MATHEMATICS PAPER 2 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS SEPTEMBER 2020 PREPARATORY EXAMINATIONS

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QUESTION 1

1.1	a = −4,1536 b = 0,958 y = −4,1536 + 0,958x Answer Only: Full Marks	✓a = −4,1536 ✓b = 0,958 ✓y = −4,1536 + 0,958x (3)
1.2	r = 0,98	✓r = 0,98 (1)
1.3	Very strong positive correlation	✓answer (1)
1.4	y = −4,1536 + 0,958(51) y = 45% Answer Only: Full Marks	✓ substitution ✓answer (2)
1.5	x̄ = 60,8 Standard deviation = 17,51 (60,8 − 17,51 ; 60,8 + 17,51) (43,29; 78,31) 6 learners	✓ Standard deviation=17,51 ✓(43,29; 78,31) ✓6 learners (3)
		[10]

QUESTION 2

2.1

Speed (km/h)	Frequency (ƒ)	Cumulative Frequency
0< x ≤ 10	10	10
10< x ≤ 20	20	30
20< x ≤ 30	45	75
30< x ≤ 40	72	147
40< x ≤ 50	23	170

✓freq column

✓ cum freq colum

(2)

2.2

✓starting pt

✓ end point

✓ shape

(3)

2.3

Q₁ = 23 (accept 22-24)
Median = 31 (accept 30-32)

✓Q₁
✓Median (2)

2.4

✓for Q₃ = 37 (accept 36-38)
✓ correct shape
(2)

2.5

170 − 110 = 60 cyclists
(accept 59-61)

✓answer (1)

[10]

QUESTION 3

3.1	m_QR = −2 − (−4) = − ¹/₃ 0 − 6	✓ substitution ✓answer (2)
3.2	m_PQ = 3 m_PQ × m_QR = 3 × ^{− 1}/₃ = − 1 ∴ PQ̂R = 90°	✓ m_PQ = 3 ✓m_PQ × m_QR = 3 × ^{− 1}/₃ = − 1 (2)
3.3	Sub: y = −x + 2 into 3x − y − 2 = 0 ∴3x − ( −x + 2 ) − 2 = 0 3x + x − 2 − 2 = 0 4x = 4 x = 1 y = 1 ∴ P(1;1)	✓substitution ✓x - coordinate ✓y = coordinate (3)
3.4	QR = √(0 − 6)² + (− 2− (− 4))² QR = 2√10 OR √40 OR 6,32 Answer Only: Full Marks	✓ substitution in correct ƒ ✓ answer (2)
3.5	PR is the diameter ( angle subtended by diameter = 90°) Midpoint of PR (⁷/₂; −³/₂) PR = √(1−6)² + (1 + 4)² PR = √50 r = ^√50/₂ (x − ⁷/₂)² + (x + ³/₂)² = (^√50/₂ )² OR (x − 3,5)² + (x + 1,5)² = (^√50/₂ )²	✓ for the statement PR is the diameter ✓✓Midpoint of PR ✓for the radius ✓equation (5)
3.6	tan PNX = −1 ∴ PNX 135° tan PMX = 3 ∴ PMX = 71,57° θ = 135° − 71,57° = 63,43°	✓ tan PNX = −1 ✓ ∴ PNX 135° ✓ tan PMX = 3 ✓ ∴ PMX = 71,57° ✓answer (5)
3.7	A = ½ × PQ × QR A_ΔPQR= ½ × √10 × √40 A_ΔPQR = 10 square units OR A_ΔPQR = ½ × PQ × QR × sin 63,43° A_ΔPQR= ½ × √10 × √40 × sin 63,43° A_ΔPQR = 10 square units	✓formula ✓ √10 ✓ answer (3)
		[22]

