GRADE/GRAAD 12 SEPTEMBER 2021
Transcript of GRADE/GRAAD 12 SEPTEMBER 2021
NATIONAL SENIOR
CERTIFICATE/NASIONALE
SENIOR SERTIFIKAAT
GRADE/GRAAD 12
SEPTEMBER 2021
TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2
MARKING GUIDELINE/NASIENRIGLYN
MARKS/PUNTE: 150
This marking guideline consists of 15 pages./
Hierdie nasienriglyn bestaan uit 15 bladsye.
2 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/SEPTEMBER 2021)
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NOTE:
β’ Continuous accuracy (CA) applies only where indicated in this marking guideline.
β’ Assuming values/answers in order to solve a problem is unacceptable.
LET WEL:
β’ Volgehoue akkuraatheid (CA) is slegs van toepassing soos aangedui in hierdie nasienriglyn.
β’ Aanvaarding van waardes/antwoorde om Ε probleem op te los, is onaanvaarbaar.
MARKING CODES / NASIENKODES
M Method / Metode
A Accuracy / Akkuraatheid
AO Answer only / Slegs antwoord
CA Consistent accuracy / Deurlopende akkuraatheid
F Formula / Formule
I Identity / Identiteit
R Rounding / Afronding
S Simplification / Vereenvoudiging
ST Statement / Bewering
RE Reason / Rede
ST RE Statement and correct reason / Bewering en korrekte rede
SF Substitution correctly in correct formula / Korrekte vervanging in die korrekte
formule
NPU No penalty for omitting units / Geen penalisering vir eenhede uitgelaat
(EC/SEPTEMBER 2021) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 3
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QUESTION/VRAAG 1
Q M
1.1 π = β4 β A
(1)
1.2 π΄π΅2 = (β2 + 5)2 + (β4 β 0)2
= 9 + 16
π΄π΅ = 5 units / eenhede
β M
β 5 CA
(2)
1.3 ππΆπ· =β4β0
3β0
= β4
3
tan π = β4
3
Ref /π£πππ€β = tanβ1 (4
3)
= 53,13Β°
β΄ π = 126,87Β°
β β4
3 A
β 53,13Β° CA
β 126,87Β° CA
(3)
1.4 π΅πΆ = 3 β (β2) = 3 + 2 = 5 units / eenhede π΄π = 0 β (β5) = 5 units / eenhede
ABCO is a parallelogram (One pair of opp. side = and β₯ / Een
paar teenoorstaande sye = en β₯)
But / maar, AO = AB (from / vanuit 1.2)
β΄ π΄π΅πΆπ is a rhombus / is 'n ruit
(parallelogram with all sides = / parallelogram met all sye =)
β π΅πΆ & π΄π M
β parm and / en
reason / rede
β rhombus / ruit
