Copyright reserved Please turn over MARKS: 150 This memorandum consists of 33 pages. MATHEMATICS P2 NOVEMBER 2010 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE 12Mathematics/P2 2 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 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. QUESTION 1 1.1 The Data Set 9 14 14 19 21 23 33 35 37 37 42 45 55 56 57 59 68 75 75 75 77 78 80 81 92 Min = 9 Max = 92 Q3 = Upper Quartile = 75 Q1 = Lower Quartile = 28 (6th number is 23. 7th number is 33. The number in position 6,25 is 23 + 5 , 25 ) 23 33 ( 41 = − ) (Accept Q1 = 25,5 ) Q2 = Median = 55 Five number summary is (9 ; 28 ; 55 ; 75 ; 92) OR (9 ; 25,5 ; 55 ; 75 ; 92) min & max upper quartile lower quartile median (4) 1.2 10 20 30 40 50 60 70 80 90 100 110 box whiskers (2) Class A Class B Q3 Q2 Q1 Note: If the candidate combines the answer of 1.1 and 1.2 by drawing the correct box and whisker diagram in 1.2 and writes the numbers on the diagram (i.e. we mark question 1.1 and 1.2 in conjunction with one another), max 5 /6 marks for question 1.1 and 1.2 If candidate writes numbers randomly in 1.1 and draws the box and whisker diagram correctly but without indicating the numbers on the diagram, then max 5 /6 marks for question 1.1 and 1.2 If candidate just draws the box and whisker in 1.2 and does not indicate values on it or answers 1.1, the max 2 /6 marks If the candidate draws two diagrams (one in the answer book and one on the diagram sheet), mark the one on the DIAGRAM SHEET. Note: Penalty 1 for not labelling or writing the numbers in ascending order Mathematics/P2 3 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 1.3 Class B Class B performed better because half of the learners got above 60% whilst half of Class A got more than 55%. Class B performed better because half of the learners got above 60% whilst half of Class A got less than 55%. Median of Class B > Median of Class A OR Class B Class B is skewed more to the left than Class A is. OR Class A 25% of class scored 75% or more in Class A while 25% of the class scored 70% or more in Class B. Highest Mark in Class A > Highest Mark in Class B. Note: If candidate answers: Cannot determine the class that does better because we have insufficient information as we do not know where the marks are clustered. Max 1 /3 Note: If candidate just answers Class A or Class B and there are no reasons, then 0 /3 marks Class B median Class B > Median Class A (3) Class B Class B skewed more left than A (3) Class A highest A > highest B 25% of A above 75% and 25% of B above 70% (2) [9] Mathematics/P2 4 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 2 2.1 EXAMINATION SCORE (x) FREQUENCY CUMULATIVE FREQUENCY 40 30 < ≤ x 12 12 50 40 < ≤ x 18 30 60 50 < ≤ x 55 85 70 60 < ≤ x 57 142 80 70 < ≤ x 43 185 90 80 < ≤ x 11 196 100 90 < ≤ x 4 200 first 3 values last 4 values (2) 2.2 shape (points must not be joined a straight line with a ruler) grounding point (30 ; 0) using the upper limit using cumulative frequencies if 4 or more points plotted correctly (5) 2.3 200 – 165 = 35 learners OR 5 , 36 5 , 163 200 5 , 163 2185 142 = − = + answer (1) answer (1) [8] Cumulative frequency graph of examination scores 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 0 10 20 30 40 50 60 70 80 90 100 110 Examination Score Cumulative Frequency Note: If learners plot the midpoint of the interval and the cumulative frequency max 1 /5 marks for shape If