REVISION MANUAL MATHEMATICS FET 2017

MATHEMATICS FET REVISION MANUAL GRADE 12 2017

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https://ultradeep.co.za 2 TABLE OF CONTENT No. TOPIC S PAGE 1  Introduction 3 2  Algebra 4 5 3  Sequences and Series and number patterns 4  Functions and Inverse functions 6 5  Calculus 8 6  Finance and Growth 10 7  Probability 10 8  Co-ordinate Geometry 11 9  Trigonometry 12 10  STATISTICS 14 11  Euclidean Geometry 16 12  Guidelines for Mathematics Grade 12 19 13  Worked out Solutions 20 https://tvetpapers.co.za

https://ultradeep.co.za 3 INTRODUCTION The purpose of this guide is to assist grade 12 learners for their 2017 end of year examination preparation. The activities in this document are meant to be a guide and not to replace any study material/ text book. To assist learners with practice model activities are provided as well as model answers. The self study guide focuses on selected challenging content topics informed by personal experience, analysis of 2016 grade 12 results and the examiner’s reports. At the end of this guide there is an indication of how marks are allocated per topic /theme. This will guide you on the amount of effort to put on each topic. https://tvetpapers.co.za

https://ultradeep.co.za 4 PAPER 1 2. GENERAL ALGEBRA Important Information to Remember This section is always question 1 in the final paper and is weighting  25 marks. This section includes factorisation and using quadratic formula to solve for x, inequalities and simultaneous equations. Solving quadratic equations by completing the square will not be examined. Solving quadratic equations using the substitution method (k-method) is examinable. Equations involving surds that lead to a quadratic equation are examinable. Solution of non-quadratic inequalities should be seen in the context of functions. Nature of the roots will be tested intuitively with the solution of quadratic equations and in all the prescribed functions. Examples: Solve for x in each of the following: 1. 2x2 + 10x =  12 2. (x  5)(x + 2) = 0 3. x(2x + 3) = 7 4. (3x  2)2  3x 5. Solve for x and y simultaneously: 2x  y = 3 x2 + 5xy + y2 = 15 3 6. x 2  2x  2 x  2x 2 4 x 1 7. Simplify, without the use of a calculator: 3 48  2 x 2 8. Given: f(x) = 3(x  1)2 + 5 and g(x) = 3 8.1 Is it possible for f(x) = g(x)? Give a reason for your answer. 8.2 Determine the value(s) of k for which f(x) = g(x) has TWO unequal roots. https://tvetpapers.co.za

https://ultradeep.co.za 5 3. NUMBER PATTERNS, SEQUENCES AND SERIES Important Information to Remember This section is always question 2 and 3 in the final paper and is weighting  25 marks. The sequence of first differences of a quadratic number pattern is linear. Therefore, knowledge of linear patterns can be tested in the context of quadratic number patterns. Recursive patterns will not be examined explicitly. Links must be clearly established between patterns done in earlier grades. EXAMPLES: 2.1 3; 9; 17; 27; ... is a quadratic sequence. 2.1.1 Write down the next term. 2.1.2 Determine an expression for the general term. 2.1.3 What is the value of the first term smaller than 269? 2.2 The sum of the first n terms of a series is given by Sn = 2n+2  4 2.2.1 Determine the first term. 2.2.2 Determine the twenty-fourth term. n 2.3 Prove that Sn = a + (a + d) + (a + 2d) +… (n terms) = [2a + (n – 1) d] 2 2.4 Consider the following sequence numbers: 1 1 1 3; ; 6; ; 9; ; 2 2 2 2.4.1 Determine the next two terms. 2.4.2 Determine the sum to 30 terms. 2.5 Determine: 15  3(2) r 0 r 1 2.6 For which value(s) of x will the series (1 x) + (1 x)2 + (1 x )3 + … converge? n 2.7 Calculate the value of n if  2  2046 k 1 k 2.8 The sum of the fifth and the ninth terms of an arithmetic sequence is 6, while the twenty-fifth term is –24. Calculate the sum of the first 50 terms. https://tvetpapers.co.za

https://ultradeep.co.za 6 4. FUNCTIONS AND INVERSE FUNCTIONS Important Information to Remember This section is always question 1 in the final paper and is weighting  35 marks. Candidates must be able to use and interpret functional notation. In the teaching process learners must be able to understand how f(x) has been transformed to generate f(x), f(x), f(x + a), f(x) + a, a f(x) and x= f(y) where aR. Trigonometric functions will ONLY be examined in Paper 2. EXAMPLES: 4 3. Given: f(x) = x3 3.1 What kind of graph is represented by f? 3.2 Write down the asymptotes of f. 3.3 Use the attached diagram sheet and sketch the graph of f clearly indicating the asymptotes and intercepts with the axes. 3.4 Write down the domain and range of f. 3.5 What will be the domain of f (x – 2)? 3.6 What is the equation of h(x) if h is the reflection of f in the x-axis? 4. 1 2 4.1 Write g(x) = x  x  3 in the form g ( x)  a( x  p) 2  q by completing the square. 4 4.2 Hence or otherwise, write down the coordinates of the turning point of g (x) 4.3 Use the attached diagram sheet and sketch g (x) on a system of axes. Clearly indicate the turning point, intercepts on the axis as well as the coordinates of any other point on the graph. 4.4 Write down the range of g (x) 4.5 Write down the equation of the new graph if g (x) is shifted 2 units to the right and 1 unit down. 4.6 Write down the equation of the axes of symmetry of the new graph. https://tvetpapers.co.za