β reason / rede
(4)
[10]
4 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/SEPTEMBER 2021)
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QUESTION/VRAAG 2
Q M
2.1.1 π₯2 + π¦2 = 25 β¦ 1 π¦ β π₯ β 1 = 0 β¦ 2
π¦ = π₯ + 1 β¦ 3
sub 3 in 1:
π₯2 + (π₯ + 1)2 = 25
π₯2 + π₯2 + 2π₯ + 1 = 25
2π₯2 + 2π₯ β 24 = 0
π₯2 + π₯ β 12 = 0
(π₯ + 4)(π₯ β 3) = 0
π₯ = β4 or / of π₯ = 3
β΄ π¦ = β3 or/of π¦ = 4
β π¦ = π₯ + 1
β π₯2 + (π₯ + 1)2 =25
β π₯2 + π₯ β 12 = 0
β π₯ = β4
β π₯ = 3
β π¦ = β3 CA
β π¦ = 4 CA
(7)
2.1.2 (3; 2): (3)2 + (2)2 = 13
β΄ π₯2 + π¦2 < π2
β΄ lies inside the circle / lΓͺ binne die sirkel
β 13
β π₯2 + π¦2 < π2 CA
βconclusion /
afleiding
(3)
2.1.3 ππππ =0β3
0β(β4)
= β3
4
β΄ ππ‘ππ Γ ππππ = β1
β΄ ππ‘ππ Γ β3
4= β1
β΄ ππ‘ππ = β1 Γ· β3
4
β΄ ππ‘ππ =4
3
π¦ = ππ₯ + π
β΄ π¦ =4
3π₯ + π
(β4; 3): 3 =4
3(β4) + π
3 =β16
3+ π
25
3= π
β΄ π¦ =4
3π₯ +
25
3
β β3
4
β 4
3
β 25
3= π
β π¦ =4
3π₯ +
25
3 CA
(4)
2.2 36π₯2 + 49π¦2 = 1764
36π₯2
1764+
49π¦2
1764= 1
π₯2
49+
π¦2
36= 1
βπ₯2
49+
π¦2
36= 1
β Shape / vorm
β π₯-intercept / π₯-
afsnit
β y-intercept /y-afsnit
(4)
[18]
π¦
π₯ β7 7
β6
6
(EC/SEPTEMBER 2021) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 5
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QUESTION/VRAAG 3
Q M
3.1.1 π‘ππ4(122.3Β°) +
2
3cos (
210,5Β°
4)
= β0,8
β SF
β β0,8
(2)
3.1.2 1
π ππ(οΏ½ΜοΏ½
3+2π΅)
=1
π ππ(210,5Β°
3+2(122,3Β°))
= β1,4
β 1
π ππ(π΄
3+2π΅)
β SF
β β1,4
(3)
3.2.1 π‘πποΏ½ΜοΏ½ =5
12=
β5
β12
β΄ πππ ππ2οΏ½ΜοΏ½ =1
π ππ2οΏ½ΜοΏ½
πππ ππ2οΏ½ΜοΏ½ =169
25
π2 = (β12)2 + (β5)2
β΄ π = 13
β π‘πππ΄ =5
12=
β5
β12 A
β correct quadrant /
regte kwadrant A
β π = 13 A
β1
π ππ2π΄ or / of
πππ ππ2οΏ½ΜοΏ½ = (13
β5)
2
CA
β169
25 CA
(5)
OR/OF
π‘πποΏ½ΜοΏ½ =5
12=
β5
β12
β΄ πππ ππ2οΏ½ΜοΏ½ = (13
β5)
2
πππ ππ2οΏ½ΜοΏ½ =169
25
3.2.2 π πποΏ½ΜοΏ½ β π πποΏ½ΜοΏ½ =
1
πππ οΏ½ΜοΏ½β π πποΏ½ΜοΏ½
=1
β12
13
ββ5
13
=13
β12β
β5
13
=β109
156
OR/OF
π πποΏ½ΜοΏ½ β π πποΏ½ΜοΏ½ =1
πππ οΏ½ΜοΏ½β π πποΏ½ΜοΏ½
=13
β12β
β5
13
=β109
156
β 1
πππ π΄π¨π« /ππββ
13
β12
β 1
β12
13
β β109
156 CA
(3)
3.2.3 π‘πποΏ½ΜοΏ½ =5
12
ref / π£πππ€β = π‘ππβ1 (5
12)
= 22,62Β°
β΄ οΏ½ΜοΏ½ = 180Β° + 22,62Β°
οΏ½ΜοΏ½ = 202,62Β°
