Grade 11 workbook paper 1 by GoMath Workbooks

Compiled by Chesley Nell

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Forward: Welcome to “ GOMATH WORKBOOKS”. This workbook is designed to be text book and class work book in one. There are sufficient exercises to ensure that learners get the required practice. A detailed memorandum booklet is available for each workbook. The statement “ You get out what you put in.” is very apt where maths is concerned. To succeed in mathematics one must be prepared to invest the time and effort to achieve that success. The partnership that you as a learner and this GO MATH WORKBOOK develop will be profitable if you allow it to be. Chesley Nell: Mathematics Educator.  Chesley Nell 2011

Grade 11 Core Mathematics

GOMATH WORKBOOKS

GRADE 11 CORE MATHEMATICS CONTENTS: Paper One: Topic

Pages

Exponents

(4 - 22)

Algebraic Factors

(23 –26 )

Quadratic Equations

(27– 35)

& Nature of Roots

(36 – 41)

Simultaneous Equations

(42 – 49)

Inequalities

(49 – 53)

Algebraic Fractions

(53 - 60)

Number Patterns

(61– 80)

Financial Math

(81 – 96)

Functions & Graphs

(97 – 149)

10.

Probability Theory

(150 –163)

Grade 11 Core Mathematics

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Exponetial Laws and Examples:

N.B. The basic laws must always be applied. Law: a m .a n  a m n

When multiplying like bases you must add the exponents. Do not multiply the bases except in the following case:

2 3.33  8.27  216.

e.g.

BUT 2 3.33  6 3  216

am Law: n = a m  n a

When dividing the bases subtract the exponents: the following exception. 3

63  6      33  27 3 2 2

Law: (a m ) n  a mn

When raising a power to a power you must multiply the exponents. 4.

Law: (a m ) 0  1

Any value raised to the power of “zero” will equal 1.

N.B. 1. Never multiply the bases 2. Never divide the bases 3. Never multiply a base by an exponent. 4. If bases are separated by plus or minus signs you MUST FACTORISE before simplifying. Prime Base Factorizing It is important to remember to use prime bases in simplifying with exponents ( especially if no calculators are allowed. NB when reducing bases by factorizing, always use the lowest possible bases. e.g for 16 rewrite as 2 4 and not as 4 2 . Use prime numbers for bases and not composite ones. Once you have factorized using prime bases then the normal laws apply.

Grade 11 Core Mathematics

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Example 1:

 

4 3

Simplify 125  53

4 3

(Now use law 3 and raise a power to a power by multiplying the exponents:)

125  5 4 3

Example 2:



4 3 3

5

3 4  1 3

 5 4  625

Simplify without the use of a calculator.

5 a  3.5 2 a 1 25a 1

Answer:

Use law 1 in the numerator

5 a  3.5 2 a 1 5 3a  2  2a 2  5 a 4 (5 2 ) a 1 5

Prime base factorise and use law 3 to simplify

Use law 2 to get the answer.

Follow the procedures as set out for any of the following types of simplifications with exponents. NB. Prime base factorizing first then apply the laws. Exercise 1.1: 2 x3 2 x 1. 2 x 1 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

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2 x 1 2 x  3 2 x  2 .2 x

4 x 1 8 x  1 3. 32 x 1 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

5 x 25 x 1 5.125 x

Grade 11 Core Mathematics

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7 x  2 49 x  2 7 3 x2 ____________________________________________________________________ 5.

6 n 1 .12 n1 .2 n 18 n  2 .8 n 1

6 n 212 2 n 1 4 2 n  3 8 3 n1 9 n1 3 n ____________________________________________________________________ 7.

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Exercise 1.2: If bases are separated using plus or minus signs one must FACTORISE FIRST.. It is easier to separate the bases as shown because it is easier to see the HCF. Factorise and simplify. NB the base with the variable exponent should always cancel leaving pure numerical values. Example:

2 x3  2 x = 2 x 1

9 2 x .2 3  2 x 2 x (2 3  1) = = 1  18 x 1 x 1 2 .2 2 .2 2

2 x 1  2 x  3 1. 2 x2  2 x ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

3 x  3 x2 3 x 1  3 x

Grade 11 Core Mathematics

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5 2  x  4 .5  x 5  x  2 .5  x  1 ____________________________________________________________________ 3.

3 n .3 4  6.3 n .3 1 7.3 n .3 2 ____________________________________________________________________ 4.

2 n .2 5  3.2 n .2 2 5 5.2 n 2 3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

EXPONENTIAL EQUATIONS There are two types of exponential equations: 1. The unknown(variable) in the exponent: 2. The unknown(variable) in the base: 1.

Variable in the exponent: Method: 1.1 Equate the bases using prime base factorizing. 1.2 If the bases are now equal then the exponents are also equal, thus simply equate the exponents and simplify further. Example 1.

2 2 x  16

Rename 16 to the base 2 and equate the bases.

22x  24 2x  4 x2 Example 2: 2 .3 3

x ( x  3)

 54

First divide by the coefficient ‘2’

 27

3 x ( x  2)  33 x 2  2x  3  0

Rename 27 to the base 3 and equate the bases.

( x  3)( x  1)  0 x3

Equate the exponents and solve for x.

or x  1

Variable in the base: Method: 2.1 Raise the power of the exponent of the variable base to its multiplicative inverse. This will give a new exponent with the value of 1. In an equation you must do the same to both sides of the equal sign. Example . 2 3

x  16  x  

2 3

3 2

   24  

x  2 4  2 3

x  26 x  64

NB. You must first remove any coefficient values prior to solving for the variable:

Grade 11 Core Mathematics

GOMATH WORKBOOKS

More advanced equations: Variables separated by + or – signs: This involves factorizing. 2 x  3.2 x  16

2 x (1  3)  16

Use 2 x as the HCF

2x  4

1.1

2 x  22 x2

Equate the bases thus and then the exponents.

NB Clues to let you know when a simple HCF must be used are: 1. Terms separated by plus and minus signs and 2.

the variable exponents have the same value. i.e the coefficients of the variables are equal in value.

Variables separated by + or – signs, BUT the values of the coefficients of the exponents are not equal. i.e. one is double the other. This involves a trinomial and needs to be factorised accordingly. 22x  2 x  8  0 (2 x  4)(2 x  2)  0

2.1

2x  4  0 2x  4 2x  2  0

2 x  2 2 or 2 x  2 x2

x 1

The exponent of the 1st term is double that of the 2nd . i.e The expression is a trinomial. The variable base is 2x

2.3 2 x  12.3 x  54  0

2.2

(2.3 x  6)(3 x  9)  0

2 .3 x  6  0 3x  3

3x  9

x 1

3 x  32 x2

The exponent of the 1st term is double that of the 2nd . i.e The expression is a trinomial. The variable base is 3x

NB a substitution method can be utilized here: Let 3 x = k [ thus 3 2 x  k 2 ] 3 x 2k 2  12k  54  0 2( k 2  6k  27  0 (k  3)(k  9)  0 k  3 or k  9 3x  9 x 3 3 or 3 x  3 2 x 1 x2

Substitute 3 x for k at this point and solve for x

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Exercise 1.3: 1. Solve for x without the use of a calculator: 1.1

3 x  81

1.2

x3 = 27

1.3

4

= 80

Grade 11 Core Mathematics

1.4

GOMATH WORKBOOKS

2 x  16 x 1

1.5

5.4 2 x  40

1.6

1 1  x2 8 (2 ) x

Grade 11 Core Mathematics

1.7

GOMATH WORKBOOKS

3.4 2 x  48 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.8. 2 x1  2 x  12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.9 3 2 x 1  2.3 2 x  45 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOď&#x201A;ˇMATH WORKBOOKS

Equations involving surds: Example: 1. â&#x2C6;&#x161;đ?&#x2018;Ľ â&#x2C6;&#x2019; 2 = 3 đ?&#x2018;Ľâ&#x2C6;&#x2019;2=9 đ?&#x2018;Ľ = 11

Isolate the surd by itself and then square both sides & simplify. Test the answers to get correct solution â&#x2C6;&#x161;2 â&#x2C6;&#x2019; đ?&#x2018;Ľ = đ?&#x2018;Ľ + 4 2 â&#x2C6;&#x2019; đ?&#x2018;Ľ = đ?&#x2018;Ľ 2 + 8đ?&#x2018;Ľ + 16 đ?&#x2018;Ľ 2 + 9đ?&#x2018;Ľ + 14 = 0 (đ?&#x2018;Ľ + 2)(đ?&#x2018;Ľ + 7) = 0

Square both sides & simplify

đ?&#x2018;Ľ = â&#x2C6;&#x2019;2 đ?&#x2018;&#x153;đ?&#x2018;&#x; đ?&#x2018;Ľ â&#x2030; â&#x2C6;&#x2019;7 Test: đ?&#x2018;Ľ == â&#x2C6;&#x2019;2 â&#x2C6;ś đ??żđ??ťđ?&#x2018;&#x2020; = 2 đ?&#x2018;&#x2026;đ??ťđ?&#x2018;&#x2020; = â&#x2C6;&#x2019;2 đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; đ?&#x2018;Ľ == â&#x2C6;&#x2019;7 â&#x2C6;ś đ??żđ??ťđ?&#x2018;&#x2020; = 3 đ?&#x2018;&#x2026;đ??ťđ?&#x2018;&#x2020; = â&#x2C6;&#x2019;3 đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;Ą đ?&#x2018;&#x17D; đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A;

Exercise: 1. â&#x2C6;&#x161;đ?&#x2018;Ľ + 4 = 7 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2. â&#x2C6;&#x161;3đ?&#x2018;Ľ â&#x2C6;&#x2019; 8 = 15 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 3. 9 â&#x2C6;&#x2019; â&#x2C6;&#x161;3 â&#x2C6;&#x2019; đ?&#x2018;Ľ = 7 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

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4. â&#x2C6;&#x161;đ?&#x2018;Ľ + 6 = đ?&#x2018;Ľ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ 5. â&#x2C6;&#x161;2 â&#x2C6;&#x2019; đ?&#x2018;Ľ = đ?&#x2018;Ľ + 4 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ 6. â&#x2C6;&#x161;3đ?&#x2018;Ľ â&#x2C6;&#x2019; 2 = đ?&#x2018;Ľ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOď&#x201A;ˇMATH WORKBOOKS

SURDS 3

Irrational numbers such as â&#x2C6;&#x161;10 ; â&#x2C6;&#x161;5 ; â&#x2C6;&#x161;17 ; â&#x20AC;Śâ&#x20AC;Śetc, are called surds. Laws of Surds: NB: Surds must always be of the same order to be able t work with them. đ?&#x2018;&#x203A; đ?&#x2018;&#x203A; đ?&#x2018;&#x203A; 1. Multiplying surds of the same order: â&#x2C6;&#x161;đ?&#x2018;&#x17D; Ă&#x2014; â&#x2C6;&#x161;đ?&#x2018;? = â&#x2C6;&#x161;đ?&#x2018;&#x17D;đ?&#x2018;? If you multiply surds of the same order you simply write their product in one root form.. đ?&#x2018;&#x2019;. đ?&#x2018;&#x201D;. 1. â&#x2C6;&#x161;4 Ă&#x2014; â&#x2C6;&#x161;9 = â&#x2C6;&#x161;4 Ă&#x2014; 9 = â&#x2C6;&#x161;36 = 6 2. â&#x2C6;&#x161;2 Ă&#x2014; â&#x2C6;&#x161;7 = â&#x2C6;&#x161;2 Ă&#x2014; 7 = â&#x2C6;&#x161;14 đ?&#x2019;?

2. Dividing surds of the same order. 3

2.1

2.2

â&#x2C6;&#x161;27 3

â&#x2C6;&#x161;3

â&#x2C6;&#x161;64 4

â&#x2C6;&#x161;4

â&#x2C6;&#x161;đ?&#x2019;&#x201A; â&#x2C6;&#x161;đ?&#x2019;&#x192;

đ?&#x2019;?

