# Mathematics Grade 12 KZN March 2018 Memo

> Grade 12 Mathematics memorandum from the KZN March 2018, 18 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/maths-march-qp-memo-2018-gr12/6c536d62-9f28-4dea-a02c-6d7b0a73732a
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: KZN March
- Curriculum: CAPS
- Pages: 18

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## About this document

Mathematics · Grade 12 · KZN March · 2018. Memorandum, 18 pages. Read online or download the PDF.

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## Document text

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### Page 1

Education

KwaZulu-Natal Department of Education I
REPUBLIC OF SOUTH AFRICA

SENIOR CERTIF
GRADE 12
go FE SRS SE SSSR ESS SSSR Es SR ESS SEES
| i MATHEMATICS '
* COMMON TEST *
* MARCH 2018 :
See ee eee
i
MARKS: 75
TIME: 1% hours
N.B. This question paper consists of 5 pages and an information sheet.
Copyright Reserved = Please Turn Over |

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Mathematics 2 March 2018 Common Test
NSC

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

ls This question paper consists of 6 questions.

2s Answer ALL questions.

3. Clearly show ALL calculations. diagrams, graphs, et cetera that you have used in
determining your answers.

4. Answers only will not necessarily be awarded full marks.

5. An approved scientific calculator (non-programmable and non-graphical) may be
used, unless stated otherwise.

6. If necessary, answers should be rounded off to TWO decimal places, unless stated
otherwise.

7. Diagrams are NOT necessarily drawn to scale.

8. Number the answers correctly according to the numbering system used in this

question paper. Write neatly and legibly.

Copyright Reserved Please Turn Over

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NSC

Mathematics

QUESTION 1
5; 12; 21 ; 32 ;... isa Quadratic Sequence.
1.1 Write down the next term of the sequence.

"term of the sequence.

1.2. Determine the n
1.3 Which term of the above sequence is 1152?

L.4 Prove that none of the terms in the sequence are perfect squares.

QUESTION 2
2;5;8;5 115... isan Arithmetic Sequence.
2.1 Determine the first term that will be greater than 2012.

2.2 Calculate the sum of the first 671 terms of the sequence.

QUESTION 3

° z in ot I-r"
3.1 Prove that the sum to n terms of a Geometric series is given by S,= ar) :
-F

3.2 The first term of a geometric series is 12, the last term is Bee and the sum of the

-. 6141 . . . .
series is a6 Determine the common ratio and the number of terms of the series.

March 2018 Common Test

a)

(4)

(4)

GB)
[12]

(3)

(G3)
[6]

(4)

(6)
[10]

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Mathematics 4 March 2018 Common Test
NSC

QUESTION 4

I 5 ‘
The graphs below are f(x) =—x° and g(x) =— | ‘i +3. A(a; b) and B(e: d) are points(s)
5 +

= x

of intersection f and g.

ax
4.1 Write down the equations of the asymptotes of g. (2)
4.2 If y = x+c isa line of symmetry to the graph of g, calculate the value(s) of c. (2)
4.3. Write down the range of fi qd)
4.4 Calculate the value(s) of x for which g(x) > 0. (4)
45 Write down the x — value(s) for which f(x) = g(x), x 2-1. (2)
4.6 Determine the value(s) of x if f(x) < g(x), for the interval x 2-1. (2)
47 If A(x) =—f (x + 2) +1, then write down the new equation of / in the
form A(x) = a(x + py +q. (2)
48 Use your graph to determine the maximum value of gids (2)
4.9 Determine the value(s) of k for which re +5) = khas one root equal to 0. (2)
. [19]

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Mathematics March 2018 Common Test

NSC
QUESTION 5
5.1 Sketch the graph of /(x)=3" ona set of axes in your answer book. (2)
5.2 Write down the equation of the inverse of f in the form aren (2)

5.3. Sketch f'' on the same set of axes indicating the intercepts with the axes and the

line of symmetry with the graph of f. (3)

5.4 Write down the equation of g(x) if g(x) = f'Cx). (2)

5.5 Determine the value(s) of x for which f~'(x) >—-1. (4)
[13]

QUESTION 6

6.1 How long will it take for a motor vehicle to depreciate to half its original value if the
rate of depreciation is 12,35% p.a., on the diminishing balance method ? (Give your
answer in years and months). (3)

6.2 On the 1 July 2017, a businessman wanted to buy a new house that costs R900 000.
He took out a loan to the value of the house with a bank at an interest rate of
10.25% p.a. compounded monthly for 20 years. The bank allowed the businessman
to start repayment towards the loan in three months time and indicated that his first
instalment will be deducted on the 1 October 2017.

6.2.1 Calculate the value of the loan on 1 September 2017. ()
6.2.2 Determine the monthly repayments if he now makes 238 monthly payments. (3)

6.2.3 Ifhe does not make the 100", 101%, and 102" payments, calculate the new
instalment that he will have to make to settle the loan in the fixed time period. (7)
[15]

TOTAL MARKS: 75

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Mathematics 6 March 2018 Common Test
NSC

INFORMATION SHEET: MATHEMATICS

_p+vn? —4ac

2a
A=P(+ni) A= P(=ni) A=P(-i)" A= P(1+i)"
T, =at(n—-ld S, = 52a)
T, =ar" Ss _ alr’ 1) ; r#l &, a4 ler
n a lr
pas +i" -1 p—al=0+i) "J
i i
fi) = lim F(xt+h)— f(x)
h>0 h
d= (x, =x)? HO, -¥,)* H(i gt 324)
a 2
v2)
ponte yoy, =m(x-x)) m= m=tan@
eK