the candidates plot the lower limit and the cumulative frequency max 1 /5 marks Note: Accept any one of 34, 35 or 36 Mathematics/P2 5 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 3 3.1 Mean 12 2404 12 184 160 249 265 185 168 161 144 239 221 211 217 = + + + + + + + + + + + = = 200,33 12 2404 answer (2) Answer only: Full marks 3.2 By means of a calculator: 37 , 37 = σ OR Pen and paper method: mean ( 33 , 200 ) = x x x x − ( )2 x x − 217 16,67 277,889 211 10,67 113,848 221 20,67 427,248 239 38,67 1495,368 144 -56,33 3173,0689 161 -39,33 1546,848 168 -32,33 1045,228 185 -15,33 235,008 265 64,67 4182,208 249 48,67 2368,768 160 -40,33 1626,508 184 -16,33 266,668 SUM 16758,666 37 , 37 126668 , 16758 = = σ answer (3) total substitution answer (3) 3.3 200,33+1(37,37) = 237,70 litres Accept any number between and including 237 and 238 litres. method answer (2) [7] Note: If candidate answers 200,33 – 1(37,37) = 162,96 litres Max 1 /2 marks Note: Penalty 1 for incorrect rounding Note: No penalty for incorrect decimal places Accept 37 Mathematics/P2 6 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 4 4.1 Fly High answer (1) 4.2 45 , 40 9 , 7 1000 5120 = × OR 96 , 40 8 1000 5120 = × OR 405 8=× Yes calculation that leads to 40 (1) 4.3 Yes. The data points suggest a straight line fit with negative gradient but Fly-High will have to be an outlier. OR Yes. Weak negative correlation. (r = – 0,2128075984) Note: If the candidate indicates “Best Air” and/or “Best Fly” and/or “Alpha” have high on time arrivals and low lost luggage max 1 /2 marks Yes negative gradient (2) Yes negative correlation (2) 4.4 Alpha, 70% on-time arrival and least luggage loss OR Best Air, best on time arrival Name of company correct justification (2) [6] Mathematics/P2 7 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 5 5.1.1 2163 ) 3 ( 3 4 7 AD == − −− == AC m m OR 21 63 ) 3 ( 3 7 4 AD = −− = − − − == AC m m substitution of A and C into correct formula answer (2) 5.1.2 25 410 ) 3 ( 1 4 6 − = − = − − − − = BC m OR 254 10 ) 1 ( 3 ) 6 ( 4− = − = − − − − = BC m answer (1) 5.2 ° ≈ ° = ° = ° − ° = ° = = − = ° = = = 2 , 85 85,24 85,236359 .. 26,56505.. 5... 409 111,801 B CˆD 095 814 , 111 tan 25 .... 56505 , 26 O DˆC O DˆC tan 21 . mmBC ADα α OR ° 57 , 26 ° 80 , 111 answer (3) x y D C(−3 ; 4) G(a ; b) A(3 ; 7) B(1 ; −6) Mα O β (x ; 0) P Note: If candidate gives x m − = 3 7 AD then 1/2 marks Mathematics/P2 8 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over ° = ° = ° + ° − ° = ° = − ° = − ° ° == 85,24 85,236359 ) ... 19859051 , 68 .. 26,56505.. ( 180 B CˆD ... 19859051 , 68 180 25 ) 180 tan( .... 56505 , 26 O DˆC 21 O DˆC tan α α OR ( )( ) ° == − + − − = + − = − = 24 , 85 B CˆD 12 1 . 1 B CˆD tan O DˆC B CˆD 21 25 21 25 CD CB CD CB m m m mα OR ° = ° − ° = ° = − = − + = − + = = = = 24 , 85 ... 76 , 94 180 B CˆD ... 76 , 94 B CˆA ... 083045 , 0 ) 116 )( 45 ( 2 173 116 45 AC.BC 2 AB BC AC B CˆA cos 173 AB 116 BC 45 AC 2 2 2 OR D(– 11 ; 0) ° == − + = − + = = = = 24 , 85 B CˆD ... 5 0830454798 , 0 ) 116 )( 80 ( 2 180 116 80 DC.BC 2 DB BC DC B CˆD cos 180 DB 116 BC 80 DC 2 2 2 OR ° 57 , 26 68,2° answer (3) CD CB CD CB m m m m . 