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MATHEMATICS N4 New Syllabus MATHEMATICS N4 New Syllabus MJJ van Rensburg ,TROUPANT l'ublljhm Preface There is no generally accepted theory of how to teach or learn; there are only guidelines which often
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MATHEMATICS N4 New Syllabus MATHEMATICS N4 New Syllabus MJJ van Rensburg ,TROUPANT l'ublljhm Preface There is no generally accepted theory of how to teach or learn; there are only guidelines which often overlap and supplement each other. It is generally accepted that when you understand something, it is much easier to learn. It is also true that if you do not initially understand something, constant repetition will eventually result in understanding. The emphasis in this book is on understanding. When you understand something, however, it does not necessarily mean that you will be able to do it. Ample provision has therefore been made for practice. One ofthe characteristics ofmathematics is that it is compact and that mathematical problems lend themselves to very short solutions. However, for students to master the subject in a meaningful way, the material must be covered in greater detail. The first examples in this book will therefore be set out properly, and will not necessarily demonstrate the shortest methods of solving problems. It is, after all, a fact oflife that before you can run, you must first learn to walk. Properly set out examples are also beneficial for exam revision, since by this time students have often forgotten the short-cut methods and are unable to work them out by themselves. THE AUTHOR Contents Subject aims and general information Important symbols and abbreviations Module : Equations, manipulation and word problems. Factorising the sum ofand difference between two cubes.. Common factors..2 Grouping..3 The difference between two squares..4 The quadratic trinomial..5 The sum ofand difference between two cubes Exercise..2 Exponents.2. The exponent laws Exercise The solution ofexponential equations Exercise.3.3 Logarithms.3. Exponential and logarithmic forms.3.2 Calculations without using a calculator Exercise Logarithmic laws Exercise Logarithms and anti-logarithms using a calculator.3.5 Changing the base xv xviii Exercise Solution of equations using logarithms Exercise.7.4 The solution of simultaneous equations.4. One variable in one equation.4.2 Two equations containing two variables.4.3 Three equations with three unknowns Exercise.8.5 Word problems.5. Compiling and solving simultaneous equations Exercise Compiling and solving quadratic equations Exercise Module 2: Determinants 2. The origin ofdeterminants 2.2 The development of a second order determinant 2.3 The solution of two equations simultaneously determinants using Exercise Third order determinants 2.5 Cramer's rule for three unknowns in three linear equations Exercise Module 3: Complex numbers Introduction The natural numbers (N) Integers (whole numbers) (Z) Rational numbers (Q) Irrational numbers Real numbers (R) Imaginary numbers Graphical representation of imaginary numbers Powers of i (or J) Multiplication of real numbers by imaginary numbers Addition and subtraction of imaginary numbers Division of imaginary numbers Exercise Complex numbers Argand diagrams The polar form of complex numbers The relation between the rectangular and