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Mathematics Telematics Resources Gr 113. Given:3.13.23.33.43.53.63.711February to October 2017f ( x) x 2 6 x 7g ( x) x 1Sketch the graph of f and g, showing clearly all intercepts with the axes and turning points.Write down the range of f.Write down the equation of the axis of symmetry.Calculate x where f(x) g(x).Hence, or otherwise, determine the value(s) of x for which f(x) t g(x).Determine the average gradient between x – 1 and x 4.Write down the axis of symmetry of f(x – 2).4. Sketched below are the graphs ofyRf ( x) ( x 2) 4g ( x) ax q2f4.14.24.34.4Write down the coordinates of R.Calculate the length of AB.Determine the equation of g.BAFor which values of x is,1) g(x) f(x)?2) f strictly increasing4.5 Write down the equation of the axis ofsymmetry of h if h(x) f(– x).4.6 Write down the range of p if p(x) – f(x).4.7 Determine the equation of a line through B, making an angle of 30q with the x-axes.xgy5. Sketched alongside are the graphs off ( x) 3 x 6 and g ( x)xLAOBx 2 8 x 20A and B are the x-intercepts of the graph of g.FPC and F are the y-intercepts of g and frespectively.QPL is a straight line with Q in f and P in g suchthat QPL A x-axis.A and D are the points of intersection of theCgraphs of f and g.E is the turning point of g.Q5.1 Write down the length of FC.5.2 Calculate the length of AB.5.3 Calculate the coordinates of D.5.4 Determine an expression for the length PQ.5.5 For which values of x, between the points Aand D, will PQ be a maximum?5.6 Calculate the maximum length of PQ, between the points A and D.gfDE

12Mathematics Telematics Resources Gr 11February to October 2017FUNCTIONS- Hyperbola, Exponential & Parabola1. Sketched below is the graph of f ( x)1.11.21.31.41.51.61.7h( x )54321x-7-6-5-4-3-2-1h( x )-2-3 2 x 12 1x 4Write down the equations of the asymptotes of h.Determine the x- and y-intercepts of the graph of h.Sketch the graph of h.Write down the domain of h.Write down the range of h.Describe the transformation of h to f if3.6.1 f(x) h(x 3)3.6.2 f(x) h(x) – 21 2 andx 3a ( x p ) 2 q are drawn. O and F arey4. The graph of f ( x)g ( x)the x-intercepts of the graph of g. E is the turningpoint of g. B is the x-intercept of the graph of f. Dis the point of intersection of the graphs of f and g.4.1 Write down the coordinates of E.4.2 Show that, g ( x)24 x2 x93fEfC4.3 Calculate the length of OF.4.4 Write down the equation of the axis ofOsymmetry of f that has a negativegradient.4.5 Write down the equation of p if p is thereflection of g in the line y 2.4.6 Find the coordinate(s) of a point on f which is closest to E.1-1Determine the y-intercept of the graph of h.Write down the equation of the asymptote of h.Draw a sketch graph of h, showing all asymptotes and intercepts.Write down the range of g(x) h(x) 3Given:3.13.23.33.43.53.662 x 1 12.5 Describe the transformation of h to g if g ( x)3.y7Write down values of p and q.Calculate the value of a if T(– 2 ; 2,5) lies on the graph of f.Write down the domain of f.Write down the equations of the lines of symmetry of f .For which value(s) of x is f(x) 0?Write down the range of f(x – 1).Describe the transformation of f to g if g(x) f(– x).2. Given:2.12.22.32.4a qx pgxBF