đ?&#x2019;&#x201A;

= â&#x2C6;&#x161;đ?&#x2019;&#x192;.

= â&#x2C6;&#x161; 3 = â&#x2C6;&#x161;9 = â&#x2C6;&#x161; 4 = â&#x2C6;&#x161;16 = 2

3. Mixed Surds & Entire Surds. â&#x2C6;&#x161;12 is an entire surd . 2â&#x2C6;&#x161;3 is a mixed surd. To convert from entire to mixed either: 1. Write in factorized form and simplify. OR and

â&#x2C6;&#x161;80 = â&#x2C6;&#x161;24 Ă&#x2014; 5 = 22 â&#x2C6;&#x161;5 = 4â&#x2C6;&#x161;5 2. Find the highest possible perfect square that divides into the given surd

then simplify further. â&#x2C6;&#x161;80 = â&#x2C6;&#x161;16 Ă&#x2014; 5 = 4â&#x2C6;&#x161;5 Reverse the procedure converting from mixed to entire surds. 4. Addition and Subtraction of Surds: In algebra one can only add or subtract LIKE TERMS i.e. 3a + 5a = 8a Surds work in the same way: The surds must be of the same order and have the same value under the root sign. e.g. 1. â&#x2C6;&#x161;7 + â&#x2C6;&#x161;7 = 2â&#x2C6;&#x161;7 2. 5â&#x2C6;&#x161;2 â&#x2C6;&#x2019; 2â&#x2C6;&#x161;2 = 3â&#x2C6;&#x161;2 3 3 3 3. 4â&#x2C6;&#x161;2 + 6â&#x2C6;&#x161;2 = 10 â&#x2C6;&#x161;2 5. Rationalizing the denominators: 1

Irrational denominator : â&#x2C6;&#x161;3 To rationalize the denominator , means one must eleiminate any root signs in the denominator.

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1. If there is only one term in the denominator then one must multiply the numerator and denominator by the value in the denominator. 1

e.g. →. we are

√3

√3 3

[we are not changing the value of the fraction as

multiplying by 1] →

2√3 √5

2√3

√5

√5 √5

2√15

2. If the denominator is a binomial then the same procedure is used as above HOWEVER use the binomial in the denominator with the opposite sign between the two terms as depicted inn the original expression. e.g.

2+√3

2(2−√3)

2−√3

= (2+

2−√3

√3)(2−√3)

4−2√3 4−3

= 4 − 2√3

6. Expressing surds as surds of the same order: First convert the surds to exponential form and then express the indices as equivalent fractions ( the same value in the denominators) 1

e.g.

= 56 = √53

√5 = 52 1

= √125

√4 = 43 = 46 = √42 = √16 It can also be seen that √5 is greater than

√4

Grade 11 Surds Worksheet: 1. Simplify: 1.1 1.2 √18 1.3 √245 √20 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.4

√24

1.5

√13 5

1.6

√96

Grade 11 Core Mathematics

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2. Simplify. 2.1. √2 + 3√2

2.2.

√3 + 3√3 − 2√3

2.4

3√8 + 5√50 − 4√32

√245 + 6√5

2.6

√54 − √16

Grade 11 Core Mathematics

2.7

2√8 + 4√32 − 3√50

GOMATH WORKBOOKS

20 2.8.

√12−√75+2√3 √3

√3(√6 − √3)

2√5(3√5 − 2√2)

3√2(2√8 − √18)

Grade 11 Core Mathematics

GOMATH WORKBOOKS

3.6 (3√2 + √5)(3√2 − √5) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 2

3.7 (√3 + √2) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 2

3.8 (√6 + 2√3) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 4. Rationalize the denominators of the following: 3 4.1 √3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 5√2

4.2 √10 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 6√18

4.3 3√12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4√3

4.4 3√12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

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4.5 2−√3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 7

4.6 3+√2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2−√3

4.7 2+√3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4.8

√7−3√2 √7−√2

2√5+√3

4.9 5√3−3√5 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Factors of Algebraic Expressions

To factorise is the process of reducing many terms to one term. It is the opposite action of distribution. Procedure: Step 1 : First look for a HCF (if possible): 1.1 Look at the constants and choose the highest possible value, that can divide equally without a remainder, into the constant values in the expression. 1.2 Look at common variables and choose the lowest power (exponential value) of these as HCF. Step 2: Place the HCF outside a bracket and then divide the HCF into each term of the expression placing the answer inside the bracket. NB there must be the same number of terms inside the bracket as there is in the original expression. Step 3:Look at the expression inside the bracket and ascertain whether it can be factorised further. It could be one of the following; 3.1

The difference of two perfect squares.

3.2

A trinomial.

3.3

A quadrinomial

3.4

The sum & or difference of two cubes.

Examples: HCF plus other expressions. 1.

2 x  4  2( x  2)

2. 3.

2 x  4 x 2  6 x 3  8 x 4  2 x(1  2 x  3 x 2  4 x 3 ) 2ax  4abx  6abcx)  2ax(1  2b  3bc)

2ax 2  8ay 2  2a( x 2  4 y 2 )  2a( x  2 y )( x  2 y )

ab4  ac 4  a(b 4  c 4 )  a(b 2  c 2 )(b 2  c 2 )  a(b 2  c 2 )(b  c )(b  c ) NB: The sum of two squares cannot be factorised at all.

Difference of two squares: Write down the product ( two brackets) of the summand difference of the roots of each term. NB look at the bracket with the difference of the 2 terms and see if it can factorise further; NB: The sum of two squares cannot be factorised at all. Examples: 1.

a 2  b 2  (a  b)(a  b)

(b 4  c 4 )  (b 2  c 2 )(b 2  c 2 )  (b 2  c 2 )(b  c )(b  c )

NB DO NOT FORGET TO BRING DOWN THE SUM OF 2 SQUARES DOWN TO THE NEXT LINE

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Trinomials: The method used here is referred to as “trial & error” In products of binomials we use the FOIL method to expand . The middle term of the expression formed is a combination of the products of the inner terms and the outer terms. i.e. the ‘OI’ of FOIL. The reverse procedure is used to factorise trinomials. We find out the correct combination of factors of the First and Last terms of the trinomial. x 2  2 x  3  ( x  3)( x  1) e.g. Method: Draw a table and use the factors of the Ist and Last terms. Cross multiply them and either subtract or add to get the middle term. The sign of the last term informs one whether to add or subtract 1x 1x

x2  2x  3  ( x  3)( x  1)

x 2  8 x  12  ( x  2)( x  6)

1x 1x

1 3

6 12

NB Trinomial + + +

Brackets ( + )( + ) ( - )( - )

+ -

( + )( - ) ( + )( - )

Not necessary to work out

Must work out which bracket is negative and which is positive

Revision Exercise 2.1: Factorise the following completely: 1.1

2x2 – 32 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

1.2

GOMATH WORKBOOKS

p2( x + y) - q2( y + x ) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.3

a2  b2  a  b ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.4

(2 x  y ) 2  ( x  2 y ) 2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.5

4( x  y) 2  9( x  y) 2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

1.6

GOMATH WORKBOOKS

x 2  7 x  12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.7

2 x 2  24x  70 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.8

9 x 2  42x  45 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

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Quadratic Equations:

1. Solve by factorizing: Method: 1. Equate to zero, factorise and solve: x2  3x  4 2 e.g. . x  3 x  4  0 ( x  4)( x  1)  0

x  4 or x  -1 2. If already factorized simply solve the equation e.g.

( x  4)( x  1)  0 x  4 or x  -1

3. If not in factorized form do the necessary steps to get the equation into factorized form before solving. x ( x  4)  12

x 2  4 x  12  0 e.g.

( x  6)( x  2)  0 x  6 or x2

Exercise 3.1 : Solve the following equations:

1. ( x  5)( x  2)  0

____________________________________________________________________ ____________________________________________________________________

2. (a  6)(a  1)  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

3. x( x  1)  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

4. ( x  2)( x  3)( x  5)  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

5. x(2 x  5)(3x  2)  0 ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

6. y 2  3 y  10  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

7. x 2  5 x  6  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

8. x 2  7 x  6  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

9. x( x  1)  6 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

10. ( x  3)( x  2)  12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

11. x 2  2 x  3  12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

12. x( x  16)  3(24  5 x) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

13. (2 x  5)(3x  2)  2(3x  11) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Solving Quadratic Equations by the method of Completing the Square : This is an alternative method that can be utilized to solve equations. NB. Only use this method when instructed to Method: Add to the equation the square of half of the value of the coefficient of x to both sides of the equation. This forms a perfect square trinomial ( a binomial squared) on the left hand side and constant values on right hand side. Mathematically zero is being added to the equation thus there is no change at all. Example 1:

Add in the square of half the coefficient of ‘x’ to both sides of the equation Mathematically adding zero as (1)2 –(1)2 = 0

x2  2x  3  0 x2  2x  3 x 2  2 x  (1) 2  3  (1) 2 ( x  1) 2  4 x 1  4

Factorise the left hand side i.e. it forms a binomial squared.

x  1 4 x  1 2 x3 OR x  1 4

Solve for x

x  1 2 x  1

Example 2. 2x2  6x  4  0 x2  3x  2  0 2

 3  3 x  3x     2     2  2 2

3 17  x   2 4  3 17  2 2 3  17 x 2 x  3,56 x

or x  0,56

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Example 3. If f ( x )  x 2  2 x , show by completing the square that f ( x  1)  ( x  2) 2  1 f ( x)  x 2  2 x f ( x  1)  ( x  1) 2  2( x  1)  x2  2x  1  2x  2  x2  4x  3 x 2  4 x  ( 2) 2  3  2 

( x  2) 2  1 ( x  2) 2  1  0

 f ( x  1)  ( x  2) 2  1 Exercise 3.2 : Use the method of completing the square to solve the following: 1.

x 2  2 x  24  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

x 2  9 x  36  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

x 2  8 x  15  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

x 2  7 x  12  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

2x 2  7x  6  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

2 x 2  11x  6  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

2 x 2  3x  8  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

SOLVING QUADRATIC EQUATIONS USING A FORMULA: The formula is derived by completing the square with the general quadratic equation: ax 2  bx  c  0

 b  b 2  4ac General Quadratic Formula: x  : 2a NB: a ; b ; & c are constant values WHERE: a = coefficient of x2 ; b = coefficient of x ; c is the constant in equations written in the form: ax 2  bx  c  0 NO x –values must be substituted into the formula. i.e. only the constant values are used. Derivation of Quadratic Formula: ax 2  bx  c  0 b c x2  x   0 a a 2

b  b   b  c x  x       a  2a   2a   a  2

b  b2 c   x   2 2a  a 4a  2

b  b 2  4ac  x   2a  4a 2  b   b 2  4ac  x  2a  2a   b  b 2  4ac x 2a

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Examples: Solving quadratic equations using the general quadratic formula. x  x  12  0 2

 b  b  4ac 2a  1  1  4(1)(12 x 2  1  49 x 2 1 7 x 2 x  4 OR 3 x

2x2  7x  6  0  b  b 2  4ac x 2a x

7  49  4( 2)(6) 2( 2)

7 1 4 8 6 x  or 4 4 3 x  2or 2 x

Exercise 3.3:

 b  b 2  4ac Solve using x  ; 2a (Answers rounded to 2 decimal places where necessary) NB: First expand if necessary and equate to zero before using the formula. 1.

x2  4x  3  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

2 x 2  x  10 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

3x2  x  2  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

x2  6x  4  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

2x2  4  7x ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

2 x ( x  3)  3  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Nature of Roots: The discriminant ,   b 2  4ac , helps us to determine what type of roots an equation will have. NB We do not solve the equation and we are not concerned about finding the specific roots of a given equation. We want to ascertain what kind of roots an equation will have classified according to the criteria: Real or Non-Real; Equal or Unequal; Rational or Irrational.

 b  b 2  4ac From the general quadratic formula, x  , the value inside the square root 2a sign is important to us as it will enable one to classify the roots. This discriminant is referred to as   b 2  4ac ( delta =)  is a perfect square then the roots are rational  is not a perfect square then the roots are irrational

  0 then the roots are Real   0 then the roots are non-real.  = 0 then the roots are equal   0 then the roots are unequal. Examples: In the following examples find the value of the discrinimant and then make a statement classifying the nature of the roots. Discuss the nature of the roots of the following equation. NB do not solve the equation.

