GRADE 11 NOVEMBER 2015 MATHEMATICS P2

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1 NATIONAL SENIOR CERTIFICATE GRADE 11 NOVEMBER 2015 MATHEMATICS P2 MARKS: 150 TIME: 3 hours *Imat2* This question paper consists of 14 pages.

2 2 MATHEMATICS P2 (EC/NOVEMBER 2015) INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 12 questions. 2. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. 3. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining the answers. 4. Answers only will NOT necessarily be awarded full marks. 5. You may use an approved scientific calculator (non-programmable and nongraphical), unless stated otherwise. 6. If necessary, round off answers to TWO decimal places, unless stated otherwise. 7. Write neatly and legibly.

3 3 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION 1 The following table represents the heights, in centimetres, of 120 boys in a school. HEIGHT (cm) FREQUENCY Complete the cumulative frequency table in the SPECIAL ANSWER BOOK. (2) 1.2 Draw an ogive, using the diagram in the SPECIAL ANSWER BOOK, to represent the information in the table. (4) 1.3 Determine, using the ogive, the five number summary. (5) 1.4 If the distribution of the data is represented by means of a box whisker diagram, comment on the spread of the data. (1) [12]

4 4 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION 2 The following is a sample of weekly wages earned by ten people working for a small printing and design company. R2 250 R2 250 R3 000 R3 300 R3 300 R3 600 R3 900 R4 350 R4 350 R Calculate the mean weekly wage. (2) 2.2 Calculate the standard deviation of the weekly wage. (1) 2.3 Determine the percentage of workers which lie within ONE standard deviation of the mean. (4) [7]

5 5 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION 3 The points A(2a 11; a +2), C(4; -1) and D(4p; p 7) are the vertices of ACD with B(-2; 3) on AC..A(2a 11; a + 2). B(-2; 3) y x..d C(4; -1) (4p; p -7) 3.1 If points A, B and C are collinear, find the value of a. (4) 3.2 Determine the equation of the line AC. (3) 3.3 Hence, determine the co-ordinates of midpoint M of AB. (3) 3.4 Determine the value of p if CD is parallel to the x-axis. (3) [13]

6 6 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION 4 In the diagram, M, N and P are vertices of MNP, with N(-6; -12). M is a point on the y-axis. The equation of the line MN is. MR = NR and NQ MP. PR and NQ intersect at the origin O. M Q R O P x N(-6; -12) y 4.1 Calculate the gradient of NQ. (1) 4.2 Calculate the gradient of MP. (1) 4.3 Calculate the angle of inclination of MP. (3) 4.4 Hence, determine the equation of the line MP. (4) 4.5 Hence, determine the coordinates of P. (4) 4.6 Determine the co-ordinates of R. (3) [16]

7 7 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION In the diagram below, P(1;3) is a point on the Cartesian plane, OP = r and = θ. y P(1;3) r O θ x Make use of the diagram to calculate the value of θ. (2) Calculate the length of OP. Leave the answer in surd form. (2) Determine the values of the following, without using a calculator: (a) sin θ (1) (b) cos (180 + θ) (2) 5.2 Determine the general solutions of: (6) 5.3 Simplify: ( ) ( ) ( ) (5) 5.4 Prove that: (5) [23]

8 8 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION COMPLETE: In ABC + A c b B a C (2) 6.2 In the diagram below, PQ is a straight line m long. RS is a vertical tower 158 m high with P, Q and S points in the same horizontal plane. The angles of elevation of R from P and Q are and θ.. R 158 m S P m θ Q Determine the length of PS. (3) Determine the length of SQ. (3) Hence, find the value of θ. (3) Determine the area of ΔSPQ. (4) [15]

9 9 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION 7 Given: ( ) ( ) 7.1 Draw the graph of f and g for x in the SPECIAL ANSWER BOOK. Show all the turning points and intercepts with the axes. Clearly show the asymptotes using dotted lines. (6) 7.2 Determine the values of x, for, for which f(x) > g(x). (6) 7.3 Write down the period of g(2x). (1) [13]

