MATHEMATICS PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
JUNE 2017

QUESTION 1

1.1	Percentages  Frequency  Cumulative Frequency  30≤𝑥<40   1  1 40≤𝑥<50  2  3 50≤𝑥<60   9  12 60≤𝑥<70   12  24 70≤𝑥<80   11  35 80≤𝑥<90   9  44 90≤𝑥<100   6  50	✓ 3, 12 ✓ 24, 35, 44 ✓ 50 (3)
1.2		✓ upper limits  ✓ cum f  ✓ shape  ✓ grounded (4)
1.3	Approx. 30 [accept between 28 – 32]	✓✓ answer/ indicated on graph.(2)

[9]

QUESTION 2

	12,4   15,1   18,9   19,7   19,7   20,0 20,9   23,7   23,8   31,1   33,6   34,5 34,9   36,5   40,1
2.1	Minimum = 12.4 Lower quartile (Q1) = 19.7 Median (Q2) = 23.7 Upper quartile (Q3) = 34.5 Maximum = 40.1	✓ min & max ✓ Q1 ✓ Q2 ✓ Q3 (4)
2.2		✓ min / max ✓ Q1 / Q3 ✓ Q2 (3)
2.3	Skewed positively to the right.	✓ positively skewed(1)
2.4	SD/SA = 8,36	✓✓ answer (2)
2.5	A small standard deviation indicates that the data is clustered around the mean. OR A large standard deviation indicates that the data is more spread out.	✓ answer (1)

[11]

QUESTION 3

3.1	M = [ 6+8 ; 9 −1]             2    2 M = (7 ; 4)	✓ x- value of M  ✓ y- value of M (2)
3.2	mFM =4−3 =  1               7−4     3  𝑦 − 𝑦1 =  1  (𝑥 − 𝑥1) m = 1                 3                      3 y – 4 =  1  (𝑥 −7) M = (7; 4)              3 ∴ y =  1 𝑥 +  5            3       3	✓substituting  ✓ value of mFM  ✓ substituting M(7; 4)  ✓ answer (4)
3.3	1 = 1 t +  5        3       3 t = -2 OR  mAF = mFM 3 − 1 = 1   4 − 𝑡    3 4 – t = 6 t = -2	✓ substitution into line equation  ✓ answer (as negative)  OR ✓ substitution into grad eqn  ✓answer as negative  (2)
3.4	mPC =3−(−1) = -1              4−8 ANSWER ONLY FULL MARKS	✓ substitution  ✓ answer  (2)
3.5	tanβ = -1 β = 135°	✓ tan β = -1 ✓ β = 135° (2)
3.6	tan α = −2 = - 1              10       5 ∴ α = 180° – 11.310 = 168.69° 𝐴𝐶 ̂𝑃 = 𝛼 − 𝛽 = 33.69°	✓ tan α = −1                  5 ✓ α = 168.69° 𝐴𝐶 ̂𝑃 = 𝛼 − 𝛽 ✓ answer (4)

[16]

QUESTION 4 

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4.1	CA =[ 7−1 ; 1−5 ]             2       2 = [3; −2] ANSWER ONLY	✓x-value ✓ y-value (2)
4.2	CA2 =(3 − 0)2 + (2 + 2)2 CA2 = 25 CA = 5 CB2 = (7 − 3)2 + (1 + 2)2 CB2 = 25 CB = 5 ∴CA = CB	✓ substitution  ✓ answer for CA ✓ answer for CB (3)
4.3		✓ substitution ✓ mAD = 7 ✓ substitution ✓ mAB  ✓ mAD mAB = -1 (5)
4.4	(𝑥 − 3)2 + (𝑦 + 2)2 = 25	✓correct centre  ✓ correct r2 (2)
4.5	mBC =  1-(-2)               7-3 =  3      4	✓ substitution ✓ mBC  (2)
4.6	mtan= −4               3 𝑦 − 1 = − 4  (𝑥 − 7)                3 𝑦 = −4 𝑥 + 31          3        3	✓ mtan ✓ subst m=−4 and B(7;1)                      3 ✓ answer (3)
4.7	AE = DB [diametersof samecircle] ∴ ABED is a rectangle [diagonals =]	✓AE = DB ✓ reason ✓ reason (3)

[20]

