GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - SENIOR CERTIFICATE EXAMINATIONS PAST PAPER (MEMORANDUM) 2016

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GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - SENIOR CERTIFICATE EXAMINATIONS PAST PAPER (MEMORANDUM) 2016

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Mathematics Paper 2
Grade 12
Senior Certificate Examinations
Past Paper (memorandum) 2016

NOTE:

If a candidate answered a question TWICE, mark only the FIRST attempt.
If a candidate has crossed out an attempt to answer a question and did not redo it, 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.

MEMORANDUM

QUESTION 1

1.1

Mean = ¹⁸⁹/₁₁

Answer only: Full marks

= 17,18

✓189
✓ answer (2)

1.2

Min = 8, max = 26
Median = 19
Q1 = 10, Q3 = 24

∴ (8 ; 10 ; 19 ; 24 ; 26)

✓ min, max
✓ median
✓ Q1 & Q3 (3)

1.3

✓ box
✓ whiskers (2)

1.4

The data is skewed to the left

Negatively skewed

✓ answer (1)
✓ answer (1)

1.5

SD/SA = 6,46

✓✓ answer (2)

1.6

17,18 + 6,46 = 23,64
∴ 3 destinations

✓ interval
✓ answer (2)
[12]

QUESTION 2

Temperature at midday (in °C)	18	21	19	26	32	35	36	40	38	30	25
Number of bottles of water (500 mℓ)	12	15	13	31	46	51	57	70	63	53	23

2.1	(30 ; 53)	✓answer (1)
2.2	a = – 38,51 b = 2,68 ∴ yˆ = 2,68x – 38,51	✓ value a ✓ value b ✓ equation (3)
2.3	∴ yˆ ≈ 36,53 bottles OR yˆ ≈ 2,68(28) – 38,51 ≈ 36,53 bottles	✓✓ answer (2) ✓substitution ✓ answer (2)
2.4	Strong The majority of the points lie close to the regression line. OR Strong r = 0,98	✓ strong ✓ reason (2) ✓strong/sterk ✓reason (2)
2.5	Temperature cannot rise beyond a certain point as this would be life threatening OR there is only so much water one can consume before it becomes a risk to your health (hyponatremia)./	✓ reason(1) [9]

