GRADE12 MATHEMATICS PAPER 2 MEMORANDUM - NSC PAST PAPERS AND MEMOS JUNE 2016

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MATHEMATICS P2
GRADE12
JUNE2016
MEMORANDUM
NATIONAL SENIOR CERTIFICATE

QUESTION

1.1

Time (hours)	Cumulative Frequency
0 ≤ t < 20	30
20 ≤ t < 40	69
40 ≤ t < 60	129
60 ≤ t < 80	157
80 ≤ t < 100	167
100 ≤ t < 120	172

(4)

1.2

40 ≤ t < 60

✓answer (1)

1.3

172

✓answer (1)

1.4

(72;148)
∴ 172-148 = 24 learners

✓148
✓24
(2)

1.5

Frequency: 30; 39; 60;28;10;5
30×10+39×30+60×50+28×70+10×90+5×110=7880
172
=7880
172
=45,81

✓frequency
✓midpoints
✓7880
172
✓answer(4)

[12]

QUESTION 2

2.1	?̅= 6772 20 ?̅=338,6 ??	✓ ?̅= 6772 20 ✓answer(2)
2.2	2,71 ml	✓✓ answer (2)
2.3	[338,6 – 2,71; 338,6 + 2,71] [335,89; 341.31]	✓✓ interval (2)

[6]

QUESTION 3

3.1	?_??= −1 3 ?_??=3 [Diagonals of a rhombus] ?− ?₁=?(?− ?₁) ?−9=3(?−3) 3?−?=0	✓S ✓R ✓ subst, m = 3 & (3;9) in eqn. (3)
3.1.2	?+3?=10……… .(1) 3?−?=0 …………(2) 3?+9?=30………(3) (1)×3 10 ?=30 (3)−(2) ? =3 ?+3(3)=10 ?=1 K(1; 3)	✓equating two eqns ✓simplification ✓y = 3 ✓x = 1 (4)
3.1.3		✓method using midpoint ✓ simplification ✓ coordinates of B (3)
3.1.4	AD =√(3−?)²+ (9−?)² ∴ √50= √9−6?+?²+81−18?+ ?² ∴ 50= ?²−6?+?²−18?+90 But:?=10−3? ∴ (10−3?)²−6(10−3?)+ ?²−18 ?+90=50 ∴ 100−60?+9?²−60+18?+?²−18?+90=50 ∴10?²−60?+80=0 ∴ ?²−6?+8 =0 ∴ (?−4)(?−2) =0 ?=4 or ?= 2 ?=10−3(4) or ?=10−3(2) ?= −2 or ?=4 A(-2;4) C(4; 2)	✓ subst into eqn dist AD ✓ subst AD = √50 ✓subst x =10 - 3y ✓ simplification ✓standard form ✓values for y ✓values for x ✓coordinates (8)
3.2.1	?PQ= 8 - 2= 6 5−(−3) 8 = ¾	✓ subs into eqn ✓ answer (2)
3.2.2	tan?= ¾ ?=36,9	✓ tan ? ✓ answer (2)
3.2.3	?=¾?+? 0= ¾(8)+? ?= −6 ?= ¾?−6	✓subst. m = 34 ✓ subst. (8:0) ✓ answer (3)

[25]

QUESTION 4

4.1	A(0; y) ∴?= 0+8=4 2 ∴D(4;4)	✓midpt equation ✓coordinates of D (2)
4.2	A_y = ?+7=4 2 A(0; 1)² ∴(?−0)²+(?−1)²= ?² ∴(4−0)²+(4−1)²= ?² ∴16+9= ?² ∴?²+(?−1)²=25 ∴?²+ ?²−2?−24=0	✓ y-coordinate of A ✓ A(0; 1) ✓✓subst (0;1)and (4;4) into equation. ✓r² (5)
4.3	?_??×?_???= −1 [tan radius]	✓S/R ✓m_AB ¾ ✓ ?_???= -4 3 ✓subst m and (4;4) into eqn ✓answer (5)
4.4	?²+ ?²= ?² (8)²+(7)²= ?² ∴?²+ ?²=113	✓subst (8;7) into eqn ✓answer (2)

[14]

QUESTION 5

5.1.1	=sin?.cos?.tan?.cos? sin?.cos?.(−????) = −cos?	✓sin x ✓cos x ✓tan x ✓cos x ✓sin x ✓cos x ✓-tan x ✓-cos x (8)
5.2	sin? + 1 + cos? = 2 1+cos? sin? sin? LHS:sin²? + (1+cos?)(1+cos?) sin?(1+cos?) sin²? + 1 + 2cos?²?) sin?(1+cos?) 2+2cos? sin?(1+cos?) 2(1+1cos?) sin?(1+1cos?) 2 sin? = RHS/RK	✓ denominator ✓ numerator ✓ simplification ✓ identity ✓ factorisation (5)
5.3	cos2?=cos(?+?) = cos?.cos?−sin?.sin? =cos²?−sin²? = cos²?−(1−cos²?) =2cos²?−1	✓ expansion ✓ identity (2)
5.4	cos2?+3sin?=2 1−2sin²?+3sin?=2 2sin²?−3sin?+1=0 (2sin?−1)(sin?−1)=0 2sin?=1 or sin?−1=0 sin?= ½ or sin?=1 ?=30°+?.360 ?=90°+?.360 ?=150+?.360 ; ? ∈?	✓ identity ✓ standard form ✓factors ✓ sin x = ½ sin x = 1 (both eqns) ✓ x = 30° ✓ x = 150° ✓ x = 90° (7)
5.5	sin?cos?+cos?sin?=sin(?+?) = sin90° = 1	✓ identity ✓ subst ✓ answer (3)