QUESTION 4

4.1	x² − 6x − y² − 4y + 9 = 0 x² − 6x + 9 + y² − 4y + 4 = −9 + 9 + 4 (x −3)² + (y −2)² = 4 C(3;2) and r = 2	✓ completing square ✓ standard form ✓ 3 ✓ 2 (4)
4.2	m_tan= −2 m_BV = ½ y − 2 = ½(x − 3) y = ½x + ½	✓ m_BV = ½ ✓ substitution ✓ answer (3)
4.3	y = 4	✓ answer (1)
4.4	TA = 4 units TB = TA (tangents from the same point) TB = 4 units	✓ length of TA ✓ S ✓ R ✓ answer (4)
4.5	T(−1;4) y = −2x +k 4 = −2( −1) + k k = 2	✓ substitution ✓ answer (2)
4.6	tan STA = −2 STA = 116,57° ∴ ACB = 116,57° (ext. angle of a.c.q) OR tan VCE = ½ ∴ VCE = 26,57° ∴ACB = 180° − (90° − 26,57°) = 116,57° (∠s on straight line)	✓ tan STA = −2 ✓ STA = 116,57° ✓ answer ✓ reason OR ✓ tan VCE = ½ ✓ ∴ VCE = 26,57° ✓ answer ✓ reason (4)
		[18]

QUESTION 5

5.1.1	cos 158° = − cos 22° =− p	✓= − cos 22° ✓=− p (2)
5.1.2	sin 112° =sin ( 90° + 22°) = cos 22° =p	✓cos 22° ✓ p (2)
5.1.3	sin 38° = sin(60° − 22°) =sin 60° cos 22° − cos 60° sin 22° = ^√3/₂p − ½√(1 − p²)	✓ sin(60° − 22°) ✓expansion ✓^√3/₂ ✓ ½√(1 − p²) (40
5.2	sin P = sin 2P sin P − sin 2P = 0 sin P − 2sin PcosP = 0 sin P(1 − 2cosP) = 0 sin P = 0 or cos P = ½ P ∈ (0° ;60° ;180° ;300° ;360°) OR P 2P + 360°k or P = 180° − 2P + 360°k , ? ∈ ℤ P = −360°k or 3P = 180° + 360°k P = 60° + 120°k P ∈ (0° ;60° ;180° ;300° ;360°)	✓standard form ✓expansion ✓factorisation ✓all correct values of P ✓P 2P + 360°k ✓P 180° − 2P + 360°k ✓P 60° + 120k ✓all correct values of P (4)
5.3	A + B + C 180° A + B = 180° − C cos(A + B) = cos(180° − C) cos(A + B) = − cosC	✓ A + B = 180° − C ✓cos(A + B) = cos(180° − C) (2)
5.4	= (2 cos x + 1 ) (cos x − 1) sin x( 2 cos x + 1) = cos x − 1 = 1 − 1 = RHS sin x sin x tan x sin x	✓1 − cos²x ✓sin x(2 cos x + 1) ✓2 cos² x − cos x − 1 ✓(2 cos x + 1 )(cos x −1) ✓ cos x − 1 sin x sin x (5)
5.5	4 + 7 cos θ = cos 2θ = 0 4 + 7cos θ + 2 cos ² − 1 = 0 2 cos² θ + 7cos θ + 3 = 0 (2 ??? ? + 1) (??? ? + 3) = 0 cos θ = −½ or cos θ = − 3 (N/A) θ = 120° = 360°.k or θ = 240° + 360.k , ? ∈ ℤ	✓ 2 cos² − 1 ✓ standard form ✓ factors ✓ cos θ = −½ or cos θ = − 3 ✓ θ = 120° = 360°.k ✓ θ = 240° + 360.k (6)
		[25]

QUESTION 6

6.1	b = ½	✓ answer (1)
6.2	A(30°;1)	✓ 30° ✓ 1 (2)
6.3	g(90°) = cos(90° − 30°) = cos 60° =½ Q(90°;½)	✓ 90° ✓ ½ (2)
6.4	x = 160°	✓ x = 160° (1)
6.5	−1 ≤ y ≤ 3 y ∈ R OR y ∈ [−1;3]	✓✓ answer (2)
		[8]

QUESTION 7

7.1	∠ LNM 180° − 2p ( angles opp. = sides)	✓answer ✓reason (2)
7.2	LM = d sin(180° − 2p ) sin p LM = d sin 2p sin p LM = d sin 2p sin p	✓for applying th sine rule ✓sin 2p (2)
7.3	tan q = h LM h = LM tan q h = d sin 2p . tan q sin p h = 2d sin p cos p tan q sin p h = 2d cos p tan q	✓tan q = h LM ✓h = d sin 2p . tan q sin p ✓h = 2d sin p cos p tan q sin p (3)
		[7]

