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MATHEMATICS
Guide to
Trigonometry, Geometry & Calculus
GRADE: 12
S.C. Nhlumayo
2
FOREWORD
Thank you for acquiring this booklet on Trigonometry,
calculus & Geometry. I just hope that you will find it
useful.
The booklet can be used in grade 11 and 12.
The focus, obviously, is on trigonometry, calculus &
Euclidean geometry. These topics are tested in paper 1 & 2
of the National senior certificate examination. One can see
that we do need to pay much attention to these topics. It is
because of this reason I have decided to write this booklet.
Albert Einstein once said that the human mind is forever
trying to master everything about something. We, mathematics
teachers, will never stop trying hard to make mathematics
accessible to all our learners.
The idea behind this project
accessible to all learners.
is
to
make
mathematics
The main focus is on using theories learnt in class to
answer typical exam questions.
For any errors you find in this booklet, please
communicate with me via email. [email protected]
do
Thank you!!!!
S.C. Nhlumayo
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3
TRIGONOMETRY
Trigonometry is the branch of mathematics which is based on
the relationship between the angles and the sides of a
triangle.
The roots of the subject lie thousands of years in the past
when early humans became interested in astronomy and used it
to predict the seasons and planting time and for navigation
and cartography.
The Babylonians understood some of the basic principles, but
it was the ancient Greeks who began to develop the concepts
of modern trigonometry systematically.
The Greek astronomer Ptolemy constructed the first table of
sines in the second century AD.
Trigonometry in its modern form is a highly evolved area of
mathematics and is used in fields such as engineering,
astronomy, architecture and surveying.
Trigonometric
equations
represented
graphically
are
extremely important in physics, chemistry, astronomy,
meteorology and even in economics
The tides of the sea, the appearance of Halley’s Comet, the
vibrations of a molecule are all examples of periodic
functions.
I am now going to give a summary of the things that you need
to know before you sit for grade 12 final examinations.
The summary will be followed by a series of activities that
you must complete as you prepare for the exam.
S.C. Nhlumayo
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4
TRIGONOMETRIC IDENTITIES
sin x
sin 2 x  cos 2 x  1
cos x
You need to know the above identities because, usually, they
are never included in the formula sheet.
tan x 
sinA  B  sin A cos B  cos A sin B
cosA  B  cos A cos B  sin A sin B
sin 2A  2 sin A cos A
cos 2A  cos2 A  sin2 A
 1  2 sin2 A
 2 cos2 A  1
THE THREE ELEMENTS( x ; y & r) IN SPACE(xy-plane)
Remember the following definitions:
If θ is the angle that the terminal ray(r) makes with the positive x-axis
then
sin  
y
r
cos  
x
r
tan  
y
x
The



above identities will help you:
Solve trigonometric equations.
Simplify trigonometric expressions.
Prove more complex identities.
Sometimes these will be used together with reduction formulae
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5
WORKING ON THE XY-PLANE USING THE THREE ELEMENTS.
What are the steps?
I explain the steps using an example.
If 5 sin   3  0
  900; 2700  with the aid of a diagram,
Show that sin2   cos2   1
STEP1:
Write the first part in the form
Trig function = fraction
3
sin  
5
This, together with the given restriction, will help
you decide on the quadrant to choose.
STEP2:
Decide on the quadrant to use.
The given restriction tells us to choose between
quadrant 2 and quadrant 3 but in STEP1 we see that
sin is positive and we know we have to go for the
2nd quadrant where sin is positive.
You can now draw your diagram.
STEP3:
Consider the signs.
‘r is always positive
In the 2nd quadrant y is positive
In the 2nd quadrant x is negative.
From trig definitions we know that sin  
y
r
 y = 3 and r = 5
STEP4:
Use the Pythagorean theorem to look for the 3rd
element, in this case x.
x 2  y2  r2
You can then answer the question.
S.C. Nhlumayo
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6
EXERCISES: Use the information
attempt the following exercises.
from
the
previous
pages
and
SCN1
1.
2.
3.
15
8
2 sin x  3 cos x
4 sin x  cos x
If tan x 
a
prove
b
sin   cos 
a b

sin   cos 
ab
If tan  
that:
P is the point in the first quadrant such that
13 cos   5  0 R is a point in the second
quadrant such that POR =900
3.1
Determine:
tan 
3.2
k
S.C. Nhlumayo
00  x  900 calculate
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7
SCN2
8
900    2700 , with the
17
aid of a diagram determine:
Given: sin  
2.1
tan 
2.2
sin(900  )
2.3
cos 2
SCN3
If
2 tan   3  0
3.1
sin(  900)
3.2
cos 2
SCN4
If sin 24 = p, determine
900    3600
the
ff
in
4.1
Cos 240
4.2
Sin 480
4.3
Cos 330
4.4
Sin 120 cos 120 – sin (-660) tan 2040
4.5
Cos 156o
S.C. Nhlumayo
determine
terms
of
Sometimes you need to apply some
clever techniques to solve these
problems.
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p
8
SCN5
Simplify
5.1
sin(180  x). tan(180  x). cos(x). sin(90  x)
cos(x  360). cos(90  x). tan(x)
5.2
tan 425. tan(295)
cos 115. cos(65)
5.3
tan 135. tan(105). tan 735
5.4
5 sin2   4 cos2 
5.5
sin(180  x). cos(90  x)  cos(180  x). cos(360  x)
6.1
If a cos   b
SCN6
and
c sin   d
show that:
a2c2  b 2c2  a2d2
6.2
6.3
Determine
Write down
the
the
maximum
value
of
minimum value of
sin x  cos x
1
3  2 sin 2
6.4
f(x)  14 cos x  5 sin x. It is also given that
f(x)  R cos(x  )
  (00 ; 900)
6.4.1
Find the value of R and .
6.4.2
Write down the minimum
S.C. Nhlumayo
value of
14cosx -5sinx
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9
SCN7
If tan(A  B) 
7.1
7.2
7.2.1
Prove
sin(A  B)
cos(A  B)
that: tan(A  B) 
hence: If
1
tan α  ;
3
0  α  90
Prove: tan 2 
tan A  tan B
1  tan A tan B
and tan β 
1
7
0  β  90
3
4
7.2.2
Evaluate: tan(2  )
7.3
Three squares are arranged as shown below:
7.3.1
S.C. Nhlumayo
Show
that
    
HINT: think in terms of the
tangents of the angles!!!!
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10
SCN8
Prove the
following
identities
8.1
2 sin x  sin 2x
2 sin x

4  3 cos x  cos 2x
5  2 cos x
8.2
sin 4t  sin 2t
 tan t
cos 4t  cos 2t
8.3
sin A  sin 3A
 2 sin A
sin2 A  cos2 A
8.4
sin 3x  sin x
 sin x
2  2 cos 2x
8.5
3 sin(x  60)  sin(x  30)  cos x
SCN9
9.1
If cos x  2 sin y cos y
0
Show that: x  2y  90
9.2
Determine the general solution:
9.2.1
1  4 sin2 A  5 sin A  cos 2A  0
9.2.2
sin 2θ  4 cos θ 
9.2.3
sin(3y  50)  cos(y  10)  0
9.3
If cos(α  β)  a
that
a  b
cos α cos β 
2
S.C. Nhlumayo
x and y are acute
3 sin θ  2 3  0
and
cos(α  β)  b
prove
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SCN10
10.1
10.2
10.3
Prove that
sin 6x
cos 6x