1. x 2  5x  3  0   b 2  4ac   ( 5) 2  4(1)( 3)   37 The roots are real, irrational and unequal.

2. 2 x 2  5x  3  0   b 2  4ac   (5) 2  4(2)(3)   25  24  1 Roots are real ; rational and unequal.

Grade 11 Core Mathematics

GOMATH WORKBOOKS

3. 4x 2  4x  1  0   b 2  4ac   (4) 2  4(4)(1)   16  16 0 The roots are real ; rational and equal. 4. x 2  5x  9  0   b 2  4ac   (5) 2  4(1)(9)   25  36   11 The roots are non-real.

Further examples involving nature of Roots. In the first 2 examples you find delta and then solve for the variable using the clue supplied. In the 3rd and 4th examples you find delta and simply make an observation in the form of statement to answer the question. NB you do not solve for any variable at all. For which values of r will 2 x 2  2 x  r , have equal roots. In this type of question find the discriminant first and then find the value of the variable in question. Example one: 2x 2  2x  r 2x 2  2x  r  0 a  2; b  2; c  r   b 2  4ac   (2) 2  4(2)(r )   4  8r

For equal roots   0 ( must equate the discriminant to zero.) 4  8r  0 8r  4 r

1 2

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Example two: For which values of h will the roots of the equation 3x 2  2hx  3 be non – real?

3 x 2  2hx  3  0 a  3; b  2h; c  3   b 2  4ac   (2h) 2  4(3)(3)   4h 2  36 For non-real roots   0 4h 2  36  0

h2  9  0

-3

(h  3)(h  3)  0

3  h  3 Example 3. Show that the roots of ax 2  (2a  1) x  (a  1)  0 are rational for all rational values of a. NB find delta first and then simply make a statement. ax 2  (2a  1) x  (a  1)  0

a  a; b  (2a  1); c  (a  1)   b  4ac   (2a  1) 2  4(a )(a  1)   4a 2  4a  1  4a 2  4a  1 Delta is a perfect square thus the roots are rational. Example 4. Prove that the roots of rx 2  x 2  4 x  r  1 are real for all real values of r. rx 2  x 2  4 x  r  1 (rx 2  x 2 )  4 x  ( r  1)  0 (r  1) x 2  4 x  (r  1)  0 a  (r  1); b  4; c  ( r  1)   b 2  4ac   (4) 2  4( r  1)(r  1)   16  4r 2  4   4r 2  12 4r 2  0 for all values of r. 4r 2  12.  12

The roots are real for all values of r.

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Exercise: 1. Determine the nature of the roots of the following without solving the equations. 1.1

x 2  4x  3  0

1.2

x2  7x  8  0

1.3

4 x 2  8 x  4

4 x 2  12 x  9

Grade 11 Core Mathematics

1.5

GOMATH WORKBOOKS

Find the value(s) of b if the roots of x 2  5 x  b  0 are equal. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 1.6 If -1 is a root of x 2  4 x  k  0 , calculate k and the other root. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

1.7

Prove that x 2  ax  a 2  1  0 has non-real roots.

Grade 11 Core Mathematics

1.8

GOMATH WORKBOOKS

Show that the roots of ax 2  bx  bx 2  a has rational roots for all rational values

of a and b ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

1.9

Determine the value(s) of h for which the graphs of y  3x 2  5h and y   x  h will not touch or intersect. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Simultaneous Equations:

One linear and one quadratic Method: Rewrite the linear equation into either the x – form or y – form and then substitute this value for x or y into the equation of the higher degree. Example: x  y  2 and x 2  y 2  52 ----- ② y  2  x ---- ① Substitute ① into ② x 2  ( 2  x ) 2  52

x 2  4  4 x  x 2  52 2 x 2  4 x  48  0 x 2  2 x  24  0 ( x  6)( x  4)  0 x = 6 or x = -4 y= -4 or y= 6 Exercise 4.1: Solve the following systems of equations simultaneously: 1.

y  x  2 and x 2  2 xy  4  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

x  y  2 and x 2  y 2  20 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ __________________________________________________________________

x  y  4 and xy  4 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ __________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

x  y  3 and xy  4 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ __________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

x  2 y  1 and x 2  2 xy  2 x  4 y  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

x  y  3 and x 2  y 2  89 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

2 x  y  1 and x 2  2 yx  2 x  y 2  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Problems involving simultaneous equations: Example: 1. The sum of two numbers is 4 and their difference is 6. 1.1 Let their numbers be x and y . Write down two equations in x and y. 1.2 Solve the equations and find the two numbers. 1.1

Let one number be x and the other y x  y  4 and x  y  6

x y  4

1.2

----(1) and y  4 x Substitute 1 into 2

x  y  6 -------(2)

x  (4  x)  6 x4 x  6 2 x  10 x5 y  1 Exercise 4.2: 1.

The sum of two numbers is 54 and their difference is 6. Find the numbers.

Grade 11 Core Mathematics

GOď&#x201A;ˇMATH WORKBOOKS

The sum of two numbers is 35 and their difference is 19. Find the numbers.

In a two digit number, the sum of the digits is 12 and their difference is 4. Find the number if the tens digit is larger than the units digit.

Grade 11 Core Mathematics

GOď&#x201A;ˇMATH WORKBOOKS

The perimeter of a rectangular flower bed is 26m. If the length exceeds the breadth by 3m, find its dimensions.

A number consisting of two digits has the following properties. When the number is added to twice the tens digit the answer is 33. If the digits are reversed, the number obtained exceeds the original number by 63. What is the original number?.

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Samantha and Warren cycle towards each other along a straight road. They start off 70km apart. Samantha cycles at 15km/h and warren at 20km/h. How far will Samantha have cycled when they meet?

Quadratic Inequalities:

Steps: 1. 2. 3. 4. 5.

Equate to zero If fractions involved then combine to make one expression Factorise all expressions fully Test for the solution with whatever method you are comfortable with. Write your solution down.

NB DO NOT CROSS MULTIPLY WITH INEQUALITIES: Examples: 1.

x2  3x  4  0 ( x  4)( x  1)  0 -1

Solution:  1  x  4

Grade 11 Core Mathematics

4 x 4 x 0 x 2 x 4 0 x ( x  2)( x  2) 0 x  2  x  0 or x  2

GOMATH WORKBOOKS

x

3 x 0 x8

-2 2

2 2

-8

x  8 or x  3 Exercise 5.1: 1.

x 2  8 x  15  0 . ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

4 x 2  49 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

x2 0 x5

2x  1 0 x4 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

 2x  5 0 3 x ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

Grade 11 Core Mathematics

GOMATH WORKBOOKS

4 x x3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

x

9 x

2 3  x2 x3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

_9.

GOMATH WORKBOOKS

1 2  x5 x7

Algebraic Fractions: Multiplication & Division:

NB. ALL TERMS MUST BE FULLY FACTORISED BEFORE ANY SIMPLIFICATION IS ATTEMPTED. Examples: 1. =

x2 x2  4

Terms are factorised and simplified

( x  2) ( x  2)( x  2)

1 x2

x2  4 x 1 x 1 x x 2 x 2x  4 x  3x  2 ( x  2)( x  2) x  1 x 1 x x ( x  1)( x  2) x 2 ( x  2) x 1  2x

All terms factorised first. And then common values are simplified.



4 x 2 x 2  8x  4 x 2  16  ( x  4) 4   ( x  4)( x  4 2 x( x  4) 4  2 x ( x  4) 2

Terms all fully factorised first and a –ve sign is used to reverse ( 4 – x ) as well as inverting the fraction after the division sign and changing the operation to multiplication.

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Algebraic Fractions Addition & Subtraction 1. NB Before you can find a LCD (lowest common denominator) you must ensure that all denominators are fully factorised. 2. The next step is to get the LCD. Write down the product of the highest power of each type of factor. Remember look at each factor in each separate term to ascertain its power. 1 2 3   Example: 2 ( x  1)( x  2) ( x  2) ( x  1) 2 The highest power of ( x  2) is in the second term and is ( x  2) 2 and the highest power of ( x  1) is in the third term and is ( x  1) 2 thus the LCD = ( x  2) 2 ( x  1) 2 3. Divide each denominator into the LCD and multiply the answer by the numerator of each fraction. 4. Simplify to get the final answer.

Examples.

2 3  x3 x2 2( x  2)  3( x  3)  ( x  3)( x  2) 2 x  4  3x  9  ( x  3)( x  2) 5x  5  ( x  3)( x  2)

LCD

Final Answer

Highest power of (x+3)

1 3 2   2 ( x  3) ( x  3) ( x  3) 3 ( x  3)( x  3)  3( x  3)  2  ( x  3)( x  3)( x  3) 

x 2  6 x  9  3x  9  2 ( x  3) 3



x 2  3x  2 ( x  3) 3

LCD Final Answer.

Grade 11 Core Mathematics

GOMATH WORKBOOKS

2 3 4  2  x  2 x  1 x  1 ( x  1) 2 2 3 4    ( x  1)( x  1) ( x  1)( x  1) ( x  1)( x  1) 2( x  1)( x  1)  3( x  1)( x  1)  4( x  1)( x  1)  ( x  1)( x  1)( x  1)( x  1)

Denominators fully factorised.



LCD = product of the highest power of each type of factor

2 x 2  4 x  2  3x 2  3  4 x 2  8 x  4 ( x  1) 2 ( x  1) 2

Expanded form of previous line

3x 2  4 x  5  ( x  1) 2 ( x  1) 2

Final answer.

Examples involving the change of signs.

2 3 4   (b  1) (1  b) (b  1)(1  b) 2 3 4    (b  1) (b  1) (b  1)(b  1) 2(b  1)  3(b  1)  4  (b  1)(b  1) 2b  2  3b  3  4  (b  1)(b  1) b5  (b  1)(b  1)

3 2 1 4    2 2 2 x 1 ( x  1) 1 x (1  x) 3 2 1 4   2   2 2 x 1 ( x  1) x  1 ( x  1) 3 2 1 4     2 2 ( x  1)( x  1) ( x  1) ( x  1) ( x  1) 

3( x  1)  2( x  1)  ( x  1)  4( x  1)( x  1) ( x  1)( x  1)( x  1)



3x  3  2 x  2  x  1  4 x 2  8 x  4 ( x  1)( x  1)( x  1)



 4 x 2  10x  2 ( x  1)( x  1)( x  1)

Use a negative to reverse the order of the denominator (1 – b) So just change the middle sign (1 + b) is the same as (b + 1) so no sign change needed.

Sign change to reverse term (1- x2)

Use a double sign change to change both brackets sign stays a +

Grade 11 Core Mathematics

GOMATH WORKBOOKS

Exercise 6.1: 1

6 x. 12

____________________________________________________________________

____________________________________________________________________ ___________________________________________________________________

4x 2 y 3 8x 3 y ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

2x  4 4

xy  y y

Grade 11 Core Mathematics

GOMATH WORKBOOKS

8x 2  4 x 4x ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

x2 1 ( x  1) 2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

x 2  x  12 x 2  7 x  12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ __________________________________________________________________

a b ba

Grade 11 Core Mathematics

GOMATH WORKBOOKS

a2 a a2 2

10.

x 2  x  12 4 x ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

11.

ab  a 2 b 2  ab X b2  a2 a2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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12.