β method / metode
β 22,62Β°
β202,62Β°
(3)
[16]
οΏ½ΜοΏ½ β12
β5 π
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QUESTION/VRAAG 4
Q M
4.1 π ππ(π β π₯). πππ ππ(2π β π₯). π‘ππ(π + π₯)
π ππ(2π β π₯). πππ (2π β π₯)
=π ππ(180Β° β π₯). πππ ππ(360Β° β π₯). π‘ππ(180Β° + π₯)
π ππ(360Β° β π₯). πππ (360Β° β π₯)
=π ππ(π₯). β
1sin (π₯)
. π‘ππ(π₯)
1cos (π₯)
. πππ (π₯)
=β1. π‘ππ(π₯)
1
= βtan (π₯)
βcorrect conversion
of rad to degrees /
korrekte herleiding
van rad na grade β π ππ(π₯)
β β1
sin (π₯)
β π‘ππ(π₯)
β 1
cos (π₯)
βπππ (π₯)
ββ1.π‘ππ(π₯)
1
ββtan (π₯)
(8) OR/OF OR/OF
π ππ(π β π₯). πππ ππ(2π β π₯). π‘ππ(π + π₯)
π ππ(2π β π₯). πππ (2π β π₯)
=π ππ(180Β° β π₯). πππ ππ(360Β° β π₯). π‘ππ(180Β° + π₯)
π ππ(360Β° β π₯). πππ (360Β° β π₯)
=π ππ(π₯). βπππ ππ(π₯). π‘ππ(π₯)
sec (π₯). πππ (π₯)
=β1. π‘ππ(π₯)
1
= βtan (π₯)
βcorrect conversion
of rad to degrees /
korrekte herleiding
van rad na grade β π ππ(π₯)
β βπππ ππ(π₯) β π‘ππ(π₯)
β π ππ(π₯) βπππ (π₯)
ββ1.π‘ππ(π₯)
1
ββtan (π₯)
(8)
4.2 πΏπ»π/πΏπΎ =πππ π
1βπ πππβ π‘πππ
=πππ π
1βπ πππβ
π πππ
πππ π
=πππ π.πππ πβπ πππ(1βπ πππ)
πππ π(1βπ πππ)
=πππ 2πβπ πππ+π ππ2π
πππ π(1βπ πππ)
=1βπ πππ
πππ π(1βπ πππ)
=1
πππ π
= π πππ = π π»S/RK
β π πππ
πππ π
β πππ π(1 β π πππ)
βπππ 2π β π πππ +π ππ2π
β1βπ πππ
πππ π(1βπ πππ)
β1
πππ π
(5)
[13]
(EC/SEPTEMBER 2021) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 7
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QUESTION/VRAAG 5
Q M
5.1 β cos start and
end point / cos
begin- en
eindpunte
β cos
turningpoints /
cos draaipunte
β cos π₯-
intercepts/cos π₯-
afsnitte
β sin start and
end point / sin
begin- en
eindpunte
β sin
turningpoints /
sin draaipunte
β sin π₯-
intercepts/sin π₯-
afsnitte
(6)
5.2.1 120Β° β 120Β°
(1)
5.2.2 (a) π₯ = 30Β° and / en π₯ = 120Β° β π₯ = 30Β°
β π₯ = 120Β°
(2)
(b) 90Β° β€ π₯ β€ 150Β° β 90Β° β€
β β€ 150Β°
(2)
[11]
π
π
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QUESTION/VRAAG 6
Q M
6.1 AB = 8 cm (opp sides of rec = / teenoorst. sye van reghoek =) β 8 cm ST
β RE
(2)
6.2 π΅πΈ
8= tan 30Β°
BE = 4,62 cm = 5 cm
β π΅πΈ
8= tan 30Β° M
β 5 cm
(2)
6.3 sin πΈοΏ½ΜοΏ½πΆ
9=
sin 30Β°
4,62
sin EBΜC =sin 30Β°