 2
sin 2x
cos 2x
Simplify:
sin3x  sin7x
cos3x  cos7x
(HINT: 3x  5x - 2x and 7x  5x  2x )
Consider the identity:
1  cos 2
 tan 
sin 2
10.3.1 For which values of θ is the identity defined?
10.3.2 Prove the identity.
10.3.3 Deduce that tan 150  2 
3
determine cos 20o in terms of w.
10.4
If sin 50  w,
10.5
If sin14 0  y determine cos 38o in terms of y.
SCN11
Determine the general solution
11.1
11.2
sin x  cos 2x  1.
6 cos x  5 
4
cos x
; cos x  0
11.3
cos 2x  1  3 cos x
11.4
sin 2 x  cos 2x  cos x  0
11.5
sin x  2 cos 2 x  1
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12
SCN12
12.1
a.
b.
12.2
12.3
Given: sin  sin  
1
cos(  )  cos(  )
2
Prove the identity
Hence, simplify:
1
 2 sin 700
0
2 sin 10
0
0
0
Prove that cos 20  cos 40  cos 80 
1
8
2
2
If cos   sin   1 then show that:
cos4   sin4   1
2

cos6   sin6   1
3
SCN13
Use compound angle identities to verify reduction formulae
13.1
sin(1800  x)  sin x
13.2
cos(1800  x)   cos x
13.3
sin(900  x)  cos x
13.4
tan(1800  x)  tan x
13.5
cos(3600  x)  cos x
13.6
cos(2700  x)   sin x
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13
SCN14
14.1
Given that P and Q are both acute, solve for P and Q if
1
sin P sin Q  cos P cos Q 
and
2
1
sin P cos Q  cos P sin Q 
2
14.2
If cos 61o = p, express the following in terms of p.
a. sin 209o
14.3
b. cos(-421o)
c. cos 10.
Prove: cos(A  B)  cos(A  B)  2 cos A cos B
Hence, evaluate cos 750  cos150
14.4
 show that sin 3x  4 sin x cos 2  sin x
sin 3x  sin x
2  2 cos 2 x
 for which values of x is the above identity valid?
 hence, or otherwise prove the identity:
14.5
If sin θ = p
15.1
If sin 230 = p, write down the following in terms of p.
and 2cos θ = 4p, find the value of p.
SCN15
 cos 1130
 cos 230
 sin 460
15.2
S.C. Nhlumayo
Simplify:
sin4 x  4 cos2 x 
cos4 x  4 sin2 x
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14
SOLUTION TO TRIANGLES
In any triangle ABC
a
b
c


sin A
sin B
sin C
1
ab sin C
2
1
 ac sin B
2
1
 bc sin A
2
(sine rule)
Area ABC 
(area rule)
a2  b2  c2  2bc cos A
b2  a2  c2  2ac cos B
(cosine rule)
c2  a2  b2  2ab cos C
Types of triangles:
 Right angled triangle
 Oblique triangle
(one angle equals 900)
(no angle equals 900)
Choosing which rule to use:
Right angled triangle
- Pythagoras theorem………..’if you know two sides and looking for the
3rd side.
- SOH-CAH-TOA………..’if you know one angle and one side.
Oblique triangle
- Area rule……. For calculating the area of a triangle.
- sine rule: (S-S-A); (A-A-S)
- cosine rule: (S-S-S) ; (S-A-S)
(S-A-S)
All these rules can be used whether you are responding to a 2D or 3D problem.
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15
In the figure below, CD is a vertical mast. The points B, C, and E are in the same
horizontal plane. BD and ED are cables joining the top of the mast to pegs on the
ground.
DE =28.1 m and BC = 20.7 m.
The angle of elevation of D from B is 43.6o. CBE = 63o and BDE = 35.7o
SCN16
16.1
Calculate the length of BD
16.2
Calculate the length of BE.
16.3
Calculate the area of ∆BEC.
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16
SCN17
In the diagram below, TQR is a horizontal plane on a sports field. PQ is
a vertical flag pole. TQ = TR = y.
QR = x units
17.1
Show that sin   sin 2
17.2
Hence, prove that in ∆PQR
2 y cos 
PR 
cos 
17.3
If TR = 2QR, prove that the area of

∆TQR = x 2 . cos 
2
17.4
Calculate the area of ∆TQR if x = 25 units and θ = 30o
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17
SCN18
The diagram below shows a rectangular block of wood. A plane cut is
made through the vertices E, G and B, revealing a triangular plane EBG.
18.1
Calculate the lengths of EB, EG and BG.
18.2
Hence, determine the magnitude of EBG.
18.3
Determine the area of ∆EBG.
SCN19
ABCD is a parallelogram with
AC =7cm
and
Calculate the numerical values of a
S.C. Nhlumayo
BD =12cm
and b
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SCN20
In the figure below, AC =e
Show
SCN21
that
and
BD = f
e2  f 2  2(a2  b2)
In the ff diagram, AB =2p, BC = 2q, DC =AC= q & AD = p
Show that
cos  
1
4
SCN22
BD = x
and AC = 6
Calculate the value of x.
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19
SCN22
22.1
22.2
22.3
SCN23
Express MC in terms of h and 
Show
If
that
BC 
h = 25mm and
 BC
 Area of ∆BCM
Consider
3h cos 
sin 
 = 43o
calculate:
∆ABC
23.1
Use Sine Rule and show that:
a  b
sin A  sin B