QUESTION 3

3.1	m_AD = y₂ - y₁ x2 - x1 = 0 - 6 -2 +8 = -6 = -1 6	✓substitution ✓–1 (2)
3.2	m_BC = -1 [BC \| \| AD] y = -x + c 10 = -8 + c c = 18 y = -x +18 OR m_BC = -1 [BC \| \| AD] y - y₁ = m(x - x₁) y - 10 = -(X - 8) Y = -X + 18	✓ gradient ✓substitute m and (8 ; 10) ✓ equation (3) ✓ gradient ✓substitute m and (8 ; 10) ✓ equation (3)
3.3	m_BD = y₂ - y₁ x2 - x1 = 10 - 0 -8 + 2 m_BD ×m_AD = 1 × -1 = -1 ∴ DB ⊥ AD OR AD² = 72 or AD = √72 or 6√2 AB² = 272 or AB = √272 or 4√17 BD² = 200 or BD = √200 or 10√2 ∴ AB² = AD² + BD² ∴ ADB = 90º	✓substitution ✓ answer ✓ ⋅ m_BD ×m_AD = 1 × -1 = -1 (3) ✓✓ calculating all 3 sides ✓ AB² = AD² + BD² (3)
3.4	tan BDM = M_BD = 1 ∴ BDM = 45º OR sin BDM = BM = 10 = 1 BD 10√2 √2 ∴ BDM = 45º	✓ tan BDM = M_BD✓ answer (2) ✓sin BDM = 1 √2 ✓ answer (2)
3.5	T (x ; y) =[ x₁ + x₂ ; y₁ + y₂] [ 2 2 ] = [-^{2 +8} /₂ ;^{0 +10}/₂] = (3;5) T symmetrical about BM ∴ distance T to BM = 5 units = distance from BM to C ∴ C (13;5) OR mDF = 3¹/₃ - 0 = 1 8 - (-2) 3 Equation of DF: y - y₁ = m(x - x₁) y - 0 = ¹/₃ (x + 2) y = ¹/₃x + ²/₃ Equation of BC: y = -x + 18 ¹/₃x + ²/₃= -x + 18 4x = 52 x = 13 ∴ y = -13 +18 = 5 ∴ C (13;5)	✓T(3 ; 5) ✓value of x ✓ value of y (3) ✓eq of DF ✓value of x ✓ value of y (3)
3.6	area ΔBDF = area ΔBDM - area ΔDFM = ½ (10)(10) - ½(10)[¹⁰/₃] = ¹⁰⁰/₃ OR 33 ¹/₃ 0R 33,3 square units OR area ΔBDF = ½. BF.DM =½ [²⁰/₃](10) = ¹⁰⁰/₃ OR 33 ¹/₃ 0R 33,3 square units OR tan FDM =m_DC = 5 - 0 = 1 or tan FDM = FM = ¹⁰/₃= 1 13 + 2 3 DM 10 3 FDM = 18,43º ∴ BDF = 103,43º [ext ∠ Δ] BF = ²⁰/3 0R 6²/3 DF² = (10)² + 3¹/3)² [pyth Δ DFM] DF = 10,54 or ^√1000/3 or ^10√10/3 ∴area = ½.BF.FD.sinBFD ½[(²⁰/3)(^10√10/3)](sin 108,43º) = ¹⁰⁰/₃ OR 33 ¹/₃ 0R 33,3 square units OR BF = ²⁰/3 or 6²/3 BD = √(10-0)² + (8 + 2)² = √200 OR 10√2 area ΔBDF = ½.BF.BD.sinDBF = ½[(²⁰/3)(√200)](sin 45º) = ¹⁰⁰/₃ OR 33 ¹/₃ 0R 33,3 square units OR area Δ BDF =area ΔBCD - areaΔBCF = ½ (10√2)(5√2) - ½(²⁰/3)(5) = ¹⁰⁰/₃ OR 33 ¹/₃ 0R 33,3 square units OR tan FDM =m_DC = 5 - 0 = 1 or tan FDM = FM = ¹⁰/₃= 1 13 + 2 3 DM 10 3 FDM = 18,43º BDF = 26,56º area Δ BDF = ½.BD.DF.sin BDF = ½.[(10√2)(^10√10/3.sin 26,56º = ¹⁰⁰/₃ OR 33 ¹/₃ 0R 33,3 square units	✓ formula/method ✓ 10 (DM) ✓ 10 (BM) ✓= ¹⁰/₃ OR 3 ¹/₃ (⊥h) ✓ answer (5) ✓ formula/method ✓ BF ✓✓ DM ✓ answer (5) ✓ gradient/ratio ✓ BFD ✓ DF ✓correct substitution into area rule ✓ answer (5) ✓ BF ✓ BD ✓ formula/method ✓correct substitution into area rule ✓ answer (5) ✓ formula/method ✓ BD = 10√2 ✓ BC = 5√2 ✓ BF = ²⁰/3 ✓ answer (5) ✓ gradient/ratio ✓ BDF ✓ DF OR BD ✓correct substitution into area rule ✓ answer (5) [18]

QUESTION 4

4.1	radius ⊥ tangent	✓ R (1)
4.2	CR² = TR² + CT² (Pyth) CR² = 20² + 10² = 500 CR = √500 or 10 √5	✓ substitution ✓ answer (2)
4.3	CR² = (X₂ - X₁)² + (Y₂ -Y1)² 500 = (k - 3)² + (21 +1)² k² - 6k + 9 + 484 = 500 k² - 6k - 7 = 0 (k-7)(k+1) = 0 k = 7 or k = -1 OR CR² = (X₂ - X₁)² + (Y₂ -Y1)² 500 = (k - 3)² + (21 +1)² (k - 3)² = 16 k - 3 = 4 or k - 3 = -4 k = 7 or k = -1	✓ substitution ✓ standard form ✓ factors ✓ k = 7 (4) ✓ substitution ✓ square form ✓ square root ✓ k = 7 (4)
4.4	(x - 3)² + (y + 1)² = 100	✓✓ answer (2)
4.5	CS = 10 and CS ⊥ PS ∴S(3; −11) ∴y = – 11	✓S(3; −11) ✓ answer (2)
4.6.1	S(3;-11) ∴ 3(-11) - 4x = 35 x = -17 ∴ P(-17;-11) OR ⁴/3x + ³⁵/3 = -11 ⁴/3x = ^-68/3 x = -17 ∴ P(-17;-11)	✓ substituting ✓ answer (2) ✓ equating ✓ answer (2)
4.6.1	PT = PS [tangents from common point] = 17 + 3 = 20 units OR = √500 or 10√5 PT² = PC² - TC² [Pyth th] = 500 - 100 = 400 ∴ PT = 20 OR = √500 or 10√5 ΔPTC = ΔRTC [90ºHS] ∴ PT = TR ∴ PT = 20	✓ S ✓R ✓ answer (3) ✓ value of PC ✓ using Pyth ✓ answer (3) ✓ value of PC ✓ S/R or proved ✓ answer (3)
4.7.1	M(3;−16)	✓answer (1)
4.7.2	Radius = 4	✓ answer (1)
4.7.3	r¹ + r² = 10 +4 = 14 distance CM = = √225 = 15 CM > r¹ + r² Therefore the two circles do not intersect or touch	✓r¹ + r² ✓ 15 ✓explanation (3) [21]

