MATHEMATICS PAPER 2
NATIONAL SENIOR CERTIFICATE
GRADE 12
MEMORANDUM
SEPTEMBER 2018
QUESTION 1 | ||||
1.1 | a = - 6 ,54 | ? value of a | (3) | |
1.2 | y = - 6 ,54 + 1,05 ( 41 ) = 36 ,51 » 37 | ? substitution ? answer | (2) | |
1.3 | On the scatter plot | ?? x-intercept 6 < x < 8 and (45;41) both correct OR ?? (69,87;66,73) and (45;41) both correct | (2) | |
1.4 | r = 0 ,88 | ? ? answer | (2) | |
1.5 | Yes. The strong positive correlation
| ? Yes ? strong positive | (2) | |
[11] |
QUESTION 2 | |||
2.1 | Range = 29 - 10 | ?answer | |
2.2 | x = 15 + 23 + 17 + 24 + 26 + 18 + 28 + 13 + 10 + 28 + 29 = 231 Answer ONLY full marks | ?? answer | |
2.3 | σ = 6 ,37 | ? min ? max | |
2.4 | (21 - 6 , 37 ; 21 + 6 , 37 ) = (14 , 63 ; 27 , 37 ) | ? 231 | |
[8] |
QUESTION 3 | |||
3.1 | (√89)2 = (-3- t)2 + (-3 - 5)2 89 = 9 + 6 t + t2 + 64 t2 + 6t - 16 = 0 ( t - 2 ) ( t + 8 ) = 0 t = 2 or t = 8 | ?substitution ? simplification ? standard form ? factors ? value of t | (5) |
3.2 | mAB = - 3 | ? mAB | (4) |
3.3 | - 3 x - 12 = 0 | ? y = 0 | (2) |
3.4 | M = (- 5 + 2 ; 3 + 5) | ? x – coordinate | (2) |
3.5 | mEM = 0 - 4 | ? correct substitution | (4) |
3.6 | tan θ = - 3 | ? size of θ | (4) |
| [21] |
QUESTION 4 | |||
4.1 | PR = √(0 - 8 )2 + (- 3 + 9 )2 | ?correct substitution | (2) |
4.2 | M = ( 0+8 ; 3+9 ) | ?x-coordinate | (2) |
4.3 | mPQ = - 3 + 6 | ?correct substitution | (4) |
4.4 | (x- 4)2 + (y + 6)2 = 25 | ?r2 = 25 | (2) |
4.5 | mrad = -6+3 4-0 =-3/4 mtan = 4/3 y = 4/3 x - 3 | ?correct subst. | |
4.6 | (√146)2 = (cosθ - 8)2 + (sinθ + 9)2 | ?correct substitution ?simplification ?sin2 θ + cos2θ = 1 ?equation ?tanθ = 8/9 | (5) |
[20] |
QUESTION 5 | |||
5.1 | sin ( A + 30º ) = sinA cos 30º + cosA sin 30º | ?x = - 2√10 | (4) |
5.2 | - sin2 (90º - x ) - tan cos (- x). sin (- x - 360º) | ?(cos x)2 | (6) |
5.3 | x2 - 2 x sin A = cos 2 A | ?standard form | (3) |
5.4 | LHS / LK = cos 3x + sin 3x | ?Simplification | (3) |
5.5.1 | sin 68º = cos 22º OR | ?cos 22º OR | (2) |
5.5.2 | cos 16º = cos (38º - 22º) | ?cos (38 0 - 22 0 ) | (4) |
[22] |
QUESTION 6 | ||||
6.1 | a = - 2 | ?a = - 2 | (2) | |
6.2 | ||||
6.2.1 | x = 60º |
| (1) | |
6.2.2 | cos( x - 60 ) = sin 3 x | ?co-function | (6) | |
6.2.3 | - 52 ,50º< x < 15º | ?both critical values | (2) | |
[11] |
QUESTION 7 | ||||
7.1 | AB = k | ? AB i.t.o | (5) | |
7.2 | AC = 139 ,5 √1 + 8 sin2 (42º) | ? correct substitution | (2) | |
[7] |
QUESTION 8 | ||
8.1 | D2 = 28º[ ∠s subt by = chords ] | ? S ? R |
8.2 | Alternate ∠s = | ? R |
8.3 | EB = EC [ radii ] | ? S ? R |
8.3.1 | D1= 28º [ ∠s opp = sides] | ? S/R |
8.3.2 | E2 = 96º [ ∠s of a Δ] | ? S ?S ? R |
QUESTION 9 | ||||
9.1 | YU = XV = YV [prop the o , UV || ZW or line to one side of a Δ ] | ? S ?R | ||
9.2 | Area of ΔXVY = ½ x 3r x 3s x sin V2 | ?substitution ?substitution ?S/R ?answer (4) | ||
9.3 | X1 = V4 [alt . ∠s , XY || WX] | ? S/R | ||
9.4 | V3 = V4 = X1 UV is a tangent to circle XYV [converse of tan chord theo.] | ?S ?R (2) |
QUESTION 10 | |||
10.1 | |||
Constr: Mark M on PQ and N on PR such that PM = AB and PN = AC Proof : In ΔABC and ΔPMN AB = PM [constr] A = P [given] AC = PN [constr] Δ ABC º ΔPMN [S∠ S] B =PMN =Q [given] MN II QR [corresp ∠s =] PM = PN [prop theo, MNQR] PQ PR but AB = PM and AC = PN [const] AB = AC PQ PR | constr SSS R S/R S/R S | ||
10.2 | |||
10.2.1 | N1 = 90º [ ∠subt by diameter /∠ in semi - circle] | ? S ? R | |
10.2.2 | P4 = L4 [ tangent chord theorem] | ? S ? R | |
10.2.3 | N1 = Q [both = 90º] | ? S | |
10.2.4 | In ΔPLR and ΔQPR | ? SSS ? R ? ratios ?substitution ?LR (5) | |
10.2.5 | NM || PR [co - int ∠s supp OR corresp ∠s =] NM = ½ PR[midpoint theorem] | ? R | |
[26] |