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2 1 1  2  2 x  3x  2 x  x  2 x  1 2

13.

x x2  2 x  y y  x2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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14.

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7x 3x  2 x  2 y 5 y  5x ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________

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Number Patterns:

3 basic types of patterns dealt with 1.

Arithmetic Sequences: General Term : Tn  a  (n  1)d These have a common difference: A 1st order sequence : d1  a

Geometric Sequence : General Term is: Tn  ar n1 These have a common ratio.

Quadratic Sequences: General Term is: Tn  an2  bn  c A 2nd order sequence. d 2  2a (n  1)(n  2) Or Tn  (n  1)T2  (n  2)T1   d2 2 NB this sequence has no constant relationship between consecutive terms. However there is a constant relationship at a higher level( 2nd level) In quadratic sequences there are two levels of differences: First Difference i.e subtract T2  T1 Second Difference ie subtract the 2nd tier of values. Example: 2

2nd Tier( level) or 2nd difference 1st Tier (level) or 1st difference Sequence

3 5 7 9 1 ; 4 ; 9 ; 16 ; 25

The first difference will be an arithmetic sequence. Example : 1. Given the number pattern 1 ; 4 ; 9 ; 16 ; 25 Find an expression for the nth term. Let 2a  d 2 and 3a  b  T2  T1 and

a  b  c  T1

SEQUENCE 1 ; 4 ; 9 ; 16 ; 25 1ST DIFF 3 5 7 9 2ND DIFF 2 2 2 2a  d 2

3a  b  T2  T1

a  b  c  T1

2a  2

3(1)  b  3

1 0  c  1

a 1

b0

c0

Tn  n 2

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Exercise 7.1: A. B. C.

Find the next two terms in the sequences below. Find an expression for the nth term of each sequence. Find the 10th term of each sequence.

1.1

1 ;3;6.;10;... _________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

1.2

15;30;48;69;...... ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

1.3

 12;7;1;12;..... ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

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1.4

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27;31;37;........ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________

1.5

12; 17; 24…… ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

1.6

-13; -4; 8….. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________

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1.7

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25; 30; 39…….. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.8 7; 12; 21…….. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ Exercise 7.2: Write down the next 3 numbers in each of these sequences. Explain the rule in words. 1.1

5; 11; 17; 23 ____________________________________________________

______________________________________________________________________ 1.2

1;4; 9 ; 16; 25_______________________________________________________

______________________________________________________________________ 1.3

10; 5; 0; -5 _______________________________________________________

______________________________________________________________________

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2.1

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Look at the pattern below Draw the next pattern.

2.2 The table below shows the shape number and the sequence of total number of black dots in each shape. Complete the table. Shape number

Number of black dots

3 4 8 10 N Hint: ‘n’ represents a general equation for the nth term of a sequence 2.3

Draw a Graph representing the data in the table in question 2.2 y

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Look at the sequence and the table below: 2; 6; 12; 20; 30; …………. Terms 1st 2nd 3rd 4th 5th 6th 7th 8th

3.1

Value of term 2 6 12 20 30 42 56 72

4 6 8 10 12 14 16

2 2 2 2 2 2

How were the answers in the third column obtained? __________________________________________________________________

________________________________________________________________________ ________________________________________________________________________

3.2 What type of pattern of pattern do the answers in 3.1 form: Give reasons: ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

3.3 What do you notice about the fourth column? ________________________________________________________________________ ________________________________________________________________________ 4. Use question 3. as a reference to complete the table below for the sequence: 5; 11; 20; 32 ; 47; 65; 86; …………… Terms 1

Value of term 5

1st Difference

2nd Difference

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Question 5: You are given a sequence where the first term is ‘a’ and the constant ratio is ‘r’. The first for terms are written down in general form as : a ; ar ; ar2 ; ar3 5.1

What is the 5th term of this sequence?

______________________________________________________________ 5.2 What is the 20th term of the sequence?

5.2

_______________________________________________________________ Now find an equation that will give the nth term of this sequence: i.e. Tn  .......... _________________________________________________________________

Exercise 7.3:

Look at each of the following sequences and : A. Find the next 3 terms. B. Find an equation that will give the nth term of the sequence.. C. Find the 20th term. 1. 0; 2; 6; 12;……..;

2. 2; -1; -6; -13;……

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3. 5; -1; -7;…….

4. 8; -4; 2. -----;

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ARITHMETIC PROGRESSIONS. 1. Sequences: An arithmetic sequence is a string of values which increase or decrease by a constant value. This value is referred to as the “common difference” ( d). The first term of the sequence is referred to as “a”.The last term (nth) in the sequence is referred to as Tn. ( N.B. There are an infinite number of terms ,however you choose which one you want to use as the last term) The formula for the nth (last) term ( Tn) in a sequence is formulated as follows. If a sequence of numbers is : 7 ; 10 ; 13 ; 16 ; …..(to the nth term). T1 = 7 T2 = 7 + 3 = 10 T3 = 7 + 3 + 3 = 13 T4 = 7 + 3 + 3 + 3 = 16 T5 = 7 + 4(3) = 19 T10 = 7 + 9(3) = 34

Tn  a  (n  1)d last term

first term

common difference

second last term

Calculations in Arithmetic Progressions (AP’s). It is a good idea to develop the formula for the nth term of a progression prior to any calculation. It is an easier formula to work with and it is relevant to the progression in question. Example: A progression is given and you want to develop the nth term formual for it. 5;9;13;17...... Simply use the general nth term formula for an arithmetic progression and substitute the ‘a’ and ‘d’ values into it and simplify. Tn  a  (n  1)d

Tn  5  (n  1)4 Tn  5  4n  4 Tn  4n  1 NOW if I need to find the 20th term simply substitute 20 for n in the formula above: i.e. Tn  4n  1 T20  4( 20)  1 T20  81

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Further Examples: 1.

If a sequence of numbers is : 7 ; 10 ; 13 ; 16 ; …..(to the nth term).

1.1

Find the 20th term in the sequence. a = 7; d = 3; n = 20 ;T20 = ? Tn = a + (n-1)d T20 = 7 + 19(3) = 64 64 is the 20th term in the sequence.

1.2

If 94 is the nth term in the sequence find out the number of terms (n). Tn = 94 ; a = 7 ; d = 3 ; n = ? Tn = a + (n-1)d 94 = 7 + (n-1)3 94 = 7 +3n – 3 90 = 3n n = 30

1.3

94 is the 30th term in the sequence. If 94 is the 30th term in the sequence Find the first term. Tn = 94 ; ; d = 3 ; n = 30 ; a = ? Tn = a + (n-1)d 94 = a + 29(3) a=7 7 is the first term in the sequence.

1.4.

If 94 is the 30th term in the sequence and 7 is the first term find the common difference. Tn = 94 ; ; n = 30 ; a = 7 ; d = ? Tn = a + (n-1)d 94 = 7 + 29d 29d = 87 d=3 the common difference is 3

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Finding a general formula that satisfies the nth term of a given sequence. i.e. you must be given or can calculate the Ist term and the common difference. Example: If the first three terms of an arithmetic sequence is 3x  1 ; 2x  3 ; 2x  1 …… NB the constant concept is the common difference. Thus T2 – T1 = T3 - T2

2x  3  3x  1  2x  1  2x  3 x= 8 T1 = 23; T2 = 19 and T3 = 15

Tn = a + (n-1)d Tn = 23 +(n-1)d Tn = 23 –4d This is the general term representing the nth term of this specific sequence. 3.

Simultaneous Equations in AP’s Given that Tn = 25 and T11 = 81 find the arithmetic progression. i. e. find the first three terms of the sequence. Start with the nth term formula and write down a specific formula for the terms supplied. Tn = a + (n-1)d T11 = a + 10d = 81 T4 = a + 3d = 25 ( subtract the 2 equations to solve for ‘d’) 7d = 56 d=8 a = 1 ( by substitution) AP = 1 ; 9 ; 17 ;….

Exercise 7.4: 1. The sixth term of an arithmetic sequence is 17 and the tenth term is 33. Determine the first term and the common difference. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

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2. x; 2x + 1; 11 are three consecutive terms of an arithmetic sequence. Calculate: 2.1 x

3.2 which term of the sequence will be equal to 57?

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GEOMETRIC PROGRESSIONS: A geometric sequence is a string of values which increase or decrease when multiplied by a constant value. This value is referred to as the “common ratio” ( r). The first term of the sequence is referred to as “a”.The last term (nth) in the sequence is referred to as Tn. ( N.B. There are an infinite number of terms ,however you choose which one you want to use as the last term) The formula for the nth (last) term ( Tn) in a sequence is formulated as follows. If a sequence of numbers is : 7 ; 14 ; 28 ; 56 ; …..(to the nth term). T1 = 7 T2 = 7 .2 = 14 T3 = 7 .2.2 = 28 T4 = 7.2.2.2 = 56 T5 = 7 .24 = 112 T10  7.2 9  3584

second last term Tn  ar n1

last term

common ratio

first term

Calculations in Geometric Progressions (GP’s) 2.

If a sequence of numbers is : 7 ; 14 ; 28 ; 56 ; …..(to the nth term).

2.1

Find the 20th term in the sequence. a = 7; r = 2; n = 20 ;T20 = ? Tn  ar n1 T20 = 7.219 = 3670016 3670016 is the 20th term in the sequence.

2.2

If 3584 is the nth term in the sequence find out the number of terms (n). Tn = 3584 ; a = 7 ; r = 2 ; n = ? Tn  ar n1 3584 = 7.2n-1 512 = 2n-1 29 = 2n-1 n–1=9 n = 10 3584 is the 10th term in the sequence.

Grade 11 Core Mathematics

2.3

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If 3584 is the 10th term in the sequence Find the first term. Tn = 3584 ; ; r = 2 ; n = 10 ; a = ? Tn  ar n1 3584 = a .29 3584 a= 512 a=7

7 is the first term in the sequence. 3.4.

If 3584 is the 10th term in the sequence and 7 is the first term find the common ratio. Tn = 84 ; ; n = 10 ; a = 7 ; r = ? Tn  ar n1 3584 = 7.r9 512 = r9 29 = r9 r=2

the common ratio is 2 Simultaneous Questions Given two terms calculate the first 3 terms of the sequence Example: Write the term that comes later in the progression first when setting the simultaneous equations up. i.e. T10 on top and T5 on the bottom. If T10  1536 and T5  48 Calculate the first 3 terms of the sequence: T10  ar 9  1536 T5  ar 4  48 r 5  32 r2 ar  48 4

48 16 a3 a

Divide to eliminate the “a” value and solve for ‘r’ NB subtract the exponents when dividing. Sequence : 3 ; 6 ; 12; …. Substitute “r” into either one of the equations to solve for “a’

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Finding a general formula that satisfies the nth term of a given sequence. i.e. you must be given or can calculate the Ist term and the common difference. Example: If the first three terms of a Geometric sequence is 3x  1 ; 2x  3 ; 2x  1 …… NB the constant concept is the common ratio. T T Thus 2 = 3 T1 T2 2x  3 2x  1  3x  1 2 x  3 (2 x  3) 2  (3x  1)(2 x  1)

4 x 2  12x  3  6 x 2  5 x  1 Tn = 23 –4d This is the general term representing the nth term of this specific sequence. Exercise 7.5: 1. 2 ; 6 ; 18 ; 54 ; …. is a geometric sequence. 1.1 Continue the sequence to the 6th term. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

1.2

Find the 20th term.

Grade 11 Core Mathematics

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2. Determine the 2nd and 3rd terms in the following sequence given that đ?&#x2018;&#x2021;1 = 5 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x2021;4 = 40 . ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 3. In the following geometric sequences: 3.1 đ?&#x2018;&#x17D; = 2 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x; = 3, đ?&#x2018;&#x201C;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018;&#x2019; 5đ?&#x2018;Ąâ&#x201E;&#x17D; đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x161;. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3.2

đ?&#x2018;&#x17D; = 1 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x; = 2 , đ?&#x2018;&#x201C;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018;&#x2019; 6đ?&#x2018;Ąâ&#x201E;&#x17D; đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x161;.

