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1 SECONDARY SCHOOL IMPROVEMENT PROGRAMME (SSIP) 2016 GRADE 12 SUBJECT: MATHEMATICS LEARNER NOTES (Page 1 of 55) © Gauteng Department of Education 2 TABLE OF CONTENTS SESSION TOPIC PAGE 10 Revision of Trigonometry (Grade 11 and12) 3 - 13 11 Revision of Calculus 13 - 21 12 Revision of Analytical Geometry 21 - 28 13 Revision of Algebra 29 - 33 14 Revision of Functions 33 - 42 15 Revision of Gr 11 Euclidean Geometry 43 - 55 © Gauteng Department of Education 3 SESSION NO: 10 (CONSOLIDATION) TOPIC: REVISION OF TRIGONOMETRY (GRADE 11 AND 12) SECTION A: TYPICAL EXAM QUESTIONS QUESTION 1: (6 MARKS) (5 MINUTES) (DoE papers pre-2014) This question revises compound and double angles. Please revise reduction formulae and special angles before attempting this question. Simplify without using a calculator: (a) cos(50 x)cos(20 x) sin(50 x)sin(20 x) (3) (b) (1 2 sin 75)( 2 sin 75 1) (3) QUESTION 2: (30 MARKS) (20 MINUTES) (DoE papers pre-2014) Revise the use of a calculator when solving a trigonometric equation. The graphical method is used in the solutions (see Section B). However, you may also use the reference angle method (see Section B). Revise identities before doing this question. Determine the general solution for the following equations: (a) 2 sin 2,5 (4) (b) cos 2 0,5 (4) (c) 2 tan (d) 3sin 2cos 0 (4) (e) sin 2 3cos 2 5sin (7) (f) cos3x sin x (7) 1 2 (4) QUESTION 3: (8 MARKS) (5 MINUTES) (DoE papers pre-2014) The equations to be solved are slightly more complicated. Specific solutions must be given in question (a). Reduction formulae, negative angles and identities feature in question (b). © Gauteng Department of Education 4 (a) Calculate the value of x, correct to one decimal digit if x 90 ; 360 and cos2x tan 23 (b) (3) Determine the general solution to the equation: cos( x 360) cos(90 x) 0 (5) QUESTION 4: (20 MARKS) (10 MINUTES) (DoE papers pre-2014) Before attempting this question, ensure that you revise sketching the graphs of trigonometric functions. The four main shifts need to be mastered before this question can be done (see Section B). Solving trigonometric equations involving double angles is dealt with in (a). Given: f ( x) 1 sin x and g ( x) cos2 x (a) (b) (c) Calculate the points of intersection of the graphs of the two functions for x 180 ; 360 (7) Draw sketch graphs of f and g for x 180 ; 360 on the same set of axes. (4) For which values of x will f ( x) g ( x) for x 180 ; 360 (3) Here are some additional level 4 challenges: (d) For which values of x will f ( x).g ( x) 0 for x 180 ; 360 (3) (e) For which values of x will f ( x).g ( x) 0 for x 180 ; 360 (3) QUESTION 5: (14 MARKS) (10 MINUTES) (DoE papers pre-2014) Before attempting this question, ensure that learners revise sketching the graphs of trigonometric functions. The four main shifts need to be mastered before this question can be done (see Section B). Solving trigonometric equations involving compound angles is dealt with in (b). (a) Draw neat sketch graphs of f ( x) cos x 30 and g ( x) sin x 30 on the same set of axes for x 120;210 . Indicate the intercepts with the axes as well as the coordinates of the endpoints of the graphs. (8) (b) Hence solve f ( x) g ( x) algebraically. © Gauteng Department of Education (6) 5 QUESTION 6: (10 MARKS) (5 MINUTES) (DoE papers pre-2014) This question focuses on three-dimensional trigonometry using the sine, cosine and area rules. First make sure that you know when to use the rules in given triangles. It might be useful to construct a model of the diagram using cardboard. A, B and L are points in the same horizontal plane, HL is a vertical pole of length 3 metres, AL = 5,2 m, the angle AL̂B 113 ° and the angle of elevation of H from B is 40. H 3m 5,2m mm L 113 40 B A (a) Calculate the length of LB. (2) (b) Hence, or otherwise, calculate the length of AB. (4) (c) Determine the area of ABL. (4) QUESTION 7: Given: (a) (9 MARKS) (5 MINUTES) (Gauteng prelim 2015) cos 2 x cos x sin x Show that cos 2 x cos x sin x cos x sin x © Gauteng Department of Education (2) 6 cos 2 x 1 can be simplified to cos 2 x sin 2 x cos x sin x 2 (b) Show that cos x (c) Hence determine the general solution of cos x cos 2 x 1 cos x sin x 2 QUESTION 8: (4 MARKS) (5 MINUTES) (4) (3) (Gauteng prelim 2015) 4x (a) Write down an expression for the area of ABC (1) (b) Calculate the value of x for which the area of the triangle will be a maximum. (2) What conclusion can you draw about the type of triangle formed when the area of the triangle is a maximum? (1) (c) QUESTION 9: (10 MARKS) (5 MINUTES) (a) Determine the maximum value of the expression (b) An equilateral triangle is inscribed in a circle. (Level 4 challenges) 2cos 2 x 7 (4) Each side of the triangle measure a units. a 2 Show that the area of the circle is 3 © Gauteng Department of Education (6) 7 SECTION B: NOTES ON CONTENT Summary of all Trigonometric Theory sin sin y r cos x r tan y x sin 90 90 cos cos tan tan ( x ; y) 180 360 180 sin sin cos cos tan tan Reduction rules sin(180 ) sin cos(180 ) cos tan(180 ) tan sin(90 ) cos cos(90 ) sin sin() sin sin(180 ) sin cos(180 ) cos tan(180 ) tan sin(90 ) cos cos(90 ) sin cos() cos sin(360 ) sin cos(360 ) cos tan(360 ) tan tan() tan Whenever the angle is greater than 360 , keep subtracting 360 from the angle until you get an angle in the interval 0 ;360 . © Gauteng Department of Education 8 Identities cos 2 sin 2 1 tan sin cos The following identities are important for tackling Grade 12 Trigonometry: Compound angle identities Double angle identities cos 2 sin 2 cos 2 sin 2 2sin cos sin(A B) sin A cos B cos A sin B sin(A B) sin A cos B cos A sin B cos(A B) cos A cos B sin A sin B cos(A B) cos A cos B sin A sin B cos 2 sin 2 cos 2 2cos 2 1 2 1 2sin Solving two-dimensional problems using the sine, cosine and area rules The sine-rule can be used when the following is known in the triangle: - more than 1 angle and a side - 2 sides and an angle (not included) sin A sin B sin C a b c The cosine-rule can be used when the following is known of the triangle: - 3 sides - 2 sides and an included angle a2 b2 c2 2bc cosA The area of any triangle can be found when at least two sides an included angle are known Area of ABC 1 ab sin C 2 Trigonometric