https://ultradeep.co.za 7 5. 1 5.1 The graph of f(x) = x(x + 3) and g ( x )   x  2 is shown below: 2 y 4 L 3 2 1 -5 -4 -3 -2 -1 O 1 2 3 4 x -1 M -2 P -3 5.1.1 Determine the value(s) of x for which f(x) = 0. -4 5.1.2 Determine the coordinates -5 of P, the turning point of f. 5.1.3 Write down the turning point of f(x)+2 5.1.4 Hence, determine the range of f(x)+2 5.1.5 Determine the average gradient between x = 1 and x = 1 on curve f. 5.1.6 Determine the maximum length of LM where LM is perpendicular to the x-axis. https://tvetpapers.co.za

https://ultradeep.co.za 8 k 6. The graphs of y  a x and y   c are shown below. x 6.1 Determine the values of a, c and k 6.2 Write down the equation of the asymptote of y  a x  2. 5. DIFFERENTIAL CALCULUS Important Information to Remember dy The following notations for differentiation can be used: f ( x), Dx , or y  dx In respect of cubic functions, candidates are expected to be able to:  Determine the equation of a cubic function from a given graph  Discuss the nature of stationary points including local maximum, local minimum and points of inflection.  Apply knowledge of transformations on a given function to obtain its image. Candidates are expected to be able to draw and interpret the graph of the derivative of a function. Surface area and volume will be examined in the context of optimisation. Candidates must know the formulae for the surface area and volume of the right prisms. These formulae will not be provided on the formula sheet If the optimisation question is based on the surface area and/or volume of the cone, sphere and/or pyramid, a list of the relevant formulae will be provided in that question. Candidates will be expected to select the correct formula from this list. https://tvetpapers.co.za

https://ultradeep.co.za 9 EXAMPLES: 1 7.1 If f ( x)  determine the derivative by using first principles. x 1 7.2 Determine the average gradient of f ( x)  between x =  1 and x = 2 x 7.3 Determine the1 following (Write your answers with positive exponents) 1 1 7.3.1 Dm [m  3 m  3m ] 3 dy x3  x2 7.3.2 if dx x2 8. The graph represents the functions f and g with f ( x)  ax 3  cx  2 and g ( x)  x  2 A and (1; 0) is the x-intercept of f. y The graphs of f and g intersect at A and C. (-1; 0) O A x -2 f C g B 8.1 Determine the coordinates of A. 8.2 Show by calculation that a = 1 and c = 3 8.3 Determine the coordinates of B, the turning point of f. 8.4 Show that line BC is parallel to the x-axis. 8.5 Find the x-coordinate of the point of inflection of f. 8.6 Write down the values of k for which f(x) = k will have only ONE root. 8.7 Write down the values of x for which f (x)  0 https://tvetpapers.co.za

https://ultradeep.co.za 10 6. FINANCE, GROWTH AND DECAY Important Information to Remember Understand the difference between nominal and effective interest rates and convert fluently between them for the following compounding periods: monthly, quarterly and half-yearly or semi-annually. With the exception of calculating for i in the Fv and Pv formulae, candidates are expected to calculate the value of any of the other variables. Pyramid schemes will not be examined in the examination EXAMPLES: 9.1 After how many years will a vehicle originally valued at R980 000 decreases to R147 000 if the rate of depreciation is 10 % p.a. on a reducing balance. 9.2 You buy a car for R180 000 and pay a deposit of 15% The balance is paid using a bank loan. Interest is charged at 13,9% p.a. compounded monthly over 5 years. 9.2.1 Determine the monthly repayments. 9.2.2 Determine the outstanding balance after 36 months. 7. PROBABILITY Important Information to Remember Dependent events are examinable but conditional probabilities are not part of the syllabus. Dependent events in which an object is not replaced is examinable. Counting principles, where the arrangement is circular and/or using combinations, is not in the spirit of the curriculum. In respect of word arrangements, letters that are repeated in the word can be treated as the same (indistinguishable) or different (distinguishable). The question will be specific in this regard. EXAMPLES: 10.1 Events A and B are mutually exclusive. It is given that: • P(B) = 2P(A) • P(A or B) = 0,57 Calculate P(B). 10.2 Two identical bags are filled with balls. Bag A contains 3 pink and 2 yellow balls. Bag B contains 5 pink and 4 yellow balls. It is equally likely that Bag A or Bag B is chosen. Each ball has an equal chance of being chosen from the bag. A bag is chosen at random and a ball is then chosen at random from the bag. https://tvetpapers.co.za