3.3

đ?&#x2018;&#x17D; = 11 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x; = â&#x2C6;&#x2019; 3 , đ?&#x2018;&#x201C;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018;&#x2019; 4đ?&#x2018;Ąâ&#x201E;&#x17D; đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x161;.

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4. In a geometric progression of which: 4.1 𝑡ℎ𝑒 6𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠 96 𝑎𝑛𝑑 𝑎 = 3, 𝑓𝑖𝑛𝑑 𝑟. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

4.2

𝑡ℎ𝑒 5𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠

7 81

𝑎𝑛𝑑 𝑎 = 7. 𝑓𝑖𝑛𝑑 𝑟.

4.3

𝑡ℎ𝑒 7𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠 192 𝑎𝑛𝑑 𝑡ℎ𝑒 2𝑛𝑑 𝑖𝑠 6, 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 3 𝑡𝑒𝑟𝑚𝑠.

4.4

𝑡ℎ𝑒 6𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠 − 32 𝑎𝑛𝑑 𝑡ℎ𝑒 9𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠

6 256

, 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 3 𝑡𝑒𝑟𝑚𝑠

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5. In a geometric progression with first 3 terms: đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; 4; đ?&#x2018;&#x2DC; + 2; 3đ?&#x2018;&#x2DC; + 1 5.1 find the value(s) of k. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

5.2

find the first term.

5.3

Find the 10th term.

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Exercise 7.6: 1.

Consider the series of numbers below. You must assume that the number of terms listed will be sufficient to conclude a rule for each sequence. Series A: 1 ; 4 ; 9 ; 16 ; 25 ; 36 ; 49 ; ------------------1 1 1 1 1 Series B: 1 ; ; ; ; ; ;-------------------2 3 4 5 6 1.1

State your observations concerning the behaviour of series A:

____________________________________________________ _____________________________________________________________________ 1.2

State your observations concerning the behaviour of series B:

_______________________________________________________ _____________________________________________________________________1.3 What will the tenth term ( 10th Term) in series A be? _______________________________________________________ _____________________________________________________________________1.4 What will the 20th term be in series B? ____________________________________________________________________ _____________________________________________________________________ 1.5 If ‘k’ represents the position of a term ( eg. Position 1 will be represents the position of a term ( eg. Position 1 will be k = 1) Write down the general rule , in an equation form , for series A and B, respectively. Series A:_________________________________________________

Series B:_________________________________________________

Find the sum ( addition) of the first 4 terms in series A. ________________________________________________________________.

Find the sum of the first 3 terms in series B. _________________________________________________________________.

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Consider the table below Terms 1st 2nd 3rd 4th 5th

Value of term -5 2 11 22 35

2.1

Complete the table.

2.2

What can you conclude about the second difference?

_____________________________________________________________________ 2.3

What can you deduce about all the values that you calculated in the column denoted “first difference”?

_____________________________________________________________________ 2.4 Determine the 10th term in the sequence ( the number occupying position ten of the sequence) _____________________________________________________________________ _____________________________________________________________________

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Financial Maths:

No business can exist without the information given by figures. Borrowing, using and making money is the heart of the commercial world. The principle of interest and interest rate calculations are extremely important. This leads into an examination of the principles involved in assessing the value of money over time and how this information can be utilized in the evaluation of alternate financial decisions. Remember that the financial decision area is a minefield in the real world, full of tax implications, depreciation allowances, investment and capital allowances. The basic principles in financial decision making are established through the concept of interest and present value: –

Definition of interest: Interest is the price paid for the use of borrowed money Interest is paid by the user of the money to the supplier of it. It is calculated as a fraction of the amount borrowed or saved over a certain period of time. This fraction is also known as interest rate and is expressed as a percentage per year (per annum).

Simple interest (Growth) is computed on the principle for the entire term of the loan and is thus due at the end of term. Si = Prt Si = is the interest paid or earned P = is the principle or Present value r = is the interest rate per annum t = is the time or term of loan

Other Formulae: Fv  Pv (1  ni) Pv 

Fv 1  ni

Pv  Fv (1  dt )

Key to Symbols Fv = Future Value = A Pv = Present Value = P i = Interest Rate n = Time period d = discounted rate

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Compound Interest (Growth) Compound interest arises when, in a transaction over an Extended period of time, interest due at the end of a payment period is not paid, but added to the principal. Thus interest also earns interest i.e. it is compounded. The amount due at the end of transaction period is referred to as the compounded amount or accrued principal. Interest periods Can vary : daily, monthly, quarterly, half-yearly or yearly.

FORMULAE: NB SOME OF THE FORMULAE USED BELOW DO NOT APPEAR ON THE FORMULAE SHEET SUPPLIED EXTERNALLY: IT WOULD BE TO YOUR ADVANTAGE TO LEARN THEM. Simple Growth A  P (1  ni)

Simple Decay A  P (1  ni)

Compound Growth:

Compound Decay: A  P(1  i) n

A  P(1  i) n

OR J   A  P 1  m  m  

A P tm t m Jm

ONLY These 4 formulae appear on the official formulae sheet. sheet

This formula can be used instead of the compound growth one above. The symbols are explained below.

= Accrued amount / Future value [ S is also used instead of A] = Initial principle / present value = the annual interest rate compounded m times per year = the number of years of investment. = the number of compounded periods per year = the nominal annual interest rate. J m  i

NB: IT IS EASIER TO USE THE DECIMAL VERSION OF % FOR CALCULATIONS: Different Compounding Periods:

Annum

 Calculated Once per year

Quartely

 Calculated 4 times per year

Half yearly

 Calculated 2 times per year

Monthly

 Calculated 12 times per year

Weekly

 Calculated 52 times per year

Daily

 Calculated 365 times per year

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Nominal Interest rates: 1.1.

In cases where interest is calculated once a year, the annual rate quoted is the nominal annual rate or simply referred to as nominal rate.

Effective Interest rates: 1.2. If the actual interest earned per year is calculated and expressed as a percentage of the relevant principal , then the so-called effective rate is obtained. The effective rate is the actual interest rate per annum taking the number of increased time periods. i.e. monthly; quarterly etc. You effectively earn more than the quoted nominal rate per annum. Converting Nominal Rate to Effective Rate: Method 1: Take the Nominal Rate and divide by the number of time periods involved and apply this to the formula: Eff Rate = [ 100(1  i) n -100]

i

nominal rate and n = number of time periods in 1 year. time periods

EG The nominal rate of interest is 22% calculated half yearly. What is the corresponding effective rate of interest: 22 2  11 % Thus R100(1.11)  R123.21 effective Interest rate is 23,21% 2 OR J eff

m   jm   100 1    1 m  

J m  no min al rate m  number of time periods J eff  the effective rate

 0.22  2  J eff  100 1    1 = 23,21% 2    J   OR 1  i   1  m  m   0.22   1  i  1   2  

= 0,2321 = 23,21 %

Converting Nominal Rate to Effective Rate:

J eff

m   jm   100 1    1 m  

J   OR 1  i   1  m  m  

J m  no min al rate m  number of time periods

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Converting Effective to Nominal Rate

Jm ď&#x20AC;˝ 100(m)(m 1 ď&#x20AC;Ť eff ď&#x20AC; 1) Finding the Rate: ď&#x192;Ś a ď&#x192;ś r ď&#x20AC;˝ 100ď&#x192;§ď&#x192;§ n ď&#x20AC; 1ď&#x192;ˇď&#x192;ˇ ď&#x192;¨ p ď&#x192;¸

Further Formulae 1. Finding Principle:

Pv ď&#x20AC;˝ Fv (1 ď&#x20AC;Ť

j m ď&#x20AC;tm ) m

jm and n ď&#x20AC;˝ tm m

NB: i ď&#x20AC;˝

đ??šđ?&#x2018;Ł = đ??šđ?&#x2018;˘đ?&#x2018;Ąđ?&#x2018;˘đ?&#x2018;&#x;đ?&#x2018;&#x2019; đ?&#x2018;&#x2030;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x2019; đ?&#x2018;&#x153;đ?&#x2018;&#x; đ??´đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;˘đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;&#x192;đ?&#x2018;Ł = đ?&#x2018;&#x192;đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;&#x2030;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x2019; đ?&#x2018;&#x153;đ?&#x2018;&#x; đ?&#x2018;&#x192;đ?&#x2018;&#x;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x2018;&#x2013;đ?&#x2018;?đ?&#x2018;&#x2122;đ?&#x2018;&#x2019; Jm = nominal interest rate. m = no of interest periods involved. n = tm = total no of time periods. t = no of years invested. Different periods of compound interest. When banks pay interest on an investment , they pay this more frequently than once a year. The effect that different periods of compounding interest can have on an investment can be quite substantial, depending on the size of the investment. Example: R100 000,00 invested at compound interest for 5 years at 12% p.a. Principle

Interest

Fv ď&#x20AC;˝ Pv (1 ď&#x20AC;Ť

i tm ) m

Time Period = 5years Yearly

m = no of compounding m = 1

Half-Yearly

Quarterly

Monthly

Daily

m=2

m =4

m = 12

m = 365

periods p.a. R100 000

I =12%

Compound Interest

R176234,17 R179084,78

R180611,12 R181669,67 R182193,91

R76234,17

R80611,12

R79084,78

R81669,67

Ci = Fv â&#x20AC;&#x201C; Pv It can be deduced from the table above that the more time periods involved in the calculation of compound interest the better the return on investment:

R82193,91

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CONVERTING EFFECTIVE TO NOMINAL CAN BE DONE AS FOLLOWS:





J m  m m 1  i   1 e.g.

Convert an effective rate of 23,21% p.a. calculated bi-annually to a nominal rate:





J m  2 2 1  0,2321  1 = 0,22 OR 22% p.a.

Calculating the rate in compound interest: Use the following formula:  A  r  100 m  1 for a compound growth  P  and :  A r  100 1  m  for a compound decay P 

Calculating time one has to use logarithms: A P t i log(1  ) m m log

Example: Thembi deposits R12000,00 into her savings account which gives an interest rate of 7,2% p.a. compounded monthly. Her savings grew to R17 181,47 over a time period. Calculate how long her money was invested for. ( answer in years) A P t i log(1  ) m m 17181,47 log 12000 t log(1  0,072) t = 5 yrs. log

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TIME LINES USED FINANCIAL CALCULATIONS Time lines are useful when dealing with complicated problems, such as changes in the interest rate during an investment period or when several deposits or withdrawals are made from a savings account. It helps to summarise the information and give a visual representation of the data in an ordered manner. Example: R7000 is deposited into a savings account , and 4 years later another R5000 is added to the savings. Calculate the value of the savings at the end of 7 years if the interest rate is 12% p.a. for the first 3 years and then increased to 13,5% for the remaining period. Solution: T0

R7000

R5000 12% p.a.

13,5 % p.a.

n (years) Interest rates

T6 etc indicates the time period of the investment.

Balance after 3 years :

Balance after 4 years :

Balance after 7 years:

A ď&#x20AC;˝ 7000(1 ď&#x20AC;Ť 0,12) 3 A ď&#x20AC;˝ R9834,496

A ď&#x20AC;˝ 9834,496(1 ď&#x20AC;Ť 0,135) ď&#x20AC;Ť 5000 A ď&#x20AC;˝ R16162,152 A ď&#x20AC;˝ 16162,152(1 ď&#x20AC;Ť 0,135) 3 A ď&#x20AC;˝ R 23631,26(nearest cent)

Alternative (shorter) Method: đ??´ = 7000(1 + 0.12)3 (1 + 0.135)4 + 5000(1 + 0.135)3 = đ?&#x2018;&#x2026;23631,26

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Example 2: In order to save for her sons University fees , Mrs Gumede deposits R8000 into a savings account at the end of January when her son is 10 years old. The rate of interest is 14% p.a. compounded annually. When her son is 18yrs old he starts a University course which has a duration of 3 years. The first fees are R9000 , payable at the end of January . The fees increase by 10% each year. Calculate: 1.

the second and third years fees.

the balance in the account after the first years fees have been paid.

the balance in the account after the second years fees are paid.

how much additional cash will be needed to cover the third years fees?

