2
11335 1 12 2 211
| | ≠ 0 –
2 31 14
20 3
22 310 143 0
2 10 43 32 4312 10 0
26393 32334 2810122 0
9 3232
9
36 61 105
925
SOLUTION
(ii) Since
LESSON 3
,
is non
singular.
Find the value of
for which the
matrix
is singular.
SOLUTION
LESSON 4
Determine
by
factoring.
A square matrix is singular if
SOLUTION
otherwise it is non
be obtained by factorizing the matrix. This is done
singular.
o
LESSON 2
The matrix
is given by
.
(i) Find the value of the determinant of
(ii) State, giving a reason, whether
or non
singular.
.
is singular
The determinant of a matrix can
by factoring out the common factor from each row
or column.
3 213 621 1055
1
6
3
35523 21 01
factoring out 3 from
factoring out 5 from
To find the determinant it is easier to use the
elements of
because it contains a zero.
1515 3 232 12121 16
1515377 1313 120
31 3 112 221
| || 5
,
–
6
4
4
1
9
3
1
6
3
5
5 9
4 26 81
5
7
9
3
1
6
54 26
2
2
3
3
2
3
3
2
||| 1 1 213 213 1
51122 11212 1 19
||| 4 69 38 2575 38 1575 69
4436590
90 2 241
41 187
87
55
16 52 46 4 2 1
16 52 46 57 9 6 38
567 6788 249
5×4 320
The transpose of a matrix is created by
interchanging the rows and columns. The
LESSON 5
The matrix
.
transpose of a matrix
is denoted
.
o
(i) Show that
(ii) Matrix
and
.
is changed to form new matrices
. Write down the determinant of EACH
o
A square matrix is symmetric if
o
A square matrix is called skew
symmetric is
.
.
of the new matrices, giving a reason for your
For example, if
answers in EACH case.
(a) Matrix
is formed by interchanging
column 1 and column 2 of matrix
and
then interchanging row 1 and row 2 of the
resulting matrix.
(b) Column 1 of matrix
then
LESSON 6
by
and
is formed by adding
column 2 to column 1 of matrix
. The
.
other columns remain unchanged.
(c) Matrix
The matrices
is formed by multiplying each
element of matrix
Calculate
by 4.
SOLUTION
(a) the determinant of
(i)
(b)
SOLUTION
(ii) (a) interchanging columns will change the
(a)
sign of the determinant to 5 and then
interchanging the rows will change the
sign of the determinant to
(b) The determinant remains as
.
(b)
since
adding a multiple of a row does not affect
the determinant.
(c) Since each element is multiplied by 4 the
value of the determinant is
and
are given
2×2
±1
|
|
|
,
∆
|||
If
then
,
are called the minors of
and
,
and
respectively. A minor of an element is obtained by
deleting the row and column containing that
element and finding the determinant of the
matrix which remains.
The cofactors of a matrix are determined by
multiplying the minor by
in the following
order
Step 1: Find the determinant of the matrix
Determinant of , det ,
The elements of any row or column can their
corresponding cofactors can be used to determine
|||
( )
the determinant. i.e.
using
the first row and its corresponding cofactors.
Step 2: Write the matrix, say
Step 3: Write the matrix,
the adjoint of
( adj)
, of cofactors.
. This matrix is called
−
|| adj
Step 4: Use the relation
to find
−
104 111 132
104 111 132
||| 11 111 324111 13
13155 4 422
11111 312 1041 321 1041 111
111 12 141 12 141 11
(5 1123 4 0 3 0 1)
3 263 31
5124 363 231
5
3
2
1
−
3 124 36 31
.
LESSON 7
SOLUTION
Let
Find the inverse of
3×1
1
0
3
73 52 01 –
− 10
128
||| 1 52 01 073 01373 52
1215 3 311
520 013 173 301 731 520
02 31 31 13 13 02
( 55 0 7 117 0 7 5 )
156 821 25
75 86 2115
11 5 26 515
− 2 71 82 215
− −
1−5 6 15 10
2 71 82 215 128
1521 2188
1 2 1751710101011010066121282281211212215
5
8
12 226 113
LESSON 8
and
are
are related by the equation
is non
Find
(a)
(b)
, when
SOLUTION
(a)
(b)
singular.
matrices and
, where
LESSON 9
Solve the equations
4 554519129
5 5 2020
, 1 5 5 2
41 55 41 2019
1 −15 30 45
− 150 28 5 06 2156
−
1501 158 306 4516 192
1 23005 0 2 5 20
150 450450
23
3
SOLUTION
Step 1: Write the system in the form
where
and
are matrices
Step 2: Find
Step 3: Multiply both sides of equation by
This equation is said to be consistent since it has a
solution.