https://ultradeep.co.za 11 10.2.1 Represent the information by means of a tree diagram. Clearly indicate the probability associated with each branch of the tree diagram and write down all the outcomes. 10.2.2 What is the probability that a yellow ball will be chosen from Bag A? 10.2.3 What is the probability that a pink ball will be chosen? PAPER 2 8. CO-ORDINATE GEOMETRY Important Information to Remember Prove the properties of polygons by using analytical methods. The concept of collinearity must be understood. Candidates are expected to be able to integrate Euclidean Geometry axioms and theorems into Analytical Geometry problems. The length of a tangent from a point outside the circle should be calculated. Concepts involved with concurrency will not be examined. y D EXAMPLES: 1. In the diagram. A (–4; 5), A(-4; 5) C (–1; – 4) and B ( 4; 1) are the vertices of a triangle in a E Cartesian plane. B(4;1) x CE  AB with E on AB. 0 E is the midpoint of straight line CD. 1.1 Determine the gradient of AB. 1.2 Determine the equation of CD. C(-1; -4) 1.3 Show that the coordinates of E is ( 2 ; 2). 1.4 Determine the coordinates of D. 1.5 Determine the equation of the straight line parallel to AC and passing through D. 1.6 Calculate the values of p if the coordinates of the point S is ( 5 ; p ) and AS = 106 . https://tvetpapers.co.za

https://ultradeep.co.za 12 2.1 In the diagram, the circle with centre C and with equation x2 – 6x + y2 – 4y = 12 cuts the y-axis at A. y BA is a tangent to the circle at A.  A B lies on the x-axis. B AC   and  2 m AC   3 .C 2.1.1 Show that the coordinates of C is ( 3 ; 2 ) O x B 2.1.2 Show that the coordinates of A is ( 0 ; 6 ). 2.1.3 Determine the equation of AB. 2.1.4 Determine the size of , round off to ONE decimal digit. 2.2 Calculate the distance between the centres of the two circles with equations given below: ( x + 4 )2 + ( y + 2 )2 = 58 ( x – 4 )2 + ( y – 6 )2 = 26 2.3 Show that the circles described in Question 5.2 intersect each other. 9. TRIGONOMETRY AND TRIG GRAPHS Important Information to Remember You should be able to: The reciprocal ratios cosec θ, sec θ and cot θ can be used by candidates in the answering of problems but will not be explicitly tested. The focus of trigonometric graphs is on the relationships, simplification and determining points of intersection by solving equations, although characteristics of the graphs should not be excluded. EXAMPLES: 3. 3.1 If 6 cos  = –5 , 0 ≤ sin and 0 ≤  ≤ 360 , Without using a calculator, calculate the value of 100 tan2  + 72 sin2  3.2 Simplify without using a calculator: 1 cos 750.  tan(360   ).cos(180   ).cos(90   ) cos(30) https://tvetpapers.co.za

https://ultradeep.co.za 13 3.3 If sin 31 = p , express the following in terms of p: 3.3.1 sin 391 3.3.2 cos 31 3.3.3 cos 59 sin x cos x 1 3.4  tan x 1  sin x  cos x 2 2 2 4. 4.1 On the same set of axes draw graphs of f(x) = sin ( x – 30 ) and g(x) = cos 2x for –180 ≤ x ≤ 180 , clearly indicate all intercepts with the axes and the turning points. (8) 4.2 If f(x) = g(x), 4.2.1 find the general solution 4.2.2 solve x for –180 ≤ x ≤ 180 4.3 For which values of x in the interval [–180 ; 0 ] will f(x) ≥ g(x)? 4.4 Consider the graph of f. If the y-axis is shifted 60 to the right, write down the equation of the graph of h which results from the shifting of the y-axis (that is, the new graph of f ). P 5. 5.1 The diagram alongside shows a top view of a part building. It is given that DR = ER = 43,4 m and DE = 79,8 m. Determine, rounded to one decimal digit, Q T  5.1.1 the size of D R E  5.1.2 the area of DRE R https://tvetpapers.co.za

https://ultradeep.co.za 14 5.2 In the figure, Q , T and R are points in the horizontal plane such that P TQ = TR = y; PQ = PR and TP represents a vertical pole positioned at T. Q T   If the angle of elevation of P from Q is  and P R Q = , R 5.2.1 express PQ in terms of y and a trigonometric ratio of  .  5.2.2 express Q P R in terms of  5.2.3 Hence, show that y sin 2  QR  cos  .sin  10. STATISTICS Important Information to Remember Candidates should be encouraged to use the calculator to calculate standard deviation, variance and the equation of the least squares regression line. The interpretation of standard deviation in terms of normal distribution is not examinable. Candidates are expected to identify outliers intuitively in both the scatter plot as well as the box and whisker diagram. In the case of the box and whisker diagram, observations that lie outside the interval (lower quartile – 1,5 IQR ; upper quartile + 1,5 IQR) are considered to be outliers. However, candidates will not be penalised if they did not make use of this formula in identifying outliers. https://tvetpapers.co.za