R8000

T10

(R9000)

n (years)

14% p.a.compounded annually NB the bracket around R9000 indicates a withdrawal. 1.

Second years fees ( fees at T9):

9000  (0.1)(9000)  R9900 Third years fees ( fees at T10) 9900  (0.1)(9900)  R10890 2.

Balance at T8:

8000(1  0.,14) 8  9000  R22820,69  9000  R13820,69 (to nearest cent) 3.

Balance at T9:

13820,69(1  0,14)  9900  R5855,59 (to nearest cent) 4.

Money in savings account at T10:

5855,59( I  0,14)  R6675,37 Additional cash required: 10890  6675,37  R 4214,63 (to nearest cent)

Grade 11 Core Mathematics

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Exercise 8.1: Revision of Grade 10 time line problems 1. Jack deposits R1000 into a savings account . One year later he adds R2000 to the savings. At the end of the second year he deposits R4000 into the same account, and finally he adds R8000 to the savings account at the end of the 3rd year. Calculate the amount (A) in Jacks account at the end of the 4th year if the interest is calculated at 11,5% compounded annually. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ _________ ___________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 2. R6500 is deposited into a savings account , and 3 years later R7400 is added to the savings. At the end of 5 years , R5800 is withdrawn from the account. How much money will be in the account at the end of 10 years if the interest rate is 11% p.a.? ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

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3. R21000 is invested where the interest rate is 7,5% p.a for the 1st 3 years. The rate is then then increased to 8,25% p.a. for the next 4 years. Calculate the value of the investment at the end of the 7 years. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

4. Mrs Smith invests R8000 for in a savings account when her two sons are 7 and 10 years old. She pays each of them R15000 in the year they turn 21. 4.1

Calculate how much money is in the savings account after she has paid her younger son. The interest rate is 14% p.a. compounded annually.

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4.2

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Is this fair on the sons? Explain your answer.

___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 5. Mr Ndlovu places R52000 in the bond market as a fixed saving for 12 years. The interest paid during first 5 years is 10,5% p.a. It is then increased to 12 % p.a. for the next 3 years, and then finally increased to 14% for the last 4 years. In each case the interest compounded annually. Calculate how much Mr Ndlovu will have in his savings account at the end of the 12 year period. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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Further ‘time line’ questions: Example 1: Jun invests R50 000,00 into an interest bearing account for 15 years. For the first 5 years he receives 10% p.a. compounded bi-annually for the next 5 years he gets 12% p.a. compounded quarterly and for the last 5 years he receives 15% compounded monthly . How much does he have in his account at the end of the 15 year period. Solution: NB: If the Principle remains the same BUT the interest rate changes: use multiplication. Pv = R50 000 Tm = 5 x 2 = 10 Tm = 5 x 4= 20 Tm = 5 x 12 = 60

10% (Half-yearly)

12% (Quarterly)

15% (Monthly)

0,10   0.12   0.15   A  500000 1   1   1   2   4   12   A  R309962,68

T1 5

0 60

Example 2: R8000 is invested into a savings account at an interest rate of 12,5 % p.a. compounded monthly. Three years later R5000 is added to the savings and after a further 3 years R6000 is withdrawn. If a final deposit of R10 000,00 is made into the account in the beginning of the 8th year how much money will be in the account after 10 years? Solution: If the principle is added to or reduced then addition and subtraction is used.

Tm = 10 x 12 = 120 Tm = 7 x 12 = 84 Tm = 4 x 12 = 48

R8000

R5000

Tm = 2 x 12 = 24

(R6000)

R10 000

15% (Monthly)

0.15 120 0.15 84 0.15 48 0.15 24 )  5000(1  )  6000(1  )  10000(1  ) 12 12 12 12 A  R52298,65 A  8000(1 

T1 0

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Exercise 8.2: 1. Jono wins R3 000 000,00 on the lottery and invests the money with Standard Bank. The bank pays 6 % p.a. compounded monthly. At the end of 3 years , Jono withdraws R500 000,00 to buy a car. Two year later he deposits R120 000,00 into his account. Calculate how much money he has in his account at the end of 8 years. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

2. James invests an amount of R100 000,00 for 15 years. He receives 10% p.a. compounded quarterly for the first 6 years and 15% compounded monthly for the last 9 years. 2.1 How much is his investment worth to the nearest rand, at the end of the full term. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

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2.2 What rate would he have to receive if his investment had the same final amount after 15 years compounded annually for the duration? (answer to 1 decimal place). ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 3. Peter invests an amount of R200 000,00 for 20 years. He receives 8% p.a. compounded monthly for the first 14 years and 12% compounded quarterly for the last 6 years. 3.1 How much is his investment worth to the nearest rand, at the end of the full term. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3.2 What rate would he have to receive if his investment had the same final amount after 20 years compounded annually for the duration? (answer to 1 decimal place).

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4. Sipho invests R150 000,00 into an interest bearing account for 10 years. For the first 2 years he receives 12% p.a. compounded bi-annually for the next 5 years he gets 15% p.a. compounded quarterly and for the last 3 years he receives 20% compounded monthly . How much does he have in his account at the end of the 10 year period. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 5. A delivery vehicle is purchased for R 1 250 000,00 in January 2001. It depreciates at 8% per annum. 5.1

Determine its book value in December 2009 if depreciation is calculated according to the straight – line method. (i.e.Simple Decay).

Determine its book value in December 2009 if depreciation is calculated according to A reducing – balance method . ( i.e compound Decay). ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

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6. R20 000 is invested into a savings account at an interest rate of 8,5 % p.a. compounded monthly. 2 years later R15000 is added to the savings and after a further 4 years R16000 is withdrawn. If a final deposit of R5 000,00 is made into the account in the beginning of the 8th year how much money will be in the account after 10 years? ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ _________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 7. Convert The following nominal interest rates to effective annual interest rates: 7.1 15,5% p.a. compounded monthly ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 7.2

32,4% p.a. compounded half-yearly.

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8. Convert an effective annual rate of 32,5 % p.a. to a nominal rate per annum compounded monthly. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 9. Sam wins R1 500 000,00 on the lottery and invests the money with Standard Bank. The bank pays 9 % p.a. compounded monthly. After 2 years , Sam withdraws R500 000,00 to buy a car. 4 year later he deposits R550 000,00 into his account. Calculate how much money he has in his account at the end of 10 years. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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FUNCTIONS & GRAPHS

General Formulae for Functions and Graphs & the effects of movement to functions: 1. Parabola: y  a( x  x1 )(x  x2 ) . Use this form of the equation to calculate the 1.1 equation of a given parabola sketch if you are supplied with the x – intercepts and one other point that the graph passes through.

y  a( x  p) 2  q : Use this form of the equation to calculate the equation 1.2 of a given parabola sketch if you are supplied with the Turning point and one other point that the graph passes through. 2.Shifting the parabola: Lateral shifts affect the axis of symmetry i.e. the p- value in the equation y  a( x  p) 2  q . Vertical shifts affect the q – value in the equation y  a( x  p) 2  q i.e the maximum or minimum value of the graph.

f ( x)  g ( x  3) is an instruction to shift the graph 3 moves to the RIGHT ( positive direction) f ( x)  g ( x  3) is an instruction to shift the graph 3 moves to the LEFT ( negative direction). f ( x)  g ( x)  3 is an instruction to shift the graph vertically upwards (positive direction) f ( x)  g ( x)  3 is an instruction to shift the graph vertically downwards (negative direction) 3.The exponential graph:

y  a x or

1 y  a  x / y    a

To get the equation of a given sketch a point must be supplied . Simply substitute into the general equation and calculate the a – value.

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4.Shifting the exponential graph:

y  a x  p is a horizontal shift to the LEFT ( negative direction) y  a x  p is a horizontal shift to the RIGHT ( positive direction) y  a x  q is a shift vertically upwards (positive direction )

y  a x  q is a shift vertically downwards (negative direction)

5.The hyperbola graph: y

k x

or y 

k x

To get the equation from a given sketch a point must be supplied. Simply substitute into general equation and calculate k –value. 6.Shifting the hyperbola graph:

y

k is a horizontal shift to the LEFT ( negative direction) x p

y

k is a horizontal shift to the RIGHT ( positive direction) x p

y

k  q is a vertical shift upwards (positive direction) x

y

k  q is a vertical shift downwards (negative direction) x

7. Trigonometric Graphs:

y  asinb(x  c)  d ; ETC.

a  affects the amplitude of the graph and also inverts the graph when negative.

b  affects the period or frequency of the graph c  shifts the graph horizontally

d  shifts the graph vertically

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Quadratic Function: Sketching a Parabola using 5 points. A parabola can be neatly sketched using the following 5 points: 1. 2. 3.1 3.2

The two roots i.e. x – intercepts. The y – intercept at x = 0

(2 points) (1 point)  sum of roots  b The axis of symmetry(a.o.s) at either  or   2 2a   The corresponding y-value by substituting (a. o. s) into the original expression. These two values are the turning point: TP(x ; y) (1 point) Axis of symmetry (ado’s) minimum value

4. The mirror image of the y- intercept.

(1 point)

Example: 1. A.

Sketch the graph of y = x2 –2x –3 x – intercepts at y = 0

x 2  2x  3  0 ( x  1)( x  3)  0 y

y – intercept = -3

a.o.s =

X = -1 or x = 3

 b  ( 2)  1 2a 2 Y – value @ f(1)= (1)2 –2(1) – 3= -4

TP ( 1 ; -4)

y= x2-2x-3

D. mirror image of the y – intercept; (2; -3) -

2; -3 1;-4

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100

Exercise 9.1: Sketch the following functions using a “5 – point method”. 1.

y  x 2  3x  4 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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101

y  x 2  4x  5 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

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102

y  x2  x  6 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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103

y  x 2  3x  10 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ 2

-12

-10

-8

-6

-4

-2

-4

-6

-8

-10

-12

-14

Grade 11 Core Mathematics

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104

y  x 2  2x  8 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

-4

-6

-8

-10

Grade 11 Core Mathematics

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105

y  x 2  4 x  12 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

-12

-10

-8

-6

-4

-2

-4

-6

-8

-10

-12

-14

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106

y  2x 2  7 x  6 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

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107

y  2 x 2  5x  3 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

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108

Parabola Graphs: Sketching Using Completing the Square: General Equation: y  a( x  p) 2  q

q is the maximum or minimum value.

p is the axis of symmetry ( a.o.s) Do not use the same setting out method for solving equations: Ensure that the coefficient of x2 is one(1). If it is larger than one place the value outside a bracket and divide each term in the equation by the value. Now complete the square in side the bracket. Finally distribute the value outside the bracket with the two terms inside. At this point the co-ordinates of the turning point can be written down. Example:

To solve equation:

2x 2  4x  6  0

[(x  1)2  4]  0

2[ x 2  2 x  3]  0

( x  1) 2  4

2[( x  1) 2  (1) 2  3]  0

x  1  2

2[(x  1) 2  4]  0

x  1 2

2( x  1) 2  8  0

x  1 or x  3

Only Use the value inside the square brackets to solve

for x pq(1 ; -8) (Turning point) To sketch the graph x-intercepts: (-1;0) and (3;0) y-intercept: -6 Turning point(pq): (1;-8) Mirror image of y-intercept for 5th point.