c
sin C
23.2
If
c2  a2  b 2  2ab cos C
show that:
2ab
1

2
1  cos C
c2  a  b 
S.C. Nhlumayo
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20
SCN24
In the figure alongside, B, C and D are 3 points on the same
horizontal plane and AB is a vertical pole of length p metres. The
angle of elevation of A from C is θ.
24.1
Express CDB in terms of θ.
24.2
Express BC in terms of p and a trigonometric function of θ.
24.3
Hence, show that p  41  3 tan  
24.4
Use the results in c to calculate the value of θ if the length of the pole is
12 m.
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21
SCN25 25.1
Nomabham’bheshe is standing on level ground, at B, a distance of
19 metres away from the foot E of a tree TE.
She measures the angle of elevation of the top of the tree at a
height of 1.55 metres above the ground as 32˚.
Calculate the height TE of the tree. Give your answer correct to
3 significant figures
25.2
When Amuh is standing 400m from the base of a mountain, the
angle of elevation from the ground to the top of the
mountain is θ. He then walks 500m straight back and measure
the angle of elevation to be .
a.
b.
c.
Express h in terms of x and θ.
Express h in terms of x and .
If θ = 25o
and  = 20o , solve for x and then find the
height of the mountain, h.
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22
EUCLIDEAN GEOMETRY
In Euclidean geometry, I concentrate a lot on applying theorems and
postulates in solving riders.
Bookwork (proving basic theorems) is not covered in this booklet. You
can check proofs in your textbook.
You are advised to know all the theorems so that you will be able to
use them in proving riders.
Although theorems won’t be proven in this booklet, we’ll look at the
proof of the theorem of Pythagoras using similar triangles.
LOGIC – The way of reasoning.
In Euclidean geometry we use the “if-then” statement when we analyse and
trying to find meaning from the theorems and postulates.
This is the same reasoning we use in the real world.
Look at this argument: All rugby players drink too much.
This means that, if John is a rugby player then he drinks too much.
This is how we work with theorems. Take the theorem, break it into substatements and then see which statements are bases and which one is the
conclusion
Look at this argument from Newton2 of motion:
An object will remain at rest or continue moving at constant velocity unless
acted upon by an unbalanced force.
Bases: the object is at rest.
The object is moving at constant velocity.
Conclusion: forces acting on the object are balanced
A theorem is just an argument which you need to break as shown above so that
you will understand it.
Trying to memorise a theorem is a total waste of time because memorizing is
not understanding.
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23
THE NINE BASIC THEOREMS ON THE CIRCLE
A line segment drawn from the centre to the midpoint of a
chord is perpendicular to that chord
The angle at the centre of a circle is twice the angle at
the circumference. (both angles subtended by the same arc)
An angle subtended by a diameter is 900.
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24
Opposite angles of a cyclic quadrilateral add up to 1800
The exterior angle of a cyclic quadrilateral equals the
interior opposite angle.
Angles in the same segment are equal
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25
An angle between a tangent and a radius is 90o.
Tangents from the same point outside the circle are equal in
length.
An angle between a tangent and a chord equals the angle
subtended by that chord in the alternate segment.
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26
THE THEOREM OF PYTHAGORAS
There are a lot of ways of proving the pythagorean theorem
but in this section we’ll use similar triangles to prove the
theorem.
The proof has got two parts.
Part one: if you drop a perpendular from a vertex of right angle of a
right angled triangle to the hypotenuse you will generate two more
triangles which are similar to each other and to the original
triangle.
GIVEN: Right angled triangle ABC with AD  BC
R.T.P: ∆ABC///∆DBA///∆DAC
PROOF: In ∆ABC and ∆DBA
D1 = A………….each equals 90o
B is common
A1 = C………..third angle
 ∆ABC /// ∆DBA
In ∆ABC and ∆DAC
C is common
A = D2
A2 = B,……third angle
ABC ///DAC
The two small triangles are both similar to the bigger triangle.
We can conclude that ∆ABC///∆DBA///∆DAC
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27
For part two, all you need to remember is that, if triangles are
similar then their corresponding sides are proportional.
(sides opposite equal angles)
In ∆ABC
and ∆DBA
AB
BC

 AB2  BD  BC
BD
AB
In ∆ABC
and ∆DAC
AC
BC

 AC2  BC  DC
DC
AC
PART TWO
2
2
2
R.T.P: BC  AB  AC
Proof: by substitution
AB2  AC2  BC  BD  BC  DC
 BC(BD  DC)
 BC  BC
 BC2
That is how prove the pythagorean theorem in grade 12. It’s the same
theorem you proved in grade 9 using four congruent triangles.
It looks like proving the
theorem of Pythagoras is
easy.
The important part is that
of proving the three
triangles to be similar.
(PART 1)
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28
REMEMBER!!!!!!!
If triangles are similar then the following
equivalent.
 Triangles are equingular
 Corresponding sides are proportional
two
statements
are
THE PROPORTINALITY THEOREM
A line parallel to a side of a triangle cuts the other two sides so as
to divide them (internally or externally) in the same proportion.
(line // one side of a∆)
Since DE // BC
then
AD
AE

BD
EC
AB
AC

AD
AE
you can use any combination
There is a special case of this theorem called the mid-point theorem.
You did it in grade 10. However I’ll remind you.
S.C. Nhlumayo
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29
THE MIDPOINT THEOREM
The line segment joining the midpoints of the two sides of a triangle
is parallel to the third side. Moreover, the line segment is half the
length of the third side.
Now, Let’s analyse the theorem and find meaning.



There is a triangle.
You join the midpoints of the two sides
You will get two results:
1. The new line segment is parallel to the third side.
2. The new line segment is half the length of the third side
YZ // WX
YZ 
1
WX
2
This was just a summary of the basic theorems that you really
have to understand before you can attempt any euclidean geometry
problem in grade 12.
You are now going to attempt the following activities using the
above theorems.
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30
SCN26
A, B, C & D are points on the circle
as shown below. AC is the diameter.
Calculate the values of x and y
SCN27
T is a point outside circle O. TA & TB are
tangents to the circle. TPQ is a secant. X
is the midpoint of PQ.
Prove:
 OXBT is a cyclic quadrilateral
 BOT = BXT
 OT bisects A
 MA//QT
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31
SCN28
PT is a tangent.
Calculate the value of x, y and z
SCN 29
Prove that PT2 = PA  PB
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32
SCN 30
Prove that:
 MKP = PKN
 SMK = KTN
SCN 31
 Prove that QS 
S.C. Nhlumayo
1
PS
2
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33
SCN 32
Calculate:
 A1
 D1
 O1
 B1
 B2
 A2
 C1
 D2
SCN 33
Determine :



S.C. Nhlumayo
C
ADB
BDC
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SCN 34
 Prove
that
ACBD
is
a
cyclic
quadrilateral
 Can you conclude that AC//DB? Discuss.
SCN 35
Determine
 AD
 BC
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SCN 36
Calculate the value of x.
SCN 37
Name the similar triangles in this figure.
Prove:
 DE2 = AE . DC
 CE2 + DE2 = BC2 + BE2 + AD2 + AE2