→ 104 111 132 100 100 001
1
1
1
1
0
0
4
4 → 00 13 36 04 10 01
– → 10 11 1310 10 00
0 13 0 6 24 10 11 0
→ 00 31 36 04 10 01
3 × 3 1 3
1
0
2
1
1
0
3
→
0
1
3
0
1
0
00 10 1 1 1 00 0 23 14 31 01
10 1
3 → 00 10 31 043 11 013
000
1
1
0
1
0
2
4
2
1
3
3
→
0
1
0
4
1
–
0 0 1 53 1 32
1
1
0
0
3
2
2 → (00 10 0144 21 131)
3
3
→ 10
5
3
2
1
−
ℎ 0 ℎ
3 124 36 33
The following elementary row operations can be
performed on a matrix.
-
The interchanging of rows
-
The multiplication of a row by a non
zero scalar.
-
The adding of the multiple of a row to
another row.
These operations can convert a
matrix (or
any other matrix) to row echelon form
or
This can be done by
1.
Beginning with the left
most column and
use row operations to make the first
element in this column a 1 and the
elements below it zeros.
2.
Ignore the row and the column with the 1
created from step 1 and repeat step 1 on
100 ℎ → 100 10
the remaining matrix.
3.
Repeat this process until the desired
matrix is obtained.
1 0 11 13
4 12
LESSON 10
SOLUTION
Find the inverse of
A system of equations can either be consistent or
inconsistent.
A consistent system can have
a unique solution
infinitely many solutions.
An inconsistent system has no solution.
.
The use of row reduction greatly assists in
To obtain the inverse of a matrix
−|
|
|
104 111 132100 010 001
we can use row operations to convert the
augmented matrix
to
.
determining the consistency of a system of
equations.
An upper triangle matrix
unique solution.
10 1
0 0 1
indicates a
100 10 00
100 10 0
3 333
4 2 2 6
Matrices of the form
Since the system has a unique solution it is said to
infinitely many solutions.
Matrices of the form
solution.
LESSON 11
be consistent.
indicate no
A system of three equations is
(i) Write the augmented matrix for the system.
(ii) Use row reduction to solve the system of
equations.
1 0 11 1333
41 11 12 36
04 11 3236
3
1
1
1
→ 04 11 32 36
4
4 → 100 131 136 363
1
1
1
3
→ 00 31 36 63
→ 001 310 362 603
1
0
2
0
3
3 → 00 10 3333
13 → 010 100 321 103
: 3 13 3
3 3011 3
2 2 3
SOLUTION
(i)
(ii)
From
From
:
:
:
From
:
2 211 20
indicate
LESSON 12
Determine the general solutions
of the system of equations.
22222 343
2 4 2 8
11 22 12 34
2 4 21 82 2 3
→ 02 44 32 78
1
2
2
3
2
2 → 00 48 36147
14 → 10 212 234374
0 81 62 142 3
8
8 → 00 10 034 740
0 0 0 0
SOLUTION
The row of zeros indicates that
,
therefore we can use a parameter for one of the
variables and express the other variables in terms
of this parameter. Furthermore, the row of zeros
indicates that the system has infinitely many
solutions.
343 774
43 47
4 4
Let
From row 2:
From row 1:
2 23 2 7 3
23 4 7 4 2 3
2 1 2 1 2 3
2 2
1
1
1 0
1
0
Therefore, if we say that
and
So
then
.
would be one solution for the
system.
LESSON 13
Discuss the solutions of the given
4 8
3 33 410
2 2 2
13 2 132 121 410
4
1
1
1
Interchange and → 3 2 32 21 10
1
1
1
4
3
2
3
→
0
6
4
2 0 0 0 22 8
04 0 0 8 44
4
08 0 0 8 8 0
equations when (i)
(ii)
SOLUTION
(i) When
, we have
This result indicates that when
, the
system of equations is inconsistent and has no
solution.
(ii) When
, we have that
This result indicates that when
8
, the
system of equations has infinitely many
solutions.