-1

-6 Y = 2x2 –4x-6

(1; -8)

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109

Exercise 9.2: a) Write the following equations in the form: y  a( x  p) 2  q i.e in the completed square form of the equation. b) Write down the coordinates of the turning point: c) Solve the equation and write down the x and y intercepts. d) Sketch the graphs of the equations. 1. y   (2 x  1)( x  1) ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

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110

2. y  x 2  2 x  3 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

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111

3. y  x 2  6 x  8 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

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112

4. y  x 2  5 x  6 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

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113

5. y  2 x 2  4 x  6 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

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114

6. y   x 2  2 x  3 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

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115

7. y  3x 2  2 x  1 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

Grade 11 Core Mathematics

116

GOMATH WORKBOOKS

Shifting parabolas: Horizontal ( Left or Right) In the completed square form of the equation simply change the “p” value and multiply the equation out for the ax 2  bx  c form ( if required) Example:

2 x 2  12x  10  0 2[ x 2  6 x  5]  0 2[( x  3) 2  4]  0 2( x  3) 2  8  0 pq(3;8)

Instruction: Shift the graph of, 2 x 2  12x  10  0 , 5 units to the left. p3 y  2( x  2) 2  8 p1  3  5 New equation y  2 x 2  8x p1  2

NB. Don’t forget to change the sign when substituting back into y  a( x  p) 2  q B:

Instruction: Shift the graph of, 2 x 2  12x  10  0 , 2 units to the right

p3 p1  3  2 p1  5

New equation

y  2( x  5) 2  8 y  2 x 2  20x  42

Exercise 9.3: 1. Shift questions 1 & 2 in Exercise 9.2 by 4 moves to the left and write down the new equation in the form of y  ax 2  bx  c ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

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Shift questions 3 & 6 in Exercise 9.2 by 5 moves to the right and write down the new equation in the form of y  ax 2  bx  c

2 x 2  12x  10  0 2[ x 2  6 x  5]  0 2[( x  3) 2  4]  0 2( x  3) 2  8  0 pq(3;8)

Instruction: Shift the graph of, 2 x 2  12x  10  0 , 5 units upwards. q  8 y  2( x  3) 2  3 q1  8  5 New equation y  2 x 2  12x  15 q1  3

Instruction: Shift the graph of, 2 x 2  12x  10  0 , 5 units downwards. q  8 y  2( x  3) 2  13 q1  8  5 New equation y  2 x 2  12 x  5 q1  13

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Exercise 9.4: 1.

Shift questions 2 & 3 in Exercise 9.2 by 3 moves upwards. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

Shift questions 5 & 6 in Exercise 9.2 by 3 moves downwards. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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Finding Equations of Parabolas: A. Given the x- intercepts and one other point. Method: Use the general form of the equation : y  a( x  r1 )(x  r2 ) and substitute the roots(xvalues) and the x & y values of the given point into this formula and solve for a. e.g. Find the equation of the parabola that has x-intercepts –3 and 4 which passes through point(1 ; -24) y  a ( x  r1 )( x  r2 )

 24  a (1  3)(1  4)  24  12a a2 y  2( x  3)( x  4) y  2 x 2  2 x  24 B. Given the turning point and one other point. Method: use the general form : y  a( x  p) 2  q i.e the completed square form of the general equation. Substitute the TP and the other point into this form to solve for a. e.g. Find the equation of a parabola that has a turning point (2 ; 3) And passes through point (1 ; 2) y  a( x  p) 2  q y  a ( x  2) 2  3 2  a (1  2) 2  3 2  a3 a  1 y  1( x  2) 2  3 y  x 2  4x  1

C: Given a sketch : Use the information supplied on the sketch to find the equation of the parabola and straight line:

y=x + 1

C(-3 ; 0)

A (0 ; y=ax2 +bx +c

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Method: Straight line : y  mx  c y0

x 1 0 x  1 B ( 1;0) Parabola: y  a( x  r1 )(x  r2 ) [Use this form as roots are known] y  a ( x  r1 )( x  r2 ) Substitute the roots into the y  a ( x  1)( x  3) equation and one other point:  3  a (0  1)(0  3) Then solve for ‘a’  3  3a a  1 y  1( x  1)( x  3)

y   x 2  4x  3

Exercise 9.5: Find the equations of the following given: 1. Turning Point (2;10) passing through (0 ;2) ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 2.

Turning point ( -1;5) passing through (1;13)

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Turning point ( -4;-1) passing through (-3;2)

x – intercepts (1;0) and (-3;0) passing through (-1;-4

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x – intercepts (2;0) and (-4;0) passing through (3;-14)

x – intercepts (1;0) and 5;0) and y –intercept (0;-5)

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Exercise 9.6: Quadratic Function: Parabolas. 1.1

Sketch the graphs of y   x 2  x  12 and y  3 x  12 on the same system of axes.

-15

-10

-5

-2

-4

1.2

Write down the co-ordinates of the points of intersection.

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2.1

124

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2 2 Write y  x  2 x  8 in the form of y  a( x  p)  q .

2.2

Write down the co-ordinates of the turning point.

2.3

Write down the roots ( x-intercepts ) of the graph.

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2 2.5 Sketch the graph of y  x  2 x  8 4

-12

-10

-8

-6

-4

-2

-4

-6

-8

-10

-12

Find the new equation if y  x 2  2 x  8 is moved 5 units to the left

2.6

3.1

Write y   x 2  4 x  5 in the form y  a( x  p) 2  q .

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3.2

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Sketch the graph of

y  x 2  4x  5 .

-12

-10

-8

-6

-4

-2

-4

-6

3.3

Sketch the graph of y  x  5 on the same system of axes.

3.3.1 Write down the co-ordinates of the points of intersection of the two graphs.

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3.4 Find the equation if y   x 2  4 x  5 is moved 4 units downwards. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ Exponential Graphs: y = ax

General formula :

a>0,a1,x  y>0

A. To sketch the graph y = ax (a) x  and y > 0 y = ax lies above the x – axis in quadrants 1 and 2. b) (i) If a > 1 , then as x increases , y increases. e.g. y = 2 x (ii) If 0 < a < 1 , then as x increases , y decreases. e.g. y = ( 12 )x. (i)

y = 2x

x y

-2

-1

1 4

1 2

(ii)

y = ( 12 )x.

x y

-2 4

-1 2

0 1

1 2

2 4

0 1

1 2

1 4

1 y    or  2 y = 0,5x

y = 2x y

1 x

These 2 graphs are mirror images of each other, the axis of symmetry being the y- axes (x = 0 ). The x- axes is a horizontal asymptote, as y will never equal zero.

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Log Graphs: Extension work ( Grade 12 Sylabus) General formula : y  loga x

y  loga x

To sketch the graph of

a) x>0 and y   thus y  loga x lies on the right of the y- axes in quadrants 1 and 4. b) If a > 1, then as x increases so y- increases c) If o < a < 1 , then as x increases so y decreases. (1)

y  log2 x ( x  2 y ) X Y

(2)

1 4

1 2

-2

-1

y  log 1 x ( x   12  ) y

1 4

1 2

-1

-2

y y = log 2x

1 1

y = logo,5x The graph y   log2 x is exactly the same as y  log 1 x 2

These two graphs are mirror images of each other, the axix of symmetry being the x – axes. The y- axes is a vertical asymptote, as x will never equal zero.

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Shifting the exponential graph: 1.

Vertical Shifts: y  a x or

y  2 x Original graph

y  ax

y  2 x  2 Graph shifted 2 units upwards. y  2 x  4 Graph shifted 4 units downwards. 3. NB: the values are added after the base (2x) for vertical movement. 2.

g x  = 2 x +2

f x  =

new horizontal asymptote for g(x) =2x + 2

2. -5

h x  = 2 x -4

-2

3. new horizontal asymptote for h(x) =2x - 4

-4

Exercise 9.7: Sketch the graph of y  3 x on a Cartesian plane.

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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1.1 Shift y  3 x by 2 units upwards and sketch this graph on the same system of axes. 1.2 Shift y  3 x by 4 units downwards and sketch this graph on the same system of axes. 2.

Write down the equations of the asymptotes after the shifts in question 1.

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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3.1 Shift y  2  x by 3 units upwards and sketch this graph on the same system of axes, 3.2 Shift y  2  x by 4 units downwards and sketch this graph on the same system of axes. Write down the equations of the asymptotes after the shifts in 3.

Horizontal Shifts: y  a x or

y  ax

y  2 x Original graph y  2 x  4 Graph shifted 4 units to the right.

2.1

y  2 x  4 Graph shifted 4 units to the left. 2.3 NB: the movement is added or subtracted in the exponent for lateral shifts. 6

s x  =

r x  = 2 x-4

2 x+4

f x  = 2 x 4

3. New vertical Asymptote x = -1

New vertical Asymptote x= 7

2. -5

-2

3. -4

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Exercise 9.8: 1. Sketch the graph of y  2 x on a Cartesian plane. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

1.1 Shift y  2 x by 4 units to the left and sketch this graph. 1.2 Shift y  2 x by 4 units to the right and sketch this graph. 2 Write down the equations of the asymptotes after the shifts in question 1. ___________________________________________________________________ __________________________________________________________________

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3 Sketch the graph of y  2  x on a Cartesian plane,

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

3.1 Shift y  4  x by 3 units to the left and sketch this graph. 3.2 Shift y  4  x by 4 units to the right and sketch this graph. 4 Write down the equations of the asymptotes after the shifts in 3.

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5 Sketch the graph of y  3 x ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

5.1

Shift the graph 4 to the right and 3 upwards and sketch the new position. i.e. the graph of y  2 x  4  3

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The Hyperbola Graph. k k or y  x x

General equation: y  Sketching the graphs:

Method 1. Table Method 4 1. y x X -4 -2 Y -1 -2

-1 -4

1 4

f x  =

2 2

4 1

2 -2

4 -1

4 x

-5

-2

-4

Sketch of y 

2. X Y

4 x

-4 1

-2 2

-1 4

1 -4

f x  =

-4 x

-5

-2

-4

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Shifting the hyperbola graph: 1.

If a constant is added to the equation after

k then this will cause a vertical x

4 4 is shifted upwards by 3 units.  3 : the graph of y  x x 4 Asymptotes of y  are y  0 ( x – axis) and x  0 ( y- axis). x

shift: e.g. y 

This shift will change the horizontal asymptote BUT not the vertical. Asymptotes are: x  0 ( y- axis). And y  3 (New horizontal asymptote)

(1;7) 6

(2;5) 4

h x  =

(4;4)

(1;4)

4 x

horizontal asymptote is y = 3 (-4;2) 2

(-2;-1)

(2;2) (4;1)

f x  =

4 x

horizontal asymptote is y = 0 -10

-5

(-4;-1)

(-1;-1) (-2;-2)

(-1;-4)

-2

-4

-6

Horizontal shifts are caused when a constant value is added to the x- value in the denominator of the hyperbola equation.

The sift is in the opposite direction of the integer in the equation. i.e if the integer is +ve then the shift is to the left and if the integer is – ve then gthe shift is to the right.

Grade 11 Core Mathematics

Shift the graph of y 

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137 4 4 as follows: y  x3 x

This shift is 3 units to the right.