polar forms Exercise Addition and subtraction ofcomplex numbers Multiplication of complex numbers A complex number multiplied by a real number Multiplication oftwo complex numbers in the rectangular form Multiplication of complex numbers in the polar form Exercise Division of complex numbers When the divisor is a real number The conjugate of a complex number (rectangular form) Division oftwo complex numbers in the rectangular form Division in the polar form Powers in the polar form Exercise Exercise 3.5 Identical complex numbers Complex roots (or zero points) Practical applications 68 Exercise Exponents and complex numbers 75 Module 4: Trigonometry Exact values General Exact values in the first quadrant Exact values in other quadrants Another figure to help you remember signs Solving easy trigonometric equations using exact values Arc functions (inverse trigonometric functions) 92 Exercise Using the calculator to determine function and arc function values The solution of 'easy' trigonometric equations More advanced trigonometric equations Exercise Exercise 4.3 Trigonometric ratios of compound angles sin (A ± B) and cos (A ± B) tan (A ± B) in terms of tan A and tan B Double angles Application of compound angles Co-functions Exercise Module 5: Sketch graphs Coordinates The domain and range Dependent and independent variables Functions and relations Function values The inverse of functions and relations Continuous and discontinuous functions or relations Symmetry Exercise Sketching the graphs Important information for identifying graphs The straight line (ax + by + c = 0) The ellipse (a,xl + by = ~ or rid 2 + yll! = ) The circle (a,xl + at = r or r + y = r or y = ±.jr - y) The rectangular hyperbola (xy = c) The hyperbola (rla 2 - ylb 2 = or ~r - dy = e) Exponential or logarithmic graphs Trigonometric graphs The parabola The cube functions Exercise Module 6: Limits and differentiation Exercise Exercise Exercise Limits A general concept of a limit Theorems on limits Limits of the form 00 Limits of the form 00/00 The binomial theorem Differentiation from first principles The gradient ofa straight line The gradient of a tangent to a curve Differentiation Introduction Symbols for differentiation Standard differential coefficients (and rules) Application of the differentiation rules Exercise The chain rule Exercise A shorter method ofdifferentiating a function ofa function Exercise The product rule Exercise The quotient rule Exercise Exercise 6.9 Maximum and minimum values Successive differentiation Maximum and minimum turning points Point of inflection Exercise Sketching graphs using maximum and minimum values Exercise Sketching graphs with the form y = axj + bxj + ex + d Exercise Module 7: Integration Introduction Standard integrals Summary of all the relevant standard integrals Exercise Integrals of trigonometric functions 7.5 Integrals of l/x. e' and a' Exercise Integrals ofthe trigonometric functions ifthe angle (e.g. x) is multiplied by a constant 7.7 Integrals of e' and ai.' Exercise Definite integrals Exercise Calculations ofareas using integration Exercise Module 8: Summary 267 Module 9: Criteria tests 289 Subject aims and general information. General subject aims Mathematics N4 for engineering students should:. equip the students with the necessary knowledge of mathematical language, concepts and operations to enable them to