1 B CˆD tan + − = substitution answer (3) cosine rule substitution into cosine rule answer (3) cosine formula substitution into cosine rule answer (3) Mathematics/P2 9 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over Equation AC: 11 2 + = x y D(– 11 ; 0) C(– 3 ; 4) 80 ) 0 4 ( ) 11 3 ( ) ( ) ( DC 2 2 2 2 2 = − + + − = − + − = D C D C y y x x Equation BC: 7 5 2 − − = x y P( ) 0 ; 57 − 25 464 ) 0 4 ( ) 3 ( PC 2 2 57 2 = − + + − = 25 2304 ) 11 ( DP 2 57 2 = + − = In ΔDCP: P CˆD cos DC.CP. 2 CP DC DP 2 2 2 − + = ° == ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − + =85,24 P CˆD . 85,23635.. P CˆD P CˆD cos . 5 464 5 2000 2 25 464 25 2000 25 2304 cosine formula substitution into cosine rule answer (3) Mathematics/P2 10 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 5.3 ( ) 0 11 2 2 11 21 3 21 7 = + − + = − = − y x x y x y OR ( ) 0 11 2 2 11 21 3 21 4 = + − + = + = − y x x y x y OR 0 11 2 2 11 212 11 ) 3 ( 21 ) 7 ( 21 = + − + = = + = + = y x x yc c c x y substitution of (3 ; 7) into ( ) 1 1 x x m y y − = − answer in any form (2) substitution of (– 3 ; 4) into ( ) 1 1 x x m y y − = − answer in any form (2) substitution of (3 ; 7) into y = mx + c answer in any form (2) 5.4 ( ) 1 ; 1 ) ; ( M 2 6 4 ; 2 1 3 ) ; ( M − − = ⎟⎠ ⎞ ⎜⎝ ⎛ − + − = y x y x substitution answer (2) Note: If candidate leaves answer as ( ) 3 21 7 − = − x y or ( ) 3 21 4 + = − x y : max 1 /3 marks Mathematics/P2 11 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 5.5 1 21 ) 1 ( 2 1 2 2 ) 1 ( 3 ) 1 ( 7 + = = ∴ + − = − + = = − − − − =x y c c c x ymAM G(a ; b) lies on the line 1 2 + = ∴ a b OR 1 21 2 4 4 8 3 3 7 7 ) 3 )( 1 ( ) 1 )( 7 ( 11 37 + = − = − − = − − + − = − − + − + = + − ++ = −− a bb a b a a ab b b ab a a b a b ab ab OR Using the point (– 1 ; – 1) 1 2 2 2 1 2 11 48 11 + = + = + = ++ = ++ a b a babab OR Using the point (3 ; 7) 1 2 2 6 7 2 37 48 37 + = − = − = −− = −− a b a babab gradient = 2 substitution (– 1 ; – 1) c = 1 conclusion (4) ab −− 37 11 ++ ab equating simplification leading to 1 2 − = − b a (4) substitution of (– 1 ; – 1) into gradient gradient = 2 equating simplification leading to 2 2 1 + = + a b (4) substitution of (3 ; 7) into gradient gradient = 2 equating simplification leading to a b 2 6 7 − = − (4) Note: If the candidate does not conclude b = 2a + 1 from y = 2x+ 1: max 3 /4 marks Mathematics/P2 12 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 5.6 3 or 57 1 or 51 0 ) 1 )( 1 5 ( 0 1 6 5 0 17 9 12 4 9 6 17 ) 4 1 2 ( ) 3 ( 17 ) 4 ( ) 3 ( 17 GC 17 GC2 2 2 2 2 2 2 2 = = ∴ = = = − − = + − = − + − + + + = − + + + = − + + = = b b a a a a a a a a a a a a b a OR 3 or 57 ) 3 )( 7 5 ( 0 21 22 5 0 89 22 5 68 4 64 32 4 25 10 17 ) 4 ( 2 5 17 ) 4 ( 3 2 1 17 ) 4 ( ) 3 ( 17 2 1 2 2 2 2 2 2 2 2 2 2 = = ∴ − − = + − = + − = + − + + + = − + ⎟⎠ ⎞ ⎜⎝ ⎛ + = − + ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + ⎟⎠ ⎞ ⎜⎝ ⎛ − = − + + = − = b b b b b b b b b b b b b b b b b a b a distance formula in terms of a and b substitution of b = 2a + 1 standard form factors or correct substitution into formula values of a values of b (6) 2 1 − = b a distance formula in terms of a and b substitution of 2 1 − = b a standard form factors or correct substitution into formula values of b (6) [20] Note: If candidate swops a and b around: max 2 /6 marks Mathematics/P2 13 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 6 x y O P M(−4 ; 4) L N Q 6.1 8 84 4 1 11 12 + = = ∴ + − = + = = −− = ∴ − = + − = x y c c c x y m m x y LN LP OR 8 ) 4 ( 1 4 + = + = − x y x y 1 − = LP m 1 = LN m equation (3) m = 1 substitution of ) ( 1 1 x x m y y − = − answer (3) 6.2 ) 5 ; 3 L( 5 8 336 2 2 8 − = + − = − = − = + − = + yyxx x x OR 5) 3; L( 3 5 10 2 2 .( .......... 8 ) 1 .( .......... 