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SSIP 2016 SEPTEMBER

MATHEMATICS GRADE 12 FACILITATORS’ MANUAL

MATHEMATICS Gr 12 SSIP SEPTEMBER 2016

Copyright

This work is protected by the Copyright Act 98 of 1978. No part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage and retrieval system, without permission in writing from Matthew Goniwe School of Leadership and Governance.

Whilst every effort has been made to ensure that the information published in this work is accurate, Matthew Goniwe School of Leadership and Governance takes no responsibility for any loss or damage suffered by any person as a result of the reliance upon the information contained therein.

Page 1


Contents Pages

1. Content that will be tested according to CAPS 3

2. Weighting of content areas 3

3. Cognitive levels of questions (TIMSS) 4

4. Patterns 5 – 13

5. Inverse functions 14 – 30

6. Solving triangles in 2D and 3D 31 – 50

7. Application of Calculus 51 – 78

8. Sequences and Series 79 – 98

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CONTENT THAT WILL BE TESTED ACCORDING TO CAPS

WEIGHTING OF CONTENT AREAS

Mark distribution for Mathematics NCS end-of-year papers: Grade 12

PAPER 1: Grade 12 bookwork: maximum 6 marksDescription Grade 12Algebra and equations (and inequalities) 25+3Patterns and sequences 25+3Finance, growth and decay 15+3Functions and graphs 35+3Differential Calculus 35+3Probability 15+3TOTAL 150PAPER 2: Grades 11 and 12: theorems and/or trig proofs: maximum 12 marksDescription Grade 12Statistics 20+3Analytical Geometry 40+3Trigonometry 40+3Euclidean Geometry and Measurement 50+3TOTAL 150

Notes: Modelling as a process will be included in all papers, thus contextual questions can be

set on any topic. Questions will not necessarily be compartmentalised in sections. Various topics can

be integrated in the same question. A formula sheet will be provided.

Page 3


Cognitive levels (TIMSS)

The questions will be levelled according to the following table:

Page 4


QUADRATIC NUMBER PATTERN

DEFINITION: Quadratic number pattern

A quadratic pattern is a pattern of numbers in which the second difference between any two consecutive terms is constant

The general terms of the form

T n=an2+bn+c

Example: Determine the general term of the pattern 4; 9; 18; 31…….if the term continues consistently.

4 9 18 31

5 9 13 1st difference

4 4 2nd difference

How can we be able to determine the general term?

By using the general term:T n=an2+bn+c

T 1=a(1)2+b(1)+c T 2=a(2)2+b(2 )+c

T 3=a(3 )2+b (3 )+c T 4=a (4 )2+b(4 )+c

a+b+c 4 a+2 b+c 9 a+3b+c 16 a+4 b+c

3 a+b 5 a+b 7 a+b 1st difference

2 a 2 a 2nd difference

NOTE: The first differences will always be a linear pattern.

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The second difference is given as an expression2 a .

then 2a=4 3 a+b=5 a+b+c=4

∴a=2 3(2 )+b=5 2−1+c=4

∴b=−1 c=3

The general term is T n=2n2−n+3

EXAMPLE 1

Consider the following number pattern 2; 7; 14; 23…..

1.1 Show that it is a quadratic number pattern.1.2 Write down the next two terms of the number pattern.1.3 Hence determine the nth term as well as the 100th term1.4 Determine which term equals 162.

EXAMPLE 2

Consider the following quadratic number pattern: 6 ; x ; 26 ; 45 ; y..

Calculate the values of x and y.

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ITS VERY IMPORTATNT TO DETERMINE THE TYPE OF PATTERN YOU ARE WORKING WITH

TO SHOW THAT A PATTERN IS A QUADRATIC NUMBER PATTERN DETERMINE THE SECOND DIFFERENCE IF IT’S COMMON THEN THE PATTERN IS QUADRATIC


EXERCISES

1 The sequence 3 ; 9 ; 17 ; 27 ; … is quadratic.

1.1 Determine an expression for the n-th term of the sequence. (4)

1.2 What is the value of the first term of the sequence that is greater than 269? (4)

2 2 ; x ; 12; y ; . . . are the first four terms of a quadratic sequence. If the second difference is 6, calculate the values of x and y. (5)

3. The quadratic pattern−3 ;4 ;x ;30 ;…………..is given. Determine the value of x. (4)

4 In the quadratic sequence4 ; x ; y ;−11; …, the first three terms of the first

differences are 2 p−4 ; p−3 andp2−1.

Determine the:

4.1 Value(s) ofp. (2)

4.2 Second difference(s). (1)

4.3 Values of x and y. (2)

4.4 General term of the quadratic sequence. (4)

5 Given the quadratic sequence 3; 5; 11; 21; …

5.1 Write down the value of the next term, if the pattern continues. (1)

5.2 Determine the value of the 48th term. (5)

5.3 Prove that the terms of this sequence will never be even numbers. (2)

5.4 If all the values of this sequence are increased by 30, determine the

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general term of the new sequence. (2)

6. A pattern of triangles is formed by increasing the base of the triangle by 2 cm and the

perpendicular height by 1cm, in each successive triangle.

The first triangle has a base of 2 cm and a height of 2 cm. The pattern continues in this manner.

TRIANGLE 1 TRIANGLE 2 TRIANGLE 3

6.1 Calculate the areas of the first four triangles. (2)

6.2 Calculate the area of the 100th triangle in the pattern (4)

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2cm

2cm

4cm 6cm

3cm 4cm


SOLUTIONS

Page 9


11.1

2a = 2 a = 13a + b = 6 b = 3c = T0 = -1Tn = n2 + 3n -1

1.2 n2+3n−1>269∴n2+3 n−270>0∴(n+18)(n−15)>0(n+18)(n−15)>0n>15=16term: 162+3 (16 )−1=303

22 ; x ; 12 ; y

x−2 12−x y−12

12−x−(x−2) y−12−(12−x)

2nd difference = 612−x− (x−2 )=6 −2 x+14=6 −2 x=−8 x=4Substitute x=4 into y−12−(12−x )=6 y−12−(12−4 )=6 y−20=6 y=26

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3−3 ;4 ;x ;30 ;…7 ; x−4 ;30−x (First difference)x−11;34−2 x (Second difference)x−11=34−2 x3 x=45x=15

44.1 2( p−3 )=2 p−4+ p

2−1

p=−2

4.2 Second difference = 34.3 x=−4 and y=−94.4 2 a=3

a=32

3 a+b=−8

3(32)+b=−8

b=−252

a+b+c=432−25

2+c=4

c=−7

T n=32

n2−252

n−7

Page 11


55.1 T 5=355.2 T n=a n2+bn+c

a=2

3(2)+b=2

b=−4

2−4+c=3

c=5

T n=2 n2−4 n+5

T 48=2(48)2−4 (48)+5

¿44215.3 2n2 even for all values of n,

4n is even for all values of n.

So T n=2n2−4n+5

is the sum of two even numbers and 5

which will not be even.

OR/OF

T n=2 n2−4 n+5

¿2 (n2−2 n+2 )+1

= (2¿

which is not even5.4 T n=2 n2−4 n+5 + 30

¿2 n2−4 n+35

Page 12


66.1 2 ; 6; 12;206.2

T 100=(100 )2+100T 100=10100 cm2

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Notation: If the original function is named f , the inverse will be f−1 .

The line y=x (y−x=0) is the line of reflection of the function and its inverse,as illustrated in the example below.

It is important to understand that every point ( x ; y ) on the graph of f becomes( y ; x ) on the graph of f−1.

According to the CAPS document learners should be able to:

Determine if a relation is a function or not Understand and be able to apply their knowledge of the concept of the inverse of a function Know how the domain of a function may need to be restricted to ensure that the inverse is a function Determine and sketch graphs of the inverses of the functions defined by:

*y=ax+q*y=a x2

*y=bx ;b>0 ,b≠ 1 Determine the following characteristics of the functions and their inverses:

*Domain and range*Intercepts with the axes*Turning points*Minima and Maxima*Asymptotes (horizontal and vertical)*Shape and symmetry*Average gradient (average rate of change)*Intervals on which the function increases/decreases.

Understand the definition of a logarithm:y=logb x⇔ x=bx where b>0∧b≠ 1

Be able to determine and sketch the function define by y=logb xfor both the cases 0<b<1∧b>1.