apply this knowledge in other engineering fields and their place ofwork..2 develop clarity of thought, logical and critical thought processes, and the higher cognitive skills of application, analysis, synthesis and evaluation in the students..3 enable the students to interpret physical problems, given in words or sketches, in mathematical terms, solve these problems, and translate the solutions into the original physical context..4 foster an appreciation for the aesthetic balance, flawless precision and limitless scope of Mathematics as a powerful instrument to solve problems in different fields of study. 2. Specific subject aims The instructional offering should be taught in such a way that: 2. the use of the correct mathematical terminology is mastered by the students. 2.2 although formal proofs of theorems are not required for examination purposes, these proofs are an integral part ofthe presentation of the instructional offering. 2.3 word problems and problem solving are the focal points of each module. 2.4 the students master the learning content by means of illustration and visual aids. 3. Duration of the course One trimester (full-time or part-time). 4. Evaluation Candidates should be evaluated on a continuous basis by means ofclass tests and assignments. 5. Examination Reproduction, application, analysis and evaluation are important aspects in determining the degree ofdifficulty ofthe subject. The division of these aspects should be as follows: Recall and reproduction ± 30% Understanding and application ± 400A Analysis, synthesis and evaluation ± 30% A three-hour examination paper totalling 00 marks will be set in April, August and November ofeach year. The pass mark is 40%. 6. General guidelines 6. Problems should be taken from the physical world and the business world so that the students can relate theory to practice. 6.2 Emphasis should be placed on the correct formulation ofdefinitions and principles and the use ofacceptable mathematical terminology. 6.3 If formulae are used in the solution of a problem, the formulae must be stated before any substitutions are done. 6.4 Students should be encouraged to memorise the basic formulae applicable to N4 Mathematics. 6.5 Calculators may be used to do mathematical calculations. 6.6 Answers to all calculations must be approximated correctly to three decimal places. Approximations may not be done during calculations. The final answer must be approximated to the stipulated degree ofaccuracy. 6.7 The weighted value of a module gives an indication of the time that should be spent on teaching the module, as well as the relative percentage of the total marks allocated to the module in the final exam paper. 6.8 The syllabus provides didactic guidelines at the end ofeach module where necessary. Numbered in accordance with the sections they refer to, these guidelines provide relevant examples, appropriate procedures and other pertinent information. 