QUESTION 5

5.1.1	sin 238° = − sin 58° = – k	✓reduction ✓ answer (2)
5.1.2	cos 58° = sin 32° = √1 − 𝑘2	✓ sin 32° ✓ answer (2)
5.2	=        tan(180°−30°).sin(360°−60°).sin10°           cos(180°+45°).sin(180°−45°).cos(90°−10°) =    (−tan30°)(−sin60°)sin10°         (−cos45°)(sin45°)sin10°	✓ - tan 30° ✓ - sin 60° ✓ - cos 45° ✓ sin 45° ✓ sin 10° ✓ simplification ✓ answer  (7)
5.3	sin(𝛼+𝛽)=cos [ 90°−(𝛼+𝛽)] =cos[(90− 𝛼)− 𝛽] =cos(90°− 𝛼)cos𝛽−sin(90°− 𝛼)𝑐𝑜𝑠𝛽 =sin𝛼cos𝛽−cos𝛼sin𝛽	✓ cos [ 90°−(𝛼+𝛽)] ✓ cos[(90− 𝛼)− 𝛽] ✓ cos(90°− 𝛼)cos𝛽−sin(90°− 𝛼)𝑐𝑜𝑠𝛽 ✓ sin𝛼cos𝛽−cos𝛼sin𝛽 (4)
5.4	cos2𝑥+1   =       2cos2𝑥−1+1      sin2𝑥.tan𝑥     2sin𝑥cos𝑥 .  sin𝑥                                          cos𝑥 2 𝑐𝑜𝑠2𝑥 2 𝑠𝑖𝑛2𝑥 =   1         𝑡𝑎𝑛2𝑥	✓ identity numerator ✓ identity denominator ✓ sin𝑥     cos𝑥 ✓cos 2𝑥 simplification     𝑠𝑖𝑛2𝑥 (4)
5.5.1	sin𝑥 =2sin𝑥  cos𝑥 sin𝑥=2sin𝑥cos𝑥 sin𝑥−2sin𝑥cos𝑥=0 sin𝑥(1−2cos𝑥)=0 sin𝑥=0 or cos𝑥= ½	✓ identity (sin𝑥)                  cos𝑥 ✓ simplification ✓ factors (3)
5.5.2	sin𝑥=0 or cos𝑥= ½ 𝑥=0°+360°𝑘,𝑘∈𝑍 OR                         𝑥= ±60°+360°𝑘 𝑥=180°+360°𝑘             𝑘 ∈ 𝑍	✓ 𝑥=0° ✓ 𝑥=180° ✓ 𝑥= ±60° ✓ 360°𝑘 ,𝑘∈𝑍(4)

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QUESTION 6 

Related Items

6.1	✓ Endpoints  ( f ) ✓ Endpoints  ( g ) ✓ (45° ; 1) ( f ) ✓ (-45° ; 0) / (135° ; 0) / x-intercepts ( g ) ✓ Asymptotes ( f ) ✓ Shape  ( g ) (6)
6.2.1	x = -45°	✓✓ -45° (2)
6.2.2	(-90°;45°] OR -90°< x ≤ 45°	✓ 90° and 45° ✓correct inequalities (2)
6.3	90°	✓ answer  (1)

[11]

QUESTION 7

7.1	In ΔABC  AC  =  d  sin 𝑘  sin 𝑧 ∴AC = 𝑑.sin 𝑘              𝑠𝑖𝑛𝑧  ANSWER ONLY	✓ proportion  ✓ answer  (2)
7.2	In ΔADC      AC        =   ℎ   sin(90°−𝑦)    sin 𝑦 AC = ℎ.cos 𝑦            sin 𝑦 AC =   h           𝑡𝑎𝑛 𝑦 OR  AC =  1    ℎ     𝑡𝑎𝑛𝑦 AC =   h           𝑡𝑎𝑛 𝑦	✓ proportion ✓ answer   AC =  1    ℎ     𝑡𝑎𝑛𝑦 AC =   h           𝑡𝑎𝑛𝑦 (2)
7.3	ℎ = AC. sin 𝑦         cos 𝑦 ℎ = 𝑑 sin 𝑘. sin 𝑦       cos 𝑦. sin 𝑧 ℎ = d. tan 𝑦. sin 𝑘             sin 𝑧 OR 𝐴𝐶 =   ℎ          tan 𝑦 𝐴𝐶 = 𝑑. sin 𝑘            sin 𝑧 ∴    ℎ   = 𝑑.sin 𝑘    tan 𝑦     sin 𝑧 ∴ ℎ = 𝑑 sin 𝑘 . 𝑡𝑎𝑛𝑦                 𝑠𝑖𝑛𝑧	✓ subst AC = 𝑑.sin 𝑘                           𝑠𝑖𝑛𝑧 OR ✓ equating AC (1)
7.4	∴ ℎ = 𝑑 sin 𝑘 . 𝑡𝑎𝑛𝑦                𝑠𝑖𝑛𝑧 ℎ = 80 . sin 38°. tan 40°                sin 125 =50,45m	✓ substitution  ✓ answer  (2)

