Mathematics P2 Grade 12 Memorandum - NSC Exams Past Papers and Memos September 2019 Preparatory Examinations

Thursday, 10 February 2022 13:52

Mathematics P2 Grade 12 Memorandum - NSC Exams Past Papers and Memos September 2019 Preparatory Examinations

More in this category: « Mathematics P1 Grade 12 Memorandum - NSC Exams Past Papers and Memos September 2019 Preparatory Examinations Mathematics P1 Grade 12 Errata - NSC Exams Past Papers and Memos September 2019 Preparatory Examinations »

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MARKING SCHEME

QUESTION 1
1.1	x = 420,8 80 = 5,26	√ 420,8 80 √ answer	(2)
1.2	total of heights = 4,86 × 7 = 34,02 error = 4,98 - 4,89 = 0,09 new total heights = 33,93 new mean = 33,93 = 4,85m 7	√ error = 0,09 √ new total √ answer	(3)
1.3	Standard deviation of individual heights of the 80 giraffes would be bigger than that of 7 males. There is a bigger spread of heights among population of juveniles and adults and babies.	√ bigger/larger standard deviation √ bigger spread of heights	(2)
			[7]
QUESTION 2
2.1		√√ all points plotted correctly	(2)
2.2	y = a + bx a = 4,64 b = 0,27 y = 4,64 + 0,27x	√ value of a √ value of b √ equation	(3)
2.3	see line in 2.1	√ correct gradient √ correct line	(2)
2.4	r = 0,78	√ r = 0,78
2.5	fairly strong positive correlation in the % of cocoa and the price	√ correct description of r	(1)
2.6.1	Brand D OR (35;24)	√ answer
2.6.2	y = 4,64 + 0,27(35) =R14,09 Overpriced by .R24,00 - R14,09 = R9,91	√ substitution √ R14,09 √ R9,91	(3)
			[13]
QUESTION 3
3.1	D(4;0)	√ answer	(1)
3.2		√ substitution √ answer	(2)
3.3	Angle of inclination of DE = tan ^-11/₃ = 18,43º ODE = 18,43º (vert.opp. ∠ s) OED = 71,57º (complementary ∠ s) ∴ PRB = 71,57º (corr. ∠ e: DE II BC)	√ tan ^-11/₃ √ 18,43º √ answer √ reason	(4)
3.4		√ substitution √ answer	(2)
3.5	AD = 3 = 1 AB 9 3	√ ratios √ simplified answer	(2)
3.6	ΔABC III ΔADE ∴ AD = DE = 1 AB BC 3 BC = 3 DE OR equation of BC y - 6 = ¹/₃ (x - 4) ∴ y = ¹/₃x + ¹⁴/₃ equation of AC	√ ratio of corresponding sides √ isolating √ substitution OR √ equation of BC √ equation of AC √ answer	(3)
3.7.1	midpoint of DE	√√ coordinates	(2)
3.7.2	m_bisector = -3 Equation of bisector	√ gradient of bisector √ substitution √ equation	(3)
3.8	LHS = -3 RHS = -3(4) + ¹⁶/₃ = - ²⁰/₃ = -6²/₃ ≈ -6,66 LHS # RHS No, bisector does not pass through A	√ substitution √ conclusion	(2)
			[21]
QUESTION 4
4.1	x² - 6x + y² = 16 (x - 3)² + (y - 0)² = 16 + 9 (x - 3)² + (y - 0)² = 25 Hence centre (3;0) Radius = √25 = 5	√ completing the square √ final form √ taking square root	(3)
4.2.1	y = 0 : hence (x - 3)² = 25 x - 3 = ± 5 x = 3 ± 5 B (8 ; 0) OR	√ equating = 0 √ solving fofr x √ coordinates of B OR √√√ use of horizontal distance	(3)
4.2.2	x = 0 : then (-3)² + y² = 25 9 + y² = 25 y² = 16 y = ± 4 C (0 ; -4) OR using pythagoras : MD² = OM² + OD² 5² = 3² + OD² 4 = OD = OC C(0;-4)	√ equating x to 0 √ coordinates of C OR √ use of pythagoras √ coordinates of C	(2)