7. Subject matter Weighted Module Theme value. Equations, manipulation and word problems (2) 2. Determinants (8) 3. Complex numbers (0) 4. Trigonometry (20) 5. Sketch graphs (0) 6. Limits and differentiation (20) 7. Integration (20) Total (00) Important symbols and abbreviations ' or = 'j f.. ~ ~ - III. II = ~ E Ii Ixl ~ orj-i e 00 is equal to is not equal to approximately is greater than (x I) is smaller than (x I) is not greater than is not smaller than is smaller than or equal to is greater than or equal to is congruent to is similar to is perpendicular to is parallel to therefore because implicates implicates and being implicated an element of not an element of absolute value is equivalent to inverse ofj 2, infinity Module Equations, manipulation and word problems Objectives and overview On completion of this section you should be able to:. factorise the sum of or difference between two cubes. 2. apply the basic operations and the laws ofindices and logarithms to manipulate formulae and solve equations. 3. solve three simultaneous equations with three unknowns by elimination and substitution. 4. translate word problems into mathematical equations and solve them.. Factorising the sum of and difference between two cubes The following four types of factors have already been done in Nl to N3:.. Common factors 2ax + 4ay + loar = 2a(x + 2y + 5x) or written out fully: 2ax + 4ay + IOar = ~: (2ax + 4ay + loar) [x ~: = ] = 2a(2ax + 4ay + loar) 2a 2a 2a = 2a (x + 2y + 5x)..2 Grouping 6ab - 4ax + 3bx - 2~ = 2a(6ab _ 4ax) + x(3bx _ 2~) 2a 2a x.x = 2a(3b - 2x) + x(3b - 2x) = (3b _ 2x) [2a(3b - 2x) + x(3b - 2X)] 3b - 2x 3b - 2x = (3b - 2x)(2a + x) or in a shorter form: 6ab - 4ax + 3bx - 2~ = 2a(3b - 2x) + x(3b - 2x) = (3b - 2x)(2a + x)..3 The difference between two squares a 2 - ~ = (a + b)(a - b)..4 The quadratic trinomial or 2~ - 5xy + 3y2 2X~-3xy (2x - 3y)(x - y) ;: y ~.'-.-./- 5xy 2~ - 5xy + 3 = ~ + (p + q)xy + 3y2 = ~ + (-3-2)xy + 3 = 2~ - 3xy - 2xy + 3 = 2x(x - y) - 3y(x - y) = (2x - 3y)(x - y) 2 a = 2, b = - 5 and c = 3 ac = 6 6=3x = p = - 3 and q = - 2 ..5 The sum of and difference between two cubes Note: This work has not been done previously. Examine the following two solutions: x 2 + ax + a 2 x - a)r _a J r - ar +ar ar - a 2 x + a 2 x _ a J a 2 x _ a J x + a)r + a J r + ar. - ar - ar - a 2 x. + a 2 x + a J a 2 x + a J... (x - a)(r + ax + a 2 ) = r - a J... (x + a)(r - ax + a 2 ) = r + a J These results can be applied to determine the factors ofall sums ofand differences between two cubes. r + a J = (x + a)(r - ax + a 2 ) r - a J = (x - a)(r + ax + a 2 ) On examination we can see that: Only the signs differ. The first and last terms of the trinomial are the same. The middle terms of the trinomials are the same but the signs differ. The signs are the opposite of the signs of the factors consisting of two terms. Example r - 8 = r - 2 J The procedure can be described as follows:. The first factor is obtained by taking the cube root ofthe two terms separately, while the sign remains unchanged.... r = (x - 2)( ) 2. The second factor is obtained by squaring the first term (r). Then the two terms are multiplied by each other and the sign is changed (+2x). The last term is squared and added to the second term.... r = (x - 2)(r + 2x + 4) 3 Example r + 27y3 = r + (3y)3 = (x + 3y)(r - 3xy + 90 Example 6r - 48Y = 6(r - 8Y') = 6 [r - (20 3 ] = 6(x - 2y)(r + 2xY + 4y ) Exercise. Factorise: I. 2tX + nr 4. a 3 - b 3 7. a X l2 - y3 3. 2p (x - y)3 _ (x + y)3 2. 2r - 5x r + y3 8. a 3-25 II. 8ry + y 4. 6X- - 54xy a3 _ b 3 3. X p a 6 + b a a - b nr - 4nr Answers I. tx(2 + x) 3. (r + I)(x + I)(x - ) 5. (x + y)(r - xy + y) 7. (a + 4)(a 2-4a + 6) 9. (a 2 + b 3 )(a 4 - aw + b 6 ) II. (2xy + y)(4ry4-2xy + y) 3. 