equations The following method (called the graphical method) is used in this session to determine the general solutions of trigonometric equations: If sin a and 1 a 1 then sin 1 (a) k .360 (k ) or 180 sin 1 (a) k .360 (k ) If cos a and 1 a 1 then cos 1 (a) k .360 If tan a and (k ) a then tan 1 (a) k .180 Note: © Gauteng Department of Education (k ) 9 Some schools use a different method which involves ignoring the negative sign when calculating the acute reference angle and then obtaining two solutions by locating quadrants. Then the general solution is obtained using these two angles. This method is acceptable as well. For example, you solve the following equation using this method: sin x 0,5 Reference angle (acute) 30 You ignored the negative sign and calculated the reference angle by working out SHIFT/sin/0.5 on your calculator Since sin x is negative in quadrant 3 and 4 x (180 30) k .360 x 210 k .360 OR x (360 30) k .360 x 330 k .360 where k Trigonometric Graphs Amplitude shifts Graphs of the form y a sin x, y a cos x and y a tan x involve amplitude shifts. The amplitude of a trigonometric graph is defined to be: 1 distance between the maximum and minimum y values 2 Vertical shifts Graphs of the form y sin x q, y cos x q and y tan x q involve vertical shifts. The basic graphs will not change shape as with amplitude shifts. The basic graph will move up or down depending on the value of q. Period shifts Graphs of the form y sin kx , y cos kx, and y tan kx involve period shifts. The central principle is that the period of these graphs is determine by taking the period of the basic graph and dividing by the value of k. Examples Period of y sin x : Period 360 © Gauteng Department of Education 10 360 180 2 360 Period 120 3 180 Period 45 4 Period of y sin 2x : Period Period of y cos 3 x : Period of y tan 4 x : Mixed shifts You may be required to sketch graphs involving more than one shift (but not more than two). Example Consider f ( x ) 2sin 1 x. 2 This graph involves an amplitude shift as well as a period shift. The amplitude is 2 and the period is 360 1 2 720 Horizontal shifts The graphs of y sin( x p), y cos( x p), y tan( x p) involve shifts to the left and right. Examples The graph of y sin( x 30) is the graph of y sin x shifted 30 to the left. The graph of y sin( x 30) is the graph of y sin x shifted 30 to the right. SECTION C: QUESTION 1: HOMEWORK QUESTIONS (13 MARKS) (5 MINUTES) (DoE papers pre-2014) Simplify without using a calculator: (a) cos(140)cos740 sin140 sin(20) (6) (b) cos2 375 cos2 (75) sin(50)sin 230 sin 40 cos310 (7) © Gauteng Department of Education 11 QUESTION 2: (a) (b) cos2x 1 0 (3) (2) 2sin x cos x 0 (4) (3) 2sin 2 x 2sin x cos x 0 (7) (4) 4cos2 x 2sin x cos x 1 (8) Solve for if sin 30 0,44 , and 0;720 (6 MARKS) (b) (5 MINUTES) (6) (DoE papers pre-2014) On the axes provided on the diagram sheet, draw neat sketch graphs of the following functions for the interval x 90 ;180 : g ( x) sin 2x (4) Hence determine graphically the values of x for which: sin 2 x tan x QUESTION 4: (a) (DoE papers pre-2014) (1) f ( x) tan x and (b) (15 MINUTES) Determine the general solutions of the following equations: QUESTION 3: (a) (28 MARKS) (14 MARKS) (2) (5 MINUTES) (DoE papers pre-2014) Solve for x if cos( x 30) sin3x where x 60 ;120 . Sketch the graphs of the following functions on the same set of axes for the interval x 60 ;120 . f ( x) cos( x 30) and g ( x) sin3x (c) (6) Determine graphically the values of x for which cos( x 30) sin3x QUESTION 5: (a) (6) (7 MARKS) (5 MINUTES) (2) (DoE papers pre-2014) Given: f ( x) sin x Sketch the graphs of the following on the same set of axes for the interval x 0 ;180 : (1) g ( x) f ( x) 3 (2) © Gauteng Department of Education 12 (2) h( x) 3 f ( x) 1 (2) (b) Determine graphically the value of x for which g ( x) h( x) (c) Explain the transformations involved in the following function: y sin(2x 60) (1) (2) QUESTION 6: (9 MARKS) (5 MINUTES) (DoE papers pre-2014) Two ships, A and B, are 120 km apart. Ship A is at a bearing of 67 from D and 97 km away from D. DN points due north. Ship B is at a bearing of 208 from D. 67 208 (a) (b) ˆ , Determine the bearing of Ship A from Ship B, i.e, MBA when BM||DN . (6) If Ship A travels due north, and Ship B travels due south, then at some instant of time, Ship A is due east of Ship B. Calculate the distance between the two ships at that instant. (3) QUESTION 7: (10 MARKS) (10 MINUTES) (Gauteng prelim 2015) In the following diagram, DE represents a vertical cell phone tower positioned on one corner of a field. The field is shaped as a cyclic quadrilateral EFGH and E, F, G and H are all in the same horizontal plane. From H, the angle of elevation to D, the top of the cell phone tower is α . EH EF p units and Ĝ β . © Gauteng Department of Education 13 α β (a) Write down DE in terms of α (b) Show that: (1) (2) (3) 1 Ĥ 2 β 2 (give reasons) FH 1 2cos β 2 FH p 2(1 cosβ) p (1) (3) (3) (3) SESSION NO: 11 (CONSOLIDATION) TOPIC: REVISION OF CALCULUS SECTION A: TYPICAL EXAM QUESTIONS QUESTION 1: (21 MARKS) (15 MINUTES) (DoE papers pre-2014) In this question first principles is revised as well as differentiation. Ensure correct layout and notation. Remember that the gradient of the tangent to a curve at a point is the derivative of the function at that point. (a) Determine f ( x) from first principles if f ( x) 4 x 2 © Gauteng Department of Education (5) 14 (b) Determine: (1) (2) (3) (c) dy 8 if y 2 x dx x 2 2p 3p 2 Dp p2 dy if xy 5 x3 dx (4) (3) (5) Determine the value of k for which the gradient of the tangent to g ( x) 3x2 kx 4 at the point where x 2 is 8. (4) QUESTION 2: (17 MARKS) (10 MINUTES) (DoE papers pre-2014) Curve sketching is revised in this question. Remember that at a turning point the derivative is zero and at that the point itself can be substituted into the equation of the function. Simultaneous equations can then be used to determine the value of the unknowns. Determining the equation of a tangent is also revised. Consider the function f ( x) ax3 bx 2 The graph of f ( x) ax3 bx 2 has a local turning point at (2 ; 4) . (a) Determine the value of a and b. (6) (b) Sketch the graph of f on a set of axes. (7) (c) Determine the equation of the tangent to f at x 1 (4) QUESTION 