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MATHEMATICS
COMMON TEST
MARCH 2020
MARKING GUIDELINES
NATIONAL
SENIOR CERTIFICATE
GRADE 12
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Mathematics NSC Common Test March 2020
2
QUESTION 1
1.1
p
5 q 19
1D 5โp qโ5 19โq
q+p-10 24-2q
2D 2 2
๐๐ โ ๐๐ = ๐
๐ = ๐๐
๐ + ๐ โ ๐๐ = ๐
๐ = ๐
A 24 โ 2๐ = 2
CAq value
A๐ + ๐ โ 10 = 2
CAp value
(4)
1.2 ๐๐ = ๐ โด ๐ = ๐
๐๐ + ๐ = ๐ โด ๐ = ๐
๐ + ๐ + ๐ = ๐ โด ๐ = โ๐
๐ป๐ = ๐๐ + ๐ โ ๐
OR
๐๐ = ๐ โด ๐ = ๐
๐๐ + ๐ = ๐ โด ๐ = ๐
โด ๐ = ๐ป๐ = โ๐
๐ป๐ = ๐๐ + ๐ โ ๐
Aa โ value
CA b โ value
CA c โ value
CAnth term
OR
Aa โ value
CA b โ value
CA c โ value
CAnth term
(4)
(4)
1.3 ๐ป๐ = ๐๐ + ๐ โ ๐ = ๐๐๐๐๐
๐๐ + ๐ โ ๐๐๐๐๐ = ๐
(๐ + ๐๐๐)(๐ โ ๐๐๐) = ๐
๐ = โ๐๐๐ ๐๐ ๐ = ๐๐๐
๐๐๐๐๐๐ term
OR
๐๐ = ๐2 + ๐ โ 1 > 10301
๐2 + ๐ โ 10302 > 0 (๐ + 102)(๐ โ 101) > 0
๐ < โ102 ๐๐ ๐ > 101
1022๐๐ term
CAequating nth term to
10301
CA factors/quadratic
formula
CA values of n
CA102
OR
CAsetting up inequality
CA factors/quad. formula
CA interval and notation
CA102
(4)
(4)
[12]
Mathematics NSC Common Test March 2020
3
QUESTION 2
๐๐ =๐
2[๐ + ๐๐]
36 =๐
2[1 + 11]
36 = 6๐
6 = ๐
11 = ๐ + (๐ โ 1)๐
11 = 1 + (6 โ 1)๐
2 = ๐
OR
๐๐ =๐
2[2๐ + (๐ โ 1)๐]
36 =๐
2[2(1) + (๐ โ 1)๐] โ (1)
11 = ๐ + (๐ โ 1)๐
11 = 1 + (๐ โ 1)๐
10 = (๐ โ 1)๐ โ (2)
36 =๐
2[2(1) + 10]
72 = 12๐
๐ = 6
10 = (6 โ 1)๐
2 = ๐
Asubstituting into sum
formula
CA36 = 6๐
CA n โ value
CAsubstituting into general
term formula
CA d โ value
OR
Asubstituting into sum
formula and forming equation
(1)
Asubstituting into general
term formula and forming
equation (2)
CA72 = 12๐
CA n โ value
CA d โ value
(5)
(5)
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Mathematics NSC Common Test March 2020
4
QUESTION 3
3.1 ๐๐ = ๐ + ๐๐ + โฏ + ๐๐๐โ1 โ (1)
๐๐๐ = ๐๐ + ๐๐2 + โฏ + ๐๐๐โ1 + ๐๐๐ โ (2)
(2) โ (1):
๐๐๐ โ ๐๐ = ๐๐๐ โ ๐
๐๐(๐๐ โ 1) = ๐(๐๐ โ 1)
๐๐ =๐(๐๐ โ 1)
๐ โ 1
Awriting equation
Amultiplying all terms by r
Asubtracting: LHS and
RHS
Afactorizing
(4)
3.2.1 3 ;
3
2
AAeach value (2)
3.2.2
๐๐ =3 [(1 โ (
12)
๐
]
1 โ12
=3069
512
6 [1 โ (1
2)
๐
] =3069
512
[1 โ (1
2)
๐
] =1023
1024
(1
2)
๐
=1
1024
(1
2)
๐
= (1
2)
10
๐ = 10
CAsubstituting into sum
formula
CA [1 โ (1
2)
๐
] =1023
1024
CA(1
2)
๐
=1
1024
CAexpressing 1024 as a
power of 2
CA m โ value
N.B. Can be solved using
logs.
