GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - AMENDED SENIOR CERTIFICATE PAST PAPERS AND MEMOS MAY/JUNE 2017
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GRADE 12
SENIOR CERTIFICATE EXAMNATIONS
2017
MEMORANDUM
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. Stop marking at the second calculation error.
- Assuming answers/values in order to solve a problem is NOT acceptable.
- Geometry:
S = a mark for a correct statement (a statement mark is independent of a reason)
R = a mark for a correct reason (a reason mark may only be awarded if the statement is correct)
S/R = award a mark if statement and reason are both correct
QUESTION 1
| TIME TAKEN (IN HOURS) | 5 | 7 | 5 | 8 | 10 | 13 | 15 | 20 | 18 | 25 | 23 |
| COST (IN THOUSANDS OF RANDS) | 10 | 10 | 15 | 12 | 20 | 25 | 28 | 32 | 28 | 40 | 30 |
| 1.1 | a = 4,806... = 4,81 b = 1,323... = 1,32 y = 4,81 + 1,32x | ✓a = 4,81 ✓b = 1,32 ✓equation(3) |
| 1.2 | Cost = 25,974… = 25,97 thousand rand (calculator) = R25 970 OR y = 4,81 + 1,32(16) y=25,93 Cost = R25 930 | ✓ 25,97 ✓ answer (in Rands) (2) ✓ substitution ✓ answer (in Rands) (2) |
| 1.3 | r = 0,949…= 0,95 | ✓ answer (1) |
| 1.4 | x = 0 y = 4,81 OR (4,80647) ∴ R4 810 OR R4806,47 | ✓ x = 0 ✓ answer (2) [8] |
QUESTION 2
| 2.1 | modal class: 80 < x ≤ 100 | ✓ correct class (1) | ||||||||||||||||||
| 2.2 |
| ✓ 13 ; 21 ✓ 31 ; 35 (2) | ||||||||||||||||||
| 2.3 |  | ✓ grounded ✓ upper limits ✓ cum frequency ✓ shape (4) | ||||||||||||||||||
| 2.4 | No. of salesmen awarded bonuses: 35 – 26 = 9 salesmen | ✓ accept (25 – 27) ✓ accept (8 – 10) (2) | ||||||||||||||||||
| 2.5 | Established Mean =(30x7) + (50x6) + (70x8) + (90x10) + (110x4) 35 = 68,86 thousand rand or R68 857,14 = R69 000 or 69 thousand rand | ✓ top line using midpts & freq ✓ 2410 ✓ answer (nearest) (3) [12] |
QUESTION 3
| 3.1 | mCD = -3 -(-5) | ✓ substitution of C & D ✓answer (2) |
| 3.2 | mAD = -5 - 3 0-(-4) = -2 mCD x mAD = ½ x -2 =-1 ∴AD⊥DC | ✓ substitution of A & D |
| 3.3 |  | ✓ correct substitution ✓ length of AB ✓ correct substitution ✓ length of BC (4) |
| 3 .4 | mCD = mBF = ½ [BF ΙΙ DC] y-4 = ½(x-3) | ✓ mBF = ½ ✓ substitution of B(3 ; 4) ✓equation (3) |
| 3.5 | tan a = -2 ∴a = 116.57º a = 90° + θ [ext∠Δ] ∴θ = 26,57º OR tan a = -2 OR mAD = -2 ∴ tan θ = ½ ∴ θ = 26,57° OR Inclination of DE is β: tan β = ½ ∴ β = 26,57° ∴ODE = 63,43 ∴θ = 90 ° - 63,43° = 26,57° | ✓tan a = -2 ✓a = 116.57º ✓θ = 26,57º (3) ✓tan a = -2 ✓tan θ = ½ ✓θ = 26,57º (3) ✓β = 26,57º ✓ODE = 63,43 ✓θ = 26,57º (3) |