3: (12 MARKS) (10 MINUTES) (DoE papers pre-2014) Problems involving area, perimeter, surface area and volume are quite important for you to master. Spend time discussing these concepts in this question. A builder wishes to construct a steel window frame in the shape of a rectangle with a semi-circular part on top. The radius of the semi-circular part is r metres and the width of the rectangular part is h metres. © Gauteng Department of Education 15 (a) (b) Write down, in terms of h and r: (1) the steel perimeter (P) of the frame. (2) (2) the area enclosed by the frame. (2) The area enclosed by the frame is to be 4 4 square metres. Show that the perimeter (P) is: P 2 r r 2 (c) (4) If the steel for the frame costs R10 per metre, calculate the value of r for which the total cost of the steel will be a minimum. (4) QUESTION 4: (2 MARKS) (5 MINUTES) (Level 4 challenge) Redraw the given diagram of y f ( x) . Now draw a possible graph of y f ( x) on the same set of axes. (2) QUESTION 5: (2 MARKS) (5 MINUTES) (DoE papers pre-2014) A stone is thrown vertically upwards and its height (in metres) above the ground at time t (in seconds) is given by: h(t ) 35 5t 2 30t (a) Determine its initial height above the ground. (2) (b) Determine the initial speed with which it was thrown. (2) (c) Determine the maximum height above the ground that the stone reached. (5) (d) How fast was the stone travelling when it reached a height of 60 metres above the ground on the way down? (5) How fast was the stone travelling when it hit the ground? (6) (e) © Gauteng Department of Education 16 QUESTION 6: (13 MARKS) (10 MINUTES) (Level 4 challenges) In the diagram alongside, the graph of y f ( x) is represented where f ( x) ax3 bx2 cx represents a cubic function. y f ( x) 15 1 5 (a) Determine the equation of f ( x) and f ( x) where f ( x) ax3 bx2 cx (b) For which values of x will the graph of f ( x) (c) (8) (1) increase? (1) (2) decrease? (2) At which values of x will the graph of f have its turning points? (2) QUESTION 7: (13 MARKS) (10 MINUTES) (Level 4 challenges) 3 2 The graph of f ( x) x 4 x 4 x k passes through the origin, has a local maximum at (b ; c) and a local minimum at (a ; 0) . (a) Explain why k 0 (b) Using this value of k , determine the values of: (c) (d) (1) (1) a (4) (2) b (4) The graph of g with equation g ( x) mx is the tangent to f at the point (0 ; 0) . Calculate the value of m. (2) Determine the values of p for which x3 4 x2 4 x 2 p will only have one negative solution. (2) © Gauteng Department of Education 17 QUESTION 8: (10 MARKS) (10 MINUTES) (Level 4 challenges) The profit (P) in rand/day which the owner of a taxi makes when the taxi is driven at an average speed of x km/hour is given by the formula: x P x 2 10 12 (a) Calculate the average speed at which the owner of the taxi will suffer a loss in revenue. (4) (b) Calculate the most economical average speed of the taxi and the corresponding daily profit. (6) SECTION B: NOTES ON CONTENT The most important fact in Calculus is that the gradient of the tangent to a curve at a given point is the gradient of the curve at that point. Other words for gradient are: Symbols for gradient are: f ( x) rate of change, derivative, slope Dx dy dx f (a ) is the gradient of f at x a f (a ) is the y -value corresponding to x a f (a) mt m f f (a ) Average gradient The average gradient (or average rate of change) of a function f between x a and x b and is defined to be the gradient of the line joining the points on the graph of the function. We say that the average gradient of f over the interval is the gradient of the line AB. Gradient of a curve at a point using first principles The formula to determine the gradient of a function from first principles is given by the following limit: f ( x) lim h 0 f ( x h) f ( x) h © Gauteng Department of Education 18 The gradient of a function using the rules of differentiation You will be required to use the following rules of differentiation to determine the gradient of a function. Rule 1 Rule 2 Rule 3 If f ( x) axn , then f ( x) a.nx n1 If f ( x) ax, then f ( x) a If f ( x) number, then f ( x) 0 Determining the equation of the tangent to a curve at a point The gradient of the tangent to a curve at a point is the derivative at that point. The equation is given by y y1 m( x x1 ) where ( x1 ; y1 ) is the point of tangency and m f ( x1 ) Rules for sketching the graph of a cubic function The graph of the form f ( x) ax3 bx 2 cx d is called a cubic function. The main concepts involved with these functions are as follows: Intercepts with the axes: For the y-intercept, let x 0 and solve for y For the x-intercepts, let y 0 and solve for x (you might have to use the factor theorem here) Stationary points: Determine f ( x) , equate it to zero and solve for x. Then substitute the x-values of the stationary points into the original equation to obtain the corresponding y-values. If the function has two stationary points, establish whether they are maximum or minimum turning points. Points of inflection: If the cubic function has only one stationary point, this inflection that is also a stationary point. For points of inflection that are not stationary points, find solve for x. Alternatively, simply add up the x-coordinates divide by 2 to get the x-coordinate of the point of inflection. point will be a point of f ( x) , equate it to 0 and of the turning points and © Gauteng Department of Education 19 Concavity: The graph of f is concave up if f ( x) 0 The graph of f is concave down if f ( x) 0 SECTION C: HOMEWORK QUESTIONS QUESTION 1: (a) (b) Given: (13 MARKS) (10 MINUTES) (DoE papers pre-2014) 1 f ( x) 2 x 2 2 (1) Determine the gradient of f at x 2 by using first principles. (6) (2) Determine the equation of the tangent to f at x 2 (3) Determine dy if dx y 6 3 6 x 1 6 x6 (4) (Write your answer with positive exponents) QUESTION 2: (4 MARKS) (5 MINUTES) (DoE papers pre-2014) 3 2 The function g defined by g ( x) ax bx cx d has the following properties. Use this information to draw a neat sketch graph of g without actually solving for a, b, c and d. g (0) 32, g (4) 0, g ( 2) 0, g (0) 0, g (4) 0 g ( x) 0 if x 0 or x 4 g ( x) 0 if 0 x 4 (4) QUESTION 3: (2 MARKS) (5 MINUTES) (DoE papers pre-2014) The graph