2(p3 + I)(P6 _ p3 + ) 5. (a - ~)(a2 + E+ ~2) 7. (~ _ ~)( ~ + 2: + ~) 2. (2x - 3)(x - ) 4. (a - b)(a 2 + ab + b 2 ) 6. (p + I)(P2 - P + ) 8. (a - 5)(a 2 + 5a + 25) 0. (X- - y)(x' + X-y + y) 2. 3(a + 3)(a 2-3a + 9) 4. 2x(2x - 3y)(4r + 6xy + 9y) 6. -2y(3r + y) 8. 4nr(i - ) 4 .2 Exponents.2. The exponent laws The following laws were explained in N3: am X an = am + n am -7 an = am - (am) = a ljxn (ambny = am b np (am)p = bn amp bnp ao = I, a =t- 0 I a- m = am m a7; is called the exponential form and~ is called the root form. Please note the following: These laws are not valid for the operations + and -. They are only valid for X, ~ and powers. Thus (a X W = a 2 X b 2 but (a + W =ta 2 + b 2 If a and b are negative, then m, nand p cannot be fractions. If the exponent laws are applied in such a case some solutions will be lost (see 3.24). Except for the above restrictions, a and b can be real numbers and m, nand p rational numbers, i.e. fractions ofthe form ~ where q and r r are whole numbers and r =t- O. Examples Simplify the following: I 2. (8XJ)3 3. (4y)-2 ISX. 4 x SX. 2 x - 5 Solutions. 2p X p3 X p-2 = 2 p l+3-2 = 2p 2 2. (8XJ)3 I = [8()]3 =2 3. (4y)-2 = (4y)2 = 6y ( 32X')~3 4. 2x-2 = (25X')~3 2x-2 = 2 3 X' = 8X' 6 I I I = [(20)2 + (45)2]4 I I I = [(4 X 5)2 + (9 X 5)2]4 I I I I I = [ ]4 I = [2J5 + 3J5]4 I = (5J5)4 I I I = 54. (52)4 I = I = = 58 =~25 I 6. 5 . 4 +2 5'. 2 - (3. 5) . 4 . (4 X 3Y- 3'. 5 . 4' . 4 X - 3' 3'. 5 . 4 '. 4'. 4-3'. 3- = 3'-'-(-). 5'-' 4,+2-,-(-) = =3 64 = 92 Exercise.2 Simplify the following without using a calculator: 3 l.ao+a+a4 2' 4'-2 3. g;=i 7. (49r)~ 9y 4 20 _ _ 2(2) ao X a X a ( 8) (3 n + I). 2(9r 8. (3n-,...,... (2n-r I 0. ~ - - l!' (-3 -.Ji)2 2. r; 5-2,,6 7 Answers.+a+a a n - n' 8 (7X) (3y)l 8. ~(~)n' e'- 5 n2 _ I _n_ x '-n'.2.2 The solution of exponential equations.2.2, Equations of the form a = ay Ifd' = a- V -+ x = y Examples. 2.8 x = J2.'. 2.(2 3 y = 22 I... 2 lx + = x+l=2... 3x=! x= x=2x3 - = x + I _ 2 x = x + = 2'... 2x + = x... x = - 8 Common factors Example 2x+2 _ 20 = 3.2 x - :. 2x+2 _ 3.2 x - = 20 :. 2 x x 2- = 20 :. 2{2 2 - ~] = x [~ - ~ ] = 20 :. 2{~J = x = 20 x 2 5 :.2' = 2 3 :. x = The quadratic trinomial Example 52 + 5' + = 50 :. 52 + 5.5' = 50 J l + 5k - 50 = 0 :. (k + O)(k - 5) = 0 [let 5 x = k] :. k + 0 = 0 or k - 5 = 0 :.k=-lo :.k =5 :. 5' = - 0 :. 5 x = 5 There is no solution because... x = x = 0g5 (- 0) and the logarithm of a negative number does not exist. Exercise.3 Solve for x if:. 4' + I = 6 2 ' - I ' - 3 x = '+ + 3'- = + _ 22. -2 = 28 9 5. 5 l ' - I = (625)' ' - 2' - = ' + = 4 -,. 52.' + 3 5'+= - 0(4') + 6 = P'- = ' + 7. e' + I + e' - = l! + 2x, = ' - 2' - = ' + 4' = \+ I - J' = '+ - 2' = ' ' - 8 = h - 5.4' = ' - 6 Jf - 27 = 0 )'+ I 8. ( 3 = 3 a 20. (a + )2' - (a + )' + = 0 Answers ; ; ; S Logarithms.3. Exponential and logarithmic forms Exponential form Logarithmic form 00 = 0 2 log lo00 = 2 6 = 4 2 log..6 = 2 25 = 52 logs25 = 2 6 = 2 4 log2 6 = 4 0, = 0- logloo,l = - N= b l login = N = b l +-+ login = I, N 0, b * and b 0 0 The above relationship is called the definition of a logarithm. If the base is lowe need not write it down. For example, log 8,76 = 0g I0 8,76. Natural logarithms are logarithms to the base e, where e = 2,783 and they are written as 0&,8 = In 2, Calculations without using a calculator Example Calculate the logarithm of 49 to the base 7. Solution Let x = 0g, = 7' = 7'... x = 2 Examples Solve for x:. x = 0g = log., g.