Page 14

INVERSE FUNCTIONS

f−1

f y=x

Introduction


INTERCHANGE (“swap”) the x and y values Make y the subject of the formula again When working with the exponential function, we need ‘logs’ to help us make y the

subject of the formula Examples:

Linear Function Quadratic Function Exponential FunctionEquation of the given function

or

INTERCHANGE the x- and y-values

Make y the subject of the formula again to determine the equation of the inverse in the form y = …..

First draw the graph of the given function Draw the line of reflection y=x (use a dotted line) Take any two known points on the given function, interchange their x- and y-values

and then plot the two new points Now draw the graph of the inverse as a reflection of the graph of the given function

about the liney=x through the two newly plotted points

Page 15

To determine the EQUATION of the inverse of a function

To draw the graph of the inverse of a function


Examples of inverses:

Linear Function

QuadraticFunction

Exponential Function

Equation ofthe originalfunction or

Graph of the original function(solid line/ curve)

Graph of the inverse(Dotted line/ curve)

The graph of the inverse of the parabola is NOT a function because it is a one-to-many mapping. (a vertical line will cut it in two places)

Page 16

Discussion:How will you explain to your learners the fact that we do not look at the inverse of the hyperbolic function?

The inverse of a the quadratic function

or


We restrict the domain of the original parabola in the following two ways so that the inverse of the parabola is a function again:

1) : the inverse is a function 2) : the inverse is a function

Restrict the domain of the negative quadratic function in two different ways so that the inverse will be a function. Draw the graphs to illustrate this.

Learners must know how to:

draw a graph of a function if the equation is given determine the equation of a function if the graph is given determine the equation of the inverse of a function and also draw the graph of

the inverse of a function determine the following of the functions (or inverses)

x- and y-intercepts turning points equations of the asymptotes equations of the axes of symmetry

Also, learners must be able to:

determine the domain and range of functions and their inverses state where the function (or its inverse) increases/decreases shift a function (or its inverse) –

o to the left or righto up or down

reflect a function (or its inverse) – o in the x-axiso in the y-axis

determine the equation of a function (or its inverse) after a shift or a reflection determine where a function (or its inverse) is positive or negative apply their knowledge of the nature of roots

Page 17

Knowledge tested with functions and inverses


The following table will assist learners in answering difficult questions on graphs of functions. Make sure they have a copy of the conditions and are able to apply it.

Condition Interval where applicablef (x)>0 The part of the graph of f that is above the x-axisf ( x )<0 The part of the graph of f that is below the x-axis

f ( x )=g (x) The point(s) where the graphs of f and g intersectf ( x )>g(x ) The part of the graph of f that lies above the graph of gf (x)≤ g (x) The part of the graph of f that lies below the graph of g, or is equal

to the graph of gf ( x )−g ( x )=k The distance between the two graphs, with the graph of f on topf ( x ) . g (x)>0 Where parts of both graphs are above the x-axis or below the x-

axisf ( x ) . g (x)<0 Part of one graph is above the x-axis and in the same interval part

of the other graph is below the x-axis

Consider the function

S H I F T SShift 3 units to the LEFT

Change in the equation

Graphical representation

Substitute the x in the equation with ( x+3 )

Original equation:

New equation:

Shift 4 units to the RIGHT



Substitute the x in the equation with ( x−4 )

Original equation:

New equation:

Page 18

Shifts and Reflections in a nutshell


Shift 3 units UPWARDS



Add 3 to the original equation

Original equation:

New equation:

Shift 4 units DOWNWARDS



Add -4 to the original equation

Original equation:

New equation:

Consider the functions and

R E F L E C T I O N SReflection in

thex-axis



Substitute y in the original equationwith –y.Then make y the subject of the formula again.

Original equation:

New equation:

− y=ax2+bx+c ∴ y=−ax2−bx−c

Original equation:

New equation:

− y=a .bx+ p+q ∴ y=−a .bx+ p−q

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Reflection in the

y-axis



Substitute x in the original equation with -x

Original equation:

New equation:

y=a (−x )2+b (−x )+c ∴ y=ax2−bx+c

Original equation:

New equation:

y=a .b−x+ p+q

Page 20


1. Given: h ( x )=2 x−3 for −2≤x≤4 . The x-intercept of h is Q.

1.1 Determine the coordinates of Q. (2)1.2 Write down the domain of h

−1. (3)

1.3 Sketch the graph of h−1

. Clearly indicating the y-intercept and the end points. (3)

1.4For which value(s) of x will h ( x )=h−1 (x ) ? (3)

Page 21

Examples


Given h ( x )=2 x−3 for −2≤x≤4 .

1.1 For x-intercepts, y = 02 x−3=0x=1,5Q (1,5 ; 0 )

x=1,5 y = 0

(2)1.2 h :

x=−2 : y=2(−2 )−3=−7x=4 : y=2( 4 )−3=5Domain of h−1 :−7≤x≤5 OR [−7 ; 5 ]

h(–2) = –7

h(4) = 5

−7≤x≤5(3)

1.3

OR

y-intercept on a straight line

line segment

accurate endpoints (x or y or both)

(3)

Page 22


1.4 h ( x )=2 x−3

For the inverse of h,x=2 y−3

y=x+32

h ( x )=h−1 (x )

2 x−3=x+32

4 x−6=x+33 x=9x=3

OR

y= x+3

2

2 x−3= x+3

2

x=3(3)

h ( x )=2 x−3h and h

−1 intersect when y=x

h ( x )=x2 x−3=xx=3

ORh ( x )=2 x−3

For the inverse of h,x=2 y−3

y=x+32

h−1 ( x )=xx+32

= x

x+3=2 xx=3

h ( x )=x

2 x−3=x

x=3(3)

y= x+3

2

x+32

= x

x=3(3)

Page 23


2. The sketch below shows the graphs of f ( x )=1x for x<0 and g ( x )=−√−x for x≤ 0.

2.1 Write down the equations of the asymptotes of the graph of f . (2)

2.2 The graphs of f and g intersect at the point A ( p ;−1 ). Calculate the value of p. (2)

2.3 Determine the equation of g−1 in the form y=¿….. (3)

2.4 Write down the equation of h if h is a reflection of g about the y-axis. (1)

Page 24


3. The graph of f ( x )=ax , where a>0 and a≠ 1 , passes through the point (3 ; 278 ).

3.1 Determine the value of a. (2)

3.2 Write down the equation of f−1 in the form y=¿….. (2)

3.3 Determine the value(s) of x for which f−1 (x )=1. (2)

3.4 If h ( x )= f (x )−5 , write down the range of h . (2)

Page 25

2.1 x=0 andy=0

x=0y=0 (2)

2.2 y=1x ( p;−1 )

−1= 1p

∴ p=−1

OR

y=−√−x ( p ;−1)

−1=−√−p

(−1 )2=(−√−p )2

1=−p ∴ p=−1

subst. x=p and y=−1p=−1

OR

subst. x=p and y=−1

p=−1 (2)

2.3g ( x )= y=−√−x

Interchange x and y

x=−√− y

x2=− y

y=−x2 ; x≤ 0 (or y ≤ 0)

x=−√− y

x2=− y

y=−x2 ; x≤ 0 (3)

2.4 g ( x )=−√−x (Given)

Reflection about the y-axis, replace x with –x:

h ( x )=−√−(−x)

∴h (x)=−√x h(x )=−√x (1)


3.1 y=ax

278=a3

( 32 )

3

=a3

a=32

substituting

(3 ; 278 )

a=32

(2)3.2

y=(32 )

x

Interchange x and y:

x=(32 )

y

∴ y=log 32

x

x=(32 )y

y=log 32

x

(2)

3.3 log 32

x=−1

x=(32 )−1

x=23

write in exponential form

x=23

(2)3.4 y∈ (−5 ;∞ ) ¿

∞ ¿ (2)

Page 26


4. The graph of f ( x )=3x is sketched alongside.

4.1 Give the coordinates of A. (1)

4.2 Write down the equation of f−1 in the form y = … (1)

4.3 For which value(s) of x will f−1≤ 0? (2)

4.4 Write down the equation of the asymptote of f (x−1) (1)

5. Sketched is the graph of f , the inverse of a restricted parabola. The pointA(8 ; 2) lies on the graph of f.

5.1 Determine the equation of f in the form y = … (2)

Page 27


5.2 Hence, write down the equation of f−1

in the form y = … (2)

5.3 Give the coordinates of the turning point of g(x) = f−1( x+3)−1 . (1)

6. The sketch below represents the inverses of g ( x )=4 x and f ( x )=a x2; x ≥ 0.

6.1 Write down the coordinates of ONE point through which both f and g will pass. (1)6.2 Determine the equation of f . (3)6.3 Calculate x if g ( x+2 )=16 . (3)6.4 If h ( x )=g−1 ( x−2 ), for which values of x will h(x )≤ 0? (2)

Page 28

y

x

(4 ; 1)

(16 ; 2)

..