4.3	MCP = 90º PM² = MC² + PC² (8¹/₃)² = 5² + PC² OR POC = 90º PC² = PO² + OC² PO = (8¹/₃) - 5 - 2 = ¹⁶/₃ and OC = 4 PC² = (¹⁶/₃)² + 4²	√ radius ⊥ tangent √ substitution into Pythag. Thm √ answer OR √ calculation of PO √ substitution into Pythag. Thm √ answer	(3)
4.4	m_BD = -½ tan-1(-½) = - 26,57º ∴ α = 180 - 26,57º ∴ θ = 180º - 153,43º = 26,57º DBC = 2 × 26,57º = 53,14º OR OR tan DBO = 4 = 1 8 2 DBO = 26,57º DBC = 2 × 26,57º = 53,14º	√ gradient of BD √ calculation of α √ angle θ √ doubling the angle OR √ sin definition √ use of arcsin √ angle θ √ doubling the angle OR √ tan ratio √ use of arctan to find angle √ angle √ doubling the angle	(4)
4.5	New circle centre -M(5 ; 1) C(2 ; -3) m_MC = ⁴/₃ m_tan = -³/₄ -3 = -³/₄(2) + c ∴ c = -³/₂ ∴ y = -³/₄x - ³/₂ OR Gradient of MC = ⁴/₃ Hence gradient of M'C' M'C' = ⁴/₃(MC ∥ M'C') Gradient of tangent at C' = −³/₄ (M'C' ⊥ tang at C') C' = (2 ; − 3) ∴𝑦 + 3= −³/₄(x - 2) ∴𝑦 = −³/₄x - ³/₂	√ coordinates of M and C √ gradient of MC √ gradient of tangent √ equation OR √ gradient of MC √ gradient of M'C' √ coordinates of C' √ equation	(4)
			[19]
QUESTION 5
5.1.1	1	√ answer	(1)
5.1.2	cos 2A	√ answer	(1)
5.2.1	3² = 2² + k² k = -√5	√ distance formula pythagoras Thm √ answer with correct sign	(2)
5.2.2 (a)	tan (θ -180º) tan θ = -√5 2	√ reduction √ answer	(2)
5.2.2 (b)	OR	√ 1 – sin²2θ = cos² 2θ √ 1 – 2sin²θ = cos 2θ √ substitution for sin √ answer OR √ 1 – sin²2θ = cos² 2θ √ 1 – 2sin²θ = cos 2θ √ substitution for cos √ answer OR √ sin²2θ = (2sinθcosθ)² √ substitution for sin θ √ substitution for cos θ √ answer	(4)
5.3	sin (−200º).cos 310º + tan (−135º).cos 380º .sin 230º = sin (20º)(cos 50º) + (tan45º)(cos 20º)(−sin50º) = sin 20º cos 50º – (1)cos 20º sin 50º = sin (20º – 50º) = sin ( − 30º) = − ½	√√√√ reductions to acute angles √ sin expansion √ answer	(6)
5.4	LHS= sin 2𝜃 + cos (2𝜃 – 90o) = sin 2𝜃 + sin 2𝜃 = 2 sin 2𝜃 = 2(2sin 𝜃 cos 𝜃) = 4 sin 𝜃 cos 𝜃 = RHS/RK	√ cos (2θ – 90º) = sin 2θ √ simplification √ expansion	(3)
5.5	10^sin𝑥 + 10^sin𝑥.10¹=110 10^sin𝑥(1+10)=110 10^sin𝑥=101 ∴ sin x = 1 ∴ x = − 270º or/of x = 90º	√ split into product of 2 bases √simplification / factorisation √division by 11 √ equating the exponents √ both solutions	(5)
			[24]
QUESTION 6
6.1		√ completing f(x) = sin 2x graph	(1)
6.2	See graph in 6.1	g(x) = cos (x – 30º) √ y intercept √ end points √ turning point √x intercepts	(4)
6.3	sin 2 x = sin (90º – ( 𝑥 – 30º)) sin 2 𝑥 = sin (120º – 𝑥) 2 𝑥 = 120º − 𝑥 + k.360º or/of 2 𝑥 = 180º – (120º − 𝑥) + k. 360º 3 𝑥 = 120º + k. 360º or/of 𝑥 = 60º + k. 360º ∴ 𝑥 = 40º + k. 120º or/of 𝑥 = 60º + k. 360º ; k 𝜖 Z k = 0 : 𝜃 = 40º ; 60º k = − 1 : 𝜃 = – 80º	√ co-ratio on RHS √ simplification √√ = 40º + k. 120º √ x = 60º + k. 360º √√ solutions	(6)
			[11]
QUESTION 7
7.1	tan y = PB k ∴ PB = k.tan y BC = PB BC ∴ BC = PB tan x Hence BC = k.tan y tan x	√ use of tan ratio √ PB = k.tan y √ BC = PB tan x	(3)
