SSIP 2016 SEPTEMBER - Sci-Bono Discovery Centre



-762000-704850004543425-70485100546735120955009600020001511302000213360SSIP 2016 SEPTEMBERMATHEMATICS GRADE 12 FACILITATORS’ MANUAL6900096000SSIP 2016 SEPTEMBERMATHEMATICS GRADE 12 FACILITATORS’ MANUALCopyright This work is protected by the Copyright Act 98 of 1978. No part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage and retrieval system, without permission in writing from Matthew Goniwe School of Leadership and Governance. Whilst every effort has been made to ensure that the information published in this work is accurate, Matthew Goniwe School of Leadership and Governance takes no responsibility for any loss or damage suffered by any person as a result of the reliance upon the information contained therein. Contents PagesContent that will be tested according to CAPS3Weighting of content areas3Cognitive levels of questions (TIMSS)4Patterns 5 – 13Inverse functions 14 – 30Solving triangles in 2D and 3D 31 – 50Application of Calculus 51 – 78Sequences and Series 79 – 98CONTENT THAT WILL BE TESTED ACCORDING TO CAPSWEIGHTING OF CONTENT AREAS Mark distribution for Mathematics NCS end-of-year papers: Grade 12PAPER 1: Grade 12 bookwork: maximum 6 marksDescriptionGrade 12Algebra and equations (and inequalities)25+3Patterns and sequences25+3Finance, growth and decay15+3Functions and graphs35+3Differential Calculus35+3Probability15+3TOTAL150PAPER 2: Grades 11 and 12: theorems and/or trig proofs: maximum 12 marksDescriptionGrade 12Statistics20+3Analytical Geometry40+3Trigonometry40+3Euclidean Geometry and Measurement50+3TOTAL150Notes:Modelling as a process will be included in all papers, thus contextual questions can be set on any topic.Questions will not necessarily be compartmentalised in sections. Various topics can be integrated in the same question.A formula sheet will be provided.Cognitive levels (TIMSS) The questions will be levelled according to the following table:QUADRATIC NUMBER PATTERNDEFINITION: Quadratic number patternA quadratic pattern is a pattern of numbers in which the second difference between any two consecutive terms is constantThe general terms of the formExample: Determine the general term of the pattern 4; 9; 18; 31…….if the term continues consistently.340995013271527813001327151981200132715138112513271557150132714006095991327140049183119812002914650027812992914650657225291464001285875291465 5 9 131st difference 4 42nd differenceHow can we be able to determine the general term?By using the general term: 3781424103505002990850103505002209800103505162877510350531432510350500904874103506002924174156210022574251371600014573251371600090487513716000 1st difference 2nd differenceNOTE: The first differences will always be a linear pattern.The second difference is given as an expression.then The general term is 02660650010477550800ITS VERY IMPORTATNT TO DETERMINE THE TYPE OF PATTERN YOU ARE WORKING WITHTO SHOW THAT A PATTERN IS A QUADRATIC NUMBER PATTERN DETERMINE THE SECOND DIFFERENCE IF IT’S COMMON THEN THE PATTERN IS QUADRATIC00ITS VERY IMPORTATNT TO DETERMINE THE TYPE OF PATTERN YOU ARE WORKING WITHTO SHOW THAT A PATTERN IS A QUADRATIC NUMBER PATTERN DETERMINE THE SECOND DIFFERENCE IF IT’S COMMON THEN THE PATTERN IS QUADRATICEXAMPLE 1Consider the following number pattern 2; 7; 14; 23…..Show that it is a quadratic number pattern.Write down the next two terms of the number pattern.Hence determine the nth term as well as the 100th termDetermine which term equals 162.EXAMPLE 2Consider the following quadratic number pattern: 6 ; x ; 26 ; 45 ; y..Calculate the values of x and y.EXERCISES1The sequence 3 ; 9 ; 17 ; 27 ; … is quadratic.1.1Determine an expression for the n-th term of the sequence.(4)1.2What is the value of the first term of the sequence that is greater than 269?(4)22 ; x ; 12; y ; . . . are the first four terms of a quadratic sequence. If the second difference is 6, calculate the values of x and y.(5)3.The quadratic pattern;…………..is given. Determine the value of x.(4) 4 In the quadratic sequence 4;x;y; -11;…, the first three terms of the first differences are 2p-4; p-3 and p2-1.Determine the: Value(s) of p.(2) Second difference(s).(1) Values of and y.(2) 4.4 General term of the quadratic sequence.(4)5Given the quadratic sequence 3; 5; 11; 21; …5.1Write down the value of the next term, if the pattern continues.(1)5.2Determine the value of the 48th term.(5)5.3Prove that the terms of this sequence will never be even numbers.(2)5.4If all the values of this sequence are increased by 30, determine the general term of the new sequence.(2)6.A pattern of triangles is formed by increasing the base of the triangle by 2 cm and the perpendicular height by 1cm, in each successive triangle. The first triangle has a base of 2 cm and a height of 2 cm. The pattern continues in this manner.539115015811453911501581154476750158114276225023431427622502343152152650234314714375234314714375234314190502343146572251377952cm02cm53244751390654cm04cm27063701365253cm03cm52781202216156cm06cm27146252216154cm04cm5810252235202cm02cm52768506159552768506159544767502235202886075615952762250615952152650223520581025615955810256159519049223520TRIANGLE 1TRIANGLE 2TRIANGLE 36.1Calculate the areas of the first four triangles.