0 1


4.1 A(0; 1) Answer (1)

4.2 f-1: y = log 3 x y = log 3 x (1)

4.3 0<x ≤1 endpoints notation (2)

4.4 y=0 y = 0 (1)

5.1f (x)=√ x

a ORf−1 (x )=a x2

(8; 2): 2 = √ 8a

a = 2 (2; 8): 8 = a(2)2 a = 2

∴ f (x )=√ x2∴ f−1 ( x )= 2x2

f ( x ) : x=2 y2

y =√ x2

a = 2

eqn (2)

5.2 f−1 (x )=a x2

(2; 8): 8 = a(2)2 a = 2∴ f−1 ( x )= 2x2

eqn(1)

5.3 (−¿3; −¿1) each value (2)

Page 29

SOLUTIONS


6.1 (1 ;4) or (2 ;16) Any one point (1)

6.2 f (x)=a x2

Substitute inverse point:

(1 ;4) OR (2 ;16)

4=a(1)2 16=a(2)2

a=4 4 a=16

a=4

∴ f ( x )=4 x2 , x≥ 0

OR

Substitute (4 ;1) or (16 ;2 ) into x=a y2:

4=a(1)2 OR 16=a(2)2

4=a 16=a .4

4=a

∴ f ( x )=4 x2 , x≥ 0

Substitute correct coordinate

Equation f ( x ) x ≥ 0

(3)

Substitute correct coordinates

Equation f ( x )

x ≥ 0 (3)

6.3 g ( x+2 )=16

4 x+2=16

4 x+2=42

∴ x+2=2

x=0

Substitution into g ( x )

Same base x=0

(3)

6.4 2<x≤ 3 x>2 x≤ 3

(2)

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SOLUTIONS

Page 43


1.1tanθ = TB

DBTB= DB tanθ

1.2 DBsin E

= DEsin B

DBsin E

= DE

sin D B¿

E

but D BΛ

E = 180ο − ( x + y )

∴ DBsin y

= 10sin [180ο −( x + y ) ]

∴ DBsin y

= 10sin ( x + y )

∴ DB = 10 sin ysin ( x + y )

1.3 TB = DB tanθ

=10 sin y . tan θsin( y + y )

=10 sin y . tan θsin2 y

=10 sin y . tan θ2 sin y . cos y

=5 tan θcos y

1.4 Area Δ BDE = 12

BD . DEsin x

= 12× 10sin y

sin( x + y )×DE × sin x

= 5sin35ο × 10 × sin 35ο

sin 70ο

¿50 sin2 35°sin 70 °

¿ 17 , 5 m2

Page 44


22.1 MQ2=k 2+(2k )2−2 k (2k )(Cos 2θ )

= k2+4 k2−4 k2 (1−2 Sin2 θ )

= 5 k2−4 k2+8 k2sin2 θ

= k2+8 k2sin2 θ

= k2(1+8 sin2 θ )

MQ=k √1+8sin2 θ2.2 MQ=139 , 5 m√1+8 sin2 420

= 299 m

33.1 In Δ DCB :

hDB

= sin y

∴ h = DB sin y −−−(1)In Δ DAB :DBsin x

= dsin D 1

but D1 = y − x

∴ DB = d sin xsin ( y − x )

−−−(2)

subst . (2 ) into (1) :

h = d sin x sin ysin ( y − x )

3.2h = 85 . sin 10° . sin 38 °

sin 28 °= 19 , 36 m

Page 45


44.1 tan 40°= LH

LB

tan 40°=3LB

LB=3 ,58 m4.2 AB2=AL2+LB2−2 AL . LB cos113 °

¿(5,2 )2+(3 .58)2−2(5,2)(3 , 58 )cos113 °AB=7 ,38 m

4.3 In Δ ABL

Area of Δ ABL=1

2. AL. BL . sin113°

=12×5,2×3 , 58sin 113

=8 ,57 m2

5AB = BC = a

AC 2=AB 2+BC 2−2 AB . BC cos { B ¿b2=a2+a2−2(a )(a )cos { B ¿ ¿b2=2 a2−2 a2cos { B ¿ ¿b2=2a2(1−cos { B ¿) ¿ b2

2 a2 =1−cos { B ¿ ¿ b2

2 a2 −1=−cos { B¿ ¿∴cos { B ¿=1− b2

2 a2 ¿¿

Page 46


66.1 cosθ=BC

2a∴BC=2 a cosθ

6.2 BO2=BC2+CO2

(2 a)2=(2 acos θ)2+CO2

∴CO=2 a sinθ

Area of Δ OCB

=12

. CO .BC

=12

. 2a sin θ. 2 acosθ

=a2 . 2sin θ cosθ¿a2 sin 2θ

6.3 sin 2 θ=12 θ=90°θ=45 °∴C (45 ° ;1)

Page 47


77.1 In Δ BAC

AC2=BC2+AB2−2 BC . AB cos { B ¿AC 2=(12)2+(20 )2−2(12 )(20)cos110°AC=26 ,6 m2

7.2 In Δ BAC

sin B ACBC

=sin {BAC

¿ sin B A C=12 sin110 °26 ,61

¿ sin B AC=0 ,4237 . .. . . . ¿∴B A C=25° ¿¿7.3 In Δ ACD

AC 2=CD2+ AD2−2CD . ADcos { D ¿26 , 612=72+282−2(7 )(28 )cos { D ¿ ¿0 ,31864 . .. ..=cos { D¿ ¿ D=71 ° ¿¿7.4 Area of ABCD

In Δ ACD

Area of Δ ACD= 1

2.CD . AD . sin { D ¿

Area of Δ ACD

=12

.7 .28 . sin 71°

=92,6608 m2

In Δ ABC

Area of

Δ ABC=12

. BC . AD . sin { D ¿=12

.12. 20 sin 110° ¿=112 .70 m2 ¿¿∴Area of ABCD = Area of Δ ABC+Area of Δ ACD =92.66 +112,70 = 205,4m2

Page 48


88.1 tan65 °= 5,9 m

PQPQ= 5,9 m

tan65 °∴PQ=2 ,75 mthenPQ≈3 m

8.2 Q P R=180 °−2 xRQ

sin(180 °−2 x )= 3 m

sin x

RQ=3 sin2 xsin x

RQ=3.2 sin xcos xsin x

RQ=6 cos x8.3 RQ=6cos (42 )

∴RQ=4 , 458868 m . . .. .

Area of

Δ PQR=12

PQ . QR sin P Q R

=12×3×4 , 4588 sin 42

=4 , 48m2

Page 49


99.1 Q R S=α

QRsin(150 °−α )

=12sin α

QR=12 sin(150 °−α )sin α

QR=12(sin 150 °cos α−cos150 °sin α )sin α

QR=12( 1

2cos α+ √3

2sin α )

sin α

QR=6 cosα+6√3sin αsin α

9.2 tan α= PQ6cosα+6√3sin αsin α

PQ=tan α . 6 cos α+6√3 sin αsin α

PQ=sin αcos α

. 6 cos α+6√3 sin α )sin α

PQ=6cosα+6√3sin αcos α

∴PQ=6+6√3 tan α9.3 PQ=6+6√3 tan α

23=6+6√3 tan α17=6√3 tan α∴α=58 .56 °

10.

Area of inner Δ =

12×50×50sin 60 °

= 1082,531755cm2

Area of outer Δ =

12×80×80sin 60 °

= 2771,281292cm2

Area of red shaded part = Area of outer Δ - Area of inner Δ = 2771,281292 – 1082,531755 = 1688,75cm2

Page 50


1111.1 Area Δ ABC=1

2mnsin 4 x

11.2 Max area if sin 4 x=14 x=90 °x=22 , 5°

11.3 Right-angled triangle

12.12.1 DE

p= tan α

DE=p tan α12.2.1 E1=180 °−β (opp s of cyclic quad)

F1=H2 (s opp equal sides; EH = EF

F1+H2=180°−(180 °−β ) (s of )=β

F1=H2=12

β

12.2.2 In EFHp

sin 12

β=

FHsin(180°−β )

p=FH.sin1

2β

sin β

=FH.sin 1

2 β

2sin 12

β .cos12

β

=FH

2cos12

β

12.2.3 FH2=p2+ p2−2 p2cos (180 °−β )¿2 p2+2 p2cos β¿2 p2(1+cos β )FH=p√2(1+cos β )

Page 51


The CAPS document states the following in terms of Differential Calculus.