2 − − = ∴ = ∴ = = − = +xy y x y x y x-value y-value Equations leading to these values must be used (2) Note: If candidate leaves it as y – 4 = x + 4 max 2 /3 marks Answer only: Full marks Note: No penalty for not leaving in coordinate form Mathematics/P2 14 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 6.3 2 ) 4 ( ) 4 ( 2 ) 4 5 ( ) 4 3 ( ) 4 ( ) 4 ( 2 2 2 2 2 2 2 2 2 = − + + = ∴ = − + + − = − + + y xr r r y x Equation can be left as: 0 30 8 8 2 2 = + − + + y y x x 2 2 2 ) 4 ( ) 4 ( r y x = − + + substitution of (– 3 ; 5) 2 2 = r (3) 6.4 Let N(x, y). Since M(– 4 ; 4) is the midpoint of LN and L(– 3 ; 5) 3 ; 5 4 2 5 ; 4 2 3 y x y x = − = ∴ = + − = − OR 3 5 5 or 3 0 ) 3 )( 5 ( 0 15 8 0 30 16 2 0 2 16 8 16 8 0 2 ) 4 8 ( ) 4 ( 2 ) 6 ( ) 4 ( 8 2 2 2 2 2 2 2 2 = = − = − = = + + = + + = + + = − + + + + + = − − + + + = − + + + = y y x x x x x x x x x x x x x x y x x y ∴N(– 5 ; 3) using the fact that M is the midpoint of LN x = −5 y = 3 (3) 0 2 ) 4 8 ( ) 4 ( 2 2 = − − + + + x x x = −5 y = 3 (3) 6.5 2 2 ) 5 ( 3 1 NQ − − = − = + − − = + − = − = x yc c c x y m OR 2 ) 5 ( 3 1 NQ − − = + − = − − = x y x ym OR Equation of LP is 2 = + y x NQ || LP ∴equation of NQ is k y x = + for some R k ∈ But N(– 5 ; 3) lies on NQ 2 3 5 − = + − = + ∴ y x gradient substitution of (– 5 ; 3) into y = mx + c c = – 2 (3) gradient substitution of (– 5 ; 3) into ( ) 1 1 x x m y y − = − equation (3) x + y = k substitution of (– 5 ; 3) equation (3) Note: Answer only: Full marks Note: If the candidate only uses the distance formula to determine the radius 2 ) 4 5 ( ) 4 3 ( 2 2 2 2 = ∴ = − + + −r r then 2 /3 marks Note: Answer only: Full marks Mathematics/P2 15 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over OR NQ is a reflection of LP (y + x = 2) in the line y = x ∴equation of NQ is 2 − = + y x 6.6 Let new radius of circle be R and centre be M/. 8 ) 4 ( ) 2 ( ) 2 2 ( ) 4 ( ) 2 (8 ) 2 ( 442 ) 4 ; 2 ( ) 4 ; 6 4 ( M 2 2 2 2 2 2 2 = − + − ∴ = − + − ∴===== + − ′ y x y xr R r R OR Let R = new radius of circle R2 = (2r)2 = 4(2) = 8 8 ) 4 ( ) 2 ( 8 ) 4 ( ) 4 6 ( 2 2 2 2 = − + − ∴ = − + + − y x y x ) 4 ; 2 ( M′ 2 2 = r equation (3) 2 ) 2 ( − x 2 ) 4 ( − y 8 or ( )2 22 (3) [17] Mathematics/P2 16 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 7 7.1 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 -8 -7 -6 -5 -4 -3 -2 -112345678 x y A(2 ; −2) B(4 ; −3) C(1 ; −4) C/(−6 ; 4) B/(0 ; 2) A/(−4 ; 0) Note: • If the candidate only draws the correct triangle with labels, full marks • If they plot the points correctly and do not draw the triangle, max 5 /6 marks • In the 3 sketches, if one vertex of the three is wrong, then 1 /2 marks for the incorrect sketch, then CA applies. • If they write down the points and do not plot the points and draw the triangle max 3 /6 marks • If the vertices are correct but not labelled and the points are joined max 5 /6 marks • If the vertices are correct, not labelled and not joined max 4 /6 marks • If a candidate finds a formula first and gets it wrong Max 1 mark for the formula Max 2 marks for the calculation of A/, B/, C/coordinates (CA) 1 mark for plotting 3 vertices 1 mark for completing the triangle and labelling 2 marks for each diagram of the transformation (6) OR If candidate works out the general rule first (x ; y) → (2x – 8; –2y – 4) /A (–4 ; 0) /B (0 ; 2) /C (–6 ; 4) 1 mark per each correct point plotted and joined (6)Mathematics/P2 17 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 