S.C. Nhlumayo
CE2
BE

2
DE
EA
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SCN 38
 Find 7 angles which are equal to Q1
 Deduce that RB is a common tangent to
circles BPT and BAQ.
 If BQ:QT = 7 : 4 and PB = 5
Find AP
Below I have included a number of theorems and postulates
that may help you. Don’t be scared. Just read through. You
will get used to them.
You are advised to attempt more geometry problems in the
previous P3 question papers.
Well, this is just for enrichment
S.C. Nhlumayo
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Postulates and Theorems
Converse of the AEA Theorem: If two lines are cut by a transversal
congruent alternate exterior angles, then the lines are parallel.
forming
Interior Supplements Theorem: If two parallel lines are cut by a
transversal, then the interior angles on the same side of the
transversal are supplementary.
Converse of the Interior Supplements Theorem: If two lines are cut by a
transversal forming interior angles on the same side of the transversal
that are supplementary, then the lines are parallel.
Parallel Transitivity Theorem: If two lines in the same plane are parallel
to a third line, then they are parallel to each other.
Perpendicular to Parallel Theorem: If two lines in the same plane are
perpendicular to a third line, then they are parallel to each other.
SAA Congruence Theorem: If two angles and a non-included side of one
triangle are congruent to the corresponding angles and side of another
triangle, then the triangles are congruent.
Angle Bisector Theorem: Any point on the bisector of an angle
equidistant from the sides of the angle.
Perpendicular Bisector Theorem: If a point
of a segment, then it is equally distant
segment.
is
is on the perpendicular bisector
from the endpoints of the
Converse of the Perpendicular Bisector Theorem: If a point is equally
distant from the end-points of a segment, then it is on the
perpendicular bisector of the segment.
Isosceles Triangle Theorem: If a
angles are congruent.
triangle
is
isosceles, then
Converse of the Isosceles Triangle: Theorem If two angles of a
congruent, then the triangle is isosceles.
its base
triangle are
Converse of the Angle Bisector Theorem: If a point is equally distant
the sides of an angle, then it is on the bisector of the angle.
Perpendicular Bisector Concurrency Theorem: The
bisectors of a triangle are concurrent.
Angle Bisector Concurrency Theorem: The
are concurrent.
S.C. Nhlumayo
from
three perpendicular
three angle bisectors of a
triangle
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Triangle Exterior Angle Theorem: The measure of an exterior angle of a
triangle is equal to the sum of the measures of the remote interior
angles.
Quadrilateral Sum Theorem: The
quadrilateral is 360°.
sum of the measures of the
Medians to the Congruent Sides Theorem: In an isosceles
medians to the congruent sides are congruent.
four angles of a
triangle, the
Angle Bisectors to the Congruent Sides Theorem: In an isosceles
the angle bisectors to the congruent sides are congruent.
Altitudes to the Congruent Sides Theorem: In an isosceles
altitudes to the congruent sides are congruent.
triangle,
triangle, the
Isosceles Triangle Vertex Angle Theorem: In an isosceles triangle, the
altitude to the base, the median to the base, and the bisector of the
vertex angle are all the same segment.
Parallelogram Diagonal Lemma: A diagonal of a parallelogram divides
parallelogram into two congruent triangles.
Opposite Sides Theorem: The opposite
the
sides of a parallelogram are congruent.
Opposite Angles Theorem: The opposite angles of a parallelogram are
congruent.
Converse of the Opposite Sides Theorem: If the opposite sides of a
quadrilateral are congruent, then the quadrilateral is a parallelogram.
Converse of the Opposite Angles Theorem: If the opposite angles of a
quadrilateral are congruent, then the quadrilateral is a parallelogram.
Opposite Sides Parallel and Congruent Theorem: If one pair of opposite
of a quadrilateral are parallel and congruent, then the quadrilateral
parallelogram.
Rhombus Angles Theorem: Each diagonal of a
angles.
rhombus bisects
sides
is a
two opposite
Parallelogram Consecutive Angles Theorem: The consecutive angles of a
parallelogram are supplementary.
Four Congruent Sides Rhombus Theorem: If a quadrilateral has
sides, then it is a rhombus.
four congruent
Four Congruent Angles Rectangle Theorem: If a quadrilateral has
congruent angles, then it is a rectangle.
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four
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Rectangle Diagonals Theorem: The diagonals of a rectangle are congruent.
Converse of the Rectangle Diagonals Theorem: If the diagonals of a
parallelogram are congruent, then the parallelogram is a rectangle.
Isosceles Trapezoid Theorem: The base angles of an
congruent.
isosceles
Isosceles Trapezoid Diagonals Theorem: The diagonals of an
trapezoid are congruent.
trapezoid are
isosceles
Converse of the Rhombus Angles Theorem: If a diagonal of a parallelogram
bisects two opposite angles, then the parallelogram is a rhombus.
Double-Edged Straightedge Theorem: If two parallel lines are intersected by
a second pair of parallel lines that are the same distance apart as the
first pair, then the parallelogram formed is a rhombus.
Tangent Theorem A tangent: to a circle
drawn to the point of tangency.
is perpendicular
to
the
radius
Converse of the Tangent Theorem: A line that is perpendicular to a
radius at its endpoint on the circle is tangent to the circle.
Perpendicular Bisector of a Chord Theorem: The perpendicular bisector of a
chord passes through the center of the circle.
Inscribed Angle Theorem: The measure of an angle
equals half the measure of its intercepted arc.
inscribed
Inscribed Angles Intercepting Arcs Theorem: Inscribed angles
the same or congruent arcs are congruent.
in a circle
that
intercept
Cyclic Quadrilateral Theorem: The opposite angles of a cyclic quadrilateral
are supplementary.
Parallel Secants Congruent Arcs Theorem: Parallel
arcs on a circle.
lines
intercept congruent
Parallelogram Inscribed in a Circle Theorem: If a parallelogram is
inscribed within a circle, then the parallelogram is a rectangle.
Tangent Segments Theorem: Tangent
congruent.
segments
from a point
to a circle are
Intersecting Chords Theorem: The measure of an angle formed by two
intersecting chords is half the sum of the measures of the two
intercepted arcs.
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Intersecting Secants Theorem: The measure of an angle formed by two
secants intersecting outside a circle is half the difference of the measure
of the larger intercepted arc and the measure of the smaller intercepted
arc.
AA Similarity Postulate: If two angles of one triangle are congruent to
two angles of another triangle, then the two triangles are similar.
SAS Similarity Theorem: If two sides of one triangle are proportional to
two sides of another triangle and the included angles are congruent, then
the triangles are similar.
SSS Similarity Theorem: If the three sides of one triangle are
proportional to the three sides of another triangle, then the
triangles are similar.
two
Corresponding Altitudes Theorem: If two triangles are similar, then
corresponding altitudes are proportional to the corresponding sides.