(2)6.2Calculate the area of the 100th triangle in the pattern(4)SOLUTIONS11.12a = 2 a = 13a + b = 6 b = 3c = T0 = -1Tn = n2 + 3n -11.2n2+3n-1>269∴n2+3n-270 >0∴(n+18)(n-15)>0(n+18)(n-15)>0n>15=16term: 162+316-1=30322 ; x ; 12 ; y4667255842001628775584201400175584201028700584208001005842026670058420 x-2 12-x y-121400175781050073342578105001095375781050466725781050 12-x-(x-2) y-12-(12-x) 2nd difference = 612-x-x-2=6 -2x+14=6 -2x=-8 x=4Substitute x=4 into y-12-12-x=6 y-12-12-4=6 y-20=6 y=263-3;4;x;30;…7;x-4;30-x (First difference)x-11;34-2x (Second difference)x-11=34-2x3x=4544.14.2Second difference = 34.34.455.15.2Tn=an2+bn+ca=23(2)+b=2b= -42-4+c=3c=5Tn=2n2-4n+5T48=2(48)2-4(48)+5 = 44215.32n2 even for all values of n, 4n is even for all values of n.So Tn=2n2-4n+5is the sum of two even numbers and 5which will not be even.OR/OFTn=2n2-4n+5 =2n2-2n+2+1 = (2(n-1)2+3 which is not even 5.4Tn=2n2-4n+5 + 30 =2n2-4n+35 66.12 ; 6; 12;206.2166687510795INVERSE FUNCTIONS00INVERSE FUNCTIONS-1968567945Introduction00IntroductionNotation: If the original function is named f, the inverse will be f-1 .The line y=x (y-x=0) is the line of reflection of the function and its inverse,as illustrated in the example below. 2990850110490f0f317182524765y=x00y=x3343275153670f-10f-1It is important to understand that every point x ; y on the graph of f becomesy ; x on the graph of f-1.571509334500According to the CAPS document learners should be able to: Determine if a relation is a function or not Understand and be able to apply their knowledge of the concept of the inverse of a function Know how the domain of a function may need to be restricted to ensure that the inverse is a function Determine and sketch graphs of the inverses of the functions defined by:*y=ax+q*y=ax2*y=bx ; b>0 , b≠1 Determine the following characteristics of the functions and their inverses:*Domain and range*Intercepts with the axes*Turning points*Minima and Maxima*Asymptotes (horizontal and vertical)*Shape and symmetry*Average gradient (average rate of change)*Intervals on which the function increases/decreases. Understand the definition of a logarithm:y=logbx ? x=bx where b>0 and b≠1 Be able to determine and sketch the function define by y=logbx for both the cases 0<b<1 and b>1.647700296545To determine the EQUATION of the inverse of a function00To determine the EQUATION of the inverse of a functionINTERCHANGE (“swap”) the x and y valuesMake y the subject of the formula againWhen working with the exponential function, we need ‘logs’ to help us make y the subject of the formulaExamples:Linear FunctionQuadratic FunctionExponential FunctionEquation of the given function or INTERCHANGE the x- and y-values Make y the subject of the formula again to determine the equation of the inverse in the form y = ….. 666750306705To draw the graph of the inverse of a function00To draw the graph of the inverse of a functionFirst draw the graph of the given functionDraw the line of reflection y=x (use a dotted line)Take any two known points on the given function, interchange their x- and y-values and then plot the two new pointsNow draw the graph of the inverse as a reflection of the graph of the given function about the line y=x through the two newly plotted pointsExamples of inverses:Linear FunctionQuadratic FunctionExponential FunctionEquation of the originalfunction or Graph of the original function(solid line/ curve) 309880381000-590556578600044831038100002095543815001352555842000400053937000Graph of the inverse(Dotted line/ curve)7175514224000527055670550022415533655004432304318000-590556578600088265-8255001206525400020193093345004197352813050025019083820001447805435600060452083820003238510477500742950247650Discussion:00Discussion:771525513715How will you explain to your learners the fact that we do not look at the inverse of the hyperbolic function?00How will you explain to your learners the fact that we do not look at the inverse of the hyperbolic function?647701316230The inverse of a the quadratic function00The inverse of a the quadratic functionThe graph of the inverse of the parabola is NOT a function because it is a one-to-many mapping. (a vertical line will cut it in two places) 31597601346202409825287020or0or We restrict the domain of the original parabola in the following two ways so that the inverse of the parabola is a function again:: the inverse is a function 2) : the inverse is a function3330575431800094996010795000 -762007810500Restrict the domain of the negative quadratic function in two different ways so that the inverse will be a function. Draw the graphs to illustrate this.695325268605Knowledge tested with functions and inverses00Knowledge tested with functions and inversesLearners must know how to:draw a graph of a function if the equation is givendetermine the equation of a function if the graph is givendetermine the equation of the inverse of a function and also draw the graph of the inverse of a functiondetermine the following of the functions (or inverses)x- and y-interceptsturning pointsequations of the asymptotesequations of the axes of symmetryAlso, learners must be able to: determine the domain and range of functions and their inversesstate where the function (or its inverse) increases/decreasesshift a function (or its inverse) –to the left or rightup or downreflect a function (or its inverse) – in the x-axisin the y-axisdetermine the equation of a function (or its inverse) after a shift or a reflection determine where a function (or its inverse) is positive or negativeapply their knowledge of the nature of roots The following table will assist learners in answering difficult questions on graphs of functions. Make sure they have a copy of the conditions and are able to apply it.ConditionInterval where applicablef(x)>0The part of the graph of f that is above the x-axisfx<0The part of the graph of f that is below the x-axisfx=g(x)The point(s) where the graphs of f and g intersectfx>g(x)The part of the graph of f that lies above the graph of gf(x)≤g(x)The part of the graph of f that lies below the graph of g, or is equal to the graph of gfx-gx=kThe distance between the two graphs, with the graph of f on topfx. g(x)>0Where parts of both graphs are above the x-axis or below the x-axisfx. g(x)<0Part of one graph is above the x-axis and in the same interval part of the other graph is below the x-axis704850317500Shifts and Reflections in a nutshell00Shifts and Reflections in a nutshellConsider