of y g ( x ) is sketched below. This graph represents the derivative graph of a quadratic function g. y g ( x ) 1 2 © Gauteng Department of Education 20 (a) (b) Determine the values of x for which the graph of g decreases. Write down the x-coordinate of the turning point of g. (1) (1) QUESTION 4: (10 MARKS) (10 MINUTES) (DoE papers pre-2014) Consider the diagram below. OCDE is an inscribed rectangle in ABO. A(0; 6) and B(8;0) are given points. A(0;6) B(8;0) (a) Determine the equation of line AB. (2) (b) By using calculus methods, determine the dimensions of rectangle OCDE so that OCDE has a maximum area. (5) (c) If rectangle OCDE is sketched such that EO = OC (thus forming a square), determine the length of the square OCDE. (3) QUESTION 5: (2 MARKS) (5 MINUTES) (DoE papers pre-2014) The graph of y g ( x ) is sketched below. This graph represents the derivative graph of a quadratic function g. y g ( x ) 1 2 (a) Determine the values of x for which the graph of g decreases. (1) (b) Write down the x-coordinate of the turning point of g. (1) © Gauteng Department of Education 21 QUESTION 6 (2 MARKS) (5 MINUTES) (Level 4 challenges) A is a town 30 km west of town B. Two athletes start walking simultaneously from the two towns. The athlete who starts from town A, walks due east in the direction of B at a constant speed of 6 km/h and reaches point P after x hours. The athlete, who starts at B, walks due north in the direction of another town C at a constant speed of 8 km/h, and reaches point Q after x hours. C . x hours A (a) . P 30km x hours Q B Prove that the distance between the athletes after x hours is given by: PQ 100 x 2 360 x 900 (5) (b) How long after they started walking were they nearest to each other? (3) (c) What was this minimum distance between them? (2) SESSION NO: 12 (CONSOLIDATION) TOPIC: REVISION OF ANALYTICAL GEOMETRY SECTION A: TYPICAL EXAM QUESTIONS QUESTION 1: (18 MARKS) (20 MINUTES) (DoE papers pre-2014) Completing the square to find the coordinates of the centre and radius is revised in this question. Make sure that you know how to determine the coordinates of the intercepts of a circle with the axes. The equation of a tangent to a circle is also revised. In the figure below, the tangent touches the circle x2 2 x y 2 4 y t 0 at D( 2; 1) . A is the centre of the circle and the circle cuts the y-axis at B and C. (a) Determine the coordinates of A, the centre of the circle. © Gauteng Department of Education (4) 22 (b) Express the radius of the circle in terms of t. (1) (c) Show that t 5 . (3) (d) Determine the coordinates of B and C. (5) (e) Write down the length of BC. (1) (f) Determine the equation of the tangent at D. (4) C D(2; 1) O A B © Gauteng Department of Education 23 QUESTION 2: (23 MARKS) (20 MINUTES) (DoE papers pre-2014) In this question, showing that a point lies on a line is revised. There is a link to trigonometry and quadrilaterals. A( 8 ; 2), B( 2 ; 6) and D(0 ; 8) are the vertices of a triangle that lies on the circumference of a circle with diameter BD and centre M, as shown in the figure below. D(0 ; 8) A( 8 ; 2) M O B( 2 ; 6) (a) Calculate the coordinates of M. (2) (b) Show that ( 8 ; 2) lies on the line y 7 x 58 (2) (c) What is the relationship between the line y 7 x 58 and the circle centred at M? (5) (d) Calculate the lengths of AD and AB. (4) (e) ˆ 90 Prove that DAB (3) (f) Write down the size of angle (1) (g) A circle, centred at a point Z inside triangle ABD, is drawn to touch sides AB, BD and DA at N, M and T respectively. Given that BMZN is a kite, calculate the radius of this circle. (6) © Gauteng Department of Education 24 QUESTION 3: (23 MARKS) (20 MINUTES) (Level 4 challenges) (a) The lines ax 3 y 5 and 2 x by 3 are parallel. Detemine ab . (b) A(1; 5) , B(3 ; 7) , C(8 ; 2) and D(x ; 0) are points in the Cartesian plane. (4) (1) ˆ 90 Show that ABC (2) (2) Show that the area of ABC 10 (5) (3) Determine the equation of AB. (3) (4) If area of ABC area of ABD , calculate the values of x. (10) SECTION B: NOTES ON CONTENT If AB is the line segment joining the points A( xA ; yA ) and B( xB ; yB ) , then the following formulas apply to line segment AB. The Distance Formula AB2 ( xB xA ) 2 ( yB yA ) 2 or AB ( xB xA )2 ( yB yA )2 The Midpoint Formula x xB yA yB M A ; where M is the midpoint of AB. 2 2 The Gradient of a line segment joining two points Gradient of AB yB yA xB xA Parallel lines Parallel lines have equal gradients. If AB||CD then mAB mCD Perpendicular lines The product of the gradients of two perpendicular lines is 1 . If AB CD, then mAB mCD 1 © Gauteng Department of Education 25 The equation of the line y yA m x xA Inclination of a line tan mAB If mAB 0, then is acute If mAB 0, then is obtuse Collinear points (A, B and C) mAB mBC or mAC mAB or mAC mBC Circles and tangents to circles The equation of a circle centre the origin is given by: x2 y 2 r 2 The equation of a circle centre (a ; b) is given by: x a 2 y b 2 r 2 mradius mtangent 1 The radius is perpendicular to the tangent: SECTION C: QUESTION 1: HOMEWORK QUESTIONS (17 MARKS) (15 MINUTES) (DoE papers pre-2014) O is the centre of the circle in the figure on the next page. P( x ; y) and Q (12 ; 5) are two points on the circle. POQ is a straight line. The point R(t ; – 1) lies on the tangent to the circle at Q. Q(12 ; 5) O R(t ; 1) P(x ; y) (a) Determine the equation of the circle. © Gauteng Department of Education (3) 26 (b) Determine the equation of the straight line through P and Q. (2) (c) Determine the coordinates of P. (2) (d) Show that the gradient of QR is (e) Determine the equation of the tangent QR in the form y ... (3) (f) Calculate the value of t. (2) (g) Determine an equation of the circle with centre Q(12 ; 5) . (3) QUESTION 2: (22 MARKS) 12 . 