Corresponding Medians Theorem: If two triangles are similar, then
corresponding medians are proportional to the corresponding sides.
Corresponding Angle Bisectors Theorem: If two triangles are similar, then
corresponding angle bisectors are proportional to the corresponding sides.
Parallel/Proportionality Theorem: If a line passes through two sides of a
triangle parallel to the third side, then it divides the two sides
proportionally.
Converse of the Parallel/Proportionality Theorem: If a line passes through
two sides of a triangle dividing them proportionally, then it is parallel
to the third side.
Three Similar Right Triangles Theorem: If you drop an altitude from the
vertex of a right angle to its hypotenuse, then it divides the right
triangle into two right triangles that are similar to each
other and to the original right triangle.
Altitude to the Hypotenuse Theorem: The length of the altitude
hypotenuse of a right triangle is the geometric mean between
of the two segments on the hypotenuse.
to the
the length
The Pythagorean Theorem: In a right triangle, the sum of the squares of
the lengths of the legs equals the square of the length of the
hypotenuse.
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Converse of the Pythagorean Theorem: If the lengths of the three sides of
a triangle satisfy the Pythagorean equation, then the triangle is a
right triangle.
Hypotenuse Leg Theorem: If the hypotenuse and one leg of a right triangle
are congruent to the hypotenuse and one leg of another right triangle,
then the two triangles are congruent.
Square Diagonals Theorem: The diagonals of a
perpendicular, and bisect each other.
square are congruent,
Converse of the Parallelogram Diagonals Theorem: If the diagonals of a
quadrilateral bisect each other, then the quadrilateral is a parallelogram.
Converse of the Kite Diagonal Bisector Theorem: If only one diagonal of a
quadrilateral is the perpendicular bisector of the other diagonal, then
the quadrilateral is a kite.
Triangle Mid-segment Theorem: A mid-segment of a triangle
the third side and half the length of the third side.
is parallel
to
Rectangle Midpoints Theorem: If the midpoints of the four sides of a
rectangle are connected to form a quadrilateral, then the quadrilateral
formed is a rhombus.
Rhombus Midpoints Theorem: If the midpoints of the four sides of a rhombus
are connected to form a quadrilateral, then the quadrilateral formed is a
rectangle.
Kite Midpoints Theorem: If the midpoints of the four sides of a kite are
connected to form a quadrilateral, then the quadrilateral formed is a
rectangle.
Product of Interior Segments Theorem: If two chords intersect in a circle,
then the product of the segment lengths on one chord is equal to the product
of the segment lengths on the other chord.
In the next section, we look at differential calculus.
DIFFERENTIAL CALCULUS
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In calculus we study the mathematics of motion and rates of
change. Calculus was developed in the seventeenth century to
meet the mathematical needs of Physicists.
It enables Scientists to define slopes of curves and to
calculate velocities and acceleration of moving bodies like
cannon balls and planets.
Calculus deals with two major problems, namely tangent and
area problems.
The study of the tangent problem is called Differentiation
and the study of the area problem is called integration.
Both these studies rest on a more fundamental concept known
as a limit.
1. THE TANGENT AND THE DERIVATIVE.
You already know how to calculate the average gradient of a
curve using the NEWTON’S QUOTIENT:
f ( x  h)  f ( x )
Where h is the horizontal distance between two
h
chosen points.
Well, we are now going to look at how we can approximate the
gradient of a curve at a certain point.
To accomplish this, we are going to use the truth that a
straight line has a constant gradient. You have already used
that when calculating the average gradient of a curve.
It is clear that a curve does not have a constant gradient
since its steepness changes as you move along the curve.
Consider the following diagram:
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PQ is the secant ( a line segment joining two points on a
curve)
RPS is the tangent to the curve of f. (a line segment that
touches the curve externally at only one point.) h is the
horizontal distance between P and Q.
If P is fixed and PQ is rotated clockwise then:
a. The horizontal distance (h) between P and Q becomes
negligible (h→0).
b. PQ will eventually coincide with RS. This means it will
become a tangent to the curve of f.
c. As you rotate further, the line will get back to being a
secant and then a tangent.
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This means that, the gradient of RS is the limiting value as
h becomes negligible (as h tends to zero.)
Calculating the gradient of RS is the same as calculating
the gradient of the curve of f at P.
We can now sum up what we have just said. Determining the
gradient of a curve includes using the limiting process to
the NEWTON’S QUOTIENT.
f ( x  h)  f ( x )
h
h 0
This can be written mathematically as: f ' ( x )  lim
This is called the derivative. You remember I said a curve
has so many gradients. A set of all gradients of a curve is
called the derivative. The derivative can be used to find
the gradient of a curve at any point.
FINDING THE DERIVATIVE OF A CURVE USING THE LIMITING PROCESS
(i.e. from first principles).
Example. Given: f ( x)  2 x 2
f ' ( x )  lim
h 0
 lim
h0
 lim
h0
 lim
h0
 lim
h0
 4x
f (x  h)  f ( x)
h
2
2 x  h   2 x 2
h
2
2  x  2 xh  h 2   2 x 2
h
2
2 x  4 xh  2 h 2  2 x 2
h
h 4 x  2 h 
h
t his results from cancelling
S.C. Nhlumayo
h with h and " forgeting 2h"
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You remember I said, in the limiting process h becomes
negligible. I did not say h becomes zero. Most learners are
tempted to replace h with zero, which is WRONG.
One other mistake most learners do is omitting
lim
h0
just
after writing the formula and continuing without writing it.
By so doing, they lose marks. You must only omit it in the
final step where you write the answer.
SCN39.
Determining the derivative from first principles
(i.e. using the limiting process)
1. f ( x)   x 2
2. f ( x)  3x 2
3. f ( x)  x 2  2
4. f ( x)  2  3x 2
5. f ( x)  1 
1 2
x
2
SCN40. Determine the derivative from first principles in
each case
1. f ( x)  2 x 3
2. f ( x)   x 3
3. f ( x)  4  2 x 3
4. f ( x ) 
2
x
5. f ( x )  
3
x
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SCN41. In each case, find the derivative of f(x) at
the point where x = 1, using the definition of the
derivative.
1. f ( x)  2 x 2  5 x  1
2
5
2
3. f ( x)  x  4 x  3
2. f ( x)   x 2  3x
1
4
5. f ( x)  3 x 2  5
4. f ( x)  x 2  6 x  14
6. f ( x)  5 x 2  2 x
2
x
7. f ( x)   4
FINDING THE DERIVATIVE USING THE POWER RULE
RULE: If y  ax
n
then
dy
 anx n1 (for any real number n)
dx
Examples :
dy
 (2  3) x 31
Given: y  2x then dx
 6x2
3
Most people fail calculus because they are not good in
algebra. It is important that you keep horning your skills
in algebra so that you will enjoy calculus.
S.C. Nhlumayo
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A LOOK AT FRACTIONS AND RADICALS
1
x2
 x 2
y 
1. If
y 
2. If
then
x
 x
 x
then
then
dy
1  43
 x
dx
3
1
2