the function S H I F T SShift 3 units to the LEFTChange in the equationGraphical representationSubstitute the x in the equation with x+3Original equation: New equation: -45720825500Shift 4 units to the RIGHTChange in the equationGraphical representationSubstitute the x in the equation with x-4Original equation: New equation: -508002349500Shift 3 units UPWARDSChange in the equationGraphical representationAdd 3 to the original equationOriginal equation: New equation: 156845146050021272520637500800107423150099885510033000Shift 4 units DOWNWARDSChange in the equationGraphical representationAdd -4 to the original equationOriginal equation: New equation: 144145-381000Consider the functions and R E F L E C T I O N SReflection in thex-axisChange in the equationGraphical representationSubstitute y in the original equation with –y.Then make y the subject of the formula again.Original equation: New equation: Original equation: New equation: 1981206350000Reflection in they-axisChange in the equationGraphical representationSubstitute x in the original equation with -xOriginal equation: New equation: Original equation: New equation: 542925287655Examples00Examples1. Given: for . The x-intercept of h is Q.1.1Determine the coordinates of Q.(2)1.2Write down the domain of .(3)1.3Sketch the graph of . Clearly indicating the y-intercept and the end points.(3)1.4For which value(s) of x will ?(3)Given for . 1.1For x-intercepts, y = 0 y = 0(2)1.2 OR h(–2) = –7 h(4) = 5(3)1.3OR y-intercept on a straight line line segment accurate endpoints (x or y or both) (3)1.4For the inverse of h, OR(3)h and intersect when ORFor the inverse of h, (3)(3)2. The sketch below shows the graphs of fx=1x for x<0 and gx=--x for x≤0.940435161925yx0gfA00yx0gfA 2.1 Write down the equations of the asymptotes of the graph of f.(2)2.2 The graphs of f and g intersect at the point Ap ; -1. Calculate the value of p.(2)2.3 Determine the equation of g-1 in the form y=…..(3)2.4 Write down the equation of h if h is a reflection of g about the y-axis.(1)2.1x=0 andy=0x=0y=0 (2)2.2 y=1x p ; -1 -1=1p∴p=-1OR y=--x (p ; -1) -1=--p-12=--p2 1=-p ∴p=-1subst. x=p and y=-1 p=-1 ORsubst. x=p and y=-1 p=-1 (2) 2.3gx=y=--xInterchange x and yx=--yx2=-yy=-x2 ; x≤0 (or y ≤0) x=--yx2=-yy=-x2 ; x≤0 (3)2.4gx=--x (Given)Reflection about the y-axis, replace x with –x: hx=--(-x) ∴h(x)= -x h(x)= -x (1) 3. The graph of fx=ax, where a>0 and a≠1, passes through the point 3 ; 278.3.1 Determine the value of a.(2)3.2 Write down the equation of f-1 in the form y=…..(2)3.3 Determine the value(s) of x for which f-1x=1.(2)3.4 If hx=fx-5, write down the range of h.(2)3.1 y=ax 278=a3 323=a3 a=32substituting 3 ; 278a=32 (2)3.2y=32xInterchange x and y: x=32y∴y=log32xx=32y y=log32x (2)3.3log32x=-1x=32-1x=23write in exponential formx=23 (2)3.4y∈-5 ; ∞(-5∞) (2)4.The graph of fx= 3x is sketched alongside. 4.1Give the coordinates of A.(1)4.2Write down the equation of f-1 in the form y = … (1)4.3For which value(s) of x will f-1≤0?(2)4.4Write down the equation of the asymptote of f(x-1)(1)5.Sketched is the graph of , the inverse of a restricted parabola. The point A(8 ; 2) lies on the graph of f.5.1Determine the equation of f in the form y = …(2)5.2Hence, write down the equation of in the form y = …(2)5.3Give the coordinates of the turning point of g(x) = .(1)6. The sketch below represents the inverses of gx=4x and fx=ax2; x≥0.49974516510yxy=f-1(x)y=g-1(x)(4 ; 1)(16 ; 2)..01yxy=f-1(x)y=g-1(x)(4 ; 1)(16 ; 2)..016.1Write down the coordinates of ONE point through which both f and g will pass. (1)6.2Determine the equation of f .(3)6.3Calculate x if gx+2=16 .(3)6.4If hx=g-1x-2, for which values of x will h(x)≤0?(2)-66675158115SOLUTIONS00SOLUTIONS4.1A(0; 1) Answer(1)4.2f-1: y = log 3 x y = log 3 x (1)4.30<x≤1 endpointsnotation(2)4.4y=0 y = 0 (1)5.1f(x)=xa OR f-1x=ax2 (8; 2): 2 = 8a a = 2 (2; 8): 8 = a(2)2 a = 2 ∴f(x)=x2 ∴f-1x= 2x2 fx: x=2y2 y =x2 a = 2 eqn(2)5.2f-1x=ax2 (2; 8): 8 = a(2)2 a = 2∴f-1x= 2x2 eqn(1)5.3(-3; -1) each value(2)6.1 (1;4) or (2;16)Any one point (1)6.2 f(x)=ax2Substitute inverse point: (1;4) OR (2;16) 4=a(1)2 16=a(2)2 a=4 4a=16 a=4 ∴ fx=4x2 , x≥0ORSubstitute (4;1) or 16;2 into x=ay2: 4=a(1)2 OR 16=a(2)2 4=a 16=a.4 4=a ∴ fx=4x2 , x≥0Substitute correct coordinateEquation fxx≥0 (3)Substitute correct coordinatesEquation fxx≥0 (3)6.3 gx+2=16 4x+2=16 4x+2=42 ∴ x+2=2 x=0 Substitution into gxSame basex=0 (3)6.4 2<x≤3x>2x≤3 (2) 7969254578350SOLUTIONS1.11.21.31.422.1 = = = = 2.2 = 299 m33.13.244.14.24.3In Area of 5AB = BC = a66.16.2Area of OCB6.377.1In 7.2In 7.3In 7.4Area of ABCDIn ACDArea of Area of In Area of Area of ABCD = Area of Area of =92.66 +112,70 = 205,488.1then8.28.3Area of 99.19.29.310.Area of inner = = 1082,531755Area of outer = = 2771,281292Area of red shaded part = Area of outer - Area of inner = 2771,281292 – 1082,531755 = 1688,751111.111.211.3Right-angled triangle12.12.112.2.1 (opp s of cyclic quad) (s opp equal sides; EH = EF (s of )12.2.2In EFH12.2.397155011430APPLICATIONS OF CALCULUS00APPLICATIONS OF CALCULUS-295275127635Introduction00Introduction18097514478000The CAPS document states the following in terms of Differential Calculus.Learners must:understand the limit concept in the context of approximating the rate of change or gradient of a function at a pointunderstand intuitively that f'a is the gradient of the tangent to the graph of f at the point a ; ybe able to determine the derivative f'x by using the definition (first principles)be able to determine the derivative f'x by making use of the rules of differentiationbe able to find the equation of the tangent to a graph of a functionunderstand that the second derivative f''x will assist in finding the point of inflection and also how it determines the concavity of a functionsketch graphs of cubic polynomial functions using differential calculus to determine coordinates of stationary points and points of inflection. Also, determine the x-intercepts of the graph using the factor theorem and other techniques (e.g. synthetic division, long division)Solve practical problems concerning optimisation and rate of change, including calculus of motionDifferential calculus is primarily about rate of change.12954009588500In Mathematics: Rate of change = gradient (m)The gradient (or rate of change) of a linear function is constant.The graph of a linear function is a straight line. When looking at a straight line graph it is clear that the gradient remains the same (or constant) at any point on the line.390525090805y0y40386001460500Example:37433259080500For fx=2x+4 the gradient is always 2. 