The horizontal asymptote stays the same BUT the vertical asymptote changes to the line x  3

vertical asymptote x = 0 vertical asymptote x = -3

(-2;4)

(-1;2)

(1;4)

(2;2) (4;1) (1;1)

f x  =

4 x

horizontal asymptote is y = 0 -10

-5

(-1;-1)

(-4;-1) (-7;-1) (-5;-2)

(-4;-4)

(-2;-2)

(-1;-4)

-2

-4

-6

h x  =

4 x+3

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Exercise 9.9 Copy and complete the following table and use it to sketch the graph of y 

1. x Y

-6 -1

-3

-2 -3

-1

2 3

6 x

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

1.1

Write down the equations of the asymptotes. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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Sketch the graph of y 

4 ( lateral move to the right) x3 6

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

1.3

Write down the equations of the asymptotes. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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140

Copy and complete the following table and use it to sketch the graph of y 

2. x y

-6 1

-3

-2 3

-1

2 -3

6 x 3

2.1 Write down the equations of the asymptotes. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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2.2 Sketch the graph of y 

6 ( lateral move to the left) x3 6

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

2.3 Write down the equations of the asymptotes. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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Complete the following table and use it to sketch the graph of y 

3. x y

-8

-4 -2

-2

-1 -8

2 4

8 x 4

-12

-10

-8

-6

-4

-2

-4

-6

-8

-10

3.1 Write down the equations of the asymptotes. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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3.2 Sketch the graph of y 

8 1 x2

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

3.3 Write down the equations of the new asymptotes. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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Trigonometric Functions: Method: Sketch the graps using the values of the special angles. Viz using 0º ; 90º ; 180º ; 270º and 360º (You do not have to set up a table) Example 1: Sketch the graph of f ( x)  sin x where x  [0  ;360 ] X y = sinx

0º 0

90º 1

180º 0

270º -1

360º 0

0

90 

180

270

360

f x  = sin x 

Example 2 Sketch the graph of f ( x)  sin x where x  [360 ;360 ]

f x  = sin x 

-360

-270

-180

-90

-1

90

180

270

360

Grade 11 Core Mathematics

Example 3:

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145

Sketch the graph of f ( x)  cos x where x  [360 ;360 ]

g x  = cos x  -360

-270

-180

-90

90

180

270

360

-1

Example 4:

Sketch the graph of f ( x)  tan x where x  [360 ;360 ]

h x  = tan x  1

-360

-270

-180

-90

-1

90

180

270

360

Grade 11 Core Mathematics

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Exercise 9.10: 1. Sketch the graphs of y  sin x and y  cos x on the same set of axes for the interval x  [0  ;360 ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  From the sketch find the following: 1.1

the period of y  sin x

1.2

the range of y  cos x

1.3

the amplitude of y  sin x

1.4

the value for x for sin x  cos x

y 2

x 0 90

-1

-2

180

270

360

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2. Sketch the graphs of y  2 sin x and y  cos x on the same set of axes for the interval x  [0  ;360 ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  From the sketch find the following: 2.1

the period of y  2 sin x

2.2

the range of y  sin x

2.3

the amplitude of y  2 sin x

2.4

the value for x for sin x  cos x

y 2

x 0 90

-1

-2

180

270

360

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3. Sketch the graphs of y  sin 2 x and y  cos x on the same set of axes for the interval x  [0  ;360 ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  From the sketch find the following: 3.1

the period of y  sin 2 x

3.2

the range of y  cos x

3.3

the amplitude of y  sin 2 x

3.4

the value for x for sin 2x  cos x

y 2

x 0 90

-1

-2

180

270

360

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4. Sketch the graphs of y   sin x and y  cos 2 x on the same set of axes for the interval x  [0  ;360 ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  From the sketch find the following: 4.1

the period of y   sin x

4.2

the range of y  cos 2 x

4.3

the amplitude of y   sin x

4.4

the value for x for  sin x  cos 2x

y 2

x 0 90

-1

-2

180

270

360

Grade 11 Core Mathematics

10.

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150

Probability Theory: Independent Events:

When looking at combining or repeated events, it is said that any event is Independent if the outcome of the event does not effect the outcome of any other event. Probabilities of events that are affected when combining them or when a probability experiment is carried out repetitively. Consider the probabilities of combinations using tables, tree diagrams and by derivation of the product rule where there are replacements. Example activity: Bandile says he is going to toss a coin three times(3) and get heads each time. Sisi says that this is not very likely and she bets him that he will not be able to get three(3) heads in a row. When he hears what his friends say, Bandile is no longer so sure he will be able to toss three(3) heads in a row. He draws a tree diagram to work out his chances of tossing 3 heads. Look at his tree diagram and discuss answers to the questions below. first toss

second toss

1 H 2

1 2

1 2 1

2 1

1 T

(HHH)

(HHT)

(HTH)

(HTT)

(THH)

(THT)

1 2

Outcomes

Third toss

2 1 2 1 2

(TTH)

(TTT)

1. For the first toss the probabilities of getting heads and tails are both

Probability 1 8 1 8 1 8 1 8

1 8 1 8 1 8 1 8

1 . Why is 2

this so? 1 again? 2 3. How many different outcomes are there when a coin is tossed three(3) times?

2. Why are the probabilities for the second toss all

4. What is the probability of Bandile tossing three(3) heads in a row? 5. What is the probability of Sisi winning the bet?

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We see from the tree diagram that we can find the probability of each of the Outcomes if we multiply the probabilities P of the outcomes of each toss (t) Together. 1 . The second toss 2 1 1 1 1 1 1 P(T) = the third toss P(T) = Thus P(TTT) =    . 2 2 2 2 2 8

e.g. to find P(TTT), we look at the first toss: P(T) =

Stated formally , we find the probability of independent events by multiplying the probability of each event together. If A and B are independent events then P(A and B) = P(A) X P(B). We call this the product rule. 6. Use the product rule to find the probability of Bandile tossing three(3) heads in a row. 7. After looking carefully at his tree diagram, Bandile decides, instead ,to bet Sisi that he will toss at least two(2) heads out of three(3) tosses. What is the probability of him doing this. Exercise 10.1: [N.B. Use a probability tree to answer the questions] 1

A bag contains 4 yellow counters and 8 white counters. Calculate the Probability that: 1.1

The first counter drawn at random is: 1.1.1 yellow

white.

Grade 11 Core Mathematics

1.2

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For a second draw you get white if first counter was white if: 1.2.1 was replaced

was not replaced before second draw was made.

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2. James cannot decide what to wear to a party. He has three(3) pairs pants â&#x20AC;&#x201C; a grey pair, black pair and blue jeans. He puts his pants on his bed, closes his eyes and chooses a pair of pants to wear. He has four(4) shirts to wear - a white shirt, a white T-shirt, a black shirt and a green T-shirt. He pulls a shirt from his drawer without looking. 2.6

Is James choice of pants independent of his choice of shirt? Explain. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.7

Complete the contingency table below to show the different possible combinations of shirt and pants that James has to choose from and answer the questions that follow. White shirt(WS)

Grey pants(G)

White T-shirt(WT)

Black shirt (BS)

Green T-shirt (GT)

G,WS

Black pants(B) Blue jeans(BJ)

2.2.1

BJ, GT

How many outcomes are there in total?

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2.2.2

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What is the probability that James will be wearing his green T-shirt?

What is the probability that James will be wearing something white

What is the probability that James chooses his black shirt?

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2.3.3

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What is the probability that James is dressed all in black?

What is the probability that James is not wearing any black at all?

Dependent Events: Events that when combined are effected when they are dependent on one another. Definition:

Events are dependent if the outcome of the second event is affected by the outcome of the first event.

Example Activity: NB P(A then B) = P(A) X P(B given A) is called the product rule for a dependent event. Sandy and her brother, John are doing their homework. Their mother puts out a plate of biscuits on the table next to Sandy. There are three(3) types of biscuits on the plate: 3 plain, 8 chocolate and 4 jam. John watches to see which biscuits Sandy is going to take. After Sandy has eaten her first one , there will be one less biscuit remaining to choose from the second time. So Sandy’s chance of taking a jam one the second time is dependent on her first choice. As the outcome of the first affects the outcome of the next event, the events are dependent.

John draws 2 venn diagrams representing the choices that Sandy has. He wants to work out the probability that Sandy’s first biscuit will be a chocolate one and her second a jam one.

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Look at John’s venn diagram and then answer the questions below. Ch

5 Pl

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

What do ‘Ch’; ‘Pl’ and ‘J’ stand for? How many biscuits are there on the plate before Sandy takes any? How many chocolate biscuits are on the plate? Calculate P(Ch), the probability of Sandy choosing a chocolate one first? How many biscuits on the plate after Sandy has taken one? How many jam biscuits would there be on the plate if Sandy took a chocolate one first? Calculate P(J and Ch), the probability of Sandy taking a jam one, given that she took a chocolate one first. Why is the denominator for P(J given Ch) 15 instead of 16? Now use the product rule to find out P(Ch , J), the probability that Sandy’s first biscuit is chocolate and her second one will be a jam one. What is the probability that Sandy chooses a jam one and then a chocolate one? Does the order in which she chooses the biscuits matter?

2.Sandy eats 2 biscuits. Each time she chooses one she reaches out to take one without looking at the plate. She takes a plain one first and a jam one second. a. Draw 2 venn diagrams to show her options for each biscuit choice. 2.2 What was the probability of her taking this combination? 2.3 Sandy’s mother says she can have another biscuit. Draw a venn diagram to show the biscuits for this 3rd choice. 2.4 What is P(Pl , J , J)? 3. How could Sandy’s choices be independent rather than dependent.

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Answers to above questions: 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

Ch stands for Chocolate biscuit, Pl stands for Plain biscuit and J stands for Jam biscuit. There 16 biscuits on the plate. There are 8 chocolate ones on the plate. 8 1 P(Ch) =  16 2 There are 15 biscuits still on the plate after Sandy has taken one. There would be 5 jam ones still on the plate if Sandy took a chocolate one first. 5 1 P(J given Ch) =  15 3 The denominator for P(J given Ch) is 15 instead of 16 as there are only 15 biscuits on the plate because Sandy ate one. 8 5 1 1 1 Using the product rule P(CH ,then J) =     16 15 2 3 6 5 8 40 1 The P(J, then Ch) =    , The order does not seem to 16 15 240 6 matter.

2.1 First Choice Ch

5 Pl

Second Choice Ch

5 Pl

2.2

P(Pl, then J) = P(Pl) x P(J given Pl) =

3 5 15 1    16 15 240 16

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Third Choice Ch

4 Pl

2.4 P(Pl, then J, then J) = P(Pl) x P(J given Pl) x P(J given Pl, given J) 3 5 4 60 1 =      0,018 16 15 14 3360 56 3

Sandy was not putting back her biscuits before making her second choice making her next choice so we say that this experiment was without replacement. For the events of choosing biscuits to be independent, there would have to be replacement. That is, each time a biscuit was taken from the plate, one of the same kind would have to be added.

Exercise 10. 2: 1.

Are these two events Dependent or Independent? 1.1 rolling a dice ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.2

taking 2 names out of a hat

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.3

the first two numbers in a lottery draw

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1.4

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tossing a coin twice ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Jack has a bag of 5 red, 7 blue, 8 white and 6 green marbles. He represents this sample space using the venn diagram below:

2.1

Jack allows some friends to choose 3 marbles each, without looking. Draw a venn diagram to show the sample space for each choice and then calculate the probability of each of these events: 2.1.1

Nick chooses first. He chooses red, then white , then green.

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2.1.2

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Vusi chooses second. He chooses green, then green, then green.

2.2

For each question below draw the relevant parts of a tree diagram to show the marble choices and their probabilities.

2.2.1

Cynthia chooses last. She wants 3 red marbles. What is the probability that she will choose 3 red ones.

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Cynthia chooses red, then white , then white. Was the probability of this choice greater than the probability of getting 3 red marbles? Explain.

Blake is getting dressed in the dark. His younger brother has been playing with his socks and they are all loose in his drawer. He has 2 grey socks, 2 black and 1 white. 3.1 He reaches into the drawer , he pulls out a sock and puts it on. 3.1.1

Draw a venn diagram to show the sample space for Blakesâ&#x20AC;&#x2122;s first choice.

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3.1.2 How many socks are in the sample space for Blake’s second choice? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3.1.3

Are the events dependent or independent?

3.2 Draw a tree diagram showing Blake’s sock choices, then answer the questions that follow: ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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3.2.1

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How many different outcomes are there altogether?

3.2.2

How many of these outcomes have two socks the same?

3.2.3

What is the probability then that Blake chooses 2 socks the same?

What is the probability that he goes to school with odd socks on?