dx
 2 x  3
dy 1  12
 x
dx 2
1
y  3
x
3. If
dy
1
3
OTHER RULES.
d
 f ( x )  g ( x )   d  f ( x )   d g ( x ) 
dx
dx
dx
d

kf ( x)  k d  f ( x) k a constant
dx
dx

SCN42 Determine
3
dy
in each case:
dx
2
1. y  2 x  4 x  6 x  3
4
2
2. y  x  3 x  2 x
1 6
x  3x 4  8x 2  7
2
3
2
4. y  4 x  x 7 x  1
3. y 
SCN43 Determine the derivative using rules
1. f ( x)   x  2 ( x  3)
2
2. y  (3  x )( x  2)
2
3. y  x 1  x 
3
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2
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SCN44.
Find the derivative using rules:
1. y  3 x  x
2. y 
3. y 
2
2
1

3
x3
x
x
 x6
3
x4
4. y 
3 x
2
SCN45: Determine the derivative using rules
x3  2 x  2
1. y 
x
1  x4 
2. y 
13
x
2
x
2
 4x
3
x
3. y 
5 x
1
x
4. y 
2x
1
2
 x
x
5. y 
2x 3
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SCN46: Determine
1. Dx 2 x  3 x  2 
2.
d  3 x  1


dx  3 x 

3. Dx 1 

4. Dx 

5.
d
dx

2
2

x  x  2  3 x 
x



x

x3  2

2
SCN47: Determine the derivative using rules
2x x2
1. y 

3 2
3
2. yx  x  2 y  8
2
3. 2 xy  2 x  7 x  6
2
4.  x  3 y  2 x  18
3
3

 1
6
x

5. y  x 
TANGENTS TO CURVES
At the beginning I said we can use the gradient of a tangent
to a curve to find the gradient of that particular curve. We
can reverse this. Given the equation of a curve and the
point of contact, we can find the equation of a tangent.
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SCN48: Determine the equation of a tangent to a
curve in each case.
1.
f ( x)  x 3  4 x 2  11x  30
at x  1.
2.
f ( x)  3 x 2  x  1
at x  2
3.
y
4.
y
6
x
at x  - 2
4
1
x
at x  2
SCN49
2
1. Find the equation of the tangent to y  x  4 x which is
parallel to y  2 x  3
2
2. Find the equation of the tangent to y  x  2 x which is
parallel to y  x  4  0
f ( x )  ax 3  bx 2
3. The gradient of the tangent to the curve
is
equal
f(1)  5.
to
13
Calculate
where
x
the values
=
-1
and
of a and b.
2
4. If y  2 x is a tangent to the curve of y  x  ax  b at
(2; 4), find the values of a and b.
5. Show
that
y  x2  x  2
y  x 1
is
a
tangent
to
at (1;2)
2
6. If y  4 x  4 is a tangent to y  x  ax  b at x= 1, find a
and b.
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APPLICATIONS OF THE DERIVATIVE.
Calculus gives you the tools you need to measure change both
qualitatively
and
quantitatively.
It
has
widespread
applications in science and engineering and is used to solve
complex and expansive problems for which algebra alone is
insufficient. Differential calculus explores and analyses
rates of change quantitatively and qualitatively.
You are now going to use the derivative to solve
many
mathematical problems.
We will start by looking at how we can use calculus in curve
sketching.
CURVE SKETCHING
Initially
I
said
tangents
are
used
to
calculate
the
gradients of curves. We all know that when a function is
increasing, its gradient is positive. If the function is
decreasing then its gradient is negative. What happens at
stationary points?
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Consider the following diagram:
You can see that the tangents (dotted lines) to f and g are
horizontal lines.
The gradient tells you by how much y changes as x changes.
With horizontal lines, as x changes y remains constant. From
the gradient formula we can see that m 
y 0

 0 . The
x x
gradient of a horizontal line is always zero. This tells us
that at stationary points, the gradient is zero. We can
therefore say that the derivative is zero. We are going to
use
this
truth
in
finding
the
x-coordinates
of
the
stationary points.
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Example: Determine co-ordinates of the turning points of
f ( x)   x 3  4 x 2  x
We have just said that at the turning points, the derivative
is zero.
f ' ( x )  3 x 2  8 x  1
f ' ( x)  0
 3x 2  8 x  1  0
then by u sin g
the quadratic
x  0.13
x  7.87
and
formula
you can then use these values of x to find the correspond ing values of y
y  -(0.13) 3  4(0.13) 2  (0.13 )  0.06
y  (7.87) 3  4(7.87) 2  (7.87)  247.57
Min(7.87; -247.57)
max(0.13; -0.06)
CONCAVITY AND POINTS OF INFLECTION
Read this several times until you understand.
If x  a is a stationary point of a function f (x) then:

f " (a)  0 means that the curve is concave up at x  a
and there is a minimum turning point there.