QUESTION 8

8.1	bisects the chord	✓ answer (1)
8.2	EB 8 − y In ΔAEB: 10² = x² + (8 − y)² ...………….(1) Eqn of the circle x² + y² = 64 x² = 64 − y²………………..(2) Subst. (2) into (1) 100 = 64 − y² + 64 −16y + y² 100 = 128 − 16y 16y = 28 y = ⁷/₄ ∴ OE = ⁷/₄	✓ for EB ✓Pythagoras in Δ AEB ✓ equation of the circle ✓substitution ✓answer (5)
8.3	Double the size of the angle subtended by the same arc.	✓ answer (1)
8.4.1	(∠ at centre = 2 x ∠ at the circumference	✓statement ✓reason (2)
8.4.2	(∠s opp = sides)	✓ statement ✓ reason (2)
8.4.3	(opp. ∠s of a cyclic quad)	✓statement ✓reason (2)
8.4.4	(∠s in the smae segment)	✓statement ✓reason (2)
		[15]

QUESTION 9

9.1	∠ PCQ = 80°( ∠ s opp = sides) ∠PCB = 100° (∠s on a straight line) ∴BC is not a diameter (angle between the tangent and BC is not equal to 90°	✓statement ✓reason ✓statement ✓reason ✓conclusion
9.2	P₁ = B (alt ∠ s, PQ \|\| AB) B = C₃ (s opp = sides; radii) C₃ = C₁ ( vert. opp. angles) ∴P₁ = C₁ ∴PQ = QC (sides opp = angles)	✓statement ✓reason ✓statement ✓reason ✓statement ✓ statement and reason
9.3	A=E₂ (ext. ∠ of a cq) D = 180° − E₂ (co–interior ∠s; BE \|\| CD) D+A 180° ˆACDF is a cq (opp ∠ s supplementary) OR D=E₁(corres. ∠ s; BE \|\| CD) E₂ = 180° − E₁ (∠ s on a straight line) A = 180° − E₁ (opp ∠ s of a cq) D+A = 180° ∴ ACDF is a cyclic quad. (opp ∠ s of a quad. supplementary)	✓statement ✓reason ✓statement and reason ✓statement and reason ✓reason ✓statement ✓reason ✓statement and reason ✓statement and reason ✓reason
		[16]

QUESTION 10

10.1	RTP : MS = MT SO TN Construction: Join SN, and OT, and construct perpendicular heights Proof : But ΔMST is common And area of ΔOST = area of ΔTNS (same base, same height) ∴ MS = MT SO TN	area ✓area of the two triangles ✓ MS SO ✓area of the two triangles ✓MT TN ✓ statement and reason (5)
10.2.1	In ΔAPS and Δ BRS P₄ = R₁ (tan – chord theorem) A = B₂ = 90° (given) / (gegee) ΔAPS \|\|\| ΔBRS (AAA)	✓statement and reason ✓statement ✓3 rd ∠ OR ✓reason for similarity (3)
10.2.2	AP = PS = AS (similar triangles) BR RS BS ∴ AP.RS = BR,PS	✓for the statement (1)
10.2.3	P₂ = 90° (∠s in a semi – circle) Let P₄ = x ∴ S₁ = 90 − x (∠s of APS) ∴ Q 90 − x (ext ∠ of a cq) ∴ R₂ = x (∠s of QPR) ∴ P₄ = R₂	✓P₂ = 90 (in a semi – circle) ✓S₁ = 90 – x ✓Q = 90 – x ✓R₂ = x (4)
10.2.4	In ΔASP and ΔPQR A = P₂ (proven) P₄ = R₂ (proven) ΔASP \|\|\| ΔPQR (AAA) AS = SP = AP (similar triangles) PQ QR PR ∴AP . QR = SP . PR ∴ AP = PR PS RQ But AP = BR PS RS (from 10.2) ∴ PR = BR RQ RS ∴ BR.RQ = RS.RP	✓statement and reason ✓statement ✓reason for similarity ✓∴ AP = PR PS RQ ✓AP = BR PS RS ✓ ∴ PR = BR RQ RS (6)
		[19]

TOTAL : 150

Last modified on Friday, 18 February 2022 07:26

Published in Grade 12 September 2020 Preparatory Examinations

MATHEMATICS PAPER 2 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS SEPTEMBER 2020 PREPARATORY EXAMINATIONS

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