(5)
3.3 ๐ = 24 (
3
4) = 18 ๐๐๐ ๐ =
3
4
๐โ =๐
1 โ ๐
=24 (
34)
1 โ34
= 72 ๐
A r โ value
CA substituting into sum to
infinity formula
CA answer
(3)
[14]
Mathematics NSC Common Test March 2020
5
QUESTION 4
4.1
4.1.1 ๐2 = ๐ฅ2 + ๐ฆ2
= (โ1)2 + (โโ3)2 = 1 + 3= 4 โด ๐ = 2
๐ ๐๐( โ ๐) = โ ๐ ๐๐ ๐
= โ (โโ3
2)
=โ3
2
A r = 2
CA substitution
CA answer
(3)
4.1.2 cos 2๐ = 1 โ 2 ๐ ๐๐2 ๐
= 1 โ 2 (โโ3
2)
2
= 1 โ 2 (3
4)
= (1 โ3
2) = โ
1
2
OR
cos 2๐ = 2 ๐๐๐ 2 ๐ โ 1
= 2 (โ1
2)
2
โ 1
= 2 (1
4) โ 1
=1
2โ 1 = โ
1
2
OR
๐๐๐ 2๐ = ๐๐๐ 2๐ โ ๐ ๐๐2๐
= (โ1
2)
2
โ (โโ3
2)
2
=1
4โ
3
4
= โ1
2
A 2sin21
CA substitution into the correct
expression
CA simplification
CA answer
OR
A 1cos22cos 2
CA substitution into the correct
expression
CAsimplification
CAanswer
OR
A ฮธsinฮธcoscos2ฮธ 22
CA substitution into the correct
expression
CA simplification
CA answer
(4)
(4)
(4)
If ๐ ๐๐( โ ๐) = ๐ ๐๐ ๐
= โโ3
2
Max. 2/3
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Mathematics NSC Common Test March 2020
6
4.2
4.2.1 sin 75o ooo 45cos15sin45cos
2
1
60cos
1545cos
45sin15sin45cos15cos
45sin15sin45cos).1590sin(
o
oo
oooo
ooo
oo
OR
sin 75o ooo 45cos15sin45cos
cos 45ยฐ(sin 75ยฐ โ sin 15ยฐ) Now
sin 75ยฐ โ sin 15ยฐ
= sin(45ยฐ + 30ยฐ) โ sin(45ยฐ โ 30ยฐ)
= 2cos 45ยฐ sin 30ยฐ
cos 45ยฐ(sin 75ยฐ โ sin 15ยฐ)
= cos 45ยฐ. 2cos 45ยฐ sin 30ยฐ
=โ2
2. 2.
โ2
2 .