7.2 ) 4 2 ; 8 2 ( ) 2 ; 4 ( ) 2 ; 4 ( ) ; ( ) ; ( ) ; ( − − − → − − − − − − → − − → y x y x y x y x y x y x OR (x ; y) → (2x – 8; –2y – 4) ) ; ( ) ; ( y x y x − → 4 − x 2 − − y ) 4 2 ; 8 2 ( − − − y x (4) 7.3 1 2 3 4 5 -4 -3 -2 -1 x y A(2 ; −2) B(4 ; −3) C(1 ; −4) Area Δ ABC = area of rectangle – sum of 3 triangle areas = ⎟⎠ ⎞ ⎜⎝ ⎛ + + − 1 23 1 6 =25 2 2 ///units 10 25 2 C B A Area = ⎟⎠ ⎞ ⎜⎝ ⎛ = Δ OR ⎟⎠ ⎞ ⎜⎝ ⎛ + + − 1 23 1 6 25 10 (4) Note: Answer only: Full marks Note: • If the candidate writes (x ; y) → (x ; –y) (x ; y) → (x – 4; y – 2) (x ; y) → (2x ; 2y) then 2 /4 marks • If the candidate writes (x ; y) → (x ; –y) (x ; y) → (x – 4; – y – 2) (x ; y) → (2x – 8 ; –2y – 4) then 4 /4 marks • If candidate writes (x ; y) → 2(x – 4; – y – 2) then 3 /4 marks Mathematics/P2 18 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 -2 -1123456 x y A(− 4 ; 0) B(0 ; 2) C(− 6 ; 4) 2 ///units 10 4 4 6 24 ) 4 )( 2 ( 21 ) 2 )( 4 ( 21 2 . 6 . 21 24 C B A Area = − − − = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + + ⎟⎠ ⎞ ⎜⎝ ⎛ − = Δ OR 2 = AC m and 21 − = AB m ∴ product = – 1 ° = ∴ 90 B AˆC AB = 5 AC = 5 ∴ Area ΔABC is ( ) 25 5 21 2 = ∴ Area Δ ///C B A is 25 4× = 10 square units OR 2 //− = C A m and 21 //= B A m ∴ product = – 1 ° = ∴ 90 B AˆC ///A/B/= 20 A/C/= 20 ∴ Area ΔA/B/C/is ( ) 10 20 21 2 = square units OR AB = 5 AC = 5 BC2 = 10 BC = 10 ⊥height = 4 10 Area of ΔABC 25 10 . 4 10 . 21 = = Area ΔA/B/C/= 25 4× = 10 square units ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ + + ⎟⎠ ⎞ ⎜⎝ ⎛ ) 4 )( 2 ( 21 ) 2 )( 4 ( 21 2 . 6 . 21 24 10 (4) AB = 5 and AC = 5 25 10 (4) A/B/= 20 A/C/= 20 10 (4) AB = 5 and AC = 5 25 answer (4) AB = 5 and AC = 5 Mathematics/P2 19 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over OR AB = 5 AC = 5 BC2 = 10 BC = 10 AC2 + AB2 = BC2 ° = ∴ 90 B AˆC Area ΔABC = ( ) 25 5 . 21 2 = = Area ΔA/B/C/= 25 4× = 10 square units OR Area ΔABC A bcsin . 21 = ( )( ) 25 90 sin 5 5 21 == Area ΔA/B/C/= 25 4× = 10 square units OR -2 -1 1 2 3 4 5 6 7 8 9 -7 -6 -5 -4 -3 -2 -11 x y A(2 ; −2) B(4 ; −3) C(1 ; −4) A/(3 ; −5) Reflect ΔABC about CB and get the square ABA//C of side 5 Area square ( )( ) 5 5 5 = = Area ΔABC 25 = Area ΔA/B/C/= 25 4× = 10 square units OR 25 answer (4) AB = 5 and AC = 5 25 10 (4) AB = 5 and AC = 5 reflection to get square 10 (4)Mathematics/P2 20 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 -2 -1123456 x y A(− 4 ; 0) B(0 ; 2) C(− 6 ; 4) A//(−2 ; 6) Reflect ΔA/B/C/about C/B/and get the square A/B/A//C/of side 20 Area square ( )( ) 20 20 20 = = Area ΔA/B/C/= 20 21 × = 10 square units A/B/= 20 and A/C/= 20 reflection to get square 10 (4) [14]Mathematics/P2 21 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 8 8.1 ) 3 3 ( 2 1 or 2 2 3 6 4 6 6 2 2 4 6 4 2 4 2 2 6 2 22 . 23 4 22 . 21 . 4 22 . 21 . 2 22 . 23 . 2 45 sin 30 cos 4 45 cos 30 sin 4 45 sin 30 sin 2 45 cos 30 cos 2 ) 45 30 sin( 4 ) 45 30 cos( 2 75 sin 4 75 cos 2 sin cos /+ − − − = − − = − − − = − − − = ° ° − ° ° − ° ° − ° ° = ° + ° − ° + ° = ° − ° = − = α α y x x OR ) 3 3 ( 2 1 or 2 2 3 6 4 6 6 2 2 4 6 4 2 4 2 2 6 2 22 . 23 4 22 . 21 . 4 22 . 21 . 2 22 . 23 . 