https://ultradeep.co.za 15 EXAMPLES: 6. A local nursery records the effect of temperature on the growth of a new plant that has recently been imported into the country. The goal of the study is to determine what temperature is ideal for maximum flowering for a particular plant. Temperature (degrees Celsius) Number of flowers 25C 1 26C 3 27C 4 28C 6 29C 7 30C 7 31C 8 32C 9 33C 10 34C 14 6.1 Draw a scatter plot on the diagram sheet to represent this data. 6.2 Describe the trends shown by this scatter plot. 6.3 Draw the line of best fit. 6.4 Predict the number of flowers if the temperature is 38C. 7. The distance travelled (in kilometres) by 10 learners to school each day were recorded as follows: 7 8 10 8 26 14 7 5 6 7 7.1 Calculate the standard deviation 7.2 Draw a box and whisker diagram for the data on the diagram sheet. 7.3 How many learners travelled distances that are outside one standard deviation? 7.4 Will the mean or the median be the most appropriate central value to use? Motivate your answer. https://tvetpapers.co.za

https://ultradeep.co.za 16 8. The following data represents the percentage of family income allocated to groceries for a sample of 50 shoppers. Percentage Frequency Class midpoint (xi) Cumulative f f.xi frequency 16 – 25 5 26 – 35 15 36 – 45 17 46 – 55 10 56 – 65 3 8.1 Complete the table on the diagram sheet. 8.2 Determine the mean of the data. 8.3 Draw an ogive of the data on the diagram sheet. 8.4 Use the graph to determine how many families use less than 50% of their income for groceries. 11. EUCLIDEAN GEOMETRY Important Information to Remember Measurement can be tested in the context of optimisation in calculus. Composite shapes could be formed by combining a maximum of TWO of the stated shapes. The following proofs of theorems are examinable: • The line drawn from the centre of a circle perpendicular to a chord bisects the chord; • The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circle (on the same side of the chord as the centre); • The opposite angles of a cyclic quadrilateral are supplementary; • The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment; • that a line drawn parallel to one side of a triangle divides the other two sides proportionally; • equiangular triangles are similar. Corollaries derived from the theorems and axioms are necessary in solving riders: • Angles in a semi-circle • Equal chords subtend equal angles at the circumference https://tvetpapers.co.za

https://ultradeep.co.za 17 • Equal chords subtend equal angles at the centre • In equal circles, equal chords subtend equal angles at the circumference • In equal circles, equal chords subtend equal angles at the centre. • The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle of the quadrilateral. • If the exterior angle of a quadrilateral is equal to the interior opposite angle of the quadrilateral, then the quadrilateral is cyclic. • Tangents drawn from a common point outside the circle are equal in length. EXAMPLES: E 9. O is the centre; BP = OB = AO; PTR is a tangent and EP  AP. Prove that: R 9.1 TEPB is a cyclic quadrilateral T 9.2  ATB ///  APE 9.3 TP = PE A  P O B 9.4  ATB is similar to  EPB 9.5 2BP2 = BE.TB 10. In the diagram PQRS is a parallelogram. W Side RS is produced to W. WQ intersect PS in X. M is a point on XQ so that MX = XW. P X Given MT // XS, PQ = 12 cm, WS = 4 cm. S M T 10.1 Determine the length of TR XM 10.2 Determine the value of Q R XQ https://tvetpapers.co.za

https://ultradeep.co.za 18 11. Two circles touch each other at point A. The smaller circle passes through O, the centre of the larger circle. A, D, B and C are points on the circumference of the larger circle. OE // CA. C 2 1 A 1 11.1 Prove, with reasons, that AE = BE 2 F 3 11.2 Prove that  AED ///  CEB. 1 2 Hence, or otherwise, show that AE 2  DE  CE 2 3 4 11.3 O 3 1 5 11.4 If AE.EB = EF.EC, show that E is the midpoint of DF E D B https://tvetpapers.co.za

https://ultradeep.co.za 19 9. GUIDELINES FOR MATHEMATICS GRADE 12 QUESTION PAPERS CONTENT P1 MARKS PATTERNS AND SEQUENCES (with 1st difference constant; with 2nd 25±3 difference constant; Arithmetic and Geometric sequence; Correctly interpret Sigma notation; Sum of series formula). ANNUITIES AND FINANCE (Straight depreciation and on a reducing 15±3 balance; finding the value of n on a compound decay formula; Solving annuity, bond repayment and sinking fund problems) FUNCTIONS AND GRAPHS (Straight line; Parabola; Exponential; 35±3 Hyperbola excluding Trigonometric functions) ALGEBRA AND EQUATIONS ( Question 1; Solving equations; 25±3 Simultaneous equation; Log and Exponential equations) CALCULUS 35±3 PROBABILITY 15±3 CONTENT P2 CO-ORDINATE GEOMETRY 40±3 DATA HANDLING( From grade 10 and 11 only; Measures of central 20±3 tendencies; Measures of dispersions; Bar and Compound bar graph; Pie chart; Five number summary; Box and Whisker diagram; Ogives; Variance and Standard deviation; Graphical representation using Histograms and Frequency polygons; Scatter plot; correlation and Regression line) EUCLIDEAN GEOMETRY 50±3 TRIGONOMETRY 40±3 https://tvetpapers.co.za