1
2
= 1
2
A ๐ ๐๐ 7 5๐ = ๐๐๐ 1 5๐ A ๐๐๐ 4 5๐ = ๐ ๐๐ 4 5๐ CA๐๐๐ (45๐ + 15๐) CA ๐๐๐ 6 0๐
CA1
2 (ACCEPT 0,5)
OR
removing common factor
sin 75ยฐ โ sin 15ยฐ = 2cos 45ยฐ sin 30ยฐ
cos 45ยฐ =โ2
2
sin 30ยฐ =1
2
answer
(5)
(5)
4.2.2
3
310sin
10sin
60tan10sin
1090cos
120tan)10(sin
100cos
2
o
o
o2
o
oo
o2o
o
OR
3
10sin
310sin
10sin
60tan80cos
)10sin(
120tan100cos
2
o
2oo
o
o2o
3
3
10sin
60tan
10sin
)1090cos(
2
oo
OR
โ ๐๐๐ 8 0๐ ; โ ๐ ๐๐ 1 0โ; ๐ก๐๐2 6 0โ
โ cos 80๐ = โ sin 10๐ ; โโ3
3
(6)
(6)
Mathematics NSC Common Test March 2020
7
4.3.1 LHS
cos
cos
sin
sin
RHS
2sin
)sin(2
cossin2
)sin(2
cossin
)(sin
2
2
cossin
)sin(
cossin
sincoscossin
Awriting as a single fraction
Asin )(
A2
2
Asimplification
(4)
4.3.2 ๐ ๐๐ 5 ๐ฝ
๐ ๐๐ ๐ฝโ
๐๐๐ 5 ๐ฝ
๐๐๐ ๐ฝ= 4 ๐๐๐ 2 ๐ฝ
๐ฟ๐ป๐ =๐ ๐๐ 5 ๐ฝ
๐ ๐๐ ๐ฝโ
๐๐๐ 5 ๐ฝ
๐๐๐ ๐ฝ
=๐ ๐๐ 5 ๐ฝ ๐๐๐ ๐ฝ โ ๐๐๐ 5 ๐ฝ ๐ ๐๐ ๐ฝ
๐ ๐๐ ๐ฝ ๐๐๐ ๐ฝ
=๐ ๐๐( 5๐ฝ โ ๐ฝ)
๐ ๐๐ ๐ฝ ๐๐๐ ๐ฝ
=๐ ๐๐ 4 ๐ฝ
12 . 2 ๐ ๐๐ ๐ฝ ๐๐๐ ๐ฝ
=2 ๐ ๐๐ 2 ๐ฝ ๐๐๐ 2 ๐ฝ
12 ๐ ๐๐ 2 ๐ฝ
= 4 ๐๐๐ 2 ๐ฝ= RHS
OR
Replace 1.3.4in5
๐ ๐ป๐ =2 ๐ ๐๐(5๐ฝ โ ๐ฝ)
๐ ๐๐ 2 ๐ฝ
=2 ๐ ๐๐ 4 ๐ฝ
๐ ๐๐ 2 ๐ฝ
=2.2 ๐ ๐๐ 2 ๐ฝ ๐๐๐ 2 ๐ฝ
๐ ๐๐ 2 ๐ฝ= 4๐๐๐ 2 ๐ฝ
Awriting as a single fraction
Asin 4 =2 sin 2 cos 2
A1
2๐ ๐๐ 2 ๐ฝ
OR
๐ผ = 5๐ฝ
sin 4ฮฒ
expansion sin 4ฮฒ
(3)
(3)
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Mathematics NSC Common Test March 2020
8
QUESTION 5
4.4
kkx
x
kkxorkx
x
xorx
xx
xx
;360.90
1sin
.;360.330360.210
2
1sin
1sin2
1sin
01sin1sin2
01sinsin2 2
OR
A factorization
CA 1sin;2
1sin xx
CA kkx ;360.210
CA kkx ;360.330
CA kkx ;360.90
[5]
5.1 ๏ฟฝฬ๏ฟฝ + 140โ = 180โ opp s of cyclic quad
โด ๏ฟฝฬ๏ฟฝ = 40โ
S/R
A
Answer only with reason 2/2
(2)
5.2 ๏ฟฝฬ๏ฟฝ1 = 2๏ฟฝฬ๏ฟฝ at centre is twice at circum.