Mathematics Telematics Resources Gr 1113February to October 2017Session 4: TrigonometryxDefinitions of trigonometric ratios:oIn a right-angled 'SinToppositehypotenuseCosTadjacenthypotenuseOn a Cartesian teadjacentTanTxopposite0 , 90 , 180 , 270 , 360 can beTxx30 , 45 and 60 can be obtainedobtained from the following unit circleT.from90qyr, the radius is1 since it is aunit circle(0 ; 1) 180q (-1 ; 0)(1 ; 0)Tx TT30qT 30qT 45qsin 30q 1 2sin 45q 13cos 45q 1cos 30q The “ASTC” rule enables you to obtain thesign of the trigonometric ratios in any of thefour quadrants.2tan 30q 1145q1T 60q32sin 60q 2cos 60q 1 2tan 45q 1tan 60q 323yT180q-TA - ALL trigratios are ve inthe first quadrantASSine veAll veTan veCos veTCxT Tan is ve in the3rdquadrantC Cos is ve inthe 4nd quadrant180q TThe trigonometric function of angles(180q T) or (360q T) or (-T)(180q T )45q2360q1270qS Sine is ve in the2nd quadranttthe following two triangles.20q360q (0 ; -1)xySpecial AnglesoTr360q-Tbecomes(180q T ) (360q T )Trigonometric function of T(360q T )The sign isdetermined bythe “ASTC” rule.( T )sin(180q T )sin Tsin(180q T ) sin Tsin(360q T ) sin Tsin(360q T )sin Tsin( T ) sin Tcos(180q T ) cos Tcos(180q T ) cos Tcos(360q T ) cos Tcos(360q T )cos Tcos( T ) cos Ttan(180q T ) tan Ttan(180q T ) tan Ttan(360q T ) tan Ttan(360q T )tan Ttan( T ) tan T

Mathematics Telematics Resources Gr 11xsin Tcos Tsin 2 T cos 2 T(cos T z 0)xCo-functions or Co-ratiosxTrigonometric Equations3.4.Determine theReference angleEstablish in whichtwo quadrants is.Calculate in theinterval [0q; 360q]Write down thegeneral solutionsin 2 T1,sin(90q T )cos Tcos(90q T )sin TrTsin T2.February to October 2017TRIGONOMETRIC IDENTITIEStan T1.14cos T0,707 11 cos 2 T ,cos 2 T1 sin 2 Tsin(90q T ) cos Tcos(90q T ) sin T90q-T yxtan T 0,866 1 1Reference sin (0,707) 45qReference cos (0,866) 30qReference tan 1 (1) 45q? ș 45qorș 180q - 45q? ș 180q- 30q or ș 180q 30q? ș 180q - 45q? ș 45qorș 135q? ș 150q orș 210q? ș r150q? ș r150 k360º where k ? ș 135q? ș 45q k360º orș 135q k360º where k ? ș 135q k180º k TRIGONOMETRY SUMMARYQuestion typeSummary of procedureExample question1. Calculate thevalue of a trigexpressionwithout using acalculator.Establish whether you need a rough sketch or specialtriangles or ASTC rules.1.1 If 13 cos D3, D [0q; 270q] and4E [0q; 180q] Determine, without using a calculator,5 and tan E a) sin cos 1.2 Calculate: a)cos( 210 ). sin 2 405 . cos 14 tan 120 . sin 104 2. Express trigratio in termsof the givenvariable.Draw a rough sketch with given angle and label 2 of thesides. The 3rd side can then be determined usingPythagoras. Express each of the angles in question interms of the angle in the rough sketch.2. If sin 27q3. Simplify atrigonometricalexpression.Use the ASTC rule to simplify the given expression ifpossible.See if any of the identities can be used to simplify it, if notsee if it can be factorized. Check again if any identity canbe used. This includes using the compound and doubleangle identities.3. Simplify:cos ( 720q x) . sin ( 360q x) . tan ( x 180q)a)sin ( x) . cos (90q x)4. Prove a givenidentity.q , express each of the following in terms of q.b) cos( 27q)a) sin 117qSimplify the one side of the equation using reductionformulae and identities until it cannot simplify any further.sin ( 90q x) . tan ( 360q x)sin (180q x) . cos (90q x) cos(540q x). cos( x)4. Prove thatb)a)tan x . cos 3 x1 sin 2 x co s 2 x1sin x2b) cos 2 (180q x) cos(90q 2 x) tan(360q x)5. Solve a trigequation.Find the reference angle by ignoring the “-“sign andfinding sin 1 (0,435)05. Solve for x [ 180q; 360q]0Write down the two solutions in the interval, x [0 ;360 ]Then write down the general solution for the given eq.From the general solution you can determine the solutionfor any specified interval by using various values of k.a)b)c)sin x 0,435cos 2 x 0,4351tan x 1 0,4352sin 2 x 1