[25]

QUESTION 6

6.1		✓shape g ✓intercepts ✓min & max value ✓shape f ✓asymptotes ✓ intercept (6)
6.2	−180°<?≤−90° or/of 0°≤?≤90°	✓critical values ✓ notation (2)

[8]

QUESTION 7

7.1

MN²=PN²+PM²−2PN.PMcos MP̂N
= 12²+10²−2(12)(10)cos126,9°
=144+100−240(−0,6)
=388,1
MN=19,7 ?
MN=AD=19,7 ?

✓ correct subst into cos rule
✓ simplification
✓ answer
(3)

7.2

½MN.PT= ½PN.PMsinMP̂N [Both equal Area of ΔPMN]
½(10)PT= ½(12)(10)sin126,9°

PT=12.sin126,9°
PT = 9,596 m
= 9,6 m

✓ Equating Area form
✓ correct subst
✓ simplification
✓ answer
(4)

[7]

QUESTION 8

8.1	Supplementary	✓ answer (1)
8.2.1	EF̂O=90° [tan radius] EĜO=90° [tan radius] EF̂O+EĜO=180° ∴FOGE is cyclic quad [opp angles supplementary]	✓ S ✓R ✓S ✓R ✓ R (5)
8.2.2	Ĝ₁= Ĥ=? [ tan chord] K̂₁= Ĥ=? [ corresp angles; EK \|\| FH] ∴Ĝ₁= K̂₁ EG is a tangent [angle between line and chord]	✓S ✓R ✓S ✓R ✓R (5)
8.2.3	Ô₁=2Ĥ [ angle at centre] = 2x ∴FÊG=180°−2? [ opp angles of cyclic quad]	✓S ✓R ✓R (3)

[14]

QUESTION 9

9.1	B̂₂ = Â₂ = x [ angles opp equal sides] Â₁ = B̂₂ = x [ tan chord] Ŝ₁ = Â₁ = x [ corresp; AD\|\|SC] ?̂₃ = ?̂₂ = x [ alt int; AD\|\| SC] AD̂C = B̂₃ = x [ ext angle of cyclic quad]	✓ S/R ✓ S/R ✓ S/R ✓ S/R ✓ S/R (5)
9.2	Â₁ = AD̂C [ from 8.1] AS \|\| DC [alt angles equal] DA \|\| CS [given] ASCD is a parallelogram [opp sides \|\|]	✓ S ✓ S/R ✓ S ✓ R (4)
9.3	ΔSAB \|\|\| ΔADB [ A,A,A]/[H,H,H]	✓ ΔSAB (1)
9.4	SA = SB [from 8.3.1] AD AB ∴AD.SB=SA.AB ButAD = SC [ ASCD is a parallelogram] and AB = SA[sides opp equal angles] = DC [ASCD is a parallelogram] ∴ SC.SB=DC²	✓ S ✓ S ✓ S/R ✓ S/R ✓ R (5)

[15]

QUESTION10

10.1.1	In proportion	✓ Answer (1)
10.1.2	RTP: PROOF: But Area ΔBDE=Area Δ CED (same base and same height) ∴ areaΔADE = area ΔADE area ΔBDE = area Δ CED ∴ AD = AE DB = EC	✓ ratio of area of ΔADE: ΔBDE ✓AD BD ✓ratio of area of ΔADE : ΔCED ✓AE EC ✓equating two areas(5)
10.2.1	QT=QW TP WR prop thm; TW \|\| VR ∴ 15 = ?+4 ? + 2 ? ?²+6?+8=15? ?²−9?+8=0 ∴(?−8)(?−1)=0 ∴?=8 or ?=1	✓QT=QW TP WR ✓R ✓ substitution ✓ std form ✓factors ✓ both values for x (6)
10.2.2	PV = PT prop thm TV \|\| QR VR = TQ PV = 10 18 = 15 PV = 12units	✓ S/R ✓ substitution ✓ answer (3)

[15]

QUESTION11

11.1

?₂=90° [ line from centre to midpoint of chord]
In Δ ABC & Δ DOC
i) Â= D₂ [ both equal to 90°]
ii)Ĉ= Ĉ [common]

∴ΔABC |||ΔDOC [AAA/???]

✓ S/R
✓ S/R
✓ S
(3)

11.2

OC = DC ΔABC |||ΔDOC
BC AC

OC = DC.BC
AC

✓ S/R
(1)

11.3

AC²=BC²−AB² [Pythagoras]
= 30²−18²
= 576
AC = 24 cm

??=??.??

= 15 × 30=18,8 ??

✓ S/R
✓ subst in eqn
✓ AC = 24
✓ subst
✓ answer
(5)

[9]

TOTAL: 150

Last modified on Tuesday, 15 June 2021 08:04

Published in Grade 12 2016 June Exams, Past Papers

GRADE12 MATHEMATICS PAPER 2 MEMORANDUM - NSC PAST PAPERS AND MEMOS JUNE 2016

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