2 45 sin 30 cos 4 45 cos 30 sin 4 45 sin 30 sin 2 45 cos 30 cos 2 ) 45 30 sin( 4 ) 45 30 cos( 2 ) 75 sin( 4 ) 75 cos( 2 ) 75 sin( 4 ) 75 cos( 2 sin cos /+ − − − = − − = − − − = − − − = ° ° − ° ° − ° ° − ° ° = ° + ° − ° + ° = ° − ° = ° − + ° − = + = α α y x x OR substitution using anti clockwise formula 75 = 30 + 45 cos expansion sin expansion substitution of special angles simplified answer for x/(6) substitution using clockwise formula 75 = 30 + 45 cos expansion sin expansion substitution of special angles simplified answer for x/(6) Note: If the candidate uses a calculator i.e. gives a decimal answer max 5 /6 marks Incorrect formula: max 5 /6 marks Mathematics/P2 22 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over First compute 4 2 6 22 . 21 22 . 23 45 sin . 30 sin 45 cos . 30 cos ) 45 30 cos( 75 cos − = − = ° ° − ° ° = ° + ° = ° and 4 6 2 22 . 23 22 . 21 45 sin . 30 cos 45 cos . 30 sin ) 45 30 sin( 75 sin + = + = ° ° + ° ° = ° + ° = ° 2 2 3 6 4 ) 2 6 6 2 4 6 2 4 4 2 6 2 75 sin 4 75 cos 2 /− − = − − = ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ + − ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − = ° − ° = x 75 = 30 + 45 cos expansion substitution of special angles in the first expansion sin expansion substitution simplified answer for x/(6)Mathematics/P2 23 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 8.2 β β sin cos /y x x − = 2 3 3 sin cos 3 − = − β β … (1) β β sin cos /x y y + = 2 3 3 1 sin 3 cos + = + β β … (1*) (1)×3 + (1)*: (1) – 3(1)*: 2 3 3 1 ( 2 3 3 3 cos 10 + + ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − = β ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − − ⎟⎠ ⎞ ⎜⎝ ⎛ = − 2 3 3 21 3 sin 10 β ( ) ⎟⎠ ⎞ ⎜⎝ ⎛ = + + − = 21 10 3 3 1 3 3 9 21 OR 2 3 10 2 ) 3 3 1 ( 3 3 3− = + − − = 21 cos = ∴ β 23 sin = ∴ β ° = ∴ 60 β ° = ∴ 60 β OR 2 3 3 sin cos 3 − = − β β … (1) β β sin 3 2 3 3 1 cos − + = … (2) Substitute (2) into (1) ° = = − = − − − − = − − = − − + − = − ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − +60 23 sin 2 3 10 sin 10 2 3 9 3 3 3 sin 10 2 3 3 sin sin 9 2 3 9 3 2 3 3 sin sin 3 2 3 3 1 3β β ββ β β β β substitution into x/ substitution into y/ simplification solving simultaneously 23 or 21 answer (6) equation (1) equation (2) substitution simplification 23 sin = β answer (6) Mathematics/P2 24 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over OR 2 3 3 1 sin 3 cos and 2 3 3 sin cos 3 + = + − = − β β β β Try ° = 60 β 3 3 3 23 3 21 3 sin cos 3 − = ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ − ⎟⎠ ⎞ ⎜⎝ ⎛ = − β β 2 3 3 1 23 3 21 sin 3 cos + = ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ + = + β β ° = ∴ 60 β OR ° = ° − ° = ° = − + = − + = ° == 60 43 , 18 43 , 78 43 , 78 3 3 3 3 1 2 3 3 2 3 3 1 tan 43 , 18 31 tan βθθαα ⎟ ⎟⎠⎞ ⎜ ⎜⎝⎛ + − 2 3 3 1 ; 2 3 3 OR ° = 60 β substitution simplification substitution simplification (6) 31 tan = α ° = 43 , 18 α 3 3 3 3 1 tan − + = θ ° = 43 , 78 θ simplification answer (6) Note: Answer only: max 2 /6 marks α β θ (3 ; 1) ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝⎛ + − 2 3 3 1 ; 2 3 3 Mathematics/P2 25 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over β β sin cos /y x x − = ° ==== − − − + + − == + = − − − = − + + = − + + = − + + = − − − = + − − 60 21 cos 2 cos 4 2 ) cos 3 ( 3 ) cos 3 ( 3 cos 3 3 cos cos 3 sin 0 sin 10 cos 3 10 6 sin 3 3 sin cos 3 cos 3 6 sin 3 3 sin 9 cos 3 9 cos 3 2 sin 3 sin 3 cos 3 3 cos 1 sin 2 3 3 cos 2 3 3 1 6 sin 3 3 sin cos 3 cos 3 3 sin 2 3 3 1 cos 2 3 3 βββ β β β β β ββ β β β β β β β β β β β β β β β β β β β β β substitution substitution simplification β β cos 3 sin − = 21 cos = β answer (6) [12] Mathematics/P2 26 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 9 9.1 5 9 16 = + = r (Pyth) 53 sin = α Accept : 0,6 any one of