https://ultradeep.co.za 20 WORKED OUT SOLUTIONS 5. 2x  y = 3 …………………….(1) x2 + 5xy + y2 = 15……………...(2) PAPER 1 from (1): y = 2x  3 x2 + 5x(2x  3) + (2x  3)2 = 15 QUESTION 1 GENERAL x2 + 10x2  15x + 4x2  12x + 9 15 = 0 ALGEBRA 15x2  27x  6 = 0 5x2  9x  2 = 0 1. 2x2 + 10x =  12 (5x + 1)(x  2) = 0 1 2x2 + 10x + 12 = 0 x =  or x = 2 5 x2 + 5x + 6 = 0 1 y = 2( )  3 y = 2(2)  3 5 (x + 3)(x + 2) = 0 y=1 17 x = 3 or x = 2 y=  5 3 2. (x  5)(x + 2) = 0 6. x 2  2 x  2 x = 5 or x =  2) x  2x 2 Let x 2  2 x  k 3  k 2 3. x(2x + 3) = 7 k k 2  2k  3  0 2x2 + 3x  7 = 0 (k  3)(k  1)  0 k  3 or k  1  (3)  (3) 2  4(2)(7) x subst. 2(2)  x 2  2 x  3 or x 2  2 x  1 x 2  2 x  3  0 or x 2  2 x  1  0  (3)  65  ( x  3)( x  1)  0 or ( x  1)( x  1)  0 4 x  3 or x  1 or x  1 x = 1, 27 or x =  2, 77 4 x 1 4. (3x  2)2  3x 7. 3 48  22x 9x2  12x + 4  3x  0 2 2 x2 = 34 3  22x 9x2  15x + 4  0 2 2 x.4 = 4 9  2x   2 (3x  1)(3x  4)  0 1 4 = 12  4 3 3 =8 1 4 x (; ) or ( ; ) 3 3 https://tvetpapers.co.za

https://ultradeep.co.za 21 8. 2.1.2 2.1.2 The second difference is constant 8.1 No, there will be no intersection between the  Tn is quadratic graphs. Min value of is 3(x  1)2 + 5 is 5  Tn  an 2  bn  c 8.2 3(x  1)2 + 5 = 3 + k 2a = 2 3(x  1)2 = k 2 k 2 > 0 for all real values of k a=1 k >2 3a + b = 6 OR 3(1) + b = 6 3x 6x + 3 + 5 = 3 + k 2 3x2  6x + 5  k = 0 b=3  = (6)2  4(3)(5  k) a+b+c=3 = 36  60 + 12k = 12k  24 1+3+c=3 For real unequal roots 12k  24 > 0 c=1 12k > 24 k >2  Tn  n 2  3n  1 2.1.3 2.1.3 n 2  3n  1  269 QUESTION 2: NUMBER PATTERNS, n 2  3n  270  0 SEQUENCES AND   SERIES ( n + 18)(n  15)  0 -18 15 2.1  14th term T14  (14) 2  3(14)  1 3 9 17 27 39 = 196 + 42  1 = 237 6 8 10 2.2 2 2 2.1.1 39 2.2.1 Sn = 2n+2  4 T1 = 21+2  4 =4 https://tvetpapers.co.za

https://ultradeep.co.za 22 2.2.2 T24 = S24  S23 15 = (2 24+2 4) (2 23+2 4) 2.5  3(2) r 1 r 0 = 67108860  33554428 = [ 3(2) 01 ]  [3(2)11 ]  [3(2) 21  ........ = 33554432 3 =  3  (6)  ............. 2.3 Sn = a + (a + d ) + …………..+ (l  d ) + l 2 Sn = l + (l  d ) + …………….+(a + d ) + a 3 a= r = 2 n = 1 2 2Sn = (a + l) + (a + l ) + ….+(a + l ) + (a + l) a(r n  1) = n (a + l) Sn = r 1 n Sn = [a+l] 3 2 [(2)16  1] S16 = 2 But l = a + (n 1)d  2 1 = 32767,5 n Sn = [ a + a + (n 1)d ] 2 2.6 Converge when 1 r  1 n Sn = [ 2a + (n 1)d ] 2 (1  x) 2 r = (1  x) 2.4 1 2.4.1 12; = (1 x) 2  1 1 x  1 1 1 2.4.2 + + ……………..(15th term) 2 2 2   x  0 1  0x2 =7 2 3 + 6 + 9 + ……….(15th term) 15 S15 = [ 2(3) + (15 1)3 ] 2 = 360 1 7 + 360 = 367,5 2 https://tvetpapers.co.za