Learners must:

understand the limit concept in the context of approximating the rate of change or gradient of a function at a pointunderstand intuitively that f ' (a ) is the gradient of the tangent to the graph of f at the point (a ; y )be able to determine the derivative f ' ( x ) by using the definition (first principles)be able to determine the derivative f ' ( x ) by making use of the rules of differentiationbe able to find the equation of the tangent to a graph of a functionunderstand that the second derivative f ' ' ( x ) will assist in finding the point of inflection and also how it determines the concavity of a functionsketch graphs of cubic polynomial functions using differential calculus to determine coordinates of stationary points and points of inflection. Also, determine the x-intercepts of the graph using the factor theorem and other techniques (e.g. synthetic division, long division)Solve practical problems concerning optimisation and rate of change, including calculus of motion

Differential calculus is primarily about rate of change.

In Mathematics: Rate of change = gradient (m)

The gradient (or rate of change) of a linear function is constant.The graph of a linear function is a straight line. When looking at a straight line graph it is clear that the gradient remains the same (or constant) at any point on the line.

Example:

For f ( x )=2 x+4 the gradient is always 2. The graph of f illustrates this fact.

Page 52

APPLICATIONS OF CALCULUS

y

4f

Introduction


For any other function the gradient changes the whole time. The gradient will change from one point to the next on the function.

Example:

For g ( x )=x2−4 the gradient will change.See the illustration alongside.

The derivative of a function gives the gradient (or rate of change) of that function at any point.(The derivative may also be referred to as the ‘gradient function’ )

Make sure that your learners clearly understand that for any function f :

f ( x )= y a y-value on the graph of f associated with a specific x-value f ' ( x )=m the gradient of f associated with a specific x-value

Differential calculus, which is primarily about rate of change, has many applications.There are mainly three types of applications:

1) In graphs2) To determine rate of change 3) Minimum and Maximum values

Page 53

0 x-2

0-2 2 x

y

g

At x=−2 ,m=−4

At x=2 , m=4


1. GRAPHS

The first and second derivative of a function is very useful to find crucial points on the graph of the function.

Turning points (Stationary points)

At the turning points: f ' ( x )=0

Minimum turning points

OR

Page 54

f ' ( x )=0

f ' ( x )=0

ff

f ' ( x )=0

f ' ( x )<0 f ' ( x )>0At a minimum turning point the sign of the gradient changes from negative to positive.

f ' ' ( x )>0

fAnother way to determine whether the turning point is a minimum is to find the second derivative. For a minimum turning point the second derivative is positive.

Basic Concepts


Maximum turning points

OR

Make sure that your learners know that a turning point can be useful in TWO ways:

Firstly, the x-value can be substituted into f ' ( x )=0 (Turning point) Secondly, the x- and y-values can be substituted into f ( x )

Points of inflection

f ' ' ( x )=0at a point of inflection.

The gradient does not change sign at the point of inflection, as illustrated in the diagrams below.

Page 55

Another way to determine whether the turning point is a maximum is to find the second derivative. For a maximum turning point the second derivative is negative.

f ' ( x )<0f ' ( x )>0

f ' ( x )=0At a maximum turning point the sign of the gradient changes from positive to negative.

f

f ' ' ( x )<0

f ' ( x )>0 f ' ( x )<0


At the point of inflection the concavity of the graphs changes.

Consider the diagram below. Along the curve PA the concavity of the graph is concave down. Along curve AQ the concavity of the curve is concave up.

Comparison between graphs of f ( x ) and f ' ( x )

Function: f ( x ) Derived function: f ' ( x )y = ax 3 + bx2 + cx + d

The x-values of the turning points are the x-intercepts of the graph of y’ =…. , which is the parabola on the right.The point of inflection’s x-value is the x-value of the turning point of the graph of y’.

y ' = 3 ax 2 + 2 bx + c

These x-intercepts, are the turning point x’s of the cubic graph, because the y- values here are 0.The turning point x-value at i, is the x-value of the point of inflection on the cubic graph.

y = ax 2 + bx + c y ' = 2 ax + b

At this x-intercept, y’ = 0. This x-intercept and the

Page 56

¿ ¿

f ' ( x )>0 f ' ( x )<0

A A

At inflection point A: f ' ' ( x )=0

P

Q

A¿

At inflection point A: f ' ' ( x )=0

x(-3 ; y)

x(-3; 0)

x x

i

(-4 ; 0) (2 ; 0)

(-1 ; y)

x

x

i

(-4 ; y)

(2; y)

(-1 ; y)

m = +

m = -

m = +

m = -


At this x-value, the gradient of the tangent = 0 x-value of the parabola on the left are the same x-values.

Function: f ( x ) Derived function: f ' ( x )

1. Given f ( x )= (x+1 )2 ( x−2 )=x3−3 x−2 and g ( x )=3 x2+3 x

1.1 Determine the x−¿intercepts of f and g. (2)

1.2 Determine the co-ordinates of the turning points f and g. (4)

1.3 Sketch the graphs showing all the relative turning points and the intercepts with the axes. (6)

1.4 Determine the solution of 2+4 x+3 x2−x3 using your graphs. (2)

1.5 For which value(s) of x will f ( x ) . g ( x )<¿0? (3)

1.6 Determine the value(s) of k for which f(x) = k has three unequal roots. (2)

Page 57

Examples

y = ax + q or y = mx + c y ' = a or my '

y = a y ' = 0

The graph of y’ = 0 is a straight line that coincides with the x-axis

m = +

m = -

m = +

m = -


1.1 f ( x )= (x+1 )2 ( x−2 )=x3−3 x−2

x=−1∨¿ofx=¿2

x−¿intercepts (0 ;1 )

& (0 ;2 )

,

g(x )=3 x2+3 x

0=x ( x+1 )

x=0, x=−1

x−¿Intercepts of f

Factors for g

x−¿Intercepts of

(3)

1.2 f ' ( x )=0

3 x2−3=0

3( x−1 )( x+1 )=0

x=−1∨x=¿1y=0∨ y=−4

TP (−1 ;0 )

(1 ;−4 )

g' (x )=6 x+3=0

x=−12

TP:(−1

2;−3

4 )

3 x2−3=0

(1 ;−4 )

x=−1

2

y=−3

4

(4)

1.3 f :

Shape

Intercepts

Both turning points g

(3)

Shape

Intercepts

TP(−12

;−34 ) (3)

1.4 2+4 x+3 x2−x3

Page 58


=3 x2+3 x − (x3−x−2 )=g (x )− f ( x )=0−(−2 )=2

Finding the differenceAnswer (2)

1.5 0<x<2∨¿of x←1 0<x<2x←1 (2)

1.6 −4<k<0 −4<k<0 (2)

2.1 Determine the points on the curve y= 4

x where the gradient of the tangent to the curve is – 1 . (5)

2.2 The graph of a cubic function with equation f ( x )=x3+ax2+bx+c is drawn.

f (1)=f (4 )=0

f has a a local maximum at B and a local minimum at x = 4.

2.2.1 Show that a=−9 , b=24 and c=−16 . (2)2.2.2 Calculate the coordinates of B. (4)2.2.3 Determine the value(s) of k for which f(x) = k has negative roots only. (2)2.2.4 Determine the value(s) of x for which f is concave up. (2)

2.1 y= 4x and the gradient of the tangent to the curve is -1

∴ y=4 x−1 exponential form

derivative

Page 59

y


dydx

=−4 x−2=−4x2

∴−4x2

=−1

x2=4

∴ x=2 or x=−2 y=2 or y=−2∴ (−2 ; −2 ) and (2 ; 2 )

derivative = -1

x-valuesy-values

(5)

2.2.1

y= (x−1 ) ( x−4 )2

y= (x−1 ) (x2−8 x+16 )y=x3−9 x2+24 x−16

y= (x−1 ) ( x−4 )2

squaring binomial (2)

2.2.2

y=x3−9 x2+24 x−16dydx

=3 x2−18 x+24=0

x2−6 x+8=0( x−2 ) ( x−4 )=0x=2 or x=4y=4 or y=0

B (2 ; 4 )

3 x2−18 x+24=0

factors

y-values

B (2 ; 4 ) (4)2.2.3

k<−16 inequality-16 (2)

2.2.4

f ''( x )=6 x−18>0x>3

6 x−18>0x>3 (2)

3. The graph of y = g/(x) is sketched below, with x-intercepts at A(− 2; 0) andB( 3; 0). The y-intercept of the sketched graph is (0; 12).

y

A B x –2 O 3

Page 60

12


3.1 Determine the gradient of at x = 0. (1)

3.2 For which value of x will the gradient of be the same as the gradient in QUESTION 7.2.1? (1)

3.3 Draw a sketch graph of . Show the x-values of the stationary points and the point of inflection on your sketch. It is not necessary to indicate the intercepts with the axes. (3)

3.1 answer (1)3.2

x = 1 (1)

3.3 y g

x -2 ½ 3

turning point at x = − 2 and at x = 3 point of inflection at x = ½ shape

(3)

1. The function defined by f ( x )=x3+ p x2+qx+30 is represented by the sketch below. A (−1 ;36) and B are the turning points of f , while g is a tangent to f at Awhich cuts f at point C.