7.2	(AC)² = (4,73)² + (3)² – 2(4,73)(3)( cos 100º) AC = 6,03 m	√ use of cosine rule √correct substitution in cosine rule √ answer	(3)
			[6]
QUESTION 8
8.1.1	BÂE = 90º ( ∠ in a semi-circle )	√ statement / S √ reason / R	(2)
8.1.2	Ê₁ = 40º ( sum of ∠ s of Δ )	√ statement / S √ reason / R	(2)
8.1.3	C₁ = 40º (∠s subtended by AB OR ∠s in the same segment)	√ statement / S √ reason / R	(2)
8.1.4	Ĉ₂ = 62º (ext. ∠ of cyclic quad.)	√ statement / S √ reason / R	(2)
8.1.5	AB̂D = 62º (∠s subtended by AD OR ∠s in the same segment OR ext. ∠ of cyclic quad.)	√ statement / S √ reason / R	(2)
8.2	equal to the angle in the alternate segment/	√ answer	(1)
8.3	R.T.P / Te Bewys: STUR cyclic Proof : Û₂ = 𝑥 (tan chord thm) R̂₃ = 𝑦 (tan chord thm) R̂₂ = 180º – (x + y) ( ∠ s on str line) OR R̂₂ = 180º – (x + y) (3 ∠ s Δ) T + R̂₂ = 180º STUR cyclic (opposite angles are suppl.)	√ statement / S √ reason / R √ S & R √ S & R √ opp. ∠ s of quad, supplementary	(5)
8.4	PN = NQ = 8 units (line from the centre ⊥ chord) M P² = 6² + 8² (Pyth. Theorem) ∴ MP = 10 units MP = RM = MT = 10 (radii) ∴ RT = 20 units	√ S & R √ S & R √ S & R √ answer	(4)
			[20]
QUESTION 9
9.1.1	AD = AC n (prop theorem; DC II EF) DE CF 12 = AC 3 5 20 = AC	√ S & R √ Substitution √ answer	(3)
9.1.2	AO = 10 (diagonals of a parallelogram bisect)	√ answer	(1)
9.1.3		√ S & R √ statement / S √ answer	(3)
9.2		(Area => Oppervlakte) √ Area Δ ADC = 16 Area Δ AEF 2 √ Area ΔADO = 1 Area ΔADC 2 √simplification OR √use of area rule √ area of ΔADO	(3)
			[10]
QUESTION 10
10.1			(6)
10.1	Construction: Mark D on PQ and E on PR such that PD = ST and PE = SU. Join DE. Proof: In ΔPDE and ΔSTU P ̂= Ŝ (given) PD = ST (construction) PE = SU (construction) ΔPDE ≡ ΔSTU (SAS) ∴ D̂₁ = T ( ≡ Δ′𝑠) But: Q̂ = T (given) ∴ D̂₁ = Q̂ ∴ DE \|\| QR (corr. ∠ s =) ∴ PD = PE (prop. theorem; DE \|\| QR) PQ PR ∴ ST = SU PQ PR	√construction √ S & R √ D̂₁ = T̂ √ D̂₁ = Q̂ √ reason for lines∥ √ prop. theorem;DE \|\| QR ;DE\|\|QR	(6)
10.2.1	Â = x (tan chord thm) Ĉ₃ = x (angles opp. = sides)	√ statement(S) √ reason(R) √ S & R	(3)
10.2.2	R.T.Ps: ΔTOC ⦀ Δ BPC ACB = 90º ( ∠ in the semi-circle) Proof : Ĉ₃= Ĉ₁ = 𝑥 (proved) T̂₂ = 90º + 𝑥 (ext. ∠ of Δ) B̂₁ = 90º + 𝑥 (ext. ∠ of Δ) ∴ T̂₂ = B̂₁ ∴ Ô₂ = P̂ (sum of ∠ s in Δ) ∴ ΔTOC ⦀ Δ BPC (∠ , ∠ ,∠ ) OR ACB = 90º ( ∠ in the semi-circle) B² = 90° − 𝑥 (sum of ∠s in Δ) ∴ P = 90° − 2𝑥 (ext. ∠of Δ) O₁ = 2𝑥 (∠ at centre = 2 × ∠ at circum.) ∴ O₂ = 90° − 2𝑥 (SO ⊥ AB) ∴ O₂ = P 𝐶₃ = 𝐶₁ = 𝑥 (proved above) ∴ ΔTOC\|\|\|ΔBPC(∠, ∠, ∠)	√ S & R √ S √ S & R √ S √ R OR √ S & R √ S √ S & R √ S √ R	(5)
10.2.3	R.T.P : TO.PO = OB.BP Proof : TO = OC BP PC But OC = OB (radii) ∴ TO = OB BP PC ∴ TO.PC = OB.BP	√sides in prop. √ S	(2)
10.2.4	in ΔOPC: OP² = OC² + PC² (pyth.theorem) But : OB = OC = BP (radii) ∴ (20C)² = OC² + PC² 4OC² = OC² + PC² ∴ PC² = 3OC²	√ S & R √ OB = OC = BP √(2OC)² = OC² + PC²	(3)
			[19]

TOTAL:			150