f " (a)  0 Means that the curve is concave down and there
is a maximum turning point there.
The concavity of a graph at any value of x is determined by
the sign of its second derivative at that value of x.
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To find the point of inflection you need to use the second
derivative to find x.
And then use the original function to find the y-coordinate.
WORKSHEETS
The only recipe for sure success in mathematics is practice,
practice, practice and more practice.
SCN50
1. For
which
values
of
b
is
the
graph
of
y  x 4  bx 3  5 x 2  6 x  8 concave down when x  2 ?
2. For which values of a is the graph of
y  x 3  ax 2  9 x
concave up when x  3?
3. Determine the x-coordinates of the stationary points of
the
graph
f ( x)  x 3  3x 2  25x  21
of
and
then
use
calculus methods to determine where the maximum and the
minimum values occur.
3
2
4. Given y  x  dx  25x  21 and the point of inflection is
(-1 ; 48). Determine the value of d.
3
2
5. The point of inflection of y  ax  3x  36 x  37 occurs
at x 
6. For
1
. Determine the value of a.
2
which
values
of
x
is
the
graph
defined
by
y  x 3  3 x 2  2 concave up?
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SCN51
3
2
Consider the function: f ( x )  2 x  3x  36 x  37
1. Show that x  1 is a factor of f.
2. Determine the x-intercepts of f.
3. Determine the stationary points of f.
4. Determine the point of inflection of f.
5. Sketch the graph of f in the xy-plane below.
y
84
77
70
63
56
49
42
35
28
21
14
7
x
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
10
-7
-14
-21
-28
-35
-42
6. Determine the turning points of
f(x+3)
7. Determine the turning points of f(x) + 1.
8. Determine the equation of the tangent to f at x = 1.
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SCN52
3
2
Consider the function: g ( x)  x  4 x  11x  30
1. Determine the coordinates of the turning points of g.
2. Determine the point of inflection of g.
3. Calculate the x and y-intercepts of g.
4. Sketch the graph of g in the Cartesian plane below.
40
y
35
30
25
20
15
10
5
x
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
10
11
-5
-10
-15
5. Determine the equation of the tangent to the graph of g
at x =1.
6. Determine the x-coordinate of the point at which the
tangent in question 5 cuts the graph of g again.
7. Determine the value(s) of p for which g(x) = p will have
only two roots.
S.C. Nhlumayo
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57
SCN53
Given: f ( x)  ( x  1)( x  1)( x  3)
1. Write down the x-intercepts of f.
2. Calculate the y-intercept of f.
3. Determine the turning points of f.
4. For which values of x is f concave down?
5. Write down the point of inflection of f.
6. In the Cartesian plane below, sketch the graph of f.
y
4
3
2
1
x
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
-1
-2
-3
-4
7. Determine the values of x for which f ( x)  0
S.C. Nhlumayo
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58
SCN54
2
Given: h( x )  ( x  1) ( x  4)
1. Determine the x and y-intercepts of h.
2. Determine the stationary points of h.
3. Sketch the graph of h.
y
x
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
-4
-8
-12
-16
4. Determine the turning points of
g if g ( x)  h( x  3)
5. Determine the turning points of h( x)  2
6. Determine the point of inflection of h.
7. For which values of x is h' ( x)  0?
S.C. Nhlumayo
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59
SCN55
3
Given: f ( x)  x  12 x  16
1. Solve for x if f ( x)  0
2. Determine the coordinates of the turning points of f.
3. Determine the inflection point of f.
4. Sketch the graph of f in the xy-plane below.
y
5
x
-6
-4
-2
2
4
6
-5
-10
-15
-20
-25
-30
5. For which values of x is f ' ( x). f ( x)  0?
S.C. Nhlumayo
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60
SCN56
3
2
Given: f ( x)  2 x  3 x  12 x  4
1. For which values of x is f concave down?
2. Write down the point of inflection of f.
3. Determine the coordinates of the turning points of f.
4. If x – 2 is a factor of f then calculate all xintercepts of f.
5. Calculate the y-intercept of f.
6. Sketch the graph of f in the xy-plane below.
y
16
14
12
10
8
6
4
2
x
-6
-4
-2
2
4
6
-2
-4
-6
-8
-10
-12
7. For which values of x is f ' ( x)  0
S.C. Nhlumayo
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61
GRAPH INTEPRETATION
You are now able to use calculus methods to sketch the cubic
function. The next thing you must practise is using calculus
to interpret graphs.
We are going to look at the graph of the cubic function and
the graph of the derivative.
I’m now starting to enjoy
Calculus.
I’m now ready to learn more
S.C. Nhlumayo
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62
SCN57
(GIVEN THE DEFINING EQUATION AND THE GRAPH)
Consider the graph of f ( x )  (5  x )(x  1) . A and B are the
2
stationary points. A and C are the x-intercepts of f.
y
B
f
x
A
C
1. For which values of x is f concave up?
2. Write down the coordinates of A and C.
3. Determine the coordinates of B.
4. Determine the equation of the tangent to f at x = 2.
5. For which values of x is f ' ( x)  0 ?
6. For which values of k will f(x) = -k have only one root?
S.C. Nhlumayo
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63
SCN58
(GIVEN THE DEFINING EQUATION AND THE GRAPH)
3
2
The graph below shows the graph of f ( x )   x  4 x  11x  30 . A
and B are the turning points.
B
y
x
O
f
A
1. Calculate the x-intercepts of f.
2. Determine the coordinates of A and B.
3. For which values of x is f decreasing?
4. Determine the gradient of f at x =-2.
5. Determine the point of inflection of f
6. Determine the turning points of f(x-1).
S.C. Nhlumayo
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64
SCN59
(GIVEN THE DEFINING EQUATION AND THE GRAPH)
Given the graph of
g ( x)  x 3  4 x 2  4 x  6 . A is the turning
point of g.
y
x
g
A
1. Determine the coordinates of
A.
2. Determine the point of inflection.
3. For which values of x is g increasing?
4. Determine the value of x where the gradient of g is -8.
5. Determine the equation of the tangent to the graph of g
at x = 4.
S.C. Nhlumayo
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65
SCN60. (Given x-cuts and one point on the graph)
3
2
Given the graph of f ( x )  ax  bx  cx  d . (1 ; 16) is the
point on the graph of f. the graph of f cuts the x-axis at x
= -3, x = -1 and x
= 2.
y
f
(1 ; 16)
x
-3
-1
2
1. Determine the values of a, b , c and d.
2. Determine the coordinates of the turning points of f.
3. Determine the equation of the tangent to f at x = 1.
4. Determine the turning points of f(x)-3.
5. For which values of x is f decreasing?
6. For which values of x is f concave up?
S.C. Nhlumayo
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66
SCN61
Sketched above is the graph of f ( x) 
1 3
x  ax 2  bx  10 . It is
2
further given that the graph of f cuts the x-axis at x =-2 ,
x = 2 and x =5.
1. Show that
a
5
and b  2
2
2. Write down the y-cut of f.
3. Determine the turning points of f.
4. Determine the equation of the tangent to f at x = -1.
5. Determine the point of inflection of f.
6. For which values of x is f ( x ). f ' ( x )  0 ?
S.C. Nhlumayo
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67
SCN62
Sketched below is the graph of h( x )  x  ax  bx  c .
3
2
y
(2 ; 0)
(-3 ; 0)
x
h
A
1. Show that a =4, b = -3 and c = -18
2. Determine the coordinates of A, the turning point of h.
3. Determine the values of x for which h is decreasing.
4. Determine the point of inflection of h.
5. Determine the equation of the tangent to h at x = -1.
3
2
6. For which values of k will x  4 x  3x  18  k have three
distinct roots?
S.C. Nhlumayo
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68
SCN63
Consider
the
graph
of
f ( x )   x 3  ax 2  bx  18 . A is the
turning point of f.
y
A
f
(2 ; 0)
x
(-3 ; 0)
1. Determine the values of a and b.
2. Calculate the x-coordinate of A.
3. For which values of x is f concave down?
4. Write down the point of inflection of f.
5. For which values of x is the gradient of f equals to 8?
6. Hence, write down the equation of the tangent to f at
this point.
S.C. Nhlumayo
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69
SCN64
3
2
Sketched below is the graph of f ( x)  x  4 x  ax  b
y
A
f
(-2 ; 0)
x
(3 ; 0)
1. Show that a = -3 and b = 18.
2. Determine the coordinates of A, the turning point of f.
3. For which values of x is f ' ( x)  0?
4. For which values of x is f ( x). f ' ( x)  0 ?
5. Determine the point of inflection of f.
S.C. Nhlumayo
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70
SCN65
3
2
Sketched below is the graph of g ( x )   x  ax  bx  18 . D is
the turning point of g.
y
(3 ; 0)
(-2 ; 0)
x
g
D
1. Write down the y-intercept of g.
2. Determine the values of a and b.
3. For which values of x is g decreasing for x  0 ?
4. Determine the coordinates of D.
5. Determine the gradient of the tangent to g at x = 2.
6. For which values of x is g ' ( x)  0 ?
S.C. Nhlumayo
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71
SOMETIMES YOU WILL BE GIVEN TURNING POINTS ONLY
SCN66
3
2
Sketched below is the graph g ( x)  x  ax  bx . A and B are the
turning points of g.
y
A(-1 ; 3.5)
x
g
B(2 ; -10)
1. Show that a  
3
2
and b  6
2. Determine the x-intercepts of g.
3. Determine the turning points of g ( x  2)
4. Determine the equation of the tangent to g at x = -2.
5. Determine the x-coordinate of the point of inflection of
g.
6. For which values of x is g concave up?
3
7. For which values of s will x 
3 2
x  6 x  s  0 have ONE
2
real root?
S.C. Nhlumayo
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72
SCN67
3
2
Sketched below is the graph of f ( x)   x  ax  bx  30 . A and
B are the turning points of f.
y
B(11/3 ; 14.8)
x
f
A(-1 ; -36)
1. Determine the values of a and b.
2. For which values of x is f ' ( x) f ( x)  0 ?
3. Determine the value(s) of k such that 16 x  y  k  0 is a
tangent to the graph of f.
4. Determine the x-coordinates of the turning points of
f ( x)  4 .
5. For which values of x is f(x) =0?
6. Write down the equation of h if h( x)   f ( x)  2
S.C. Nhlumayo
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73
GIVEN A TURNING POINT AND AN INFLECTION POINT
SCN68
3
2
The diagram below shows the graph of f ( x)  2 x  ax  bx  37 .
A and C are the turning points of f. B is the point of
inflection of the graph of f.
y
A(-2 ; 81)
B(0.5 ; 18.5)
f
x
C
1. Show that a  3
and b  -36
2. For which values of x is f concave up?
3. Determine the coordinates of C.
4. Determine
the
x-coordinate
of
the
point
where
the
tangent to f at C cuts the graph of f.
S.C. Nhlumayo
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74
SCN69
3
2
Sketched below is the graph of g ( x)  ax  11x  bx . B and C
are stationary points of g. A is the point of inflection of
g.
y
C
g
x
A(11/6 : -2.65)
B(3 : -9)
1. Determine the values of a and b.
2. Determine the x-intercepts of g.
3. Determine the coordinates of C.
4. Find the equation of the tangent to the graph of g at
the point (2 ; -4).
5. Calculate
the
co-ordinates
of
the
point
where
this
tangent cuts the curve again
S.C. Nhlumayo
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75
THE GRAPH OF THE DERIVATIVE.
SCN70
2
Sketched below is the graph of f ' ( x )  ax  bx  c . The x-cuts
are given as shown in the diagram. f is a cubic function.
y
f'(x)
x
-2
5
1. Write down the x-coordinates of the stationary points of
f.
2. State whether each stationary point is a local minimum
or a local maximum. Support your answer.
3. Determine the x-coordinate of the point of inflection of
f.
4. For which values of x is f decreasing?
5. Draw a sketch graph of f.
S.C. Nhlumayo
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76
SCN71
Consider the graph of f ' where f is a cubic function.
y
f'(x)
x
-2
5
1. For which values of x is f increasing?
2. For which values of x is f decreasing?
3. Write down the x-coordinate of the point of inflection.
4. Write down the x-coordinates of the turning points of f
and state whether each stationary point is a local
maximum or a local minimum.
5. Sketch the graph of f.
S.C. Nhlumayo
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77
SCN72
Sketched below is the graph of g ' ( x)  
1
 x  52  4 where g is
2
a cubic function
y
C
A
x
B
g'
D
1. Determine the x-coordinates of A, B and C.
2. Write down the x-coordinates of the stationary point of
g.
3. For which values of x is g decreasing?
4. For which values of x is g increasing?
5. For which values of x is g concave up?
6. For which values of x is g concave down?
7. Write down the x-coordinate of the point of inflection
of g.
S.C. Nhlumayo
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78
OPTIMISATION
 Apply the derivative of a function to solve optimization
problems
 Apply the derivative of a function to determine the rate
of change of physical quantities.
OPTIMIZATION: Looking for the best possible outcome.
Seeking the best possible outcome could mean one of the following:

Maximisation (seeking the largest possible outcome) or

Minimisation (seeking the smallest possible outcome)
Some physical quantities which can be maximized or minimized:

Lengths and distances

Areas and surface areas

Volumes

Velocity

Acceleration
Steps in Optimisation

Draw a diagram, if necessary

Set up a function for the object/ quantity to be optimised.

Take note of the restrictions that could be connected to the practical
situation. (e.g. it would be wrong to find something like -3 seconds)


Use the first derivative to determine the critical numbers.
Use the second derivative to determine whether maximums or minimums
occur at the critical numbers.
S.C. Nhlumayo
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79
SCN73.
(NOV 2010: DIAGRAM MODIFIED)
The object below is made of a cylinder with a hemisphere at
each end. The radius of the cylinder is r and its height is
h.
The volume of the object is
1. Show that
h

.
6
1
4r

2
6r
3
2. Hence, show that the outer surface area of the object is
given by S 
4 r 2


3
3r
3. Calculate the minimum outer surface area of this object.
V  r 2 h (Cylinder)
A  2rh (curved surface area of a cylinder)
4
V  r 3 (Sphere)
3
TSA  4r 2 (Sphere)
S.C. Nhlumayo
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80
SCN74
Ayabonga
wants to make a bird-cage. She has 180 cm of wire
and bends it to form a rectangular prism with an extra band
around its middle.
1. Show that L  45  3 x
2. Show that the equation for the volume
3
enclosed by the frame is V  3x  45x
2
3. Calculate the dimensions of the bird-cage
that will give the maximum volume.
S.C. Nhlumayo
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81
SCN75
A cylinder fits exactly into one half of a spherical
container as shown. The radius of the sphere is 2 3 cm. The
height of the cylinder is x.
1. Show that
r
2
 12  x 2
2. Calculate the value of x that will maximize the volume of
the cylinder.
3. Calculate the maximum volume
2
Volume of a Cylinder  r h
S.C. Nhlumayo
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82
SCN76
Gubevu wants to make an open box (i.e. a box has no lid)
with a square base and a volume of 2.5 m3.
1. Determine the height of the box in terms of x.
2. He wants to paint the inner and the outer surface area
of the box, including the base. Show that the area to be
2
painted is given by the equation: A  2 x 
20
x
3. Hence, determine the value of x that will give the
minimum area to be painted.
S.C. Nhlumayo
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83
SCN77
A cylinder fits in a cone as shown in the diagram below. The
length from the rim of the cylinder to the center of the
base is 2 3 .
The radius of the cylinder is r and its height is x.
2 3
2
1. Show that r  12  x
2
2. Write down the equation of the volume of the cylinder in
terms of x.
3. Calculate the value of x that will maximize the volume
of the cylinder.
4. Calculate the radius of the cylinder if it is of maximum
volume.
S.C. Nhlumayo
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84
SCN78.
An object is made of a cylinder and a hemisphere. The radius
of the cylinder is x mm. the total length of the object is
to be (600 –x) mm as shown,
1. Show that the volume of the object is V  600x 2 
4 3
x
3
2. Find the value of x at which maximum volume occurs.
SCN79
1
4
3
The equation d (t )   t 
1 2
t  4 shows the depth of water in
2
metres in a water tank. The time, t, is the number of hours
that have passed since 8 a.m.
1. What is the depth of water at 10:30 a.m?
2. At what rate is the depth of water changing at 11:00
a.m?
3. At what time will the inflow of water be the same as the
outflow?
S.C. Nhlumayo
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85
SCN80
The relationship between the number of employees working for
your uncle and the profit he makes is given by the formula
P( x)  2 x 3  600 x  10000
1. Determine the number of employees he needs for the
business to make maximum profit.
2. Calculate maximum profit.
SCN81
It is given that, the number of bacteria at any moment
be represented by the formula
can
f (t )  5t 2  50t  1000 where
f(t) is the number of bacteria (in millions) present and t
is
the
number
of
hours
after
the
beginning
of
the
experiment.
1. Calculate the rate of change of the number of bacteria
after 2 hours.
2. After how many hours does the population of bacteria
start decreasing?
3. How long will it take for all the bacteria to die.
S.C. Nhlumayo
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86
CALCULUS OF MOTION.
In calculus of motion you need to know that the motion of an
object is defined by its position relative to a fixed point,
its velocity and acceleration.
The term used for position is displacement.
Velocity is the rate of change of displacement with respect
to time.
Acceleration is the rate of change of velocity with respect
to time.
If
s  f (t ) is the equation of motion, which gives position at time t
Then v 
And a 
ds
gives the velocity at time t.
dt
dv
dt
is the acceleration at any time t.
____________________________________________________________
S.C. Nhlumayo
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87
SCN82
The height that a stone is above ground level at any time, t
seconds after being catapulted up is given by S (t )  25t  5t
2
1. Determine the velocity of the stone at the instant that
it is catapulted.
2. Determine the maximum height that the stone reaches.
3. What is the acceleration of the stone while it is in the
air?
SCN83.
The stone is thrown vertically upwards. Its height, s metres
at any given time(sec) is given by S (t )  20t  5t
2
1. Find what height the stone is after (i) 1sec
(ii) 2 sec
(iii) 3sec
2. At what time will the stone again hit the ground?
3. What is the greatest height reached by the stone?
4. Find an expression for the acceleration of the stone.
S.C. Nhlumayo
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88
SCN84
The distance in metres covered by two objects in t seconds
is
given
by
the
f (t )  t 3  2t 2  3t  4
functions
and
g (t )  3t 2  5t  2 .
1. After how
many seconds is
the velocity of the
two
objects the same?
2. How far has each object travelled after 2 seconds?
3. What
is
the
acceleration
of
an
object
that
has
a
velocity given by r (t )  6t  5 ?
SCN85
Let the displacement (in metres) of an object be represented
by the function
S (t )  2t 2  7t  4 where t is the time in
seconds.
1. Determine the velocity of the object at t = 3 seconds.
2. Determine the acceleration of the object at any time.
S.C. Nhlumayo
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89
This marks the end of my worksheets on
Trigonometry,
Geometry
and
Calculus.
Unfortunately I could not provide solutions to my
worksheets. I trust teachers to help learners in that
case. If you get stuck, you may still communicate
with me via email: [email protected] or call
me: 082 291 7325/ 072 922 5920.
S.C. Nhlumayo
[email protected]