3762375125730f0f40074856350404The graph of f illustrates this fact.4600575154305x00x399097533655000379222025400-20-23409315112395For any other function the gradient changes the whole time. The gradient will change from one point to the next on the function.4337685116840y0y483679527305At x=2, m=400At x=2, m=444481751606550Example:49720507112000396240033020g0g4591050768350039624001428753981450577850For gx=x2-4 the gradient will change.5133975100330x00x47948851657352024403725164465000See the illustration alongside. 3505200121285038862000-20-23743324641352981325117475At x=-2, m=-400At x=-2, m=-4The derivative of a function gives the gradient (or rate of change) of that function at any point.(The derivative may also be referred to as the ‘gradient function’ ) 151765-330200Make sure that your learners clearly understand that for any function f:fx=y a y-value on the graph of f associated with a specific x-valuef'x=m the gradient of f associated with a specific x-valueDifferential calculus, which is primarily about rate of change, has many applications.There are mainly three types of applications:In graphsTo determine rate of change Minimum and Maximum values1.GRAPHS542925339090Basic Concepts00Basic ConceptsThe first and second derivative of a function is very useful to find crucial points on the graph of the function.Turning points (Stationary points)At the turning points: f'x=02628900103505f'x=000f'x=0105727523495838200105410f00f24384003048028657565207003381375101600f00f866775118110f'x=000f'x=072390099060Minimum turning points1314450234952990850153035At a minimum turning point the sign of the gradient changes from negative to positive.00At a minimum turning point the sign of the gradient changes from negative to positive.1752600110490f'x>000f'x>0466725110490f'x<000f'x<0 981075990601076325118110f'x=000f'x=0 OR299085074930Another way to determine whether the turning point is a minimum is to find the second derivative. For a minimum turning point the second derivative is positive.00Another way to determine whether the turning point is a minimum is to find the second derivative. For a minimum turning point the second derivative is positive.13144502349570485062865f00f 1076325118110f''x>000f''x>097155099060Maximum turning points126682588900f'x=000f'x=03143250148590At a maximum turning point the sign of the gradient changes from positive to negative.00At a maximum turning point the sign of the gradient changes from positive to negative.1437005520700108585030480187642573660f'x<000f'x<060960092710f'x>000f'x>0 OR3200400156210Another way to determine whether the turning point is a maximum is to find the second derivative. For a maximum turning point the second derivative is negative.00Another way to determine whether the turning point is a maximum is to find the second derivative. For a maximum turning point the second derivative is negative.126682588900f''x<000f''x<0143700552070010858503048087630092710f00f-139700933450Make sure that your learners know that a turning point can be useful in TWO ways:Firstly, the x-value can be substituted into f'x=0 (Turning point)Secondly, the x- and y-values can be substituted into fxPoints of inflectionf''x=0 at a point of inflection.The gradient does not change sign at the point of inflection, as illustrated in the diagrams below.35433001397009239251016003337560135255f'x<000f'x<01062990135255f'x>000f'x>0137160055880A00A378079013716000387667555880A00A145669012763500348043546990f'x<000f'x<087630040640f'x>000f'x>0156210012065At inflection point A: f''x=000At inflection point A: f''x=0At the point of inflection the concavity of the graphs changes.3457575321945Q00QConsider the diagram below. Along the curve PA the concavity of the graph is concave down. Along curve AQ the concavity of the curve is concave up.1447800723903409950142875At inflection point A: f''x=000At inflection point A: f''x=02705100161925A00A26092156223000120967534290P00PComparison between graphs of fx and f'xFunction: Derived function: 50546015240xxi(-4 ; y)(2; y)(-1 ; y)00xxi(-4 ; y)(2; y)(-1 ; y)The x-values of the turning points are the x-intercepts of the graph of y’ =…. , which is the parabola on the right. The point of inflection’s x-value is the x-value of the turning point of the graph of y’.201930094783m = +0m = +81026041275xxi(-4 ; 0)(2 ; 0)(-1 ; y)00xxi(-4 ; 0)(2 ; 0)(-1 ; y)2038350123190m = -0m = -These x-intercepts, are the turning point x’s of the cubic graph, because the y- values here are 0.The turning point x-value at i, is the x-value of the point of inflection on the cubic graph. 610235247015x(-3 ; y)00x(-3 ; y) At this x-value, the gradient of the tangent = 0276860208915x(-3; 0)00x(-3; 0) 190500052691m = +0m = +190500080010m = -0m = -At this x-intercept, y’ = 0. This x-intercept and the x-value of the parabola on the left are the same x-values. Function: Derived function: or 581025495300052387519494500 or 187071062865m = +0m = +1932940428625m = -0m = -2971801739900066167021590001893570207010m = +0m = + 1946275-6350m = -0m = -The graph of y’ = 0 is a straight line that coincides with the x-axis561340286385Examples00ExamplesGiven fx=x+12x-2=x3-3x-2 and gx=3x2+3x1.1 Determine the x-intercepts of f and g.