MATHEMATICS PAPER 2 ERRATA GRADE 12 - 2018 SEPTEMBER PREPARATORY EXAM PAPERS AND MEMOS

MATHEMATICS PAPER 2
GRADE 12
ERRATA
SEPTEMBER 2018

QUESTION 1

1.3

QUESTION 2

[8]

QUESTION 3

3.2	OR 3 = -3(p) -12 15 = -3p -5 = p	?subst. A(p ; 3) in equation of line AB ?simplification ?value p	(3)
3.5	OR AB =√(-3 + 5)² + (-3-3)² =2√10 AE = √(-4 + 5)² + (0 + 3)² = √10 Eis the midpoint of AB [AE=EB=10] Eis the midpoint of AB and M the midpoint of AC from3.4 EM II BC [midpoint theorem]	?correct subst ?AB = 2√10 ?AE = √10 ?R	(4)
3.6	tan β = -3 β = 108,4349488º tan θ = ⁸/₅ θ = 57,99461679º ABC = 50,44	?size of θ ?size of a ?size of ABC	(4)

QUESTION 4

4.3	m_PQ= -3 + 6 0 - 9 = - ¹/₃m_QR = -6 + 9 9 - 8 = 3 PQR = 90º [m_PQ x m_QR = -¹/₃ x 3 = -1]	?correct substitution ?m_PQ ?m_QR ?m_PQ x m_QR = -¹/₃ x 3 (4)
	OR PQ² = (0 - 9)² + (-3 + 6)² = 90 QR² = (9 - 8)² + (-6 + 9)² = 10 PR² = (0 - 8)² + (-3 + 9)² =100 but PQ ² + QR² = 90 + 10 = 100 ∴ PR² = PQ² + QR² [conv] PQR= 90º	?PQ² ?QR² ?PR ² ?R (4)

QUESTION 6

6.1	OR a = -2 and p = -150º

QUESTION 7

7.1

AB = 2k
AC =√2k + k² - 2.2k.k.cos 2θ
= √5k² 4k² .cos 2θ
= √k² (5 - 4(1 - 2 sin ² θ))
= √k²(5 - 4 + 8sin² θ)
= k√1 + 8sin² θ

? AB i.t.o
? cosine rule formula in ∆ ABC
? correct subst.
? cos 2θ = 1 - 2sin² θ
? simplification

(5)

QUESTION 9

9.3

X₁ = V₄ [corresp∠s, XY II UV]
V₃ = V₄[given]
V₃ = W₂ [corresp∠s, WZ II UV]
X₁ = W₂
WXYZ is cyclic quad [converse ∠s same segment/ line subt ∠s]

?S/R

? S/R

? R

(3)

QUESTION 10

10.2
10.2.1	N₁ = 90 [∠subt by diameter /∠ in semi - circle] LN = NP [line from centre ⊥ to chord]	? S ? R ? R	(3)
	OR NM = NM [common] N₁ = N₂ = 90º PM = LM ΔPMN = LMN [RHS] LN = NP	?SSS ?S ?R
10.2.2	P₄ = L [tangent chord theorem] LPR = 90º [∠ subt by diameter] R₂ = 90º - P₄ [∠s of Δ LPR] R₁ = 90º - P₄ [∠s of Δ RPQ]	S ? R ? S/R ? S	(4)
10.2.3	N₁ = Qˆ [both = 90º ] P₂ = L [∠s opp.= sides] = P₄ M₂ = R₁ [3^rd ∠] ΔPNM\|\|\|Δ PQR [∠∠∠]	? S ?S ? R ?R	(4)
10.3.1	In ΔPLR and/en ΔQPR LPR = Qˆ [both = 90º ] R₂ = R₁ [proved] L = P₄ [3^rd] Δ PLR\|\|\|ΔQPR [∠∠∠] LR = PR PR QR LR = 30² 15 = 60	? SSS ? R ? ratios ?substitution ?LR	(5)
10.3.2	NM PR [co - int ∠s supp OR corresp ∠s =] ∴ NM =½PR [midpoint theorem] sin x = 30√3 15 x = 60º	? R ? R ? ratio ? value of x	(4)
	OR NM PR [co - int ∠s supp OR corresp∠s =] ∴ NM = ½ PR [midpoint theorem] cos x = ¹⁵/₃₀ x = 600	? R ? R ? ratio ? value of x

Last modified on Tuesday, 07 September 2021 12:14