66 44 2222
113 23
11 2 44
31 13 4
3 3
Let
From
:
From
:
DIFFERENTIAL
DIFFERENT
IAL EQUA
EQUATIONS
TIONS
At the end of this section, students should be able to:
0 a, ,, ∈ ℝ
1. solve first order linear differential equations
using an integrating factor,
given that is a real constant or a function of , and is a function;
2. solve first order linear differential equations given boundary conditions;
3. solve second order ordinary differential equations with constant coefficients of the form
, where
and
is:
(a) a polynomial,
(b) an exponential function,
(c) a trigonometric function;
and the complementary function may consist of
(a) real and distinct root,
(b) 2 equal roots,
(c) 2 complex roots;
4. solve second order ordinary differential equation given boundary conditions;
5. use substitution to reduce a second order ordinary differential equation to a suitable form.
DIFFERENTIAL EQUATIONS
A differential equation is an equation which contains derivatives of a function or functions.
For a first order differential equation the highest derivative is the first derivative.
For a second order differential equation the highest derivative is the second derivative.
sin
where
The solution of this type of equation can be
achieved by separating the variables and
integrating both sides of the equation with respect
These solutions are called general solutions of the
differential equation because the value of the
constant is unknown.
to the relative variables.
LESSON 3
LESSON 1
SOLUTION
5 3
Solve the differential equations
53 3
55 3 3
55 3 3
2 53 3
103 6
coscos tan
cost
o
s
t
a
n
1tan coscos ln
cossin coscos
lnsisin sisin+
sisin
LESSON 2
SOLUTION
Solve the differential equation
Find the particular solution of the
csc
4,4,
1 cscsc;c1 ; 3 when 4
csc1 −
csc
sisin −
coscos 1 1
cos1 31 4 1
2 14 1 → 4
coscos 4
differential equation
when
SOLUTION
.
INTRODUCTION
∫
Linear differential equations of the form
can be solved by multiplying throughout by the
Integrating Factor,
LESSON 4
.
Solve the following differential
3 3 7
3 7
. ∫3
. ∫− − −
− 3 3− 7−
− 3 3− 7−
27
equations
SOLUTION
Step 1: Write the DE in the form
Step 2: Calculate the I.F using
Step 3: Multiply each term in the equation by I.F
L.H.S resembles the product rule where I.F is
actually
Step 4: Integrate both sides of the equation w.r.t.
72
cotcot 2 coscos
2
cotcot 2 coscos
cotcot cos
cotcot sin lnsin
. sicoscosn
sin s sin.n . sin . 2sincn cos
sin cos
cos 2c2 cosss sin
si nsinsin csc
+
+
1 1
2
2 sin2 sin2
2 1 1
s1in cs csc
LESSON 5
Solve the differential equation
given that
when
.
SOLUTION
when
LESSON 6
Determine the particular solution
of the differential equation
−
4
4
3 0
4
4 −
4
4 −
given that
SOLUTION
when
.
I.F ∫
4 4 −
4 4
−
3,
0
32
− 2 2
When
0 andand d 0
,,
are first and second order homogenous equations
where
and
are constants. The solution of
these equations is called the complementary
function (C.F).
LESSON 7
5 2 2
1 5 2
1 5 2
5
ln 25
+
SOLUTION
5 2
2 0
Solve the differential equation
In general, the solution (complementary function)
0
−
of a first order differential equation is of the form
is
Auxiliary Quadratic Equation
Given the equation
0
If the auxiliary quadratic equation has a
repeated root , ,
The complementary function is determined by the
is the general solution of the differential
equation.
roots of the nature of the roots of the quadratic
auxiliary equation which is
0
5 5 4 0
5 5 4
4 0
51,15,4 4 0
LESSON 8
Solve the equation
SOLUTION
Auxiliary equation
If the auxiliary quadratic equation has
real and distinct roots, and then
2 2 5
5 0
2 2 5 5 0
212± 25 0
cos
cos22 sin 22
±
cos
cos s sin
LESSON 10
SOLUTION
Auxiliary equation
If the auxiliary quadratic equation has
complex roots of the form
,
is the general solution of the differential
equation.
is the general solution of the differential
equation.
