MATHEMATICS PAPER 2
GRADE 12
ERRATA
SEPTEMBER 2018

QUESTION 1

1.3 graph


QUESTION 2

2.1 Range = 29 - 10
= 19 
🗸 answer  (1) 
2.2

x=  15 + 23 + 17 + 24 + 26 + 18 + 28 + 13 + 10 + 28 + 29
                                                   11
= 231
    11
= 21
Answer ONLY full marks

🗸 231

🗸 answer 
(2)

(2)
2.3 σ = 6.37 🗸answer  (2)
2.4  (21 6,37;21 6,37) = (14,63;27,37)
5weeks 
🗸min 🗸 max
answer
(3) 

[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 

mPQ=  -3 + 6 
             0 - 9
= - 1/3
mQR =  -6 + 9 
              9 - 8
= 3
PQR = 90º [mPQ x mQR = -1/3 x 3 = -1]

🗸correct substitution
🗸mPQ
🗸mQR
🗸mPQ x mQR = -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)

 

Related Items

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 qn  
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 [3rd ∠]
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 [3rd]
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