| 3.6 | x2 + y2 = r2 (4) 2 +(-3) 2 = 25 x2 + y2 = 25 | ✓ r2 = 25 ✓ equation (2) [17] |
QUESTION 4
| 4.1 |  | ✓ substitution M & P ✓ x-value of N ✓ y-value of N (3) |
| 4.2 |  | ✓ substitution N & P ✓ r = √13 ✓ LHS of eq ✓ RHS of eq (4) ✓ substitution N & M ✓ r = √13 ✓ LHS of eq ✓ RHS of eq (4) |
| 4.3 |  | ✓ correct substitution ✓ NMm ✓ MRm ✓ substitution of MRm&(–3 ;–2) ✓ equation (5) |
| 4.4 | Symmetry of a kite: S(–3 ; 4) OR PSM = 90º P [∠ in semi circle] PS ⊥ SM ∴ S(–3 ; 4) OR (NS)2 = (radius)2 (-3+1)2 + (y-1)2 = 13 (y-1)2 = 9 y-1 = ± 3 y=4 OR y≠-2 ∴ S(–3 ; 4) | ✓ x-value of S ✓ y-value of S (2) ✓ x-value of S ✓ y-value of S (2) ✓ x-value of S ✓ y-value of S (2) |
| 4.5 |
| ✓ equating lengths ✓ simplification ✓ y-value of R ✓ x-value of R (4) ✓ yR = 1 ✓ horizontal line OR R lies on y = 1 ✓ equating ✓ x-value of R (x < −4,6) (4) ✓ y = 2 x + 6 3 ✓ equating ✓ x-value of R (x < −4,6) ✓ y-value of R (4) |
| 4.6 |
|
|
| [22] |
QUESTION 5
| 5.1.1 | tanA = sinA
| ✓identity ✓value of tan A (2) ✓ y x ✓ value of tan A (2) |
| 5.1.2 | sin2 A + cos2 A = 1 (2p)2 + p2 = 1 4p2 + p2 = 1 5p2 = 1 p2 = 1 5 ∴p = -√ 1 5 | ✓ (2p)2 + p2 = 1 ✓ simplification of LHS ✓ answer (3) |
| 5.2 | 2sin2 x -5sin + 2 = 0 (2sinx -1)(sinx-2) = 0 sinx = ½ or sinx =2(no solution) ref ∠ = 30° ∴x= 30° + k360° or x=150° + k360°; k∈Z | ✓ factors or formula ✓ both equations ✓ no solution/geen opl ✓30°+k.360.30° ✓ 150°+k360°; ✓ k∈Z (6) |
| 5.3.1 | sin(x+300°) = sinxcos300° + cosxsin300° | ✓ expansion (1) |
| 5.3.2 | sin(x+300°) - cos(x-150°) | ✓ 2nd expansion 2nd expansion |
| 5.4 |  | ✓ identity of tan x ✓sinx + cosx cosx ✓sin 2x + cos 2x cosx ✓ sin 2x + cos 2x = 1 ✓ simplify (5) ✓ identity of tan x ✓sin 2x + cos 2x cosx ✓ sin 2x + cos 2x = 1 ✓ simplify ✓ multiplication (5) |
| 5.5.1 |  | ✓ square both sides ✓ sin2x+cos2x=1 ✓ sin 2x (3) |
| 5.5.2 | From 5.5.1 sin x + cosx = √1+sin2x ∴ max value: sinx + cosx = √1+1 =√2 OR Maximum value of 1 + sin 2x = 1 + 1 = 2 ∴ maximum value of sinx + cosx = √2 OR (sin x cos x)2 = sin2 x+2sinxcosx + cos2 x =1+sin2x ∴ max value: (sinx + cosx)2 = 1+1 =2 ∴ max value of sinx + cosx = √2 | ✓ max of sin 2x =1 ✓ answer (2)
|
| [27] |
QUESTION 6
| 6.1 | Period = 180° | ✓ answer (1) |
| 6.2 | °−75 | ✓ answer (1) |
| 6.3 |  | ✓cos45°.cosx+sin45°.sinx ✓ cos(x-45°) ✓✓ answer (4) |
| [6] |
QUESTION 7
| 7.1 | KC = 6cm | ✓ answer (1) |
| 7.2 | Let P be the point of intersection of KL and CB