Page 61

f

g

0 x

y

A(–1;36)

B

C

30


1.1 Show that p=−4 and q=−11. (7)1.2 Determine the coordinates of C. (3)1.3 Write down the coordinates of a turning point of k , if

k ( x )= f ( x )−10. (2)

2. The turning points of the graph of a cubic polynomial h ( x ) are (2 ;−3 )and (5 ;4 ). Draw a sketch graph of the derivative function h '( x), clearly showing the x-intercepts. (3)

3. Given: h( x )=−x3+ax2+bx and

g( x )=−12 x . P and Q(2 ; 10) are the turning points of h. The graph of h passes through the origin.

3.1 Show that a=3

2 and b=6 .

(5)

3.2 Calculate the average gradient of h between P and Q, if it is given that x = –1 at P. (4)

3.3 Show that the concavity of h changes at x=1

2 .(3)

3.4 Explain the significance of the change in QUESTION 9.3 with respect to h. (1)

3.5 Determine the value of x, given x < 0, at which the tangent to h is parallel to g. (4)

4. The diagram below shows the graph of f ' ( x ), the derivative of f ( x )=a x3+b x2+cx+d . The graph of y=f ' ( x )intersects the x- axis at 1 and 5. A (4 ;−9 ) is a point on the graph of f ' .

Page 62

y

0 1 5x

y=f ' (x )


4.1 Write down the gradient of the tangent to f at x=4. (1)4.2 Draw a rough sketch of the graph of f . (3)4.3 Write down the x-coordinate of the point of inflection of f . (2)4.4 For which value(s) of x is f strictly decreasing? (2)

2. TO DETERMINE RATE OF CHANGE

Remember that rate of change refers to the speed at which something is changing.

Make sure your learners memorise the following:

If an equation s ( t ) is given with distance (s) and time (t), then

s ( t ) represents distance at time t s' (t ) represents speed (velocity) at time t s' ' (t ) represents acceleration at time t

Learners struggle with this content. Use an easy example like the following, to explain to them the different scenario’s relevant in this situation:

Given a height (or distance) equation: s (t )=50 t=5t 2 with distance (s) in meters and time (t) in seconds. (Like a stone thrown up into the air)

s(2) is the distance after 2 seconds s(5) is the distance after 5 seconds s' (2 ) is the speed after 2 seconds s' (5 ) is the speed after 5 seconds s(0) is the distance EVEN BEFORE THE MOTION HAS STARTED Maximum height will be at s' (t )=0 (the turning point of s) Maximum height after how many seconds?

Page 63

A (4 ;−9 )


Solve for: s' (t )=0 ∴50−10 t=0

∴t=5 seconds What is the maximum height? (Substitute t = 5 into s)

s (5 )=50 (5 )−5 (5 )2

∴ s (5 )=125 meter When (at what time) does it hit the ground again? (When speed = 0 again,

thus s (t )=0) 50 t−5 t2=0⇒ t=0 s∨t=10 s

Use the bigger value because it will be the end of the ‘journey’ of the stone.

Speed with which it hits the ground? (Calculate s' (10 ))s' (10 )=50−10 (10 ) ¿−50 m /s

The negative indicates that the stone hits the ground when in DOWNWARD motion.

SPECIAL CASES

Suppose TWO moving bodies are involved, one going up and on coming down, and the ask:

When (at what time) will they pass each other?This will happen when the two heights are equal.Therefore, equate the two height equations of the bodies and solve for t.

When (at what time) will their speeds be equal?Now you equate the two speed equations (s’ = s’ ) and solve for t.

1. A stone falls from a cliff 320 metres high. The distance covered by the stone while falling is given by the function s ( t )=5 t2, where s is the distance covered in metres and t is the time in seconds.

1.1 Determine the speed of the falling stone after 3 seconds. (3)1.2 How long will it take before the stone hits the bottom of the cliff? (3)1.3 What will the speed of the stone be when it hits the bottom of the cliff? (2)

1.1 s ( t )=5 t2

Speed: s' (t )=10 tAfter 3 seconds: s' (3 )=10 (3 )=30 m / s

s' (t )=10 ts' (3 )=10 (3 )30 m /s (3)

1.2 Distance covered by stone = 320 m5 t2=320 5 t2=320

Page 64

Examples


t 2=64t=± 8

It takes 8 seconds to hit the bottom of the cliff

t=± 8Choose t = 8 s (3)

1.3 Speed after 8 seconds: s' (8 )=10(8) ¿80m /s

s' (8 )=10(8)¿80m /s

2. Sales of a new product grow rapidly and then level off with time. This situation is represented by the equation S ( t )=−t 3−4 t2+80 t where t represents time in months and S(t ) represents sales.

2.1 Determine the rate of change of sales during the third month. (3)2.2 Determine after how many months a maximum sale is obtained. (5)

2.1 S (t )=−t 3−4 t2+80 tRate of change: S' (t )=−3 t2−8t+80After 3 months: S' (3 )=−3 (3 )2−8 (3 )+80 ¿29 sales per month

S' (t )=−3 t2−8t+80Subst. t = 329 (3)

2.2 Max sales if: s' (t )=0−3 t 2−8 t+80=0

3 t 2+8 t−80=0(3 t+20 ) (t−4 )=0

∴t=−203

∨t=4

After 4 months a maximum sale will be obtained

s' (t )=0

Standard formfactorst-values

Choose t = 4 (5)

1. The mass of a baby in the first 30 days of life is given by M ( t )=t 3−9 t2+3 000

; 0≤t≤30

. t is the time in days and M is the mass of the baby in grams.

1.1 Write down the mass of the baby at birth. (1)1.2 A baby's mass usually decreases in the first few days after birth. On which day will the baby's mass return to its birth mass? (4)

Page 65


1.3 On which day will this baby have a minimum mass? (4)1.4 On which day will the baby's mass be decreasing the fastest? (2)

2. A Petrol tank at BP Depot has both the inlet and the outlet pipes which are usedto control the amount of petrol it contains. The depth of the tank is given by

D (t )=6+ t2

4− t3

8 where D is in metres and t is in hours that are measured

from 9h00.

2.1 Determine the rate at which the depth is changing at 12h00, and then tell whether there is and increase or decrease in depth. Answer correct to two decimal digits.

(3)

2.2 At what time will the inflow of petrol be the same as the outflow?(4)

3. A stone is thrown vertically upwards. Its height (in metres) above the ground at time t

(in seconds) is given by: .

3.1 Determine the initial height of the stone above the ground. (1) 3.2 Determine the time taken to reach the maximum height. (3) 3.3 How fast was the stone travelling when it hit the ground? (5) 3.4 Determine the acceleration of the stone. (1)

3. Maximum and minimum values

We use calculus to find the x-value that will minimise or maximise a quantity like area, volume, etc.

It is important that the learners will remember that the gradient of a function is zero at the minimum and maximum values.

Follow these steps:

Page 66


If necessary calculate the missing dimensions (e.g. height in terms of the radius)

Determine a formula in terms of one variable (e.g. x) for what must be minimised or maximised, like

Area………...A ( x ) or Cost………...C ( x ) or Volume…….V ( x )

If your formula contains a value like π, then keep it as π and only in the final answer convert it to a value/number

Now determine the first derivative of your formula (e.g. A' ( x )) Solve for x in A' ( x )=0 Test the x-values obtained. They will maximise or minimise the area, volume,

cost and so forth. REJECT the x-value that is not valid (e.g. a negative area etc.)