Mathematics Telematics Resources Gr 1115February to October 2017TRIGONOMETRY QUESTIONS1In the diagram below, P(–8 ; t) is a point in the Cartesian plane such that OP 17 units and reflexXÔP T .yTOx17P(– 8 ; t )231.1Calculate the value of t.1.2Determine the value of each of the following WITHOUT using a calculator:If sin 17q(2)(a)cos( T )(2)(b)1 sin T(2)a , WITHOUT using a calculator, express the following in terms of a :2.1tan 17q(3)2.2sin 107q(2)2.3cos 2 253q sin 2 557q(4)Simplify fully, WITHOUT the use of a calculator:cos( 225q). sin 135q sin 330qtan 225q1(cos x 1)(cos x 1)(6) 1tan x.cos 2 x4Prove the identity:5Determine the general solution for 2sin x. cos x2cos x.(4)(6)[31]

Mathematics Telematics Resources Gr 1116February to October 2017Session 5: Grade 11 geometryBelow are Grade 11 Theorems, Converse Theorems and their Corollaries which you must know. The proofs ofthe theorems marked with (**) must be studied because it could be examined.1Theorem** The line drawn from the centre of a circle perpendicular to a chord bisects the chord;(line from centre ٣ to chord)ConverseThe line from the centre of a circle to the midpoint of a chord is perpendicular to the chord.(line from centre to midpt of chord)The perpendicular bisector of a chord passes through the centre ofthe circle; (perp bisector of chord)2Theorem** The angle subtended by an arc at the centre of a circle is double the size of the anglesubtended by the same arc at the circle (on the same side of the chord as the centre); ס at centre 2 ס ס at circumference)( ס Corollary1. Angle in a semi-circle is 900( ס ס s in semi circle)2. Angles subtended by a chord of the circle, on the same side of the chord, are equal ס s in the same seg)( ס 3. Equal chords subtend equal angles at the circumference (equal chords; equal ס s)4. Equal chords subtend equal angles at the centre (equal chords; equal ס s)CorollaryConverse35. Equal chords in equal circles subtend equal angles at the circumference of the circles.(equal circles; equal chords; equal ס s)1. If the angle subtended by a chord at the circumference of the circleis 900, then the chord is a diameter. (converse ס s in semi circle)2. If a line segment joining two points subtends equal angles at two points on the same sideof the line segment, then the four points are concyclic.Theorem** The opposite angles of a cyclic quadrilateral are supplementary; (opp ס s of cyclic quad)ConverseCorollaryIf the opposite angles of a quadrilateral are supplementary then the quadrilateral is a cyclicquadrilateral. (opp ס s quad sup OR converse opp ס s of cyclic quad)The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle of thequadrilateral. (ext ס of cyclic quad)If the exterior angle of a quadrilateral is equal to the interior opposite angle of thequadrilateral, then the quadrilateral is cyclic.Converse(ext ס int opp ס OR converse ext ס of cyclic quad)The tangent to a circle is perpendicular to the radius/diameter of theTheoremcircle at the point of contact.(tan ٣ radius)If a line is drawn perpendicular to a radius/diameter at the point where the radius/diameterConversemeets the circle, then the line is a tangent to the circle. (line ٣ radius)Two tangents drawn to a circle from the same point outside the circle are equal in length.Theorem(Tans from common pt OR Tans from same pt)Theorem** The angle between the tangent to a circle and the chord drawn from the point of contact isequal to the angle in the alternate segment. (tan chord theorem)If a line is drawn through the end-point of a chord, making with the chord an angle equal toConversean angle in the alternate segment, then the line is a tangent to the circle.(converse tan chord theorem OR ס between line and chord)Corollary456