4,62Γ 9
EBΜC = 76,91Β° = 77Β°
β SF
β sin πΈοΏ½ΜοΏ½πΆ =sin 30Β°
4,62Γ
9
β 77Β°
(3)
6.4 π΅οΏ½ΜοΏ½πΆ = 180Β° β 30Β° β 77Β°
= 73Β° (Int. β βs of βBEC/ Binne. β βe van βBEC)
π΄πππ βBCE =1
2π. π. π πποΏ½ΜοΏ½
=1
2(5)(9)π ππ73Β°
= 21,52 cm2
β ST RE
β 1
2(5)(9)π ππ73Β° SF
β 21,52 cm2 CA
(3)
6.5 πΆπΉ2 = πΆπΈ2 + πΉπΈ2 β 2(πΆπΈ)(πΉπΈ) cos πΆοΏ½ΜοΏ½πΉ
πΆπΉ2 = (9)2 + (10)2 β 2(9)(10) cos(25Β°)
= 17,86 β¦
πΆπΉ = β17,86 β¦ πΆπΉ = 4,22 β¦ = 4 cm
β (9)2 + (10)2 β2(9)(10) cos(25Β°)
SF
β 17,86 β¦ S
β 4 cm CA
(3)
[13]
(EC/SEPTEMBER 2021) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 9
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QUESTION/VRAAG 7
Q M
7.1 double the size of the angle subtended by the same arc
at the circumference of a circle / is dubbel die grootte van die
hoek wat dit by dieselfde koord die omtrek van Ε sirkel
onderspan
β
(1)
7.2
7.2.1 π΄οΏ½ΜοΏ½π΅ = 180Β° β 48Β° β¦(suppl. β βs/ suppl. β βe)
= 132Β°
β ST RE
(1)
7.2.2 οΏ½ΜοΏ½ =1
2(132Β°)β¦( β at centre = 2 x β at circum./ β by midpt = 2 x
β by omtrek)
= 66Β°
β RE
β 66Β° ST
(2)
7.2.3 ποΏ½ΜοΏ½πΈ = 90Β° β¦( tan β₯ rad)
π΄οΏ½ΜοΏ½π· = 180Β° β (90Β° + 48Β°)
= 42Β° (int β β²s of triangle / binne β β²e van driekhoek)
β ST
β RE
β ST RE
(3)
7.2.4 οΏ½ΜοΏ½ = π΄οΏ½ΜοΏ½π· = 42Β° β¦(β β²s opp.= sides /β β²e teenoor = sye)
π΄οΏ½ΜοΏ½π· = 84 β¦.(ext β of β / buite β van β
β ST RE
β ST RE
(2)
7.2.5 π΅πΈ2 = ππΈ2 β ππ΅2 β¦.(Pythagoras)
= 72 β 52
= 24
π΅πΈ = β24 β 4,9 cm
β M Pythagoras
β ST π΅πΈ2 = 24
β ST π΅πΈ = β24
(3)
[12]
10 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/SEPTEMBER 2021)
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QUESTION/VRAAG 8
Q M
8.1 supplementary / supplementΓͺr β
(1)
8.2
8.2.1 (a) ποΏ½ΜοΏ½π = οΏ½ΜοΏ½ = π₯ (β β²s subt.chord QR / (β β²e onderspan deur krd QR)
β ST
β RE
(2)
8.2.2 (b) ποΏ½ΜοΏ½π = 180Β° β (58Β° + π₯) (opp. β β²s quad/teenoorts. β β²e kvh)
= 122Β° β π₯
β ST βRE
β 122Β° β π₯ ST
(3)
8.2.2 οΏ½ΜοΏ½ + ποΏ½ΜοΏ½π + ποΏ½ΜοΏ½π = 180Β° (int. β β²s β/binne β β²e β)
22Β° + 72Β° + π₯ + 58Β° = 180Β°
π₯ = 28Β°
β ST RE
β π₯ = 28Β° ST
(2)
OR/OF OR/OF
οΏ½ΜοΏ½ + οΏ½ΜοΏ½ + οΏ½ΜοΏ½ = 180Β° (int. β β²s β/ binne β β²e β)