(2) Determine the co-ordinates of the turning points and g.(4) Sketch the graphs showing all the relative turning points and the intercepts with the axes.(6)1.4 Determine the solution of using your graphs.(2)1.5 For which value(s) of x will fx . gx<0?(3)1.6 Determine the value(s) of k for which f(x) = k has three unequal roots. (2)1.1fx=x+12x-2=x3-3x-2 x=-1 or/ofx=2x-intercepts & ,g(x)=3x2+3x0=xx+1x=0, x-Intercepts of fFactors for g x-Intercepts of (3)1.2x=-1 or x=1y=0 or y=-4TP g'x=6x+3=0 TP: (4)1.3f:ShapeIntercepts Both turning points g (3)ShapeIntercepts TP(3)1.4Finding the differenceAnswer(2)1.50<x<2 or/of x<-10<x<2 x<-1(2)1.6 (2)2.1Determine the points on the curve where the gradient of the tangent to the curve is – 1 .(5)2.2The graph of a cubic function with equation is drawn.1905000405765y00yf has a a local maximum at B and a local minimum at x = 4.2.2.1Show that (2)2.2.2Calculate the coordinates of B.(4)2.2.3Determine the value(s) of k for which f(x) = k has negative roots only.(2)2.2.4Determine the value(s) of x for which f is concave up.(2)2.1 and the gradient of the tangent to the curve is -1 exponential formderivativederivative = -1x-valuesy-values (5)2.2.1squaring binomial (2)2.2.2factorsy-values (4)2.2.3inequality-16 (2)2.2.4 (2)3.The graph of y = g/(x) is sketched below, with x-intercepts at A(? 2; 0) and B( 3; 0). The y-intercept of the sketched graph is (0; 12). 16719551187450 y98615595814137731512700120012 A B x6140455715 –2 O 3 3.1Determine the gradient of at x = 0.(1)3.2For which value of x will the gradient of be the same as the gradient in QUESTION 7.2.1?(1)3.3Draw a sketch graph of . Show the x-values of the stationary points and the point of inflection on your sketch. It is not necessary to indicate the intercepts with the axes.(3)3.1 answer (1)3.2 x = 1 (1)3.3 158178564770 y212471053340049593555872 g169608558420381635158750 x1096010-95250 -2 ? 3 turning point at x = ? 2 and at x = 3 point of inflection at x = ? shape (3)1.The function defined by fx=x3+px2+qx+30 is represented by the sketch below. A (-1 ;36) and B are the turning points of f , while g is a tangent to f at Awhich cuts f at point C.8235959144030CBA(–1;36)yx0gf30CBA(–1;36)yx0gf1.1Show that p=-4 and q=-11.(7)1.2Determine the coordinates of C.(3)1.3Write down the coordinates of a turning point of k , ifkx=fx-10.(2)2.The turning points of the graph of a cubic polynomial hx are 2 ; -3and (5 ;4). Draw a sketch graph of the derivative function h'(x), clearly showing the x-intercepts.(3)3. Given: and . P and Q(2 ; 10) are the turning points of h. The graph of h passes through the origin. 3.1 Show that and .(5) 3.2 Calculate the average gradient of h between P and Q, if it is given that x = –1 at P.(4) 3.3 Show that the concavity of h changes at .(3) 3.4 Explain the significance of the change in QUESTION 9.3 with respect to h.(1) 3.5 Determine the value of x, given x < 0, at which the tangent to h is parallel to g.(4)4.The diagram below shows the graph of f'x, the derivative of fx=ax3+bx2+cx+d. 2237740327849yyThe graph of y=f'x intersects the x - axis at 1 and 5. A4 ; -9 is a point on the graph of f'.144208510984180310814948002941320125284y=f'(x)y=f'(x)198149214287511124777513671400281241514306455408876538289xx 901065-446 255270098133A4 ; -9A4 ; -9 268258318415 4.1 Write down the gradient of the tangent to f at x=4.(1)4.2 Draw a rough sketch of the graph of f.(3)4.3 Write down the x-coordinate of the point of inflection of f.(2)4.4 For which value(s) of x is f strictly decreasing?(2)2.TO DETERMINE RATE OF CHANGERemember that rate of change refers to the speed at which something is changing.Make sure your learners memorise the following:78105010985500If an equation st is given with distance (s) and time (t), thenst represents distance at time t s't represents speed (velocity) at time ts''t represents acceleration at time tLearners struggle with this content. Use an easy example like the following, to explain to them the different scenario’s relevant in this situation:Given a height (or distance) equation: st=50t=5t2 with distance (s) in meters and time (t) in seconds. (Like a stone thrown up into the air)s(2) is the distance after 2 secondss(5) is the distance after 5 secondss'2 is the speed after 2 secondss'5 is the speed after 5 secondss(0) is the distance EVEN BEFORE THE MOTION HAS STARTEDMaximum height will be at s't=0 (the turning point of s)Maximum height after how many seconds? Solve for: s't=0 ∴50-10t=0 ∴t=5 secondsWhat is the maximum height? (Substitute t = 5 into s) s5=505-552∴s5=125 meterWhen (at what time) does it hit the ground again? (When speed = 0 again, thus st=0) 50t-5t2=0?t=0s or t=10sUse the bigger value because it will be the end of the ‘journey’ of the stone.Speed with which it hits the ground? (Calculate s'10)s'10=50-1010 =-50m/sThe negative indicates that the stone hits the ground when in DOWNWARD motion.SPECIAL CASESSuppose TWO moving bodies are involved, one going up and on coming down, and the ask:When (at what time) will they pass each other?This will happen when the two heights are equal.Therefore, equate the two height equations of the bodies and solve for t.When (at what time) will their speeds be equal?Now you equate the two speed equations (s’ = s’ ) and solve for t.594995240665Examples00ExamplesA stone falls from a cliff 320 metres high. The distance covered by the stone while falling is given by the function st=5t2, where s is the distance covered in metres and t is the time in seconds.1.1 Determine the speed of the falling stone after 3 seconds.(3)1.2 How long will it take before the stone hits the bottom of the cliff?(3)1.3 What will the speed of the stone be when it hits the bottom of the cliff?