Remember that these questions could be asked about distances and shapes on the Cartesian plane. Then height is given by the y-values and the horizontal distances are given by the x-values

Make sure your learners KNOW the formulas for the surface areas and volumes of right prisms. For other shapes (e.g. cones and pyramids), formulas will be given and learners should be able to select the applicable formula.

1. PQRS is a rectangle with P on the curve h(x )=x2and with the x-axis and the line x=6 as boundaries.

Page 67

Examples

h

6 RQ

SP

0x

y


1.1 Show that the area of rectangle PQRS can be expressed as: A¿6 x2−x3 . (3)

1.2 Determine the largest possible area for rectangle PQRS. Show all your calculations. (4)

2. A wooden block is made as shown in the diagram below. The ends are right-angled triangles having sides 3 x , 4 x∧5 x. the length of the block is y. The total surface area of the block is 3 600 cm2.

2.1 Show that: y=300−x2

x(3)

2.2 Determine the value of x for which the block will have a maximum volume. (6)

1.1

Page 68

y

x0

P S

Q R6

h

3 x4 x

y

5 x


P(x ;x2) and Q (x ;0)QR ¿6−xPQ ¿ x2

Area of PQRS = PQ × QR = x2×(6−x) = 6 x2−x3

QR ¿6−x PQ ¿ x2

Method (3)

1.2 For maximum area: dAdx ¿0

12 x−3 x2=0 3 x ( 4−x )=0 x=0 or x=4 N/A

Maximum area ¿6 (4 )2−( 4 )3

¿32 units2

12 x−3 x2

dAdx

¿0

x=4

Answer (4)

2.1 SA=2. 12

.3 x .4 x+3 x . y+4 x . y+5 x . y

3600=12 x2+12xy 300=x2+xy 300−x2=xy

300−x2

x= y

formula

3 600=12 x2+12xy300−x2=xy

(3)

2.2 Volume=12

×b × h× H

V (x )=12

(3 x ) (4 x )( 300−x2

x ) V ( x )=1 800 x−6x3

For max: V ' (x )=0

formula

Correct substitution

V ( x )=1 800 x−6 x3

1 800−18 x2=0

Page 69

x


1800−18 x2=01800=18 x2

100= x2

∴ x=± 10 but x>0

For a max volume, x = 10 cm

x-values

Choose x = 10 (6)

1. A rectangular box has a length of 5x units, breadth of (9−2 x )units and its height of x units.

1.1 Show that the volume (V) of the box is given by V=45 x2−10 x3(2)

1.2 Determine the value of x for which the box will have maximum volume. (5)

2. A 340 ml can of cool drink with height h and radius r is shown below.

Page 70

(9 – 2x) 5x

r

h


2.1 Determine the height of the can in terms of the radius r. (3)

2.2 Show that the surface area of the can be written as SA=2π r2+680r . (2)

2.3 Determine the radius of the can in cm, if the surface area of the can has to be as small as possible. (4)

1. ABCD is a square with sides 20 mm each. PQRS is a rectangle that fits inside the square such that QB=BR=DS=DP=k mm.

3.1 Prove that the area of PQRS=−2k (k−20 )=40 k−2 k2 (4)3.2 Determine the value of k for which the area of PQRS is a maximum. (4)

Page 71

B

CD

P

Q

R

S

A


1. GRAPHS

QUESTION 91.1 f ( x )=x3+ p x2+qx+30

At turning point f ' ( x )=0: 3 x2+2 px+q=0Substitute x=−1 : 3 (−1 )2+2 p (−1 )+q=0 3−2 p+q=0 −2 p+q=−3 - - - (i)

Substitute (−1 ;36) in f ( x ) : (−1)3+ p (−1)2+q (−1 )+30=36 −1+ p−q+30=36 p−q=7 - - - (ii)

(i) + (ii): −p=4 p=−4Substitute p=−4 into (ii): −4−q=7 q=−11

f ' (x)

Substitute x=−1 Equation (i)

Substitute (−1 ;36)

Equation (ii)

p=−4

q=−11 (7)

1.2 At C: f ( x )=g ( x )=36

Page 72

SOLUTIONS


x3−4 x2−11 x+30=36 x3−4 x2−11 x−6=0 But x=−1 at A ∴ ( x+1 ) (x2+kx−6 )=0 k+1=−4 OR touching point at A k=−5 ∴ ( x+1 ) ( x+1 ) ( x−6 )=0 ∴ ( x+1 ) (x2−5 x−6 )=0 ∴ ( x+1 ) ( x+1 ) ( x−6 )=0 x=6 C (6 ;36)

f (x)=36

x+1 factor

x=6 (3)

1.3 Turning point of k: (−1 ;26) x=−1 y=26

(2)

2.

Shape x=2 x=5

(3)

3.1 Substitute Q(2; 10) intoh( x )=−x3+ax2+bx−23+a(22 )+b(2 )=10−8+4 a+2b=102 a+b=9 .. . .. .. . .. .. . .. .. . line 1

substitute Q intoh

findingderivativeh' (2)equating

Page 73

(5; 4)

(2; -3)

x

2 5 x


h' ( x )=−3x2+2 ax+bAt Q: h' (2)=0

−3(2 )2+2 a(2)+b=0−12+4 a+b=0

4 a+b=12 . .. . .. .. . .. .. line 2line 2 − line 1: 2 a=3

a=32

Substitute in line 1: b=6

derivative to 0

solving simultaneously for a and b

(5)

3.2 f (−1 )=−(−1 )3+32

(−1 )2+6 (−1 )

¿−3,5

Average gradient=

f ( xQ )− f ( x P )xQ−xP

Average gradient =10−(−3,5 )2−(−1)

=4,5

f (−1 )= –3,5

formula

substitution

answer (4)

3.3 h' ( x )=−3 x2+3 x+6h' ' ( x )=−6 x+3=−3(2 x−1)

For x<1

2 , h is concave up and for x>1

2 , h is concave down

h' ( x )=−3 x2+3 x+6

h' ' ( x )=−6 x+3

explanation

(3)

3.4The graph of h has a point of inflection at

x=12 answer

(1)

Page 74

x


OR

The graph of h changes from concave up to concave down at

x=12

OR

The graph of h changes concavity at x=1

2

answer

(1)

answer

(1)

3.5 Gradient of g is – 12Gradient of tangent is:

h' ( x )=−3 x2+3 x+6

{h' ( x )=−12 ¿−3 x2+3 x+6=−12 ¿3 x2−3 x+18=0 ¿x2−x+6=0 ¿(x−3)( x+2)=0 ¿ x=−2only ¿¿h' ( x )=−3 x2+3 x+6

h' ( x )=−12

factors

selection of x-value (4)

4.1 Gradient ¿−9 answer (1)

4.2

correct shape

TP at x=1

TP at x=5

(3)4.3 x=3 x=3

(2)4.4 x∈ (1;5 ) ¿

5¿

Page 75

x=1 x=5


(2)

2. RATE OF CHANGE

Given: M ( t )=t3−9 t2+3000 ; 0≤t≤301.1 M (0 )=03−9 (0 )2+3000

¿3000 ganswer

(1)1.2 t3−9 t2+3000=3000

t3−9 t2=0t2 ( t−9 )=0t=0 or t=9Baby’s mass will return to the birth mass on the 9th day

M ( t )=3000

t3−9 t=0factors

t=9(4)

1.3 M ' ( t )=03 t2−18 t=03 t ( t−6 )=0t=0 or t=6Baby’s mass will be a minimum on the 6th day

M ' (t )=0

3 t2−18 tfactors

t=6(4)

1.4 M ' ( t )=3 t2−18 tM ' ' (t )=6 t−180=6 t−18t=3OR / OF

Using symmetry:

t=0+62

=3

0=6 t−18answer

(2)

0+62

answer(2)

2.1D ( t )=6+ t2

4− t3

8

D' (t)= t2−3 t2

8

D' (3)= 32−

3 (3 )2

8=−15

8m. h−1

¿−1,875 decreasing in depth/verlaging in diepte

Derivative

Substitution

Answer

Page 76


2.2 D' (t)=0

t2−3 t 2

8=0

t ( 12−3 t

8 )=0

t=0 or t=43

At 9h00 and at 43

×60=80minutes later. i.e. at 10h20

Derivative

factors

answers

Answer (4)

3.1

answer (1)

3.2

OR

answer (3)

correct formula

substitution into the correct formula answer (3)