5 (2) (15 MINUTES) (DoE papers pre-2014) Given below is a circle with centre M passing through R(3; 2), T(5; 4) and V on the y-axis. Q( 2; 2) lies on the line RP. TV is drawn and then extended to meet RP at P. T(5;4) R( 3;2) Q( 2; 2) (a) Determine the equation of the circle with centre M. (4) (b) Show that PR is a tangent to the circle at R. (3) (c) Determine the equations of the lines: (1) (2) PR TP (3) (6) (d) If the equations of PR and TP are y 4 x 10 and y x 1 respectively, determine the coordinates of P. (3) (e) Calculate the angle of inclination of the line TVP (refer to the angle as θ ). © Gauteng Department of Education (3) 27 QUESTION 3: (a) (13 MARKS) (11 MINUTES) (DoE papers pre-2014) Determine the centre and radius of the circle with the equation defined by x2 y 2 8x 4 y 38 0 (4) A second circle has the equation ( x 4)2 ( y 6)2 26 . Calculate the distance between the centres of the two circles. (2) (c) Hence, show that the circles intersect each other. (3) (d) Show that the circles intersect along the line y x 4 . (4) (b) QUESTION 4 (13 MARKS) (11 MINUTES) (DoE papers pre-2014) The straight line through PQ with equation 2 x y 1 cuts the y-axis at Q. The straight line through PR with equation 3x 2 y 12 cuts the x-axis at R. The coordinates of P are (2;3) . P(2;3) Determine: (a) the coordinates of Q and R. (2) (b) the equation of QR. (3) (c) the coordinates of point S, the midpoint of QR. (2) (d) the equation of PS. (1) © Gauteng Department of Education 28 (e) the size of angle rounded off to the nearest degree. (6) (f) the area of PQR rounded off to two decimal places. (6) QUESTION 5: (9 MARKS) (10 MINUTES) (Level 4 challenge) The coordinates of trapezium ABCD are A(4 ; 3), B( x ; 6), C(4 ; y ) and D(2 ; 1) AD||BC and BC 2AD . Determine the values of x and y. (9) B(x ; 6) A( 4 ; 3) C(4 ; y ) D( 2 ; 1) © Gauteng Department of Education 29 SESSION NO: 13 (CONSOLIDATION) TOPIC: REVISION OF ALGEBRA SECTION A: TYPICAL EXAM QUESTIONS QUESTION 1: (34 MARKS) (25 MINUTES) (DoE papers pre-2014) This question revises Grade 11 Algebra including algebraic fractions, the quadratic formula, quadratic inequalities, surd equations, simultaneous equations, the concept of real and non-real and the nature of roots. (a) Solve for x rounded off to two decimal places where necessary: x2 x 2 1 0 (1) x 1 (4) (2) 5x( x 3) 2 (4) (3) 20 ( x 1)( x 2) (4) (4) (b) 2 7 x 2x 0 (5) Solve simultaneously for x and y in the following set of equations: (1) y 2 x 2 and 2 x2 2 y 2 (2) (3 x y ) 2 ( x 5) 2 0 (6) (4) 4 x be non-real? x6 (c) For which value(s) of x will the expression (d) Show that the roots of the equation 5 p 2 x( x 3 p) are non-real if p 0. (3) (4) QUESTION 2: (20 MARKS) (20 MINUTES) (DoE papers pre-2014) Simplifying exponential and surd expressions and solving exponential equations is dealt with in this question. The use of the k-method is ideal for 2(b). (a) Simplify the following without the use of a calculator: (1) 3 81 . 3 3 4 (4) © Gauteng Department of Education 30 3 (2) (b) 16 3 54 3 128 3 (5) 128 Solve for x: 32 x1 4.3x2 81 0 (1) (5) 1 x x4 2 0 (2) (6) 2 x3 4 (3) QUESTION 3: (10 MARKS) (a) Show that a2 (b) Determine the values of p for which the equation 3 1 p will have real solutions. (3) (c) For which real values of x will (d) If r 0 and p 0 and the equation p( x 2) x rx( x 3) p has equal 1 can be written as a (Level 4 challenges) a (a 1) a ( x 3)2 be a real number? 16 p and hence determine the equal roots. 9 Exponential and surd laws Laws a . a a m n m m 2. 3. (3) (8) If P x 2 5 x 6 and x 2 5 x 6 0 , determine the possible values for P. (5) SECTION B: NOTES ON CONTENT 1. (4) x roots prove that r (e) (15 MINUTES) n a a mn n a (a m ) n a mn (a n ) m 1. Definitions a 1 2. 1a 1 , with aR 3. xn 0 1 xn © Gauteng Department of Education 31 4. ab m ambm 5. am a m b b 4. 1 xn n x m 5. n a m m an Rules for exponential and surd equations (a) 1 even number x (b) Whenever you have equations of the form x n number , always rewrite the negative number m m 1 expression x n in the form ( x n ) m . A negative number (c) Quadratic formula and nature of roots x b b 2 4ac 2a b 2 4ac Roots 0 0 Non-real Real 0 Real, rational and unequal Real, irrational and unequal Real, rational and equal 0 and perfect square 0 and a perfect square SECTION C: HOMEWORK QUESTIONS QUESTION 1: (41 MARKS) (30 MINUTES) (DoE papers pre-2014) (a) Solve for x rounded off to two decimal places where necessary: (1) x2 4 3x x2 (4) (2) ( x 3)( x 2) 8 (4) (3) 7 x 2 18 x 9 0 (4) (4) 5 x 1 x 0 (5) © Gauteng Department of Education 32 (b) Solve for x and y simultaneously: 2x y 7 (1) (2) (c) x2 xy 21 y 2 (7) ( x 2 8 x 16) ( y 2 2 y 1) 0 (6) 2 a and If 3 b , express the following in terms of a and b: 108 18 (d) (e) (3) For which values of p will the equation x(4 x 3) p have equal roots? Show that the roots of ( k 1) x kx 1 0 are non-real for all real 2 2 values of k. QUESTION 2: (a) (4) (23 MARKS) 9 x 1 6.32 x Show that: 3 4 x 1 (15 MINUTES) (DoE papers pre-2014) 3 for all values of x. (5) 10 2 x 20 7 8 x 20 (b) Simplify: (c) Solve for x: (5) 18 x 20 3x 1 3x 1 8 (1) (3) 2 3 6 x 54 (2) (d) (4) (3) If 4 . 6 48 , then find the value of x y x y 12 QUESTION 3: (28 MARKS) (20 MINUTES) (4) , then find the value of b in terms of a (4) Without using a calculator, show that: (b) If a 1 2 and b 1 2 n (Level 4 challenges) 4 8 16 648 250 (a) n (7) 24 16 12 © Gauteng Department of Education 33 If x 3 2 2 3 2 2 , then determine the value of x without using a calculator. (6) (d) Solve for x x 2 2 x 2 0 leaving answers in surd form. (4) (e) If ( x y 1)( x y 3) 0 and x y 3 , find the value of x y (3) (f) For which real values of k will (c) 1 k have real roots? x 1 2 SESSION NO: 14 TOPIC: REVISION OF FUNCTIONS (7) (CONSOLIDATION) SECTION A: TYPICAL EXAM QUESTIONS QUESTION 1: (14 MARKS) (10 MINUTES) (DoE papers pre-2014) In this question, the focus is on revising quadratic functions. Revise the two different forms of a quadratic function and how to determine the coordinates of the turning point for each form. (a) (b) Consider the function f ( x) ( x 1)2 4 (1) Draw a neat sketch graph indicating the coordinates of the intercepts with the axes, the coordinates of the turning point and the equation of the axis of symmetry. (6) (2) Write down the range. (2) (3) Determine the values of x for which the graph increases. (1) Consider the function g ( x) x2 4 x 3 (1) Determine the maximum value of g. (2) Determine the coordinates of the turning point of the graph of g ( x 1) (3) (2) © Gauteng Department of Education 34 QUESTION 2: (13 MARKS) (10 MINUTES) (DoE papers pre-2014) In this question, you are required to determine the equation of a quadratic and hyperbolic function. This is extremely important for Paper 1. In the diagram