https://ultradeep.co.za 23 n QUESTION 3: FUNCTIONS AND 2.7  2 k  21  2 2  23  ....  2 n  2046 k 1 INVERSE FUNCTIONS a r n  1)  Sn  r 1 22 n  1 3.1 Hyperbola 2046  2 1 3.2 x = 3, y=0 1023  2  1 n 3.3 y 1024  2 n 210  2 n OR n  log 2 1024 n  10 n  10 y=0 0 x 4  3 x=3 2.8 T5 + T9 = 6 a + 4d + a + 8d = 6 a + 6d = 3 . . . (1) 3.4 Domain: x R, x  3 T25 = - 24 Range: y  R, y  0 a + 24d = - 24 . . .(2) (2)-(1) : 18d = -27 3.5 x R, x  5 3 d=  4 2 3.6 y = x3 a = 12 4 50   3  y= x3 S50=  2(12)  49    2  2  4. = –1 237,5 1 2 4.1 g(x) = x  x3 4 1 2 = ( x  4 x  12) 4 1 2 = ( x  4 x  4  4  12) 4 1 = [( x  2) 2  16] 4 1 = ( x  2) 2  4 4 https://tvetpapers.co.za

https://ultradeep.co.za 24 4.2 TP( 2; 4) 4.3 y OR b x =  2a  6 3 0 x =  -2 2(1) 3 =  -3  (4; -3) 2  3 2 3 y = ( ) + 3(  ) (2; -4) 2 2 9 4.4 y  [4 ; ) =  4 1 4.5 y ( x  4) 2  5 3 9 4 P(  ; ) 2 4 4.6 x=4 3 1 5.1.3 (  ; ) 2 4 5. 1 5.1.4 [ ;) 4 5.1.1 0 = x(x + 3) OR x=0 or x = 3 1 y  ;yR 4 5.1.2 5.1.5 f(1) = 1(1 + 3) f ( x )  x 2  3x = -2 9 9  x 2  3x   f(1) = 1(1 + 3) 4 4 2  3 9 =4  x     2 4 4  ( 2) Ave. gradient =  3 9 1  ( 1)  P  :    2 4 = 3 https://tvetpapers.co.za

https://ultradeep.co.za 25 1 QUESTION 7: DIFFERENTIAL 5.1.6 LM = (  x  2)  ( x 2  3x) 2 CALCULUS 7 1 1 =  x2  x2  2 7.1 f (x)  lim x  h x h 0 h 7  2x  0 b 2 x  ( x  h) x =  OR 2a 7 x ( x  h) x  lim h0 h 4 x  ( x  h) 7  lim h h  0 x ( x  h) ( ) = 2 h 1 2(1)  lim  h  0 x ( x  h) h 7 1 =  lim h  0 x ( x  h) 4 1 7 2 7 7  2 y = ( )  ( )2 x 4 2 4 1 81 7.2 (1; 1) (2; ) = 2 16 1 81 1  maximum length of LM is units av. gr. = 2 16 2  (1) 6. 1 = 6.1 y= ax 6 4 = a2 7.3 1 a=2 1 1 7.3.1 Dm [m  3 m  3m ] 3 c=1 2 1 3 1 k = m   3m 2 y = 1 3 3 x 1 1 3 k =   2 4= 1 2 3 m 2 3m 3 k=6 6.2 y = -2 https://tvetpapers.co.za

https://ultradeep.co.za 26 x3 x2 = 4 7.3.2 y = 2  2 x x B(1; 4) =x1 8.4 f ( x)  x 3  3x  2 and y  x  2 dy  1 x 3  3x  2  x  2 dx 8. x 3  4x  0 8.1 g(x) = 0 x( x 2  4)  0 x–2=0 x( x  2)( x  2)  0 x = 2 x  0 or x = 2 or x = 2 A(2; 0) NA NA yc = 2 2 8.2 f ( x)  a( x  x1 )( x  x2 )( x  x3 ) =4  a( x  1)( x  1)( x  2) (0; 2) C(2; 4) and B (1; 4) 2 = a (1)(1)( 2) Eq. of BC: y =  4 2 = 2a Eq. of x – axis: y = 0 a=1  BC  x – axis f ( x)  1( x 2  2 x  1)( x  2) 8.5 f (x) = 0  x 3  2x 2  x  2x 2  4x  2 6x = 0  x 3  3x  2 x=0  a = 1 and c = 3 8.6 x 3  3x  2  k x 3  3x  2  k  0 8.3 f ( x)  x 3  3 x  2 y-int.:  2 k f (x)  3x 2  3   2 k   2 0  x2 1 k 0 (x  1)(x +1) = 0 or  2 k  2 x = 1 or x = 1 4k f (1)  (1)  3(1)  2 3 https://tvetpapers.co.za