4 4 4 0
4 4 4 0
4 41 4 1 0
2twicee
LESSON 9
Solve the equation
SOLUTION
Auxiliary equation
Solve the equation
Non
–
–
–
, ≠ 0
homogeneous first
order and second
order differential equation are of the form
The particular integral is any solution of these
types of differential equations.
is a Polynomial
5 5 6
→ → 0
0665 5 5 66
6 11
6
55 6
5 6 0
36
LESSON 11
SOLUTION
Solve the equation
It seems sensible to think that the
solution is of the form
Substituting into original equation
So
is a solution of the given equation,
but it cannot be the complete solution since it
does not contain any arbitrary constant. However,
it must be part of the complete solution, and is
called a particular integral (P.I). The remainder of
the solution can be found by considering the
5 5 6 0
16 365
simpler differential equation
4 4 3 2 2 1
4±24 0
−
SOLUTION
Auxiliary equation
C.F:
Particular Integral:
→ 22 → 2
24 4 44
4 22 4 3 3 22221 1 1
4 3 → 341
4 2 → 2 5
2 4 1 → 8
−
34 12 58
Substituting into original equation
P.I is
General solution:
is a Trigonometric Function
whose solution is
LESSON 13
Thus the complete solution is
differential equation
which is the combination of the complementary
SOLUTION
function and the particular integral.
LESSON 12
4
4 3 2 2 1
Solve the differential equation
Find the complete solution of the
4 5 5 cos sisi n
4 5 5 cos s sin
44 51 1511 10 0 sisi n cos
Auxiliary equation
Trial Solution
14 , 1
cos
cos s
sinn
sin coscos
coscos sisin
cos csos sisinsinncoscos5sisisnin c
4co
cos
44 5
5 coscoscosos sisi44n5 5 sin
333 5
55 11
1741
17
174 coscos 171 sin
174 coscos 171 sisin
C.F is
Substituting into original equation
Particular Integral is
The complete solution is
LESSON 14
(i) Solve the D.E
4
4 3 65sin2
3, 7 when 0
(ii) Hence, find the particular solution for which
SOLUTION
Auxiliary equation
41,4 33 0
−
−
sisi n cos
sisi n 2 coscos 2
22 coscos 2 2sin2
44sisin 2 4cos2
334sisisinn22 4co
cos2
4cos
s
2
4
2
c
o
s
2
2si
2si
n
2
cos2 6565 sisin 2
C.F is
Trial Solution
Substituting into original equation
465si4 8
n82 33 sin2 44 8 8 33 cos2
88
8 si65n2 88 coscos 2 65sisi65 n 2
8 0 → 1si& n 2 8 8 coscos 2
−
− sisi n 2 8 cos 2
7
0,
3,
−
−− sisi−n 2 8 cos 2
3 2 2 coscos 2 16sin 2
3
33
8 sisi n0 8co
8coss0
11
3
77 3
332 2co
2coss0 16si16sin0
3
3 9
Particular Integral is
General solution is
When
3
3 119 → 10 & 21
21− 10
10− s sin2 8 cos 2
2 2 10
10
26
5 11 0
2222±1010240411010
21
1 ± 3
−cos
cos33 sin 33
Particular Solution is
is an Exponential Function
LESSON 15
Solve the differential equation
given that
and
when
your answer in the form
.
SOLUTION
Auxiliary equation:
Complementary function:
Particular Integral:
. Give
53 cos
cos00 sin 0 2
11;11; 0
−cos
cos33− sin 33
2 33 sisin 3 3cos3
3cos3
cos
3si
11111
cos0
0
s
i
n
0
3si
n
03cos0
03cos0
2
1
3
1
1
3
3
2
2
12 43
−3 cos 3 4 sisin 3 2
2 2 3
3 22−
→
0
→
∞
3
0
32 3231 10 0
1,1, 3−
−
−
−− −
LESSON 16
Solve the differential equation
given that
as
and that
when
.
SOLUTION
Auxiliary equation:
Trial Solution
Both the trial
and the
original
have the same
form. Coefficients differ.
2 2 10 10 26
26
13
22
−cos
cos33 sin 33 2
5; 0
General solution:
Particular solution:
C.F:
In this case, the particular integral would be of the
form
but since this is already included in
the complementary function we have to use
.
Trial Solution
Both the trial
and the
original
have the same
form. Coefficients differ.
