NATIONAL SENIOR CERTIFICATE MEMORANDUM MATHEMATICS MEMORANDUM P2 SEPTEMBER 2016 GRADE 12

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1 NATIONAL SENIOR CERTIFICATE MEMORANDUM MATHEMATICS MEMORANDUM P SEPTEMBER 06 GRADE This memo consists of 5 pages

2 Income Maths Memo / P September 06 QUESTION answer () answer ().. 45 minutes answer ()..4 min / max median ().. 40 p 54 values notation () 40 p p 00 p 9 equation answer () [] QUESTION y Advertising points plotted correctly 4 points plotted correctly 6 points plotted correctly. y 7,69x 0, 49 7,69x 0,49 (). r 0, 95 answer ().4 It is a very strong positive correlation. answer ().5 y 7,69(,5) 0,49 7, 405 substitution Income = R answer () x [9] ()

3 Maths Memo / P September 06 QUESTION.. x 5 y 4 x 0 y 8 x 8 y 5 B(8;5) x 8 y 5 ().. m AM 4 5 m y 4 ( x 5) y 4 x 5 gradient of AM perp gradient correct sub into equation y x x 9 9 x 9 C ;0 answer y = 0 coordinates of C (4) m AC 0 9 gradient of AC 9 tan A CX 9 A C X 80 4, 7 45,º = 45, 4, 7 answer (5)

4 Maths Memo / P 4 September BC is a diameter converse of in a semi circle S R () D ; D (;6) coordinates of D AD sub in distance formula.. 5 D ;6 E x;4 AD DE answer (4) 5 x x 4x x 4x 04 0 x 4x 96 x x 8 0 x or x 8 NA. mab mac y y 4 8 y 40 0 y sub in distance formula simplification standard form factors x 8 only (5) equating gradients simplification answer () [5]

5 Maths Memo / P 5 September 06 QUESTION 4 4. x x y 4y x y 49 M ; equation of circle answer () Line AK : y x 5 leta m AM a; bthen : Aa; a 5 a 5 a a a a b 4 A ;4 r AM 4 8 x y Distance = radius of bigger circle radius of smaller circle = K 5;0 AK = 7 - or 4, Area AMK AM. AK 8 8 standard form coordinates of A substitution sub and r value equation method () () answer () x 5 y 0 length of AK sub in area formula answer (5) [6]

6 Maths Memo / P 6 September 06 QUESTION sin cos Pythag correct sketch in correct quadrant r 8 5. cos45. tan x. sin x tan x sin x.tan x answer (5) reduction formulae value of special angle answer (6) [] QUESTION cos sin 0 4cos cos sin cos0 cos sin0 sin sin cos sin cos expansion substitution of special angles 4cos sin cos ()

7 Maths Memo / P 7 September cos sin sin cos tan n division 0 or g : y-intercept and x- intercepts shape tan answers (4) O ( 0; ) f : y-intercept and x-intercepts turning points 6.. O 0 90 shape values (5) notation () values notation () [8]

8 Maths Memo / P 8 September 06 QUESTION7 H 50 A 6,6 C P,8 B In AHC; AHC 6,4 AC 50 sin6,4 sin 6,6 AC 64,4 In CHB : CHB 68, BC 50 sin68, sin,8 BC 65,04 In ABC AB AB 987,94 64,4 65,04 64,465, ,65 sin A sin04,5 In ABC : 65,04 987,94 cos04,5 method value of AC method value of BC sub in cos-rule answer (6) method sin A 0,67 A 9, sin A 0, 67 answer () [9]

9 Maths Memo / P 9 September 06 QUESTION 8 8. S P T 4 N O W 0 R 8.. P P 90 in semi circle S P P 80 ' s in 8.. R Ŝ Ŝ 60 S N N PR subt = ' s N N 60 N N T R80 ' s in R 80 R 0 OR T 4 P S 80 ' s in 90 P P 0 P R NR subtends = ' s R 0 S R answer S R answer P 0 S/ R answer () () ()

10 Maths Memo / P 0 September O R at centre =. on circumference O (0) 60 N T 80 ' s in N O N S R answer () OR 8..4 O (0) 60 at center =. at circumference N O R 0 adj. on str. line N R ' s opp sides N O R N 80 ' s in 0 N 80 N 80 0 N 0 But N 60 N N 0 NW = WR given W 90 line from centre, midpoint chord T 90 W T 80 TNWO is a cyclic quad opp ' s suppl S/R S/R answer S S/R S R () (4)

11 Maths Memo / P September P 0 T O 8 Q R P T Q 90 in semi-circle PQ QT PT Pythagoras theorem 0 8 PT PT = 6 P T Q 90 rad tangent PR PQ QR Pythagoras theorem ( x 6) 0 QR x x x 64 QR x 64 QR 8.. QR QT TR Pythagoras theorem QR 8 x ( x 6) 0 8 x x x x x 8 x S R S S/R substitution S substituting QR answer () () () []

12 Maths Memo / P September 06 QUESTION 9 9. E C D O B Draw diameter OBE and EC A constr. B 90 radius tangent B Ĉ 90 in semi circle E 90 s in B E 90 B but B B E D E subt BD D construction S/R S/R S S/R (5) OR E D O C B A Draw diameter BOE and ED constr. B 90 radius tangent B D 90 in semi circle D But D B EC subtends = s But D E B D or ABC D construction S/R S/R S/R S (5)

13 Maths Memo / P September AB 6 CD 4 D 90 C B 4 E D F A 9.. In DBC and ADC, 9.. D A tan-chord DCA DCA common DBC /// ADC,, DC AC DBC /// ADC BC DC DC BC. AC 4 BC( AB BC) S/R S R correct ratio simplification substitution () 6 6BC BC BC 6BC 6 0 ( BC 8)( BC ) 0 BC 8 or BC BC 9.. DEC /// AFC given DC EC 4 EC AC FC 8 FC EC FC std form factors BC only S correct ratio AC = 8 simplification (6) (4) EC EF

14 Maths Memo / P 4 September CDF CFA FA FD given CDF CFA S R but CDF CFA ADC CDF CDF ADC S CDF ADC S and CDE CDF CDE ACD CDF CDF 6 answer (5) []

15 Maths Memo / P 5 September 06 QUESTION 0 Q R S W T 0. P T tan chord theorem P V S R P Q tan chord theorem T Q S RS // ST corresp s are equal R (4) 0. PS PW lines // one side of SR WT PS PT SR TQ PW PT WT TQ S R S () [7] TOTAL: 50

NATIONAL SENIOR CERTIFICATE GRADE 12

NATIONAL SENIOR CERTIFICATE GRADE 12 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 04 MEMANDUM MARKS: 50 This memorandum consists of pages. Mathematics/P DBE/04 NSC Grade Exemplar Memorandum NOTE: If a candidate answers a question