https://ultradeep.co.za 27 8.7 f (x)  0 QUESTION 10: PROBABILITY  f decreasing 10.1 P(A or B) = P(A) + P(B) 0,57 = P(A) + 2P(A)  x (1; 1) 0,57 = 3P(A) P(A) = 0,19 QUESTION 9: FINANCE, GROWTH ∴ P(B) = 2(0,19) AND DECAY = 0,38 9.1 A  P(1  i ) n 3 10.2 AP 147000  980000(1  i ) n 5 P 10.2.1 1 0,15  (0,90) n A 2 2 Y AY n  log 0,9 0,15 5 5 n  18 years 9 P BP 1 B 9.2 2 4 15 Y BY 9.2.1  180000  R27000 9 100 180000  27000 = 153 000 x[1  (1  i)  n ] 1 2 P 10.2.2 P(AY) = ( )( ) i 2 5 1 = 0,139 60 5 x[1  (1  ) ] 153000  12 0,139 1 3 1 5 10.2.3 P(P) = ( )( )  ( )( ) 12 2 5 2 9 3 5 = x(43,0729.....) =  10 18 26 x = R3552,12 = 45 9.2.2 Outstanding balance x[(1  i) n  1]  A(1  i) n  i 0,139 36 3552,12[(1  )  1] 0,139 36 12  153000(1  )  12 0,139 12 = 231606,8416  157551,6627 = R74055,18 https://tvetpapers.co.za

https://ultradeep.co.za 28 1.4 PAPER 2 Dx  C x Dy  C y CO-ORDINATE GEOMETRY Ex  Ey  2 2 y y 1.1 A(– 4;5) ; B(4;1), m AB  2 1 x  (1) y  (4) x x 2 1 2 2 1 5 2 2  4  ( 4) x 1  4 y4 4   4 1  x5 y 8 8 2 D(5; 8) 1.2 mCD = 2, CE AB C(–1; –4) y – y1 = x – x1 1.5 A(-4;5) and C(-1;-4) y – (–4) = 2( x – (–1)) y + 4 = 2( x + 1 ) y y m AC  2 1 y = 2x + 2 – 4 x x 2 1 y = 2x – 2 45   1  ( 4) m AB   1 9 1.3 A( -4;5)  2 3 1 y  5   ( x  (4))  3 2 1 Line//AC mline = –3, D(5;8) y 5   x 2 2 1 y –y1 = x – x1 y   x3 2 y – 8 = –3( x – 5 ) 1  x  3  2x  2 2 y = –3x + 15 + 8 5 x5 2 y = –3x + 23 x=2  y = 2(2) – 2 = 2 E(2; 2) 1.6 A(-4 ; 5 ) S(5; p) AS  ( p  5) 2  (5  (4)) 2 B( 4;1)  p 2  10 p  25  81 1 m AB   2  p 2  10 p  106 1 y  1   ( x  4)  2 p 2  10 p  106  106 1 y 1   x  2 p 2  10 p  106 106 2 1 y   x3 p2 – 10p + 106 = 106 2 p2 – 10p = 0 p(p – 10 ) = 0 p = 0 or p = 10 https://tvetpapers.co.za

https://ultradeep.co.za 29 2. 2.2 Centre of circles: (-4; -2) and (4; 6) 2.1 Distance between centres = 2.1.1 2 2 x – 6x + y – 4y = 12 (4  4) 2  (6  2) 2 x2 – 6x + 9 + y2 – 4y + 4 = 12 + 9 + 4 = 11,31 ( x – 3 )2 + ( y – 2 )2 = 25 C( 3 ; 2 ) 2.3 Sum of radii = 58  26  12,71 Distance between centres is 11,31 2.1.2 y- intercept  x = 0 Sum of the radii > distance between the centres (0)2 – 6(0) + y2 – 4y = 12  the circles intersect as they overlap y2 – 4y – 12 = 0 ( y – 6 )( y + 2 ) = 0 y = 6 or y = –2 TRIGONOMETRIC A( 0 ; 6 ) FUNCTIONS AND GRAPHS 2.1.3 2 m AC   3 mAB  mAC  1........AC  AB 3. 3  m AB  3.1 2 A(0;6) y  y  m( x  x ) 1 1 3 y  6  ( x  0) 2 3 y  x6 2 OR 6 cos  = –5 y = mx + c 5  cos   6 3 6 (0)  c y = 6 – (–5)2 2 2 2  y  11 c=6 100 tan2  + 72 sin2  2 2 3  11   11  y  x6  100   72  2  5   6   11   11  2.1.4 tan ABO= mAB = 3/2  100   72  ABO = 56,3  25   36   = 90+ ABO………. ext. Angle ∆ABO =44 + 22 = 66 = 146,3 https://tvetpapers.co.za