= 2(40ยฐ)
= 80ยฐ
SR
80ยฐ CA
Answer only with reason 3/3
(3)
5.3 ๏ฟฝฬ๏ฟฝ3 =1
2(180ยฐ โ 80ยฐ) s opp = sides
= 50ยฐ
SR
A
Answer only with reason 3/3
(3)
5.4 ๏ฟฝฬ๏ฟฝ5 = ๏ฟฝฬ๏ฟฝ3 + 28โ tan โ chord theorem
= 50โ + 28โ = 78โ
S/R
CA Answer
Answer only with reason 2/2
(2)
[10]
MaDownloaded from Stanmorephysics.com9
QUESTION 6
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QUESTION 7
6.1 ๏ฟฝฬ๏ฟฝ1 = ๐ฅ corr s ; KP// LN
๏ฟฝฬ๏ฟฝ1 = ๐ฅ Tan-chord theorem
๏ฟฝฬ๏ฟฝ1 = ๏ฟฝฬ๏ฟฝ1 NMLP is a cyclic quad (Converse s in the same
segment)
SR
S/R
R (4)
6.2 ๏ฟฝฬ๏ฟฝ1 = ๐ฅ Tan-chord theorem
๏ฟฝฬ๏ฟฝ2 = ๏ฟฝฬ๏ฟฝ1 = ๐ฅ s in same segment
๏ฟฝฬ๏ฟฝ3 = ๏ฟฝฬ๏ฟฝ2 = ๐ฅ alt s KP//MN
๏ฟฝฬ๏ฟฝ = ๏ฟฝฬ๏ฟฝ3
โด โ๐พ๐ฟ๐ is isosceles (Sides opp equal angles)
S/R
SR
SR
R (6)
[10]
Construction: Mark off PA = FG and
PB = FH and join AB.
๐ฅ๐๐ด๐ต โก ๐ฅ๐น๐บ๐ป Congruency s s
๐๏ฟฝฬ๏ฟฝ๐ต = ๏ฟฝฬ๏ฟฝ๏ฟฝฬ๏ฟฝ = ๏ฟฝฬ๏ฟฝ given
โด ๐๏ฟฝฬ๏ฟฝ๐ต = ๏ฟฝฬ๏ฟฝ โด ๐ด๐ต//๐๐ corresp. s equal
๐๐
๐๐ด=
๐๐
๐๐ต line // to one side of ฮ
โด๐๐
๐น๐บ=
๐๐
๐น๐ป
construction
S R
S
R
S/R
[6]
P
* H
Q โ
G โ
F
* R
A B
Mathematics NSC Common Test March 2020
10
QUESTION 8
8.1 ๏ฟฝฬ๏ฟฝ2 = 30โ s opp = sides
๐^
4 = 30โ tan-chord theorem
๏ฟฝฬ๏ฟฝ = 30โ corresponding s NP//MO
S/R
S/R
S/R (3)
8.2 ๏ฟฝฬ๏ฟฝ2 + ๏ฟฝฬ๏ฟฝ3 = 90๐ ( in the semi-circle) โด ๐ ๏ฟฝฬ๏ฟฝ๐ = 90โ + 30โ
= 120โ
S/R
CA (2)
8.3.1 ๐ผ๐๐ฅ๐ ๐๐๐๐๐๐ฅ๐ ๐๐๏ฟฝฬ๏ฟฝ = ๏ฟฝฬ๏ฟฝ common
๏ฟฝฬ๏ฟฝ4 = ๏ฟฝฬ๏ฟฝ = 30โ proven
๏ฟฝฬ๏ฟฝ2 = ๐ ๏ฟฝฬ๏ฟฝ๐ rem โด ๐ฅ๐ ๐๐///๐ฅ๐ ๐๐ (๐ด๐ด๐ด)
S/R
S/R
R (3)
8.3.2 RN
W
RT=
NP
TN (/// triangles)
RN. TN = RT. NP ฬ
2 = 90โ = ๐๏ฟฝฬ๏ฟฝ๐ corr ; NP//MO NW = WT line from centre โฅ chord โด NT = 2WT โด 2WT. RN = RT. NP
TW. RN =1
2RT. NP
SR
S/R
S/R
substituting NT = 2WT
(5)
[13]
TOTAL MARKS: 100