(2)1.1st=5t2Speed: s't=10tAfter 3 seconds: s'3=103=30m/ss't=10ts'3=10330m/s (3)1.2Distance covered by stone = 320 m5t2=320t2=64t=±8It takes 8 seconds to hit the bottom of the cliff5t2=320t=±8Choose t = 8 s (3)1.3Speed after 8 seconds: s'8=10(8) =80m/ss'8=10(8)=80m/sSales of a new product grow rapidly and then level off with time. This situation is represented by the equation St=-t3-4t2+80t where t represents time in months and S(t) represents sales. 2.1 Determine the rate of change of sales during the third month.(3)2.2 Determine after how many months a maximum sale is obtained.(5)2.1St=-t3-4t2+80tRate of change: S't=-3t2-8t+80After 3 months: S'3=-332-83+80 =29 sales per monthS't=-3t2-8t+80Subst. t = 329 (3)2.2Max sales if: s't=0-3t2-8t+80=0 3t2+8t-80=03t+20t-4=0∴t=-203 or t=4After 4 months a maximum sale will be obtaineds't=0Standard formfactorst-valuesChoose t = 4 (5)The mass of a baby in the first 30 days of life is given by ; . t is the time in days and M is the mass of the baby in grams.1.1 Write down the mass of the baby at birth.(1)1.2 A baby's mass usually decreases in the first few days after birth. On which day will the baby's mass return to its birth mass? (4)1.3 On which day will this baby have a minimum mass?(4)1.4 On which day will the baby's mass be decreasing the fastest?(2)A Petrol tank at BP Depot has both the inlet and the outlet pipes which are usedto control the amount of petrol it contains. The depth of the tank is given byDt=6+t24-t38 where D is in metres and t is in hours that are measured from 9h00. 2.1 Determine the rate at which the depth is changing at 12h00, and then tell whether there is and increase or decrease in depth. Answer correct to two decimal digits. (3) At what time will the inflow of petrol be the same as the outflow?(4)A stone is thrown vertically upwards. Its height (in metres) above the ground at time t (in seconds) is given by: . 3.1 Determine the initial height of the stone above the ground.(1) 3.2 Determine the time taken to reach the maximum height.(3) 3.3 How fast was the stone travelling when it hit the ground? (5) 3.4 Determine the acceleration of the stone.(1)3.Maximum and minimum valuesWe use calculus to find the x-value that will minimise or maximise a quantity like area, volume, etc.It is important that the learners will remember that the gradient of a function is zero at the minimum and maximum values.Follow these steps:If necessary calculate the missing dimensions (e.g. height in terms of the radius)Determine a formula in terms of one variable (e.g. x) for what must be minimised or maximised, likeArea………...Ax orCost………...Cx orVolume…….VxIf your formula contains a value like π, then keep it as π and only in the final answer convert it to a value/numberNow determine the first derivative of your formula (e.g. A'x)Solve for x in A'x=0Test the x-values obtained. They will maximise or minimise the area, volume, cost and so forth. REJECT the x-value that is not valid (e.g. a negative area etc.)Remember that these questions could be asked about distances and shapes on the Cartesian plane. Then height is given by the y-values and the horizontal distances are given by the x-valuesMake sure your learners KNOW the formulas for the surface areas and volumes of right prisms. For other shapes (e.g. cones and pyramids), formulas will be given and learners should be able to select the applicable formula.594995234950Examples00ExamplesPQRS is a rectangle with P on the curve h(x)=x2 and with the x-axis and the line x=6 as boundaries.21399559690yx0PSQR6hyx0PSQR6h 1.1 Show that the area of rectangle PQRS can be expressed as: A =6x2-x3 .(3) 1.2 Determine the largest possible area for rectangle PQRS. Show all your calculations.(4)A wooden block is made as shown in the diagram below. The ends are right-angled triangles having sides 3x, 4x and 5x. the length of the block is y. The total surface area of the block is 3?600 cm2.25755601422404x04x2143125161290002257425142240002228850444501635760552453x03x1854901266712204085438155x05x167640018732502924175193675y0y1870710152400167640064135001857375546102.1 Show that: y=300-x2x(3)2.2 Determine the value of x for which the block will have a maximum volume.(6)1.12857576200yx0PSQR6hyx0PSQR6hP(x;x2) and Q (x;0)QR =6-xPQ =x2Area of PQRS = PQ × QR = x2×(6-x) = 6x2-x3QR =6-xPQ =x2Method (3)1.2For maximum area: dAdx =0 12x-3x2=0 3x4-x=0 x=0 or x=4 N/AMaximum area =642-43 = 32 units212x-3x2dAdx =0x=4Answer (4)2.1 SA=2.12.3x.4x+3x.y+4x.y+5x.y 3 600=12x2+12xy 300=x2+xy 300-x2=xy 300-x2x=yformula3 600=12x2+12xy 300-x2=xy (3)2.2Volume= 12×b×h×H V(x)=123x4x300-x2x Vx=1 800x-6x3For max: V'x=01?800-18x2=01?800=18x2100=x2∴x=±10 but x>0For a max volume, x = 10 cmformulaCorrect substitution Vx=1 800x-6x31?800-18x2=0x-valuesChoose x = 10 (6)A rectangular box has a length of 5x units, breadth of units and its height of x units.1608455349255x(9 – 2x) 005x(9 – 2x) 3855085224155x00x1.1 Show that the volume (V) of the box is given by (2)1.2 Determine the value of x for which the box will have maximum volume.(5)1242060337185hr00hrA 340 ml can of cool drink with height h and radius r is shown below.2.1 Determine the height of the can in terms of the radius r.(3)2.2 Show that the surface area of the can be written as SA=2πr2+680r.(2)2.3 Determine the radius of the can in cm, if the surface area of the can has to be as small as possible.(4)ABCD is a square with sides 20 mm each. PQRS is a rectangle that fits inside the square such that QB=BR=DS=DP=k mm.156210036830BCDPQRSA00BCDPQRSA3.1 Prove that the area of PQRS=-2kk-20=40k-2k2(4)3.2 Determine the value of k for which the area of PQRS is a maximum.