− − −
−
−
2
4
2
2−−
2−− 2−
− 3− 2−
4 12
2
12 −
− 12 −
→→ ∞0 − →→∞0 3 0
→ 0:0: 0
3,3, 0
0
− 12 − 12 −
3 5 12 12 0
2
52 − 12 −
LESSON 17
(i) Show that by using the substitution
may be written in the form
Particular Integral:
(ii) Find the general solution of
General Solution:
as
and that
Therefore we have
From general solution (
Particular solution:
and hence find the general solution of
when
:
):
2
2
2 2
2 2
2
2
differential equation
SOLUTION
(i)
×
1 ×
2
2
1
1
1
2
2 2
2 2 0
2
(ii) Auxiliary equation:
Complementary function:
Particular Integral
Let
, the
∴ 2
2 2 22
21 1
2
12 2 0
14
1 → 1
1 12 14
1 4 24 2 1
4 24 2 1
≠0
2 233 1 3 33 2 2 18
6 6 9 18
2 233 1 3 33 2 2 18
Particular Integral:
Complete solution:
LESSON 18
It is given that
the differential equation
(a) Show that the substitution
this differential equation into
,
satisfies
transforms
(b) Hence find the general solution of the
differential equation
giving your answer in the form
SOLUTION
(a)
.
2
2 23 1 1 3 333 2 18
6 6 2 2 9 9 6 6 1818
2 2 6 99 1818
6 6 9
9 1818
6 6 9 0
3,3
−
0
0 66 9 18
6 9
9 9 9 18
9 18
2
6 9 0
9 124
3
2 43
(b) Auxiliary equation
Complementary function:
Particular Integral
Let
− 2 2
− 2 2 43
− 2 34
General solution:
(b)
LESSON 19
, > 0,0, > 0
2
4 2
2
88 1 1 12
12 12
4 4 3 3
88 1 1 12
12 12
×
12 − → 2
2
2
(a) Given that
function of , show that:
and
is a
(i)
(ii)
(b) Hence show that the substitution
transforms the differential equation
into
(c) Hence find the general solution of the
differential equation
giving your answer in the form
SOLUTION
(a) (i)
Now,
(ii)
− 2 2
−
2 2 2
2 4 4
88 1 1 12
12 12
8 8 1 1 12 12
4 2 2 8 8 −2 12 12
4 2 16 2 12
12 1212
4 16
16 12
12 1212
4 4 3
3 3
4 4 3 0
1,1, 3
0
0 44 3 3
44 3
3 3
3 33
3 3
1
44 3
3 0
4 3 0
3 4
.
(c) Auxiliary equation
Complementary function:
43
43
, 4
3
General Solution:
Since
LESSON 20
(a) A pond is initially empty and is then filled
5√41 5 5
gradually with water. After
depth of the water,
minutes, the
metres, satisfies the
differential equation
Solve this differential equation to find
terms of .
in
(b) Another pond is gradually filling with water,
after minutes, the surface of the water forms
a circle of radius
metres. The rate of change
of the radius is inversely proportional to the
area of the surface of the water.
(i)
Write down a differential equation, in
the variables
and and a constant of
proportionality, which represents
how the radius of the surface of the
water is changing with time.
(You are not required to solve your
differential equation.)
(ii)
When the radius of the pond is 1
metre, the radius is increasing at a
rate of 4.5 metres per second. Find
the radius of the pond when the
radius is increasing at a rate of 0.5
metres per second.
SOLUTION
(a)
√+
+
+
1√45 51 1
45
4 5− 15 1 −−
45
4551 15 1 −−
2
25 √4 5 15 1 1
0,
0
25 4 50 15 1 1 0
12 1 1
5 √4 5 5 1 1 5 11
√4 5 2 2 1
4 5 52 211
5 52 211 4 4
15 52 211 45
9 1, 4.5
29 1
2 9
2
29
When
(b) (i)
(ii)
When
12 29
39
LESSON 2
The number of bacteria in a
liquid culture is observed to grow at a rate
proportional to the number of cells present. At the
beginning of the experiment there are 10,000 cells
and after three hours there are 500,000. How
many will there be after one day of growth if this
unlimited growth continues? What is the
of the bacteria?
SOLUTION
1
ln1
+
0,
50050 000 10 000
ln50 1ln350 ≅ 1.304
103 000.
22 ..
ln2 ln1.2304≅ 0.532 hourhourss
1.304
Let
represent the number of bacteria present
at time . Then the rate of change is
where
is the constant of proportionality
where
When
Also,
Therefore,
(initial population)
Using the information from the problem we now
determine
Doubling time refers to the amount of time for the
bacteria to double in number from its original
number.
0
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