below, the graphs of the following functions are represented: a q x p f ( x) a( x p)2 q and g ( x) The turning point of f is (1; 4) and the asymptotes of g intersect at the turning point of f. Both graphs cut the y-axis at 3. (1 ; 4) 3 (a) Determine the equation of f. (4) (b) Determine the equation of g. (4) (c) Determine the coordinates of the x-intercept of g. (3) (d) For which values of x will g ( x) 0 ? (2) © Gauteng Department of Education 35 QUESTION 3: (19 MARKS) (15 MINUTES) (DoE Nov 2015) This is quite a challenging question with some level 4 type questions. Given: h( x) 2x 3 for 2 x 4 and the x-intercept of the graph is Q. 2 4 P( x ; y) 3 (a) Determine the coordinates of Q. (2) (b) Write down the domain of h 1 (3) (c) Sketch the graph of h 1 , clearly indicating the intercepts and endpoints. (3) (d) For which value(s) of x will h( x) h1( x) ? (3) (e) P(x ; y) is the point on the graph of h that is closest to the origin. Calculate the distance OP. (5) (f) Given: h( x) f ( x) where f is a function defined for 2 x 4 (1) Explain why f has a local minimum. (2) Write down the value of the maximum gradient of the tangent to the graph of f. (1) © Gauteng Department of Education (2) 36 QUESTION 4: (4 MARKS) The function defined as y The domain is x (5 MINUTES) (DoE Nov 2015 level 4) a q has the following properties: x p where x 2 y x 6 is an axis of symmetry The function is increasing for all x where x 2 Draw a neat sketch graph showing asymptotes, if any. (4) QUESTION 5: (6 MARKS) (5 MINUTES) (Gauteng prelim 2015) The straight line y mx 32 is a tangent to the graph of y 2 x2 4 x 30 . Calculate the possible values of m. (6) QUESTION 6: (7 MARKS) (10 MINUTES) (Gauteng prelim 2015) 2 The sketch below shows the graphs of f ( x) 2 x 5x 3 and g ( x) ax q The angle of inclination of graph g is 135 in the direction of the positive x-axis. P is the tangent point of f and g such that g is a tangent to the graph of f at P. 135 (a) Calculate the coordinates of the point of contact between the two graphs. (4) (b) Hence, or otherwise, determine the equation of g (2) (c) Determine the values of d for which the line y x d will not intersect the graph of f. (1) © Gauteng Department of Education 37 QUESTION 7: (12 MARKS) (10 MINUTES) (Exemplar 2014) In this question, the focus is on exponential functions. It is important for you to understand the concept of reflection about the axes as well as translating graphs horizontally and vertically (see (c) and (d)). In the diagram below (not drawn to scale), the graph of f ( x) 2a x . The graph of f passes through the point (1; 4) and cuts the y-axis at A. ( 1 ; 4) 1 and hence write down the equation of f. 2 (a) Show that a (b) Determine the coordinates of A. (c) Show that the equation of g, the reflection of f about the y-axis, can (2) be written as g ( x) 2 x 1 (d) (3) (3) Draw a neat sketch graph of y f ( x 1) 2 indicating the intercepts with the axes as well as the equation of the asymptote. (4) QUESTION 8: (10 MARKS) (5 MINUTES) (DoE papers pre-2014) This question revises exponential and logarithmic functions. The graph of f ( x) a x 1 passes through the point 0 ; 1 . 2 (a) Calculate the value of a. (2) (b) Write down the equation of the inverse of f in the form f 1 ( x) .... (2) (c) Sketch the graph of f and its inverse on the same set of axes. (5) (d) Write down the domain of f 1 (1) © Gauteng Department of Education 38 SECTION B: NOTES ON CONTENT QUADRATIC FUNCTIONS (a) The general equation for the quadratic function is y a( x p) 2 q . (remember that the exponent of x is always squared). (b) Draw the “mother graph” y ax 2 by using a table of selected x-values. (c) The value of a tells us if the graph is concave (happy) or convex (sad). (d) If a 0 , the shape is concave If a 0 , the shape is convex. 2 The graph of y a( x p) q is obtained by shifting the graph of y ax 2 by p units to the left or right and then q units up or down. If p 0 , the shift is left If p 0 , the shift is right If q 0 , the shift is upwards If q 0 , the shift is downwards (e) For graphs of the form y ax 2 q , the value of p is 0 meaning that the mother graph will shift up or down but not left or right. The equation of the axis of symmetry of the graph y a( x p) 2 q is obtained by putting the expression x p 0 and solving for x. The equation of the axis of symmetry for the graph of a parabola of the form y ax 2 q is x 0 (the y-axis). The axis of symmetry passes through the x- (f) coordinate of turning point of the parabola. The value of q is the y-coordinate of the turning point. To determine the coordinates of the turning point of parabolas of the form y ax 2 bx c , express y ax 2 bx c in the form y a( x p)2 q by completing the square or use the formula x (g) b to find the x-coordinate and 2a hence substitute this x-value into the original equation to find the y-coordinate. The y-intercept can be determined by letting x 0 in the equation of y a( x p)2 q . The y-intercept of the graph of y ax 2 q is the value of q. (h) If the parabola cuts the x-axis, the x-intercepts can be determined by letting (j) y 0 in the equation of y a( x p)2 q and solving for x. If a 0 the parabola has a minimum value at q. If a 0 the parabola has a maximum value at q. If a 0 then the graph of the parabola decreases for all x p and increases for all x p. If a 0 then the graph of the parabola increases for all x p and decreases for all x p . (k) For any quadratic function of the form y ax 2 q or y a( x p) 2 q : (i) Domain: x ; © Gauteng Department of Education 39 Range: y q ; y ; q if a 0 if a 0 HYPERBOLIC FUNCTIONS (a) (b) (b) (c) (d) The general equation for the hyperbola is y (remember that x is in the denominator). a The hyperbola y q has two asymptotes: x Vertical asymptote: x 0 (y-axis) Horizontal asymptote: a The hyperbola y q has two asymptotes: x p Vertical asymptote: x p 0 Horizontal asymptote: Shape: a0 a0 (g) yq yq (The dotted lines are the asymptotes) a Start by drawing the graph of y by using a table of selected x-values (preferably x two positive and two negative factors of a). a a The graph of y q is obtained by shifting the graph of y q units up or down. x x a If q 0 , then the graph of y shifts q units upwards. x a If q 0 , then the graph of y shifts