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QUESTION 8 

8.1	Line from centre perpendicular to chord, bisects the chord.	✓ answer (1)
8.2.	AĈB = 90° [ angle in semi-circle]  AĈB = D̂1 [ both = 90°] ∴OE || AC [corresp ∠'s equal]	✓ S ✓R ✓ R (3)
8.3	Â = x [ tan chord]  EÔB = x [corresp ∠s' ; AC || OE]	✓S ✓R ✓S ✓R (4)
8.4	EÔB = EĈB [ both = x] ∴ OBEC is cyclic quad [converse angles in same segment]	✓S ✓R (2)

[10]

QUESTION 9

9.1	Â= 49° [ ∠ at centre=2 ∠ at circumf.	✓S ✓R (2)
9.2	Ĉ1 = B̂1 [angles opp equal sides] B̂1 =180°−98°   [ angles of Δ]             2 𝐵1 = 41°	✓R ✓R ✓ S (3)
9.3	BĈD = 90° [ ∠’s in semi-circle]  B̂2 = Ĉ3 = 26° [ ∠’s in same segment]  Ĉ2 = 23°	✓ S/R ✓ S/R ✓ S (3)

[8]

QUESTION 10

10.1.1	QR2 = PQ2 + PR2             Pyth.Theo = 62 + 82 ∴UR =  5    RQ    10 =½	✓ subst. in Pyth ✓ QR = 10  UR = ½ RQ     (3)
10.1.2	PM = 4    [diagonals bisecteachother] QM2 = 62 + 42    [Pyth.Theo] QM = 2√13  MS = QM = 2√13 [Diagonals bisecteachother] ∴MV = √13 ∴VM = √13    MQ   2√13  = ½	✓ R ✓ QM = 2√13  ✓ MV = √13 VM = ½ MQ     (4)
10.2	UR  =  VM      [both = ½]  RQ      MQ ∴MR = VU [line divides two sides of Δ in prop]	✓ S ✓ R (2)

[9]

QUESTION 11 

11.1	Constr. On AB mark off AG = DE  On AC mark off AH = DF  Join GH.  Proof : In ΔAGH & ΔDEF: i) AG = DE (constr)  ii) Â = D̂ (given)  iii) AH = DF (constr)  ∴ ΔAGH ||| ΔDEF (SAS)  ∴ Ĝ1 = Ê But 𝐵 ̂= 𝐸 ̂ given ∴ Ĝ1 = B̂ ∴ GH || BC (corresp angles equal)  ∴ 𝐴𝐵 = 𝐴𝐶    𝐴𝐺   𝐴𝐻 ∴ 𝐴𝐵 = 𝐴𝐶    𝐷𝐸   𝐷𝐹   ( AG= DE, AH = DF) ∴ AB = AC= BC    DE   DF    EF	✓constr  ✓S ✓ S/R ✓S ✓R ✓S ✓R (7)
11.2
11.2.1	R2 = 𝑥 [ tan chord]  T1 = 𝑥 [ ∠’s opp equal sides]  Q3 = 𝑥 [ tan chord]  R1 = 𝑥 [ tan from same point]	✓ S/R ✓ S/R ✓ S/R ✓ S/R (any three) (3)
11.2.2	Q2 = 180° − 2𝑥 [ angles of Δ]	✓S ✓R (2)
11.2.3	𝑃 ̂ = 180° − 2𝑥 [sum of angles of Δ PQR] R3 = 𝑄2 = 180° − 2𝑥 [ tan chord] ∴ TR || QP [ corresp ∠’s =]	✓ S ✓ S/R ✓ R (3)
11.2.4	In Δ STR & Δ SRQ Ŝ = Ŝ common  R̂3 = Q̂2 tan chord  ∴ ΔSTR |||ΔSRQ [ AAA] / [HHH]	✓ S ✓ S ✓ R (3)
11.2.5	ST = SR SR    SQ ΔSTR |||ΔSRQ RS2 = ST. SQ	✓ S ✓R (2)
11.2.6	SP = SQ       [line toonesideof a Δ] PR    TQ = 5     3 PQ = PR [tan fromsamepoint] SP =  5  PQ    3	✓ S/R ✓ R valueof SP =  5              PQ     3 (3)

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TOTAL: 150