10 10 MATHEMATICS P2 (EC/NOVEMBER 2015) GIVE REASONS FOR YOUR STATEMENTS AND CALCULATIONS IN QUESTIONS 8, 9, 10 AND 11. QUESTION Complete: The line drawn from the centre of the circle perpendicular to the chord (1) 8.2 In the figure below, AB and CD are chords of the circle with centre O. OE AB. CF = FD. OE = 4 cm, OF = 3 cm and CD = 8 cm. B E. O F D A C Calculate the length of OD. (3) Hence calculate the length of AB. (4) [8]

11 11 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION In the diagram O is the centre of the circle and ABC are points on the circle. Use the diagram in your SPECIAL ANSWER BOOK to prove that:. C. O A B (6) 9.2 In the figure below, and O is the centre of the circle. A, B, E C and D are points on the circumference. Calculate, giving reasons, the sizes of: A 1 B O 1 E D 1 25 C (2) (2) (2) (2) [14]

12 12 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION 10 A, B, C and D are points on the circumference of the circle in the diagram below. ECF is a tangent at C, B 1 = B 2. D A F 1 2 B C 10.1 If B 1 = x, find, with reasons, TWO other angles equal to x. (4) 10.2 Hence, show that DC bisects. (2) [6] E

13 13 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION Complete: Opposite angles of a cyclic quadrilateral (1) 11.2 In the figure, ABCD is a cyclic quadrilateral. AB DC in circle with centre O. BC and AD produced meet at M. D 3 = x B 1 C 2 O 1 A 1 2 D 3 M Show that MC = MD. (5) If D 3 = x, determine the value of, in terms of x. (2) Hence, show that BODM is a cyclic quadrilateral. (3) [11]

14 14 MATHEMATICS P2 (EC/NOVEMBER 2015) QUESTION 12 The solid in the diagram is made up of a right prism with square base, and a right pyramid on top of the prism. The length of the prism is 12 cm, the side of the base is 6 cm and the height of the pyramid is 8 cm Calculate the slant height of the triangular face of the pyramid. (3) 12.2 Calculate the area of one of the triangular faces. (3) 12.3 Calculate the total surface area of the solid. TSA = area of slanted faces + area of right prism (5) [11] TOTAL: 150

17 LEARNER NAME NATIONAL SENIOR CERTIFICATE MATHEMATICS P2 GRADE 11 NOVEMBER 2015 SPECIAL ANSWER BOOK QUESTION MARK INITIAL MOD TOTAL This answer book consists of 22 pages. *Iwisab2*

18 2 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) QUESTION Solution Cumlative Frequency Table Marks HEIGHT (cm) FREQUENCY CUMULATIVE FREQUENCY (2) Ogive Heights (in centimetres) (4) (5) (1) [12]

19 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 3 QUESTION Solution Marks (2) (1) (4) [7]

20 4 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) QUESTION 3.A(2a 11; a + 2). B(-2; 3) y x..d C(4; -1) (4p; p -7) 3.1 Solution Marks 3.2 (4) 3.3 (3) (3)

21 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK (3) [13]

22 6 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) QUESTION 4 M Q R O P x N(-6; -12) y 4.1 Solution Marks 4.2 (1) 4.3 (1) 4.4 (3) (4)

23 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK (4) (3) [16]

24 8 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) QUESTION 5 y P(1;3) r O θ x Solution Marks (2) 5.1.3a (2) 5.1.3b (1) 5.2 (2) (6)

25 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK (5) (5) [23]

26 10 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) QUESTION 6 A c b B a C 6.1 Solution Marks 6.2 R (2) 158 m S P m θ Q (3) (3)

27 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK (3) (4) [15]

28 12 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) QUESTION Solution Marks 7.2 (6) 7.3 (6) (1) [13]

29 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 13 QUESTION 8 B E. O F D A 8.1 Solution C Marks (1) (3) (4) [8]

30 14 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) QUESTION 9 C. O A B 9.1 Solution Marks (6)

31 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK A 1 B O 1 E D 1 25 C (2) (2) (2) (2) [14]

32 16 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) QUESTION 10 D A F 1 2 B C E 10.1 Solution Marks 10.2 (4) (2) [6]

33 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 17 QUESTION Solution Marks 11.2 B (1) 1 C 2 O 1 A 1 2 D 3 M (5)

34 18 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) (2) (3) [11]

35 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 19 QUESTION 12 8 cm 12 cm 12.1 Solution 6 cm Marks 12.2 (3) 12.3 (3) (5) [11]

36 20 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015) ADDITIONAL PAGES

37 (EC/NOVEMBER 2015) MATHEMATICS P2 SPECIAL ANSWER BOOK 21

38 22 MATHEMATICS P2 SPECIAL ANSWER BOOK (EC/NOVEMBER 2015)

41 NATIONAL SENIOR CERTIFICATE GRADE 11 NOVEMBER 2015 MATHEMATICS P2/WISKUNDE V2 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 8 pages. Hierdie memorandum bestaan uit 8 bladsye.