q units downwards. x a a q is obtained by shifting the graph of y The graph of y p units left or x p x right and then q units up or down. If p 0 , the shift is left If q 0 , the shift is upwards (e) (f) a q x p If p 0 , the shift is right If q 0 , the shift is downwards The y-intercept can be determined by letting x 0 . The x-intercept can be determined by letting y 0 . a For any hyperbolic function of the form y q : x Domain: x ; where x 0 Range: y ; where y q For any hyperbolic function of the form y a q: x p Domain: x ; where x p 0 Range: y ; where y q © Gauteng Department of Education 40 EXPONENTIAL FUNCTIONS (a) The general equation for the exponential function is y ab x p q where b 0 and (b) b 1 (remember that the exponent is x). The exponential graph has one asymptote: Horizontal asymptote: y q (c) Draw the basic graph y ab x by using a table of selected x-values (best to use x 1; 0 ;1 ). (d) The graph of y ab x p q is formed by shifting the graph of the function y ab x p units horizontally (left or right) and then q units vertically (up or down). If p 0 , the shift is left If q 0 , the shift is right If q 0 , the shift is upwards If q 0 , the shift is downwards (e) Determine the x-intercept of the graph of y ab x p q by letting y 0 . (f) (g) The y-intercept can be determined by letting x 0 in the equation of y ab x p q . For any exponential function of the form: y ab x p q [ a 0, b 0 and b 1 ] x ; Domain: y q ; Range: Summary of the exponential and logarithmic functions Increasing functions Decreasing functions f ( x) a x a 1 yx f ( x) a x 0 a 1 f 1 ( x ) log a x (0 ;1) a 1 yx (0 ;1) (1;0) (1;0) f 1 ( x ) log a x 0 a 1 Some important theory: The expression log a x is defined only if: 0 a 1 or a 1 (a cannot be negative, 0 or 1) x 0 (x cannot be negative or 0) © Gauteng Department of Education 41 SECTION C: QUESTION 1: HOMEWORK QUESTIONS (10 MARKS) (5 MINUTES) (DoE papers pre-2014) 2 x Functions y f ( x ) bx and y g ( x) a are not sketched according to scale. 9 2 8 (a) Determine the values of a and b. (b) Write down a possible restriction for the domain of f ( x) so that the inverse of f will be a function. (1) (c) Hence determine the equation of the inverse function of f ( x) . (3) (d) Write down the equation of the inverse of g ( x) in the form g 1( x) ..... (2) (e) If h( x) g ( x 2) 3 , state the transformation that was applied to g ( x) to form h( x) and hence or otherwise determine the equation of h( x) (3) QUESTION 2: Given: (3) f ( x) (23 MARKS) (15 MINUTES) (DoE papers pre-2014) 2 x 1 (a) Write down the equations of the asymptotes. (2) (b) Sketch the graph of f indicating the coordinates of the y-intercept as well as the asymptotes. (6) © Gauteng Department of Education 42 (c) Determine graphically the values of x for which (d) State the domain of f. 2 1 x 1 (2) (2) QUESTION 3: (23 MARKS) (15 MINUTES) (DoE papers pre-2014) The straight line f ( x) 4 x 32 passes through the turning point of the parabola with equation g ( x) a( x p) 2 q . The line and parabola intersect at the point (7 ; 4) . (7 ; 4) (a) Determine the coordinates of D. (2) (b) Determine the equation of g. (4) (c) State the range of g. (1) QUESTION 4: (6 MARKS) (5 MINUTES) (DoE Exemplar 2014) x An increasing exponential function with equation f ( x) a . b q has the following properties: Range is y 3 The points (0 ; 2) and (1; 1) lie on the graph (a) Determine the equation that defines f (4) (b) Describe the transformation from f ( x) to h( x) 2 . 2 x 1 (2) © Gauteng Department of Education 43 SESSION NO: 15 (CONSOLIDATION) TOPIC: REVISION OF GRADE 11 EUCLIDEAN GEOMETRY SECTION A: TYPICAL EXAM QUESTIONS QUESTION 1: (a) (b) (10 MARKS) (10 MINUTES) (DoE papers pre-2014) Complete: (1) The angle in a semi-circle is ……. (1) (2) The perpendicular from the centre of a circle to a chord …….. (1) In the diagram below, AOB is a diameter of the circle centre O. OD||BC. The length of the radius is 10 units. (1) What is the size of Ĉ ? State a reason. (2) (2) What is the size of Ê1 ? State a reason. (2) (3) Why is AE EC ? State a reason. (1) (4) If AC 16 units, calculate the length of ED. (3) © Gauteng Department of Education 44 QUESTION 2: (15 MARKS) (10 MINUTES) (DoE papers pre-2014) In the figure below, RDS is a tangent to the circle centre O at D. BC = DC and C D S = 40° 40 (a) What is the size of B̂1 . State a reason. (2) (b) What is the size of D̂ 2 . State a reason. (2) (c) (d) What is the size of Ĉ . State a reason. Calculate the size of Ô 2 State a reason. (2) (2) (e) Calculate the size of Ô1 . State a reason. (2) (f) Calculate the size of D̂3 State reasons. (3) (g) Calculate the size of  State a reason. (2) QUESTION 3: (9 MARKS) (5 MINUTES) (DoE papers pre-2014) In the diagram, O is the centre of the circle passing through A, B, C and D. AB||CD and B̂ 20 20 (a) Calculate the size of Ĉ1 ? State a reason. (2) (b) Calculate the size of Ô1 ? State a reason. (2) (c) (d) (e) Calculate the size of D̂ ? State a reason. Calculate the size of Ê1 ? State a reason. Why is AOEC a cyclic quadrilateral? (2) (2) (1) © Gauteng Department of Education 45 QUESTION 4: (9 MARKS) (5 MINUTES) (DoE papers pre-2014) LOM is a diameter of circle LMT. The centre is O. TN is a tangent at T. LN NP . MT is a chord. LT is a chord produced to P. L . O M 1 2 N 1 2 3 4 T 2 1 P Prove that: (a) MNPT is a cyclic quadrilateral (4) (b) NP = NT (5) QUESTION 5: (9 MARKS) (5 MINUTES) (DoE papers pre-2014) In the diagram below, AB is a diameter of the circle ABCD. AE is a tangent to the circle at A. B̂1 x . (a) Prove that AB is a tangent to the circle through A, D and E. (7) (b) ˆ Eˆ Prove that C 1 (2) © Gauteng Department of Education 46 QUESTION 6: (4 MARKS) (5 MINUTES) (DoE Nov 2014) In the diagram, AB and AE are tangents to the circle at B and E respectively. BC is a diameter of the circle. AC 13, AE x and BC x 7 . Calculate the length of AB. (4) x 13 x7 QUESTION 7: (9 MARKS) (10 MINUTES) (Level 4 challenge) ˆ . T lies on PR is a common chord of circles PQSR and PXRY. PR bisects XRY chord PS such that ST SR SQ . MPY touches the larger circle at P. Prove that: (a) ˆ Q ˆ Q 1 2 (6) (b) QP is a tangent to the smaller circle. (7) © Gauteng Department of Education 47 SECTION B: NOTES ON CONTENT REVISION OF GRADE 11 THEOREMS THEOREM 1 The line segment joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. If . O AM MB then OM AB which means that ˆ M ˆ 90 M 1 2 A || 1 2 || M B THEOREM 1 CONVERSE The perpendicular drawn from the centre of a circle to a chord bisects the chord. If OM AB then . O AM MB A THEOREM 2 || 1 2 || M B The angle which an arc of a circle subtends at the centre of a circle is twice the angle it subtends at the circumference of the circle. C C x x C . . O 1 2x A B A A 1 . 