(4)28575148590SOLUTIONS00SOLUTIONS1.GRAPHSQUESTION 91.1 fx=x3+px2+qx+30 At turning point f'x=0: 3x2+2px+q=0Substitute x=-1: 3-12+2p-1+q=0 3-2p+q=0 -2p+q=-3 - - - (i)Substitute (-1;36) in fx: (-1)3+p(-1)2+q-1+30=36 -1+p-q+30=36 p-q=7 - - - (ii)(i) + (ii): -p=4 p=-4Substitute p=-4 into (ii): -4-q=7 q=-11 f'(x)Substitute x=-1Equation (i)Substitute (-1;36)Equation (ii)p=-4q=-11 (7)1.2At C: fx=gx=36 x3-4x2-11x+30=36 x3-4x2-11x-6=0 But x=-1 at A ∴x+1x2+kx-6=0 k+1=-4 OR touching point at A k=-5 ∴x+1x+1x-6=0 ∴x+1x2-5x-6=0 ∴x+1x+1x-6=0 x=6 C (6;36)f(x)=36x+1 factorx=6 (3) 1.3Turning point of k: (-1;26)x=-1y=26 (2)2.141218-7040(5; 4)(2; -3)hx(5; 4)(2; -3)hx369818-770325y=h'(x)x25y=h'(x)xShapex=2x=5 (3)3.1Substitute Q(2; 10) into substitute Q into hfinding derivativeequating derivative to 0solving simultaneously for a and b (5)3.2Average gradient= –3,5formulasubstitutionanswer (4)3.313970073660x00xFor , h is concave up and for , h is concave downexplanation(3)3.4The graph of h has a point of inflection at ORThe graph of h changes from concave up to concave down at ORThe graph of h changes concavity at answer (1)answer (1)answer (1)3.5Gradient of g is – 12Gradient of tangent is: factorsselection of x-value (4)4.11700530316865x=500x=51022350318135x=100x=1Gradient =-9answer (1)4.2134937516446500203136538100049847510287000correct shapeTP at x=1TP at x=5 (3)4.3x=3x=3 (2)4.4x∈1 ; 5(15) (2)2.RATE OF CHANGEGiven: ; 1.1answer(1)1.2Baby’s mass will return to the birth mass on the 9th dayfactors(4)1.3Baby’s mass will be a minimum on the 6th day factors(4)1.4OR / OFanswer(2)answer(2)2.1Dt=6+t24-t38D'(t)=t2-3t28D'(3)=32-3328=-158m.h-1=-1,875 decreasing in depth/verlaging in diepteDerivativeSubstitutionAnswer (3)2.2D'(t)=0t2-3t28=0t12-3t8=0t=0 or t=43At 9h00 and at 43×60=80 minutes later. i.e. at 10h20DerivativefactorsanswersAnswer (4)3.1 answer (1)3.2OR answer (3) correct formula substitution into the correct formula answer (3) 3.3 h(t) = 0 factors t = 8 –20m/s or 20 m/s downwards (5)3.4 OR answer (1) answer (1) 3.MINIMA & MAXIMA1.1V=l×b×h =5x9-2xx =45x2-10x3formulasubstitution (2)1.2V'x=90x-30x290x-30x2=0 30x3-x=0x=0 or x=3Therefore the box will have a maximum at x=3derivativederivative = 0factorsx-valueschoosing x=3 (5)2.1V=πr2h = 340 cm3h=340πr2???V=πr2hV=340cm3answer(3)2.2total surface area = 2πr2+2πrh = 2πr2+2πr ( 340πr2 ) SA=2πr2+680r??2πr2+2πrhsubst. of h in the total SA formula(2)2.3SA=2πr2+680rdSAdr=4πr- 680r2For the surface area to be as small as possibledSAdr=04πr- 680r2=04πr3-680=0r3=1704π=13, 52817…r=2, 38cm????derivative.dSAdr=0r3=1704π=13, 52817…r=2, 38cm(4)87414598710BCDPQRSAkk20-k2k2(20-k)20-k0BCDPQRSAkk20-k2k2(20-k)20-k3.1QR=2k PythagorasSR= 2(20-k) PythagorasArea PQRS= 2k2(20-k) = 2k20-k=40k-2k2 QR=2k SR= 2(20-k) Area PQRS= 2k2(20-k)(4)3.2A=-2k2+40kdAdk= -4k+40=0 k=10-4k+40=0 10(4)TOPIC: SEQUENCES AND SERIES INTRODUCTIONYou have already had some experience of working with number sequences and number patterns. In grade 11 you have dealt with quadratic or second difference sequences. In grade 12 there are two specific types of sequences which both have quite unique properties. These are Arithmetic and Geometric sequences.4322445447675What do I do? It’s not arithmetic or geometric Mr K didn’t give me the formula00What do I do? It’s not arithmetic or geometric Mr K didn’t give me the formula1895475190500201; 2001; 20001; …Work out this, what sequence is this? Find the formula00201; 2001; 20001; …Work out this, what sequence is this? Find the formulaArithmetic sequenceArithmetic sequence is a sequence of numbers in which the difference between two consecutive numbers is constant.The numbers in the sequence are called terms i.e. T1;T2; T3;T4…Tn and Tn the general term.T2-T1=T3-T2= T4-T3 = Tn –Tn-1 = d.149987085090Check:Is this Arithmetic?-1; 1; 3; 6;…00Check:Is this Arithmetic?-1; 1; 3; 6;…d is the constant difference .To determine whether the sequence is arithmetic or not we need to find the difference between two consecutive terms. Given -1; 1; 3; 6; …find the difference1--1=3-1 2= 2 But T4-T3=3 ≠ T3-T2The difference between two consecutive numbers is not always the same in this sequence. “So the sequence is not arithmetic”In arithmetic sequences the difference between every two consecutive terms should be the same. And when you add the constant difference to the previous term you get the next term.112395073660Can the procedure given above be used every time?00Can the procedure given above be used every time?247650294640 Consider the sequence -3; -1; 1; 3; …to find T135. Yes, but it will take long to keep on adding the constant difference in order to get to T135, hence we need the formula that can help us to get the answer.T1=aT2=a+dT3=a+2dT25=a+24dTn =a+n-1d 4867275403860General term??00General term??1494155-421005Tn=a+n-1dis the general termfor Arithmetic Sequence00Tn=a+n-1dis the general termfor Arithmetic SequenceGeometric sequence Geometric sequence is a?sequence?of numbers where each term after the first is found by multiplying the previous one by a non-zero number called the common ratio(r).?Note: T2T1= T3T2=r T1 =aT2=a.rT3=a.r2T4=a.r3Tn=a.rn-1166687598425Tn=a.rn-1Is the general term for Geometric sequence00Tn=a.rn-1Is the general term for Geometric sequence923925-577215General term again??00General term again??1181100937895The general term (nth term) is a formula that you can use to determine any term within the sequence up to infinity!!!00The general term (nth term) is a formula that you can use to determine any term within the sequence up to infinity!!! Activity 11.1Find the numerical values of the first three terms in an arithmetic sequence given by x;4x+5 and 10x-5. (3)1.2Given the sequence 4; -2; 1; …Find:1.2.1 the next two terms(2)1.2.2 the nth term(3)1.3The 10th term of an arithmetic sequence is 17 and the 16th term is 44. Find the first three terms in the sequence.(4)1.4The first three terms of arithmetic sequence are Calculate the 60th term in terms of(3)[15]Activity 22.1Given 6; 3; 2; 1; 23; -1;… if the sequence behaves the same find:2.1.1The next two terms(4)2.1.2Explain your answer in 2.1.1(4)2.1.3What will be the general term of the sequence if it exists?