https://ultradeep.co.za 30 3.2 4. 1 4.1 cos 750.  tan(360   ).cos(180   ).cos(90   ) cos(30) 1  cos(2.360  30).  ( tan  ).( cos  ).( sin  ) cos 30 1 sin   cos 30.  .(cos  ).(sin  ) cos 30 cos   1  sin 2   cos 2  3.3 3.3.1 sin 391 = sin ( 360 + 31 ) = sin 31 4.2 =p 4.2.1 sin( x - 30) = cos 2x = sin( 90 - 2x) x - 30 = 90 – 2x + n.360 3x = 120 + n.360 3.3.2 cos2 31 + sin2 31 = 1 x = 40 + n.120 ; n Є Z cos2 31 = 1 – p2  cos 31  1  p 2 OR x - 30 = 180 – (90 – 2x) + n.360 3.3.3 cos 59 = sin 31 x – 30 = 180 - 90 + 2x + n.360 =p –x = 120 + n. 360 3.4 x = –120 – n. 360 ; n Є Z sin x cos x 4.2.2 x = –120 ; –80 ; 40; 160 LHS  1  sin 2 x  cos 2 x  sin x cos x 4.3 –120 ≤ x ≤ –80 cos 2 x  cos 2 x sin x cos x  4.4 y = sin( x + 60 - 30 ) = sin ( x + 30) 2 cos 2 x sin x  5. 2 cos x 1  tan x 5.1.1 DE2 =DR2 + ER2 – 2DRERcos DRˆ E 2 79,82 = 43,42 + 43,42 – 243,443,4.cos DRˆ E 6368,04 - 3767,12 = –3767,12cos DRˆ E -2600,92 = –3767,12cos DRˆ E cos DRˆ E = -0,69042… ref angle = 46,3 https://tvetpapers.co.za

31 DRˆ E = 180 – 46,3 STATISTICS = 133,7 5.1.2 Area of DRE = 0,5 DRERsin DRˆ E = 0,5 43,443,4sin133,7 6. = 680,9 m2 6.1 5.2 5.2.1 cos  y PQ y PQ  cos 5.2.2 In ∆PQR, PQ = PR PQˆ R  PRˆ Q   PQˆ R  PRˆ Q  QRˆ P  180 0 QRˆ P = 180– 2 5.2.3 QR PQ  sin QPR sin QRˆ P ˆ y sin(180  2 ). QR  cos  6.2 As the temperature increases, the number of sin  flowers increases. y sin 2 . QR  cos  6.3 On graph ( y = 1,19x – 28 ) sin  y sin 2 QR PQ 6.4 17 flower QR   cos  . sin  sin QPˆ R sin QRˆ P y sin(180  2 ). QR  cos  7. sin  y 7.1 Standard deviation = 5,90 (with calculator) sin 2 . QR  sin  cos  4 .7.2 . .8 10. 12 14 16 18 20 22 24 26. 2 6 y sin 2 QR  cos  . sin  7.3 Mean = 9,8 One standard deviation: (9,8 – 5,9; 9,8 + 5,9) = (3,9; 15,7)  one learner (26km)

32 7.4 Median. The mean is influenced by the outlier. 9. 9.1 ATB  90 0 Subtended by diameter AB ATB  90 0  EPB 8. TEPB is a cyclic quadrilateral 8.1 Exterior angle = opp interior angle 9.2 In ATB and APE Percentage F (xi) Cumulative f.xi frequency A  A Common 16 – 25 5 20,5 102,5 5 26 – 35 15 30,5 457,5 20 ATB  APE  90 0 36 – 45 17 40,5 688,5 37 46 – 55 10 50,5 505 47 ΔATB /// ΔAPE they are 56 – 65 3 60,5 181,5 50 equiangular 9.3 RTA  TBA Tangent chord 8.2 1935 mean   38,7 RTA  ETP Vertically 50 opposite angles 8.3 TBA  TEP Exterior angle of cyclic quardrilateral TEPB TPE is Isosceles. ETP  TEP Hence TP = PE 9.4 ATB  EPB  90 0 PTB  A Tangent chord PTB  PEB Subtended by BP A  PEB Hence ATB /// EPB They are equiangular BP BE 8.4 42 families 9.5  ATB /// EPB TB AB AB.BP = BE.TB 2BP.BP = BE.TB AO = OB = BP 2BP2 = BE.TB

33 10. 11. 10.1 SR = PQ opp sides //m 11.1 In ABC : CO = OB radii  SR = 12 given CA//OE given AE = EB midpoint theorem ST XM  SW XW ΔWMT: XS // MT. line // div. propor 11.2 In ADE and CBE ST = XW = 4 given Aˆ3  Cˆ1 subt BD TR = SR – ST = 12 – 4 = 8 cm Dˆ  Bˆ subt AC Eˆ  Eˆ  Eˆ 4 1 2 vert. opp  ADE /// CBE 3 s XM ST 10.2  XS // MT // QR XQ SR 11.3 AE  DE (from 7.2) CE BE 4 1  AE.BE  CE.DE   12 3  AE.AE  CE.DE (AE=BE)  AE  CE.DE 2 11.4 AE.EB  EF.EC given and AE.EB  CE.DE  EF.CE  CE.DE  EF  DE E is the midpoint of DF