2x 2x O x B O 1 B ˆ 2C ˆ For all three diagrams: O 1 C THEOREM 3 The angle subtended at the circle by a diameter is a A right angle. We say that the angle in a semi-circle is 90 . In the diagram Ĉ 90 © Gauteng Department of Education . B 48 C THEOREM 3 CONVERSE If the angle subtended by a chord at a point on the circle is 90 , then the chord is a diameter. If Ĉ 90 , then the chord subtending Ĉ is a diameter. . A B THEOREM 4 D An arc or line segment of a circle subtends equal angles at the circumference of the circle. We say that the angles in the same segment of the circle are equal. . ˆ B ˆ ˆ and Cˆ D In the diagram, A C A B THEOREM 5 D The opposite angles of a cyclic quadrilateral are supplementary (add up to 180 ) . ˆ C ˆ 180 and B ˆ D ˆ 180 In the diagram, A A 180 C 180 B THEOREM 5 CONVERSE If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is a cyclic quadrilateral. D D . C A 180 B A C 180 B © Gauteng Department of Education 49 THEOREM 6 D An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. 1 . If ABCD is a cyclic quadrilateral, then Ĉ1 A . C A B THEOREM 6 CONVERSE If an exterior angle of a quadrilateral is equal to the interior opposite angle, then the quadrilateral is a cyclic quadrilateral. D D 1 1 . C C A A B B THEOREM 7 A tangent to a circle is perpendicular to the radius at the point of contact. . O If ABC is a tangent to the circle at the point B, ˆ B ˆ 90 . then the radius OB ABC , i.e. B 1 2 A . 1 2 B C THEOREM 7 CONVERSE If a line is drawn perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle. If line ABC OB and if OB is a radius, then ABC is a tangent to the circle at B. . O A © Gauteng Department of Education . 1 2 B C 50 THEOREM 8 B 1 If two tangents are drawn from the same point outside a circle, then they are equal in length. || A . O || 1 C THEOREM 9 The angle between a tangent to a circle and a chord drawn from the point of contact in equal to an angle in the alternate segment. E E . 1 . D 2 A C B D 1 2 B A C ˆ Eˆ and B ˆ D ˆ. In both diagrams, B 2 1 THEOREM 9 CONVERSE If a line is drawn through the endpoint of a chord, making with the chord an angle equal to an angle in the alternate segment, then the line is a tangent to the circle. E E D 1 A . D 2 B C 1 A 2 B C ˆ Eˆ or if B ˆ D ˆ , then ABC is a tangent to the circle passing through the points If B 2 1 B, D and E. © Gauteng Department of Education 51 How to prove that a quadrilateral is cyclic ABCD will be a cyclic quadrilateral if one of the following conditions is satisfied. Condition 1 ˆ A ˆ ) (C ˆ C ˆ ) 180 or (B ˆ B ˆ ) (D ˆ D ˆ ) 180 (A 1 2 2 3 1 2 1 2 Condition 2 ˆ A ˆ A ˆ C 1 1 2 Condition 3 ˆ or D̂ A ˆ B ˆ or B̂ A ˆ ˆ or C D̂1 C 3 2 1 2 1 2 2 How to prove that a line is a tangent to a circle ABC would be a tangent to the “imaginary” circle drawn through ˆ Eˆ EBD if B 1 PROOFS OF THEOREMS REQUIRED FOR EXAMS THEOREM 1 The line drawn from the centre of a circle perpendicular to a chord bisects the chord. Proof Join OA and OB. In OAM and OBM : (a) (b) (c) OA = OB ˆ M ˆ 90 M 1 2 OM = OM OAM OBM radii given common RHS AM = MB © Gauteng Department of Education 52 THEOREM 2 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). Proof Join CO and produce. For all three diagrams: ˆ C ˆ A ˆ ext of OAC O 1 1 ˆ A ˆ OA = OC , radii But C 1 ˆ 2C ˆ O 1 1 ˆ 2C ˆ Similarly, in OCB O 2 2 ˆ ˆ ˆ ˆ O O 2C 2C 1 2 1 2 ˆ O ˆ 2(C ˆ C ˆ ) O 1 2 1 2 ˆ ˆ AOB 2ACB For the third diagram: ˆ O ˆ 2C ˆ 2C ˆ O 2 1 2 1 ˆ ˆ ˆ ˆ O O 2(C C ) 2 1 2 1 ˆ 2ACB ˆ AOB THEOREM 5 The opposite angles of a cyclic quadrilateral are supplementary (add up to 180 ). Proof Join AO and OC. ˆ 2D ˆ O 1 ˆ 2B ˆ O 2 at centre = 2 at circ at centre = 2 at circ ˆ O ˆ 2D ˆ 2B ˆ O 1 2 ˆ O ˆ 360 's round a point and O 1 2 ˆ B) ˆ 360 2(D ˆ B ˆ 180 D ˆ C ˆ 180 Similarly, by joining BO and DO, it can be proven that A © Gauteng Department of Education 53 THEOREM 9 The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle in the alternate segment. Proof Draw diameter BOF and join EF ˆ B ˆ 90 B tan rad 1 2 in a semi-circle Eˆ Eˆ 90 1 2 ˆ Eˆ But B 1 1 ˆ Eˆ B 2 arc FD subt = 's 2 ˆ BED ˆ CBD Draw diameter BOF and join FD B̂1 90 tan rad D̂1 90 ˆ D ˆ B 2 in a semi-circle arc FE subt = 's 2 ˆ D ˆ 90 and B ˆ D ˆ Now B 1 1 2 2 ˆ ˆ ˆ ˆ B B D D 1 2 1 2 ˆ BDE ˆ ABE SECTION C: HOMEWORK QUESTIONS QUESTION 1: (10 MARKS) (5 MINUTES) (DoE papers pre-2014) In the diagram below, M is the centre of the circle. D, E, F and G are points on the circle. If F̂1 10 and D̂2 50 , calculate, with reasons, the size of: (a) D̂1 D (2) 1 2 50 (b) M̂1 (2) (c) F̂2 (2) (d) Ĝ (2) . M 1 2 E (e) Ê1 1 2 (2) 10 2 1 F © Gauteng Department of Education G 54 QUESTION 2: (17 MARKS) (10 MINUTES) (DoE papers pre-2014) In the diagram below, QP is a tangent to a circle with centre O. RS is a diameter of ˆ and the circle and RQ is a straight line. T is a point on the circle. PS bisects TPQ ˆ 22 . SPQ Calculate the following, giving reasons: (a) P̂2 (2) (b) R̂ 2 (2) (c) Pˆ3 Pˆ4 (3) (d) R̂1 (4) (e) Ô1 (3) (f) Q̂ 2 (3) QUESTION 3: (12 MARKS) (10 MINUTES) (DoE papers pre-2014) ALB is a tangent to circle LMNP. ALB||MP. Prove that: (a) LM = LP (4) (b) ˆ LN bisects MNP (4) (c) LM is a tangent to circle MNQ (4) © Gauteng Department of Education 55 QUESTION 4: (16 MARKS) (10 MINUTES) (DoE papers pre-2014) EC is a diameter of circle DEC. EC is produced to B. BD is a tangent at D. ED is produced to A and AB BE . Prove that: (a) ABCD is a cyclic quadrilateral. (4) (b) ˆ Eˆ A 1 (3) (c) BD BA (5) (d) ˆ C ˆ C 2 3 (4) QUESTION 5: (16 MARKS) (10 MINUTES) (Level 4 challenge) In the diagram, PA is a tangent to the circle at A. AC and AB are chords and PM is produced to K such that AK AM . K and M lie on AC and AB respectively. Chord CB is produced to P. ˆ Prove that KP bisects APC (7) © Gauteng Department of Education