Mathematics Telematics Resources Gr 1117February to October 2017QUESTION 1O is the centre of the circle PTR. N is a point on chord RP such that ON A PR. RSand PS are tangents to the circle at R and P respectively.RS 15 units; TS 9 units; RP̂S 42,83q.P42,83qON9TS15R 1Calculate the size of NÔR. 2Calculate the length of the radius of the circle.QUESTION In the diagram, PQRS is a cyclic quadrilateral. PS and QR are produced and meet at T. PRbisects Q P̂ S . Also, PŜR 92q and QP̂S 68 .P12S192 68 TM2311Q2RCalculate the size of the following angles: .1RP̂T .2TQ̂S .3PQ̂S .4T̂

Mathematics Telematics Resources Gr 1118February to October 2017QUESTION In the diagram, M is the centre of the circle. A, B, C, K and T lie on the circle.AT produced and CK produced meet in N. Also NA NC and B̂ 38q .B .1Calculate, with reasons, the size of the 38 following angles:(a) KM̂ A ˆ 2(b) TCM(c) Ĉˆ(d) K A1 2 .2Show that NK NT. .3Prove that AMKN is a cyclic quadrilateral.3K412TQUESTION In the diagram M is the centre of the circle passing through points L, N and P.PM is produced to K. KLMN is a cyclic quadrilateral in the larger circle having KL MN.LP is joined. KM̂L 20qK1 2 .1LWrite down, with a reason, the size2of NK̂ M .1 .2Give a reason why KN LM. .3Prove that KL LM. .4Calculate, with reasons, the size of: 4. .4.2KN̂MLP̂N20 1M2 4311N22P

Mathematics Telematics Resources Gr 1119February to October 2017QUESTION P is a common chord of the two circles. The centre, M, o t rger . In the diagram, Qcircle lies on the circumference of the smaller circle. PMNQ is a cyclic quadrilateral in thesmaller circle. QN is produced to R, a point on the larger circle. NM produced meets thechord PR at S. P̂2 x.P21xS1 2M1R211 2NQ .1Give a reason why N̂ x.2 . 2Write down another angle equal in size to x. Give a reason. .3 4Determine the size of R̂ n termsiof x.Prove that PS SR.K .2In the diagram O isthe centre of thecircle. KM and LMare tangents to thecircle at K and Lrespectively. T is apointonthecircumference of thecircle. KT andTL are joined. Ô1 106 .O 1 106 21TL .2.1Calculate, with reasons, the size of T̂1 . .2.2Prove that quadrilateral OKML is a kite. .2.3Prove that quadrilateral OKML is a cyclic quadrilatera Calculate, with reasons, the size of M̂ . .2.4M

Mathematics Telematics Resources Gr 1120February to October 2017QUESTION In the diagram, PN is a diameter of the circle with centre O. RT is a tangent to the circle atR. RT produced and PN produced meet at M. OT is perpendicular to NR. NT andOR are drawn.MN1O13322PS2 31T1 2 3R .2Prove that TO RP. .2It is further given thatTR̂ N x . Name TWO o ther angles each equal [ .3Prove that NTRO is a cyclic quadrilateral. .4Calculate the size of M̂in terms of x. .5Show that NT is a tangent to the circle at N.QUESTION BIn the diagram, the vertices A, Band C of 'ABC are concyclic.EB and EC are tangents to thecircle at B and C respectively.T is a point on AB such thatTE AC. BC cuts TE in F.2 1T .1Prove that Bˆ .2Prove that TBEC is a cyclicquadrilateral.1T̂3 .1322F1E321 .3Prove that ET bisects BT̂ C . .4If it is given that TB is a tangent to the circle through B, F and E, prove that TB TC. .5Hence, prove that T is the centre of the circle through A, B and C.AC

Mathematics Telematics Resources Gr 11 6 February to October 2017. Session 2: Equations & Inequalities In this session we will be solving quadratic equations and quadratic inequalities. The standard form of a quadratic equ