(4)2.2The first term of an arithmetic sequence is 2. The 1st; 3rd and 11th terms of an arithmetic sequence are the first three terms of a Geometric sequence. Find the 7th term of the Geometric Sequence.(6)[18]Arithmetic seriesWhen the terms of a sequence are added together, the series is formed.A story of a historical event or of a contrived situation can motivate pupils. To introduce the sum of an arithmetic series, tell the story about young Carl Freidrich Gauss (10 years) in a class that was asked by its instructor to add the numbers from 1 to 100. Much to the astonishment of the instructor, young Gauss produced the correct answers immediately. When asked how he arrived at the answer he quickly explained that:1 + 100 = 1012 + 99 = 1013 + 98 = 101Since there are 50 such pairs, the answer is 50 × 101 = 5050. This scheme can be used to develop the sum of an arithmetic series.-371475-40640000Proof for the sum of terms of arithmetic series( Examinable) S1=T1 S2=T1+T2S3=T1+T2+T3Sn= T1+T2+T3+…+TnSn=a+ a+d+a+2d+…++n-2d+[a+n-1d…………(1)Sn=a+n-1d+a+n-2d+…+a+2d+a+d+a………(2)Add equation (1) and (2)2Sn=2a+n-1d+2a+n-1d+…+2a+n-1d+2a+n-1d2Sn=n[2a+n-1d]?Sn=n2[2a+n-1d]Alternatively: Sn=n2[a+l] ; l is the last term=a+n-1d-45783524638000Geometric Series (Examinable)S1=T1S2=T1+T2S3=T1+T2+T3Sn=a+ar+ar2+…+arn-2+arn-1 …………………..(1)(×r):rSn=ar+ar2+ar3+…+arn-2+arn-1+arn………..(2)(1) – (2) Sn-rSn=a-arnSn1-r=a(1-rn) Sn=a1-rn1-r;r<1 or Sn=a(rn-1)r-1 ;r>1Activity 33.1The sum of n terms of a sequence is given as Sn=n25n+93.1.1Calculate the sum to 50 terms.(2)3.1.2Calculate the 50th term of the sequence.(3)3.2The sum of the first 20 terms of the following arithmetic series is 3?360.Calculate the value of (4)3.3Given the arithmetic series: 2+9+16+…to 251 terms3.3.1Write down the fourth term of the series(1)3.3.2Calculate the 251st term(3)3.3.3Calculate the sum of the series(2)3.3.4How many terms in the series are divisible by 4?(4)3.4Calculate the sum of the first 6 terms of a geometric series if the second term is 8 and the sixth term is 648.(4)3.5Find the smallest natural number n for which the first n terms of the geometric sequence1; 1, 1; 1, 21; 1, 331; … will have a sum greater than 70.(5)[28]Sigma notationThe mathematical symbol Σ is the capital letter S in Greek alphabet. It is used as the symbol for the sum of a series. Note that the last value you substitute is not always giving the number of terms that will be added. The number of terms = Top – Bottom +1 What is the number of terms in this series?Ans: It will be 20-3+1=18, so we will calculate the sum of 18 terms in the series.Activity 44.1Expand and then calculate(5)4.2Determine the value of n:(5)4.3Write the following series in the sigma notation:2+6+18+…+162(3)4.4In the arithmetic series 4.4.1Prove that a=6 and b=20(2)4.4.2Determine the sum of the first 20 terms of the series above.(3)4.4.3Write down the series in 4.4.2 in the sigma notation(3)[21]Infinite geometric seriesA Geometric series can also have?smaller and smaller?values. What happens when? .Therefore the series will converge to a number referred to as the sum to infinity. This is the case only if Example:9; 3; 1…; ; the series converges and the sum to infinity existsFormula for sum to infinity Activity 55.1Given: Determine for which values of x the series will be convergent.(4)5.2115570015938500 1998593798170172824979817328642974930C0C92235174930B0B The mid- points of the sides of are joined to form a smaller triangle. The process is repeated on the smaller triangle, and then continued indefinitely. The perimeters of all the triangles including are added to obtain 44 units. Calculate the perimeter of (5)[9]2447925-3581400A0ARecurring decimalIt’s a number that keeps repeating after a comma. A recurring decimal is represented by placing a dot above the number or numbers that repeat.All recurring decimals can be written as common fractions.How do we convert recurring decimal to common fractions?e.g. Convert the recurring decimal in to a common fraction.Ans: the series converge = Activity 66.1 Use the formula for of a geometric series to write as common fraction(3) 184151651000 The application of all formulae is examinableDescriptionArithmeticGeometricFirst termaaNumber of termsnnCommon differenced=T2-T1=T3-T2-Common ratio-r=T2T1=T3T2General termTn=a+n-1dTn=a.rn-1Sum of n termsSn=n2[a+l]Sn=n2[2a+n-1d]Sn=a1-rn1-r;r<1Sn=a(rn-1)r-1;r>1Sum to infinity-S∞=a1-r if-1<r<1The sum of a number of termsr=1310+r=11+12+13r=13110r= 110+1102+1103SOLUTIONSActivity 11.14x+5-x=10x-5-4x+54x+5-x=10x-5-4x-5 3x+5=6x-10 -3x=-15 x=5∴5;25;45;…(3)1.2 .1(2)1.2.2(3) 1.3………………….(1)………………...(2)(2) – (1) ∴d=92Subst in eq (1) ∴ a=-472 ∴ -472; -19; -292;…(4) 1.4 (3)[15]Activity 2 2.1.1(4)2.1.2The sequence consists of arithmetic and geometric sequences, therefore you work out the terms from two separate sequences.(4)2.1.3For geometric sequence: For arithmetic sequence: (4)2.2 ÷each equation by 2Subst in to (6)[18]Activity 33.1.1 (2)3.1.2 (3)3.2(4)3.3.1(1)3.3.2(3)3.3.3Or(3)3.3.4The new series is 16+44+72+…+1752OrT3 is divisible by 4, then T7; T11; T15;…are divisible by 4 thus each 4th termis divisible by 4.Number of divisible terms by 4 will be== 63(4)3.4 (4)3.5 The smallest natural number is (5)[29]Activity 44.1 (5)4.2 (5)4.3 OR (3)4.4.127-b=b -13 b=13+272 = 2027 – 20 = 13 – aa=6OR27 – 13 = 2dd=7b=13+7=20∴a = 13 – 7 = 64.4.2a=6 ;d=7S20=n2 2a+(n-1d) = 202(26+19(7) = 1450(2)4.4.3n=1206+7n-1 or n=120(7n -1)(3)[21]Activity 55.1 For convergent series -1<x2-3x+1<1 x2-3x+1>-1 or x2-3x+1<1 x2-3x+2>0 or x2-3x<0 x-2x-1>0 or xx-3<0 x>2 and x<1 or x<3 and x>01>x>2 or 0<x<3 (4)5.2Let perimeter of be x the next term will be 12xx; 12x;… the perimeter is 22 units (5)6.1 (3) ................
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