Page 77


3.3 h(t) = 0

factors

t = 8

–20m/s or 20 m/s downwards (5)

3.4OR

answer (1)

answer (1)

3. MINIMA & MAXIMA

1.1 V=l × b× h ¿5 x (9−2 x ) ( x ) ¿45 x2−10 x3

formulasubstitution (2)

1.2 V ' (x )=90 x−30 x2

90 x−30 x2=0 30 x (3−x )=0x=0∨x=3Therefore the box will have a maximum at x=3

derivativederivative = 0factorsx-valueschoosing x=3 (5)

2.1 V=π r2 h = 340cm3 ✓✓✓

V=π r2 h

V=340 cm3

answer

(3)

Page 78


h=340π r2

2.2 total surface area = 2π r2+2 πrh

= 2 π r2+2 πr (340π r2 )

SA=2π r2+680r

✓✓

2 π r2+2 πrh

subst. of h in the

total SA formula

(2)

2.3 SA=2π r2+680r

dSAdr

=4 πr−680r2

For the surface area to be as small as possible

dSAdr

=0

4 πr−680r2 =0

4π r3−680=0

r3=1704 π

=13 , 52817 …

r=2 ,38cm

✓

✓

✓

✓

derivative.

dSAdr

=0

r3=1704 π

=13 , 52817 …

r=2 ,38 cm

(4)

3.1 QR=√2k Pythagoras QR=√2k

Page 79

B

CD

P

Q

R

S

Ak

k


SR=√2(20−k ) PythagorasArea PQRS=√2k [√2(20−k )] = 2k (20−k )=40k−2 k2

SR=√2(20−k ) Area PQRS=√2k [√2(20−k )](4)

3.2 A=−2 k2+40 kdAdk

=−4 k+40=0

k=10

−4k+40¿0 10 (4)

TOPIC: SEQUENCES AND SERIES

INTRODUCTION

You have already had some experience of working with number sequences and number patterns. In grade 11 you have dealt with quadratic or second difference sequences. In grade 12 there are two specific types of sequences which both have quite unique properties. These are Arithmetic and Geometric sequences.

Page 80

201; 2001; 20001; …

Work out this, what sequence is this? Find the formula

What do I do? It’s not arithmetic or geometric Mr K didn’t give me the formula


Arithmetic sequenceArithmetic sequence is a sequence of numbers in which the difference between two consecutive numbers is constant.The numbers in the sequence are called terms i.e. T 1;T 2;T 3;T 4 …T n

and Tn the general term.T 2−T 1=T3−T2=T4−T 3 = Tn –Tn-1 = d.d is the constant difference .

To determine whether the sequence is arithmetic or not we need to find the difference between two consecutive terms. Given -1; 1; 3; 6; …find the difference

1− (−1 )=3−1 2=2But T 4−T 3=3 ≠ T 3−T 2

The difference between two consecutive numbers is not always the same in this sequence. “So the sequence is not arithmetic”In arithmetic sequences the difference between every two consecutive terms should be the same. And when you add the constant difference to the previous term you get the

Page 81

Check:

Is this Arithmetic?

-1; 1; 3; 6;…


next term.

Consider the sequence -3; -1; 1; 3; …to find T135.

Yes, but it will take long to keep on adding the constant difference in order to get to T135,

hence we need the formula that can help us to get the answer.

T 1=a

T 2=a+d

T 3=a+2d

T 25=a+24 d

T n=a+ (n−1 ) d

Page 82

Can the procedure

given above be used every

time?


Geometric sequenceGeometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a non-zero number called the common ratio(r).Note:T2

T1=

T3

T2=r

T 1=aT 2=a . rT 3=a . r2

T 4=a .r 3

T n=a . rn−1

Page 83

T n=a+ (n−1 ) d

is the general term

for Arithmetic Sequence

T n=a . rn−1

Is the general term for Geometric sequence

General term??


Activity 1

1.

1

Find the numerical values of the first three terms in an arithmetic sequence given

by x ; 4 x+5∧10 x−5.

(3)

1.

2

Given the sequence 4; -2; 1; …Find:

1.2.1 the next two terms (2)

1.2.2 the nth term (3)

1.

3

The 10th term of an arithmetic sequence is 17 and the 16th term is 44. Find the first

three terms in the sequence.

(4)

1.

4The first three terms of arithmetic sequence are 2 x ;3 x+4 ; 4 x+8 ; .. .Calculate the

60th term in terms ofx

(3)

[15]

Activity 2

2.1 Given 6; 3; 2; 1; 23 ; −1 ;… if the sequence behaves the same find:

2.1.1 The next two terms (4)

Page 84

The general term (nth term) is a formula that you can use to determine any term within the sequence up to

infinity!!!

General term again??


2.1.2 Explain your answer in 2.1.1 (4)

2.1.3 What will be the general term of the sequence if it exists? (4)

2.2 The first term of an arithmetic sequence is 2. The 1st; 3rd and 11th terms of an

arithmetic sequence are the first three terms of a Geometric sequence. Find the

7th term of the Geometric Sequence.

(6)

[18]

Arithmetic series

When the terms of a sequence are added together, the series is formed.

A story of a historical event or of a contrived situation can motivate pupils. To introduce the sum of an arithmetic series, tell the story about young Carl Freidrich Gauss (10 years) in a class that was asked by its instructor to add the numbers from 1 to 100. Much to the astonishment of the instructor, young Gauss produced the correct answers immediately. When asked how he arrived at the answer he quickly explained that:

1 + 100 = 101

2 + 99 = 101

3 + 98 = 101

Since there are 50 such pairs, the answer is 50 × 101 = 5050. This scheme can be used to develop the sum of an arithmetic series.

Proof for the sum of terms of arithmetic series( Examinable)

S1=T 1

S2=T 1+T2

S3=T 1+T 2+T 3

Sn=T1+T 2+T 3+…+T n

Sn=a+(a+d )+ (a+2 d )+…+[+( n−2 ) d ]+¿Sn= [a+(n−1 ) d ]+[ a+(n−2 )d ]+…+ (a+2d )+(a+d )+a ………(2)Add equation (1) and (2)2 Sn=[2a+(n−1 ) d ]+ [2 a+(n−1 )d ]+…+ [2a+ (n−1 ) d ]+[2 a+(n−1 ) d ]2Sn=n ¿]

Page 85


⁖Sn=n2[2a+(n−1 ) d]

Alternatively:Sn=n2[a+l ] ; l is the last term=a+(n−1 )d

Geometric Series (Examinable)

S1=T 1

S2=T 1+T2

S3=T 1+T 2+T 3

Sn=a+ar+a r2+…+a rn−2+ar n−1…………………..(1)(× r):r Sn=ar+ar2+a r3+…+a rn−2+a rn−1+a rn………..(2)(1) – (2)

Sn−rSn=a−a rn

Sn (1−r )=a(1−rn)

Sn=a (1−rn )

1−r; r<1 or Sn=

a (rn−1)r−1

;r>1

Activity 3

3.1 The sum of n terms of a sequence is given as Sn=n2

(5n+9 )

3.1.1 Calculate the sum to 50 terms. (2)

3.1.2 Calculate the 50th term of the sequence. (3)

3.2 The sum of the first 20 terms of the following arithmetic series is

3 360.

2x+2 .2x+3 . 2x+. . .

Calculate the value of x

(4)

Given the arithmetic series: 2+9+16+…to 251 terms

Page 86


3.3

3.3.1 Write down the fourth term of the series (1)

3.3.2 Calculate the 251st term (3)

3.3.3 Calculate the sum of the series (2)

3.3.4 How many terms in the series are divisible by 4? (4)

3.4 Calculate the sum of the first 6 terms of a geometric series if the

second term is 8 and the sixth term is 648.

(4)

3.5 Find the smallest natural number n for which the first n terms of the

geometric sequence1; 1, 1; 1, 21; 1, 331; … will have a sum greater than

70.

(5)

[28]

Sigma notationThe mathematical symbol Σ is the capital letter S in Greek alphabet. It is used as the symbol for the sum of a series.

∑m=1

n

T m=T 1+T 2+T 3+. . .+Tn

Note that the last value you substitute is not always giving the number of terms that will be added.

The number of terms = Top – Bottom +1

∑t=3

20

(3 t+2 ) What is the number of terms in this series?

Ans: It will be 20-3+1=18, so we will calculate the sum of 18 terms in the series.