| ✓ trig ratio ✓ length of KP ✓ answer (3) |
| 7.3 | DK2=62 + 122 DK = √180 or 6√5 or 13,42cm sinKDL = sinDLK KL DK sinKDL = KL sinDLK DK = 8+3√3 or 13,20 or 0,98 6√5 13,42 | ✓ DK=6√5 ✓ use of sine rule ✓sinKDL = KL sinDLK DK ✓ answer (4) |
| [8] |
QUESTION 8
| 8.1 | L-100º [ext∠cyclic quad = int opp∠] OR N1=80° [∠s on straight line] L = 100° [opp ∠s of cyclic quad] | ✓S ✓R (2) ✓S ✓R (2) |
| 8.2 | N1=80° [∠s on straight line] O1=160º [∠ at centre=2 × ∠ at circumference] OR refelx KOM=200º [∠ at centre=2 × ∠ at circumference] O1=160º [∠s around a pt] | ✓S ✓S ✓R (3) ✓S ✓R ✓S (3) |
| 8.3 | M1 = 360º-(100º+55º+160º) ∴M1 = 45º | ✓S ✓S (2) [7] |
QUESTION 9
| 9.1.1 | ∠ in semi-circle | ✓answer (1) |
| 9.1.2 | Opp ∠s of quad = 180° | ✓answer (1) |
| 9.2.1 | OF ⊥ AC [line from centre bisects chord] ∴ AC || GO [co-interior∠s = 180° OR/ corresp ∠s =] | ✓ S ✓ R ✓ R (3) |
| 9.2.2 | G1 = A2 [corresp ∠s AC ΙΙ GO] A2 = B1 [∠s in same segment] ∴G1 = B1 OR G1 = B2 [ext ∠cyclic quad] but ΔABF = ΔCBF [s,∠,s] ∴B2 = B1 ∴G1 = B1 | ✓ S ✓ R ✓ S ✓ R |
| 9.3 | OF:FB = 3:2 DB=2r ∴DO = 5k and DF =8k OR DF= 2r - 2 r = 8 r 5 5 ∴DG = DO = r [line ΙΙ side of Δ] DA DF 8 r 5 ∴DG = 5 DA 8 | ✓ S ✓ R ✓ S (3) |
| [12] |
QUESTION 10
| 10.1 | Tangent-chord theorem | ✓ R (1) |
| 10.2.1 | A2 + A3 =B1 + B2 [∠s opp = sides] | S ✓ R ✓ S ✓ R ✓ R (5) ✓ S ✓ R ✓ S ✓ R ✓ R (5) |
| 10.2.2 | B2 = x [tan chord theorem] x + T4 = B1 + B2 [corresp ∠s AB ΙΙ ST] T4 = B1 B1 = A1 [tan chord theorem] T4 = A1 | ✓ S ✓ R ✓ S ✓ R ✓ R (5) |
| 10.2.3 | T4 = A1 [proven in 10.2.2] ∴RTAP is a cyclic quadrilateral [line subtends =∠s ] | ✓ S ✓ R (2) |
| [13] |
QUESTION 11
| 11.1 | Constr: On sides AB and AC of ΔABC, mark points G and H respectively such that AG = DE and AH = DF. Draw GH Proof: ΔAGH = ΔDEF [s,∠,s] ∴ AGH = E =B [B=E given] ∴ GH ΙΙ BC [corresp ∠s =] ∴ AG = AH [line ΙΙ side of Δ] AB AC ∴ DE = DF [constr] AB AC | ✓ construction ✓ S/R ✓ S ✓ S /R ✓ S ✓ R (6) |
| 11.2.1(a) |  AP = PC [diag ΙΙm bisect each other] But TP = PS [given] AP - TP = PC - PS ∴ AT = SC( | ✓S ✓S OR S (2) |
| 11.2.1(b) | In PSR and PBA: | ✓S ✓R ✓S ✓R ✓R (5) ✓S ✓R ✓S ✓R ✓S (5) |
| 11.2.2(a) | PR = PS [ΙΙΙ Δs] PA PB ∴PR = TR = PS [Given PR = TR ] PA AD PB PA AD ∴PR = TR = TP [PS = TP;PB=PD] PA AD PD ∴ΔRPT ΙΙΙ ΔAPD [sides of Δ in prop] | ✓ S (all 3 ratios) ✓ S ✓ R (3) |
| 11.2.2(b) | T1 = D2 [ΙΙΙ Δs] ∴ATRD is a cyclic quad [converse:ext ∠ of cyclic quad] | ✓ S ✓ R (2) |
| [18] |
TOTAL: 150