the diagram value of r answer (3) 9.2 64 , 0 25 16 25 25 2 53 1 α 2 sin 1 α) (90 2 cos − = − = − ⎟⎠ ⎞ ⎜⎝ ⎛ = − = − − ° OR 64 , 0 25 16 2 54 2 cos ) 2 cos 2 (sin α 2 sin 1 α) (90 2 cos − = − = ⎟⎠ ⎞ ⎜⎝ ⎛ − = − = + − = − − ° α α α α sin α) cos(90 = − ° substitution of sin α = 53 (2) 9.3 25 1 25 24 1 54 53 2 1 cos sin 2 1 2 sin 1= − = ⎟⎠ ⎞ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ ⎛ − = − =− α α α OR 25 1 51 54 53 ) cos (sin cos cos sin 2 sin 2 sin 1 2 2 2 2 2 = ⎟⎠ ⎞ ⎜⎝ ⎛ − = ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ ⎛ − ⎟⎠ ⎞ ⎜⎝ ⎛ = − = + − =− α α α α α αα sin 2α = 2sin α.cosα 25 24 answer (3) 1 cos sin 2 2 = + α α 2 54 53 ⎟ ⎟⎠ ⎞ ⎜ ⎜⎝ ⎛ ⎟⎠ ⎞ ⎜⎝ ⎛ − ⎟⎠ ⎞ ⎜⎝ ⎛ answer (3) [8] 3 4 α 5 α 3 4 r (4 ; 3) r α Mathematics/P2 27 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 10 10.1 ° − ° θ − θ + ° + θ + ° 135 tan 180 sin ) sin( ) 180 cos( ) 90 sin( ) sin 1 ( cos sin . cos cos 1 0 ) sin )( cos ( cos θ θ θ θ θ θ θ θ + = + = + − − + = θ cos θ cos − θ sin − 0 + 1 answer (5) 10.2 2 2 6 42 46 2 21 . 22 23 . 22 2 ] 30 sin 45 cos 30 cos 45 [sin 2 ) 30 45 sin( 2 15 sin 2 2 cos 15 sin . 2 cos 2 ) sin 2 1 ( cos sin 2 15 sin . 2 cos cos sin 4 ) sin 2 1 ( 2 sin 15 sin . 2 cos cos sin 4 2 2 − = ⎥⎦ ⎤ ⎢⎣ ⎡ − = ⎥⎦ ⎤ ⎢⎣ ⎡ − = ° ° − ° ° = ° − ° = ° = ° = − ° = − ° A A A A A A A A A A A A A OR RHS A A A A A A A A = − = ⎥⎦ ⎤ ⎢⎣ ⎡ − = ⎥⎦ ⎤ ⎢⎣ ⎡ − = ° ° − ° ° = ° − ° = ° = ° = − ° = 2 2 6 42 46 2 22 . 21 22 . 23 2 ] 45 sin 60 cos 45 cos 60 [sin 2 ) 45 60 sin( 2 15 sin 2 2 cos 15 sin . 2 cos 2 ) sin 2 1 ( cos sin 2 15 sin . 2 cos cos sin 4 Side Hand Left 2 A Acos sin 2 A A 2 cos sin 2 1 2 = − ° 15 sin 2 15 = 45 – 30 or 15 = 60 – 45 substitution ⎥⎦ ⎤ ⎢⎣ ⎡ − 42 46 2 (6) A Acos sin 2 A A 2 cos sin 2 1 2 = − ° 15 sin 2 15 = 45 – 30 or 15 = 60 – 45 substitution ⎥⎦ ⎤ ⎢⎣ ⎡ − 42 46 2 (6)Mathematics/P2 28 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over OR RHS A A A A A A A A A A = − = ⎥⎦ ⎤ ⎢⎣ ⎡ − = ⎥⎦ ⎤ ⎢⎣ ⎡ − = ° ° − ° ° = ° − ° = ° = ° = − ° = 2 2 6 42 46 2 21 . 22 23 . 22 2 ] 30 sin 45 cos 30 cos 45 [sin 2 ) 30 45 sin( 2 15 sin 2 2 cos 2 sin 15 sin . 2 cos 2 sin 2 ) sin 2 1 ( cos sin 2 15 sin . 2 cos cos sin 4 Side Hand Left 2 10.3 , 360 . 240 or , 360 . 120 or solution no 21 cos 34 cos 0 ) 1 cos 2 )( 4 cos 3 ( 0 4 cos 5 cos 6 4 cos 5 6cos cos 4 5 cos 6 2 2 Z k k x Z k k xx or x x x x x x x x x ∈ ° + ° = ∈ ° + ° = − = = = + − = − − = − = − Alternative solution for 21 cos − = x ° ± ° = 120 360 . k x Z k ∈ standard form factors both equations ° + ° 360 . 240 k ° + ° 360 . 120 k Z k ∈ (6) [17] Note: If candidate puts ±k.360 then k ∈ N0 Mathematics/P2 29 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 11 11.1 m 1 2 , 117 75 , 64 cos 50 AC 50 75 , 64 cos = ° == ° AC OR m 21 , 117 AC m 2144026 , 117 AC 75 , 64 cos 50 AC = ∴ = ∴ ° = OR m 21 , 117 AC 25 25 sin 90 50sin AC 90 sinAC 25 25 sin 50= ∴ °° = ∴ ° = ° , , substitution in ratio AC subject of the formula answer (3) substitution into ratio AC subject of the formula answer (3) substitution into sine rule AC subject of the formula answer (3) 11.2 PC is given to be m 32 ) 64 ( 21 = 117,21 32 C AˆP tan = ° = 27 , 15 θ (15,27042173...) Note: If the candidate takes the unrounded answer for AC, then the answer is 15,27° (15,26987495...) C AˆP tan 117,21 32 answer (3) A C P 117,21 32 θ C Q ° 75 , 64 A P MD40 ●●Mathematics/P2 30 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 11.3 m 81 , 82 CD 289092 , 6857 25 , 25 cos ) 40 )( 21 , 