42 2 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015) NOTE: If a candidate answers a question TWICE, only mark the FIRST attempt. If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. Consistent accuracy applies in ALL aspects of the marking memorandum. Assuming answers/values in order to solve a problem is NOT acceptable. LET WEL: Indien ŉ kandidaat ŉ vraag twee keer beantwoord, merk slegs die eerste poging. Indien ŉ kandidaat ŉ antwoord doodgetrek het, maar nie oorgedoen het nie, merk die doodgetrekte antwoord. Volgehoue akkuraatheid geld in ALLE aspekte van die memorandum. Aanname van antwoorde/waardes om ŉ probleem op te los, is Onaanvaarbaar. QUESTION/VRAAG 1: [12] 1.1 Height (cm) Hoogte (cm) Frequency Frekwensie Cumulative frequency Kumulatiewe frekwensie (2) 1.2 grounding at(150;0)/anker by (150;0) Ogive/Ogief plotting (155;4)/plot van 150 (155;4) 100 plotting with upper limits/plot by boonste 50 limiete 0 joining the points to form a smooth curve/verbind Heights/Hoogte (in cm) punte om glade kurwe te vorm (4) Cumlative Frequency / Kumulatiewe Frekwensie 1.3 (150, 160,5, 163, 166, 175) OR/OF Min = 150 Q 1 = 160,5 Q 2 = 163 Q 3 = 166 Max = , (5) 1.4 Skewed to the left / Skeefgetrek na links. answer/antwoord (1) [12]

43 (EC/NOVEMBER 2015) MATHEMATICS P2/WISKUNDE V2 3 QUESTION/VRAAG 2 [7] 2.1 = R3 555 answer/antwoord (2) 2.2 answer/antwoord (1) 2.3 ( ,12 ; ,12) = (2654,88 ; 4455,12) interval 7 answer/antwoord werkers lê binne een standaardafwyking of workers lie within one standard deviation. van die werkers lê binne een standaardafwyking. QUESTION/VRAAG 3 [13] 3.1 m AB = m BC = m AC ANSWER ONLY : FULL MARKS SLEGS ANTWOORD : VOLPUNTE m AB = m BC = m AC (4) [7] substitute into equation/vervang in vgl ( ) ( ) substituting into m AC /vervang in vgl m AC 3.2 ( ) m CD = 0 OR/OF p 7 = -1 p = 6 Kopiereg voorbehou ( ) [ ] [ ] = [ ] answer/antw (4) equation/vgl subst of m and(-5;5) into form/vervang van m en (4;-1) in formule equation/vgl (3) subst into correct formula/vervang in korrekte formule coordinates/coordinate (3) correct m = 0 substitution into eqn/vervang in vgl answer/antw equating/ answer (3) [13] Blaai om asseblief

44 4 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015) QUESTION/VRAAG 4: [13] 4.1 answer/antw = M NQ M MP = -1 [NQ MP] answer/antw 4.3 value of /waarde van (1) (1) (3) 4.4 ( ) 4.5 ( ) equation/vgl subst of m = and (0;6) into eqn/vervang m = en (0;6) in vgl answer/antw (4) subst into eqn/vervang in vgl P (2;5) 4.6 MR = NR R = [ ] P (2;5) (4) Substitution/vervang. R = [ ] R = [-3;-3] QUESTION/VRAAG 5: [23] tan θ ( ) ( ) 5.1.3a 5.1.3b ( ) x value/waarde y value/waarde (3) [16] answer/antwoord (2) OP 2 answer/antwoord (2) answer/antwoord -cosθ (1) ANSWER ONLY FULL MARKS/ SLEGS ANTWOORD:VOLPUNTE answer/antwoord (2)