72Β° + 58Β° + π₯ + 22Β° = 180Β°
π₯ = 28Β°
β ST RE
β π₯ = 28Β° ST
(2)
[8]
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QUESTION/VRAAG 9
Q M
9.1 All angles of the one triangle is equal to the angles in the other
triangle / Alle hoeke van die een driehoek is gelyk aan die hoeke
van die ander driehoek
Sides of triangles are in proportion / Sye van die driehoek is in
verhouding
β ST
β ST
(2)
9.2
9.2.1 In βπΆπ·πΉ and/en βπΈπΆπΉ:
πΉοΏ½ΜοΏ½π· = πΆοΏ½ΜοΏ½π· = 52Β° (tanchord / raaklyn koord)
πΆοΏ½ΜοΏ½π· = πΆοΏ½ΜοΏ½πΈ (common β / gemene β )
πΆοΏ½ΜοΏ½πΉ = π·οΏ½ΜοΏ½πΈ (int. β β²s β/ binne β β²e β))
β΄ βπΆπ·πΉβ¦βπΈπΆπΉ (β β β )
β ST RE
β ST RE
β ST βπΆπ·πΉβ¦βπΈπΆπΉ
OR β RE (β β β )
(4)
9.2.2 πΆπ·
πΈπΆ=
π·πΉ
πΆπΉ=
πΆπΉ
πΈπΉ (βπΆπ·πΉβ¦βπΈπΆπΉ)
β΄ πΆπΉ2 = πΈπΉ. πΉπ·
β πΆπ·
πΈπΆ=
π·πΉ
πΆπΉ=
πΆπΉ
πΈπΉ ST
β πΆπΉ2 = πΈπΉ. πΉπ· ST
(2)
9.2.3 π·πΉ = 15 β 6 = 9 πΆπ·
πΈπΆ=
π·πΉ
πΆπΉ=
πΆπΉ
πΈπΉ (from 9.2.2 / vanuit 9.2.2)
β΄9
πΆπΉ=
πΆπΉ
15
β΄ πΆπΉ2 = 135
β΄ πΆπΉ β 12 cm
βπ·πΉ = 9 A
β 9
πΆπΉ=
πΆπΉ
15 SF
β πΆπΉ2 = 135 CA
β 12 cm CA R
(4)
OR/OF OR/OF
πΆπΉ2 = 15 Γ 9 (from 9.2.2 / vanuit 9.2.2)
πΆπΉ = β135
πΆπΉ = 11,619 β¦
β΄ πΆπΉ = 12 cm
βπΆπΉ2 = 15 Γ 9 A
β πΆπΉ = β135 S
β πΆπΉ = 11,619 β¦ CA
β 12 cm CA R
(4)
9.2.4 πΆπ·
πΈπΆ=
12
15=
4
5
β 12
15 SF A
β 4
5 A
(2)
9.2.5 πΈοΏ½ΜοΏ½πΉ = 44Β° + 52Β° = 96Β° β 90Β°
β΄ CE not a diameter / πΆπΈ πππ πβ² ππππππππ¦π πππ (Diameter not
perpendicular to tangent/middellyn nie loodreg met raaklyn nie)
β ST
β ST RE
(2)
12 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/SEPTEMBER 2021)
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OR/OF OR/OF
πΆοΏ½ΜοΏ½πΈ = 180Β° β 44Β° β 52Β° = 86Β° β 90Β°
β΄ πΆπΈ πππ‘ π ππππππ‘ππ / πΆπΈ πππ πβ² ππππππππ¦π πππ (not Converse
β β²s in semi-circle / nie-omgekeerde β β²e in semi-sirkel)
β ST
β ST RE
(2)
[16]
QUESTION/VRAAG 10
Q M
10.1
10.1.1 οΏ½ΜοΏ½ = 30Β° Γπ
180Β°
οΏ½ΜοΏ½ =π
6 radians / radiale
β Γπ
180Β°
β π
6 radians/radiale
(2)
10.1.2 π = ππ
π΅πΊ = (4) (π
6)
π΅πΊ =2π
3 cm
β F
β SF
β2π
3 cm
(3)
10.1.3 Area of a sector / πππ π£ππ πβ² π πππ‘ππ =π2π
2
β΄ Area of sector AEC / πππ π£ππ π πππ‘ππ π΄πΈπΆ =(9)2π