Activity 4

4.1

Expand and then calculate∑

3

10

( r (r+2 )2 ) (5)

4.2 Determine the value of n: (5)

Page 87


∑k=1

n

22 k−1=682

4.3 Write the following series in the sigma notation:

2+6+18+…+162

(3)

4.4 In the arithmetic series a+13+b+27+. . .

4.4.1 Prove that a=6∧b=20 (2)

4.4.2 Determine the sum of the first 20 terms of the series above. (3)

4.4.3 Write down the series in 4.4.2 in the sigma notation (3)

[21]

Infinite geometric series

A Geometric series can also have smaller and smaller values.

What happens whenn→∞ ? rn→0 .Therefore the series will converge to a number referred

to as the sum to infinity. This is the case only if −1<r<1

Example:

9; 3; 1…

r=13 ;

T n=9( 13 )

n−1

; −1< 1

3<1

∴ the series converges and the sum to infinity exists

Formula for sum to infinity

Page 88


Sn=a (1−rn )1−r

Sn=a−ar n

1−r

Sn=a1−r

−arn

1−rIf −1<r<1 , n→∞∴r n→0∴arn→0

∴arn

1−r→0

∴Sn=a1−r

;n→∞

S∞=a1−r

;−1<r<1

Activity 55.1

Given: ∑n=1

∞(x2−3 x+1 )n

Determine for which values of x

the series will be convergent.

(4)

5.2

The mid- points of the sides of Δ ABC are joined to form a smaller triangle. The

process is repeated on the smaller triangle, and then continued indefinitely. The

perimeters of all the triangles including Δ ABC are added to obtain 44 units.

Calculate the perimeter of Δ ABC (5)[9]

Page 89

B C


Recurring decimal

It’s a number that keeps repeating after a comma. A recurring decimal is represented by

placing a dot above the number or numbers that repeat.

All recurring decimals can be written as common fractions.

How do we convert recurring decimal to common fractions?

e.g. Convert the recurring decimal 0 ,4¿

5¿

in to a common fraction.

Ans: 0 ,4¿

5¿

=0 , 45454545 . .. =0 , 45+0 , 0045+0 , 000045+. ..a=0 , 45r=0 ,01−1<0 , 01<1

the series converge

S∞=0 ,45

1−0 ,01

=

511

∴0 , 4¿

5¿= 5

11

Activity 6

6.1 Use the formula for S∞of a geometric series to write 0,2 5¿

as common fraction (3)

Page 90


The application of all formulae is examinable

Description Arithmetic GeometricFirst term a aNumber of terms n nCommon difference d=T2−T 1=T 3−T 2 -Common ratio

- r=T 2

T 1=

T 3

T 2

General term T n=a+ (n−1 ) d T n=a . rn−1

Sum of n terms Sn=n2[a+l ]

Sn=n2[2a+(n−1 ) d]

Sn=a (1−rn )

1−r; r<1

Sn=a (rn−1)

r−1; r>1

Sum to infinity - S∞=a

1−rif−1<r<1

The sum of a number of terms ∑

r=1

3

10+r=11+12+13 ∑r=1

3

( 110 )

r

= 110

+( 110 )

2

+( 110 )

3

Page 91


SOLUTIONS

Activity 11.1 4 x+5−x=10 x−5−(4 x+5 )

4 x+5−x=10 x−5−4 x−53 x+5=6 x−10−3 x=−15x=5∴5 ;25 ;45 ;…

(3)

1.2 .1 −12

; 14

(2)

1.2.2 a=4 ;r=−12

T n=4−(12 )n−1

(3)

1.3 a+9 d=17 ………………….(1)a+15d=44 ………………...(2)(2) – (1)6d=27∴d=9

2Subst d in eq (1)

a+9( 92 )=17

∴a=−472

∴−472

;−19 ;−292

;…

(4)

1.4 a=2 x ;d=3x+4−2 xd= x+4T 60=a+59 d

=2 x+59( x+4 ) =2 x+59 x+236 =61 x+236

(3)

[15]

Page 92


Activity 2 2.1.1 2

9;−3 (4)

2.1.2 The sequence consists of arithmetic and geometric sequences, therefore you work out the terms from two separate sequences.

(4)

2.1.3

For geometric sequence: 6 ;2 ; 2

3; .. .

a=6 ;r=13

T n=6(13 )n−1

For arithmetic sequence: a=3 ;d=−2T n=3+(n−1)(−2)T n=3−2n+2∴T n=5−2n

(4)

2.2 a=2a+2d=ara+10 d=ar 2

2+2d=2 r2+10 d=2 r2

÷ each equation by 2

1+d=r . .. . .. .. . .. .. . .. .(1)1+5 d=r 2. . .. .. . .. .. . ..(2 )Subst (1) in to (2)1+5 d=(1+d )2

1+5 d=1+2 d+d2

d2−3 d=0d (d−3)=0d≠0∴d=3∴r=4T 7=2 ( 4 )6

=8192

(6)

Page 93


[18]

Activity 3

3.1.1S50=

502 [5(50 )+9 ]

=25 (259) =6475

(2)

3.1.2 T 50=S51−S50

S51=512 [5(51)+9 ]

=6732∴T 50=6732−6475 =257

(3)

3.2 a=2x

d=2x

S20=202 [2(2x )+(20−1)2x ]

3360=10 (2. 2x+19.2x )¿by 10336=2x (2+19)16=2x

24=2x

∴ x=4

(4)

3.3.1 (1)

3.3.2 (3)

3.3.3

Or

(3)

Page 94


3.3.4 The new series is 16+44+72+…+1752

Or

T3 is divisible by 4, then T7; T11; T15;…are divisible by 4 thus each 4th termis divisible by 4.Number of divisible terms by 4 will be

=

251−34

+1

= 63

(4)

3.4S6=

a(r n−1 )r−1

ar=8 .. . .. .. . .. .. .. . .. .(1)ar 5=648 .. .. . .. .. . ..(2 )(2)÷(1)r4=81r4=34

r=3

a=83

S6=

83 [ (36−1 ) ]

3−1

∴S6=29123

=970 ,67

(4)

Page 95


3.5 a=1 ;r=1,1Sn>70

Sn=1[ 1,1n−11,1−1

=1,1n−1110

¿10(1 .1n−1 )

Sn>70⇒

10(1,1n−1)>701,1n−1>7∴1,1n=8

∴n>1og1,1 8∴n>21,8

∴The smallest natural number is n=22

(5)

[29]

Activity 44.1 15+18+21+. ..

a=15 ;d=3S8=

82 [2(15 )+(8−1)(3 )]

=4 (51 )=204

(5)

4.2 2+8+32+. ..=682

Sn=2(4n−1)

4−1

682= 2(4n−1)3

32×682=2(4n−1)

3×3

2

1023=4n−1 1024=4n

45=4n

∴n=5

(5)

Page 96


4.3 a=2r=3T n=2. 3n−1

T n=162162=2 .3n−1

81=3n−1

35=3n−1

∴n=6

∑n=1

6

2 .3n−1

OR ∑n=0

5

2 .3n−1

(3)

4.4.1

27−b=b−13

b=13+272 = 20

27 – 20 = 13 – a

a=6

OR

27 – 13=2 d

d=7

b=13+7=20

∴a=13 –7¿6

4.4.2

a=6 ;d=7

S20=n2¿)

= 202¿

= 1450

(2)

4.4.3 ∑

n=1

20( 6+7 (n−1 ) )

¿¿

or

∑n=1

20

(7 n−1¿)¿

(3)

Page 97


[21]

Activity 5

5.1 ( x2−3 x+1)+( x2−3 x+1)2+( x2−3 x+1)3+.. .

r=x2−3 x+1

For convergent series −1<r<1

−1<x2−3 x+1<1

x2−3 x+1>−1∨x2−3 x+1<1

x2−3 x+2>0∨x2−3 x<0

( x−2 ) ( x−1 )>0∨x ( x−3 )<0

x>2∧x<1∨x<3∧x>0

1>x>2∨0< x<3

(4)

5.2Let perimeter of

Δ ABC be

x the next term will be

12

x

x ; 12

x ;…

S∞=44

r=12

S∞=a

1−r

S∞=x

1−12

(5)

Page 98


44= x12

2 x=44

x=22

∴

the perimeter is 22 units

6.1 0,25¿

=0,2+0 ,05+0 ,005+0 , 0005+.. . .a=0 , 05r=0,1

S∞=0 , 051−0,1

= 118

∴

0,2 5¿=0,2+ 1

18

=2390

(3)

Page 99


Page 100

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