QUESTION 5

5.1.1(a)	sin T = ¹/_√5 = ^√5/₅ = 0,45	✓value (1)
5.1.1(b)	cos S = ³/_√10 = ^3√10/₁₀ = 0,95	✓value (1)
5.1.2	cos (T + S) = cosTcosS - sinTsinS = ⁶/_√50 - ¹/_√₅₀= ⁵/_√50 or ¹/_√2 or ^√2/₂	✓expansion ✓²/_√5 ✓¹/_√10 ✓ simplification ✓ answer (5)
5.2	1 - tan²(180º + θ) cos(360º - θ)sin(90º - θ) = 1 - tan² θ (cos θ)(cos θ) = 1 - (sin² θ) (cos² θ) (cos² θ) = 1 - sin² θ cos² θ = cos² θ or 1 - sin² θ cos² θ 1 - sin² θ	✓cosθ ✓cosθ ✓tan² θ ✓sin² θ cos² θ ✓identity ✓ answer (6)
5.3	(sin x - cos x)² = [³/₄]² sin²x - 2sinxcosx + cos²x = ⁹/₁₆1 - 2sinxcosx = ⁹/₁₆ 2sinxcosx = ⁷/_{16 ∴sin2x = ⁷/₁₆}	✓ squaring both sides ✓expanding LHS ✓ using identity ✓ simplifying ✓answer (5) [18]

QUESTION 6

6.1	4sinx + 2 cos 2x = 2 2sinx + cos 2x - 1 = 0 2sinx + (1 - 2sin²x ) - 1 = 0 2sin²x - 2sinx = 0 2sinx (sin x - 1) = 0 2sin x = 0 or sin x - 1 = 0 sin x = 0 sin x = 1 x = k.180º or x = 90º +k.360,k∈Z	✓using identity ✓ standard form ✓ factors ✓sin x = 0 or sin x =1 ✓ k.180° ✓ x = k.180º or x = 90º +k.360,k∈Z (6)
6.2.1		✓ turning point (–90° ; –3) ✓ turning point (90° ; 1) ✓ (–180° ; –1) & (0° ; –1)
6.2.2	(−90°;0°) OR − 90°< x < 0°	✓ ✓ answer (2) ✓ ✓ answer (2)
6.2.3	f(x) = g(x) ∴ -180º ; 0º; 90º; 180º f(x +30º) = g(x + 30º) ∴x = -30º; 60º ; 150º	✓ any ONE correct ✓ other 2 correct (2) [13]

QUESTION 7

7.1

ABD = θ [alternate∠s ; || lines]
cos θ = BD = 64
AB 81
θ = 38º

OR
sin BAD = 64
81
BAD = 52,18º
θ = 38º

✓correct trig ratio
✓substitution into
correct ratio
✓ answer (to the nearest degree) (3)

✓correct trig ratio
✓substitution into correct ratio
✓ answer (to the nearest degree) (3)

7.2

BC² = AB² + AC² - 2(AB)(AC)cosBAC
=81² + 87² - 2(81)(87)cos82,6º
=12312,754.....
BC = 110,97m

✓use cosine rule
✓correct substitution into cosine rule

✓answer (3)

7.3

sinDCB = sinBDC
BD BC
sinDCB = BD.sinBDC
BC
sinDCB = 64.sin110º
110,97
∴DCB = 32,82º

✓ use sine rule
✓ substitution
✓ answer (3)
[9]