117 ( 2 40 21 , 117 CD 2 2 2 = ∴ = − + = OR AM = ACsin 64,75° OR AM = CM tan 64,75° OR AM = AC cos 25,25° = 106,0111876 = 50 tan 64,75° = 117,21.cos25,25° = 106,01 = 106,01 = 106,01 DM = 106,01 – 40 = 66,01 CD2 = CM2 + DM2 = (50)2 +(66,01)2 = 6857,3201 CD = 82,81 metres OR AM = ACsin 64,75° OR AM = CM tan 64,75° OR AM = AC cos 25,25° = 106,0111876 = 50 tan 64,75° = 117,21.cos25,25° = 106,01 = 106,01 = 106,01 DM = 106,01 – 40 = 66,01 DC2 = (50)2 + (66,01)2 – 2(50)(66,01).cos 90° = 6857,3201 CD = 82,81 metres OR 01 , 34 21 , 117 32 40 75 , 64 sin = + + = °x x CD2 = CM2 + DM2 = (50)2 +(32 + 34,01)2 = 6857,3201 CD = 82,81 metres cos rule substitution answer (4) AM = 106,01 DM = 66,01 Pythagoras answer (4) AM = 106,01 DM = 66,01 cos rule answer (4) [10] C M D A 64,75° 25,25° 40 117,21 50 x 32 Note: If don’t use the rounded off then CD = 82,81 m. Accept this answer. Mathematics/P2 31 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over QUESTION 12 12.1 -120 -105 -90 -75 -60 -45 -30 -15 15 30 45 60 75 90 105 120 -5 -4 -3 -2 -112345 x y f g g g tan 2x: asymptotes ) 45 & 45 ( ° − ° x-intercepts middle curve end curves Note: The curve must tend toward the asymptotes 2 cos x: y-intercept x-intercepts (6) Note: If the candidate draws outside of the interval, NO PENALTY. Mathematics/P2 32 DBE/November 2010 NSC – Memorandum Copyright reserved Please turn over 12.2 For these values of x, cos 2x ≠ 0 ° = ° ± = = = − == − + = = − + = = − − == 30 90 0 cos or 21 sin or 1 sin 0 ) 1 sin 2 )( 1 (sin 0 cos or 0 1 sin sin 2 0 cos or sin sin 2 1 sin 2 1 cos sin 2 2 cos2 sin cos 2 2 2 2 x or x x x x x x x x x x x x xx xxx x OR For these values of x, cos 2x ≠ 0 ° = ° ± = = = − == − + = = − + = − + = − − = − − = − − == 30 90 0 cos or 21 sin or 1 sin 0 ) 1 sin 2 )( 1 (sin 0 cos or 0 1 sin sin 2 0 ) 1 sin sin 2 ( cos 2 0 cos sin 2 ) sin 2 1 ( cos 2 cos sin 2 ) 1 ) sin 1 ( 2 ( cos 2 cos sin 2 ) 1 cos 2 ( cos 2 1 cos 2 cos sin 2 2 cos2 sin cos 2 2 2 2 22 2 x or x x x x x x x x x x x x x x x x x x x x x x x x x x xxx x OR ° = ° ± = = = − == − + = − + = − = == − = − == 30 90 0 cos or 21 sin or 1 sin 0 ) 1 sin 2 )( 1 (sin 0 1 sin sin 2 sin sin 2 1 0 cos 2 or sin 2 cos 0 ) sin 2 (cos cos 2 0 cos . sin 2 2 cos . cos 2 2 sin 2 cos . cos 2 2 cos2 sin cos 2 2 2 x or x x x x x x x x x x x x x x x x x x x x x x x xx x xx 2 cos2 sin x x cos sin 2 x 2 sin 2 1− 0 cos = x factors equations ° ± 90 300 (8) xx 2 cos2 sin x x cos sin 2 1 cos 2 2 − x 0 cos = x factors equations ° ± 90 300 (8) xx 2 cos2 sin x x cos sin 2 x 2 sin 2 1− 0 cos = x factors equations ° ± 90 300 (8) Mathematics/P2 33 DBE/November 2010 NSC – Memorandum Copyright reserved OR ° − = ° + ° − = ° = ° − = ° + ° = ° + ° = ∈ ° + − ° ± = ° = − ° = = == − = − == 90 360 . 90 30 90 120 . 30 360 . 90 3 360 . ) 90 ( 2 90 ) 90 cos( 2 cos 0 cos 2 or sin 2 cos 0 ) sin 2 (cos cos 2 0 cos . sin 2 2 cos . cos 2 2 sin 2 cos . cos 2 2 cos2 sin cos 2 x k x or x or x k x k x Z k k x x x x x x x x x x x x x x x x x x xx x xx 2 cos2 sin x x cos sin 2 ) 90 cos( sin x x − ° = 0 cos = x factors equations ° ± 90 300 (8) 12.3 ° < < ° 45 0 x Or ° − < < ° − 45 90 x critical points notation (4) 12.4 Period ° = ° = 720 ) 360 (2 answer (2) 12.5 x = – 45° + 25° = – 20° x = 45° + 25° = 70° OR ° − = ° − = ° − = ° − ° − = ° − 20 40 2 90 50 2 90 ) 25 ( 2 xx x x and ° = ° = ° = ° − ° = ° − 70 140 2 90 50 2 90 ) 25 ( 2 xx x x x = – 20° x = 70° (2) [22] TOTAL : 150 Note: Answer only: full marks