6
2
=3π
2 cm2
β F
β SF
β 3π
2 cm2
(3)
10.1.4 Area of a sector / πππ π£ππ πβ² π πππ‘ππ =π2π
2
β΄ Area of sector ABG / πππ π£ππ π πππ‘ππ π΄π΅πΊ =(4)2π
6
2
=4π
3 cm2
β΄ Area of BECG / πππ π£ππ π΅πΈπΆπΊ =27π
4β
4π
3
β΄ Area of BECG / πππ π£ππ π΅πΈπΆπΊ =65π
12 cm2
β SF
β 4π
3
β M
β 65π
12
(4)
(EC/SEPTEMBER 2021) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 13
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10.2
10.2.1 π = 2ππ
π = 2π(120)
π = 240π rad/min
β F
β SF
β 240π rad/min
(3)
10.2.2 π£ = ππ·π
π£ = π(3 Γ 2)(120)
π£ = 720π cm/min
β F
β SF
β 720π cm/min
OR/OF OR/OF
π£ = ππ
π£ = (3)(240π)
π£ = 720π cm/min
β F
β SF
β 720π cm/min
(3)
10.2.3 Linear speed of large pulley / reglynige spoed van groter katrol =720π cm / min
π£ = ππ
720π = π(15)
48π rad / min = π
β F
β SF
β 48π rad/min
(3)
14 TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 (EC/SEPTEMBER 2021)
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10.3
4β2 β 4πβ + π₯2 = 0
4β2 β 4(10)β + (8)2 = 0
4β2 β 40β + 64 = 0
β2 β 10β + 16 = 0 (β β 8)(β β 2) = 0
β΄ β = 8 cm or/of β = 2 cm
β΄ β = 2 cm
β F
β SF
β Standard
form/standaard
vorm
β M
ββ = 2 cm
OR/OF OR/OF
πΆπΈ = 4 (line from centre perpendicular to chord / lyn vanaf
middelpunt loodreg op koord)
πΉπ = ππΊ (radii)
ππΆ2 = ππΈ2 + πΆπΈ2 (Pyth)
β΄ ππΈ2 = ππΆ2 β πΆπΈ2
β΄ ππΈ2 = (5)2 β (4)2
β΄ ππΈ2 = 25 β 16
β΄ ππΈ2 = 9
β΄ ππΈ = Β±β9
β΄ ππΈ = Β± 3
β΄ ππΈ = 3 cm
β΄ πΈπΊ = 5 β 3
β΄ πΈπΊ = 2 cm
β ST RE
β ST RE
β ππΈ2 =(5)2 β (4)2 SF
β ππΈ = 3 cm
CA
β πΈπΊ = 2 cm
CA
(5)
[26]
(EC/SEPTEMBER 2021) TECHNICAL MATHEMATICS P2/TEGNIESE WISKUNDE V2 15
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
QUESTION/VRAAG 11
Q M
11.1
π΄π = π(π1 + π2 + π3+. . . +ππ)
π΄π = 3(6,4 + 5,45 + 4,7 + 4,25 + 4,25)
π΄π = 3(25,05)
π΄π = 75,15 cm2
β F
β SF
β S
β 75,15 cm2
OR/OF OR/OF
π΄π = π (π1+ππ
2+ π2 + π3 + π4 + β― + ππβ1)
π΄π = 3 (5,9+5,4
2+ 6,9 + 4 + 5,4 + 3,1)
π΄π = 3(25,05)
π΄π = 75,15 cm2
β F
β SF
β S
β 75,15 cm2
(4)
11.2
πππ΄ = 2ππ2 + 2ππβ β ππ2 β ππ2
πππ΄ = 2π (1,85
2)
2
+ 2π (1,85
2) (2,5) β π (
1,85
2)
2
β π (1
2)
2
πππ΄ = 16,43 m2
β F
β SF
β17,22 m2
(3)
[7]
TOTAL/TOTAAL: 150