QUESTION 8
8.1

8.1.1	P = 32º [opp ∠ s of cyclic quad	✓ S ✓ R (2)
8.1.2	O₁ = 2(32º) = 64º [∠centre = 2∠ at circum ] OR reflex O = 296º [∠centre = 2∠ at circum] O₁ = 64º [∠s around a point]	✓ S ✓R (2) ✓ S and R ✓ S (2)
8.1.3	OMS = 180º -(32º +18º +43º) [sum ∠ sΔ] = 87º	✓ S ✓ S (2)
8.1.4	R3 = TMP = 87º + 18º - 6º =99º	✓ R ✓ S (2)

8.2

8.2.1

corres ∠s; AB| | DC

✓ R (1)

8.2.2

E₂ = x [tan - chord theorem]
B₂ = x [∠s opp = sides]
E₃ = x [alt ∠s ; AB|| DC]
DAB = x [opp ∠s ||^m OR alternate ∠ BC || AD]

Any 3 ∠s correct

✓S ✓R
✓ S ✓ R
✓ S ✓ R (6

8.2.3

D = 180º - x [co - int ∠s suppl ; AD || BC]
∴ B2 + D = 180º
∴ ABED a cyc quad [converse opp ∠s of cyclic quad]

DAB = x [OPP ∠s ||^m] OR [alt ∠s ; BC|| AD]
E₃ = DAB = x
∴ ABED a cyc quad [converse ext ∠ of cyclic quad]

✓S ✓ R
✓ R (3)

✓S ✓ R
✓ R (3)
[18]

QUESTION 9

9.1

… in the alternate segment

✓ answer (1)

9.2

9.2.1	A₁ = D₁[tan chord theorem] B₄ = A₁ + D₂ [ext ∠ Δ] =D₁ + D₂	✓S ✓ R ✓ S ✓ R (4)
9.2.2	B₄ = B₂ [vert opp ∠s] D₁ + D₂ = B₂ [proven] =G₂ [∠s in same segment] ∴ AGCD is cyc quad [converse ext ∠ cyc quad]	✓ S ✓ S ✓ R ✓ R (4)
9.2.3	D₁ = A2 [∠s in same segment] A₂ = F [∠s in same segment] ∴ D₁ = F ∴ DC = CF [side opp ∠s ]	✓ S ✓ R ✓ S ✓ R (4) [13]

QUESTION 10

10.1

Constr:
Draw line BC such that MB = AK and MC = AF
Proof:
In ΔBMC and ΔKAF
MB = AK [constr]
M = A [given]
MC = AF [constr]
ΔBMC = ΔKAF [s∠s]
∴MBC = AKF or MCB = AFK [ΞΔ]
but V = K or T = F [given]
∴MBC = V or MCB = T
But these are corresponding ∠s
∴ BC || VT [corr∠s≡]
∴ MV = MT [prop theorem; BC || VT]
MB MC
but MB = AK and MC = AF [const]
∴ MV = MT
AK AF

✓ constr/konstr

✓ S / R
✓ S

✓ S

✓ S / R

✓S ✓R (7)

10.2

10.2.1 (a)	In ΔKGH and ΔKEF K is common H₂ = F [ext ∠ CYCLIC QUAD] G₃ = E [sum∠s Δ or ext∠cyclic quad] ∴ ΔKGH \|\|\| ΔKEF [∠∠∠]	✓ S ✓S ✓R ✓ naming third angle OR ∠∠∠ (4)
10.2.1 (b)	EF = KE [\|\|\| Δs] GH KG ∴ EF = KE GH EF [KG = EF] ∴ EF² = KE.GH	✓ S ✓ S (2)
10.2.1 (c)	KG = EM [prop theorem ; MG \|\| EK] KF EF but EF = KG [given] KG = EM KF KG KG² = EM.KF	✓ S ✓ R ✓ S (3)
10.2.2	KE.GH = EM.KF EM = 20 × 4 16 =5 units	✓ KE.GH = EM.KF ✓ substitution ✓ answer (3) [19]

TOTAL: 150

Last modified on Tuesday, 15 June 2021 08:15

Published in Grade 12 Amended Senior Certificate Exam Past Papers and Memos 2016

GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - SENIOR CERTIFICATE EXAMINATIONS PAST PAPER (MEMORANDUM) 2016

MEMORANDUM

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