45 (EC/NOVEMBER 2015) MATHEMATICS P2/WISKUNDE V ( ) ( )( )= 0 standard form/st vorm 1 sin 2 x 5.3 = = -tan x 5.4 ( ) ( ) sin x = ½ or no soln (sin x = 2) 30 ; 150 k.360, sin x -sin x cos 2 x tan x numerator / teller denominator / noemer (6) (5) ( ) ( ) ( ) ( ) expansion / uitbreiding simplification / vereenvoudiging factorisation / faktorisering (5) QUESTION/VRAAG 6: [11] 6.1 b 2 = a 2 + c 2 2acCos B a 2 + c PS = 338, In PQS = ( )( ) = ,606 SQ = 1218,40 m 2 [23] 2acCos B (2) Ratio for tan/verhou. vir tan substitution 65 / verv van 65 answer/antwoord (3) use of cos formula/gebruik van cos formule Substitution/vervang SQ (3) Kopiereg voorbehou Blaai om asseblief

46 6 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015) In RSQ: Area of ΔSPQ = ½ SP.PQ.sin = ½ (338,83)(1 500)sin 30 =127061,25 m 2 value of /waarde van (3) correct formula / korrekte formule substitute / vervang (338,83), (1500) Sin 30 0 answer / antwoord (4) [15] QUESTION/VRAAG 7: [12] 7.1 f asymtotes / asimptote min value / min waarde max value / maks waarde g (-45 ; -1) (45 ; 1) (135 ; -1) 7.2 ( ) ( ) ( ) answer / antwoord (6) (6) (1) [13]

47 (EC/NOVEMBER 2015) MATHEMATICS P2/WISKUNDE V2 7 QUESTION/VRAAG 8: [8] 8.1 bisects the chord. answer/antwoord (Pythagoras) (substitution/vervang) method/metode (1) (Pythagoras) (substitution/vervang) answer/antwoord method/metode (3) cm cm But AB = 2AE AB = 2(3) = 6 cm QUESTION/VRAAG 9: [13] 9.1 (OE AB) C S/R answer/antwoord (4) [8] construction/konstr A. O D B S/R S/R S/R S/R conclusion / gevolgtrekking (6) CONSTR: Join CO, extend to D PROOF: In AOC i) (ext of /buitehoek van ) ii) (ext of /buitehoek van ) iii) (AO = OC) iv) (BO = OC) = 2( ) ( radii equal/radiusse gelyk) S R (ext of /buitehoek van ) S R (angles in same segment/ hoeke in dieselfde segment) (opp angles of cyclic quad/ teenoorstaande hoeke van ŉ koordevierhoek) S R S R (2) (2) (2) (2) [14] Kopiereg voorbehou Blaai om asseblief

48 8 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2015) QUESTION/VRAAG 10: [6] 10.1 = x (angles in the same segment/ hoeke in dieselfde segment) S R = x (tan chord/tan koord) 10.2 (both equal to x/albei gelyk aan x) S R S/R conclusion / gevolgtrekking QUESTION/VRAAG 11 [11] 11.1 Are supplementary OR add to 180. answer/antwoord (Ext of cyclic quad/buitehoek van koordevhk) S R = (corresponding angles, AB DC/ Ooreenkomstige hoeke AB DC) (base angles of equal/basis hoeke van gelyk) S R R (angles of /hoeke van ) S R ( at centre = 2 at circumference/ by middle = 2 by omtreks) S R S (4) (2) [6] (1) (5) (2) QUESTION/VRAAG 12 [11] 12.1 Sh 2 = (Pythagoras) Sh 2 = (substitution/vervang) Sh = = 8,54 cm 12.2 Area of face = = ( )( ) = 25,63 cm TSA = area of slanted faces + area of right prism TBO = oppervlakte van skuinsvlakke + oppervlakte van reghoekige prisma = 3(25,63) (6 x 12) = 76, = 400,89 cm 2 S method/metode answer/antwoord formula/formule substitution/vervang answer/antwoord TSA substitute/vervang answer/antwoord (3) [11] (3) (3) (5) [11] TOTAL/TOTAAL: 150

GRAAD 11 NOVEMBER 2015 WISKUNDE V2

NASIONALE SENIOR SERTIFIKAAT GRAAD 11 NOVEMBER 2015 WISKUNDE V2 PUNTE: 150 TYD: 3 uur *Iwis2* Hierdie vraestel bestaan uit 14 bladsye. 2 WISKUNDE V2 (EC/NOVEMBER 2015) INSTRUKSIES EN INLIGTING Lees die