{3,2 x 0-4 ) = = 0g7x g362 = x Solutions. 3. x = 0g = 2' = 2'... x = 7 5 = log,l .IOOOOO=x s... los = x S... x = = 0g7x... X = X = 49 5. log..(3,2 x 0-4 ) = ,2 X 0-4 = x 5.'. 0,00032 = x 5.'. (0,2)5 = X 5 :. X = 0,2 Exercise.4 Solve for x without using a calculator if: 3. x = log..e 5. x = log~8i 3 7. x = log~ log 0,25 = 3. log,.e3 = x 3. logqx = 2 2. log) 9 = x 4. logx256 = = log x logx 25 = 3 0. logxo,008l = 4 2. log..x = Answers 5 9. 0,5 0. 0, e Logarithmic laws The following laws are very useful and they are easy to prove. Logarithmic laws were dealt with in N3. a. log,il. b = log,il + log,b 2. log,.b = log,il - log,b 3. log,il m = m log,il 4. log a = log,il 5. lo~ =-- 6. log b =- log,b log b 2 Examples Simplify: I. IOg2(8 x 32) IOg2 8 3. log,a8 I 5. 2 log log 2-2 log log 5 Solutions I. IOg2(8 x 32) = IOg2(2 3 x 2 5 ) 8 = log22 = 8og22 = 8 x I = 8 3. Let x = log_8, = CJ2)' 5. I... 2' = 22' I.'. 2x = 3... x = log2 8 = loga = IOg222 = 2og22 = 2 x I =2 IOg82 + IOg832 = IOg82 x 32 = IOg864 = IOg882 = 2og88 =2xl =2 I 2 log log 2-2 log log 5 I = log log 2' - log 82 + log 53 = log 3 2 X 2 3 X I = log 9 x 8 x 25 9 = log 000 = log lo 0 3 = 3 log lo 0 = 3() = 3 3 Examples Calculate the following without using a calculator:. log log,. ~ 3. log e. log 0 4. (e l)ln log 4 Solutions. log828 _ log log2 8 log2 27 = log223 7 log22 = 30g [law 5] 3. log e x In 0 log 0 = log e x-- [law 5] log e = log 0 I 4. Let x = (~) = log e3 6 = - log e 3 = 2() =2.'. loglx = In 6 [definition of a logarithm] e. log x.. -- = 0~6 log ē log x log 6 --I = log e log ē [law 5]. log 6 I.. log x = -I2 x log - og e log 6... log x = -- x log e- I log e 4 log 6 x-i log e., ogx = I og e... log x = - log 6... log x = log X = 6- =6 5. Let x = '. 0g,oX = 2 log 4 [definition of a logarithm].'. 0g,oX = log X = 4 2 Exercise.5 = 6 Simplify without using a calculator:. log38 2. log83 3. log2h 4. 0& ,30, ~26 6. log 20 + log log log 2 - log g2m - log3m 9. log22 - log2 6 + IOg24 0. log 0,00 + log , II. log2(log38) _ log 2 + log 27-3 log 6 + log In~ lnlo 9. In 0,00 6. en 4 Solve for x: 20. lo~ + lo~(x + I) = 2. 2 log2(x - I) = 0gA 22. log3( - x) - log3(x - I) = log~ log x + log 2 - log (3x + 2) = X + 084(X + 3) - I = Solve for x and y if: 2' = 8 2y - and 9 x = 27 Y 26. Solve for x if: 3 log~ + - = 4 o~ 27. Solve for x if: log2 64 = -2 5 Answers ; ; ± 3 I 24. ; (; j) ; Logarithms and anti-logarithms using a calculator Examples Calculate the value ofx if:. x = log 0,46 4. x = log (0g,.5) 7. 0g,x = 4,2 2. x = In 42,6 5. log x = 0,06 3. x = 0g,.526,3 6. log x = -3,2 Solutions. x = -0,38 2. x = 3,74 3. x = 6, x = 0, x =, x = 0,00 7. x = 66,686 press D [!] IlogI press [!] ~ D ~ press rn D Q] ~ press [!] rn [ ] IlogI press D [Q] It YI press Q] D rn [!] 0 Y press [!] D ~ ~.3.5 Changing the base' Examples Calculate the following:. 0g g 3 e 4. 0~25-0g,8 3. 0g 2 s3,7 6 Solutions. 3. log27 log07 = log lo 2 0,845 = 0,30 = 2,807 log2.83,7 In 3,7 = In 2,8,308 =,030 =,27 [law 5] [taw 5] gJe = 0g..3 [law 6] =,099 = 0,90 0~25 - log8 log 25 log 8 = log 4 - log 7,398,255 = 0,602-0,845 = 2,322 -,485 = 0,836 Exercise.6 Calculate the following:. log29 4. log30, &.s42, log43 - log '0,'0 - i og / 4. log~25 + log & &2 5. 0g..3,5 6. 0gJ.se 8. log,, ' 9. log2.48,3. 2 0gIOe - 3 0g log2jr + 0gn2 Answers. 3, , , , , ,825 5., , , , , , , , Solution of equations
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