(x-ay +( s-by =r?
f é £ 2 =? 4.6? —2he.cos A area AABC = asin C

= = — a
sn A sm Bo smc

In AABC:
sin(a+ B) = sin a.cos 8 + cosa.sin £ sin(@ — B) = sin a.cos f — cosa@.sin #

cos(a + £)= cosa@.cos f -sin asin B cos(a — 2)=cosa.cos # +sin asin 8

2 es
cos’ a@—sin” @
cos2@ =41—2sin° a sin 2a = 2sin a.cosa@

2cos’a-l

s (x, = x)

ves n #

P(A) = ae P(A or B) = P(A) + P(B) - P(A and B)
Me

Dsatbx p= KID)

Copyright Reserved

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Education

KwaZulu-Natal Department of Education
REPUBLIC OF SOUTH AFRICA

" MATHEMATICS

" COMMON TEST

" MARCH 2018

« MEMORANDUM
NATIONAL

| SENIOR CERTIFICATE
GRADE 12
MARKS: 75

TIME: L.5 hours

This memorandum consists of 9 pages.

Copyright Reserved

Please Turn Over |

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Mathematics 2 March Common Test 2018

QUESTION 1

[il 145 }Av (1)
1.2 5 \ A “32 Oo ;
1D 7 \ / \ /
2D 2 2
2a=2 val AY avalue
Batb=7  ..b=4 CAM value
atb+c=5 ..c=0 CAY c value
T= w+4n CAM answer
OR
(4)
OR
T, =T, +(n—Vd, _ hada 2), - “ =) d, AY formula
=54(n—-I(1)+ (n a =2) (2) AY substitution into correct formula
=547n-T+H -3n+2 CAY simplifying
ata (4)
. owen “ CAY answer
13 } n+4n=1152 ~ i /M¥ equating to 1152
n’ +4n—-1152=0 my CA” standard form
(n+36)n-32)=0 we lias | CAY factors
n=-36 or n=32
nla CAV n values and rejecting (4)
14 Suppose
i 4An= 1? ho MY equating 1" term to n?
n=0 ’ i
4 AY value of 1
Bute N Kas |
: Ves |
Therefore no such 1 exists and no terms are, <j fh
perfect squares in the sequence. '| AYfeason (3)
[12]

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Mathematics 8
QUESTION 2

March Common Test 2018

\ 5, = a(l=r")
l-r

=

2.1 T, =3n-1 AY n term=3n—1
3n-1> 2012 or 3n-1= 2012 AY setting up inequality /equation
3n > 2013 3a = 2013
n> 67 n=671
CAY answer
The 672™ term will be the first term to be
greater than 2012. (3)
2.2 = AY Sq formule
8, == Pat (n-Dd] onmre
2 AY substitution of a, n and d into
671 a -
Sen = -—[2(2) + 670(3)] correct formula
2
_ CAY answer
= 675697 G)
OR OR
T, =a+(n—l)d
Tey, = 2+ (671-3) = 2012
n AY Sn formula
8, = 5k +7] AY substitution of a, n and 7, into
671 correct formula
Son = se +2012] CAY answer
= 675697
(3)
[6]
UESTION 3
3.1 S,satartar +..+ ar tar > AY setting up equation |
nS, =artar tar +....+ar"| tar" > (2) | AY setting up equation 2
Fd () -(2):S, -1S, =a-ar" AY subtraction LHS and RHS
f a
\ Sir) = aly AV factorizing LHS and RHS 4)

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Mathematics 4 : March Common Test 2018
3.2 a=12
3
ar” a 2r't=— >() . .
256 256 AY setting up equation
S = I2d-r') = 6141 > (2) AY setting up equation
" l-r 256 iil
From (1):
lar" = =—_ Opt = 3r >) AY making 12r the subject
r 256
Substituting (3) into (2)
3r
56 _ 6141
I-r 256 CAY substitution of 127” into (2)

3r 6141614 Ir

256-256-256
3072 3r _ 614i G14Ir

256 256 256 ©6256
61387 = 3069

CAY r value

CAY nr value
(6)

OR

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Mathematics

March Common Test 2018

a=12
3
ar"! =—— [2r"'= 2. >)
256 256
120—r" 14]
5, - A)
l-r 256
From (\):
12" 3 t 2%

= Sr"= r
r 256 1024
Substituting (3) into (2)

ae
1024) 6141

l-r 256
it»
1024. _ O14)
l-r 3072

1 6141 6141

1-—_r= ———7
1024 3072 3072
6141 | 1 6141

, _
3072 1024" 3072

6141 3 6141 3072
p= r= aa
3072 3072 »=—- 3072-3072

6138 _ 3069
3072 3072

I
pet

2

1y\" 1V1 1
2) ae ha}™ a
n=11

AY setting up equation

AY setting up equation

AY making 7” the subject

CAY substitution of ” into (2)

CAY r value

CAY n value

(6)

(10) |

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J \

| Mathematics : 6
9 QUESTION 4

March Common Test 2018

4.1) x=-l , 5} p AY vertical asymptote
Doe cA
y=3 ‘ ry AY horizontal asymptote (2)
42 yoxte AY substituting (-1 ; 3 )
8 elie CAY c— value (2)
c=4
43 y20. or yel[0;0) AY answer “(D
4.4
—— sh a 0 Ifa candidate
atl . AY determining LCD and
=14+3%43 50 cross multiplies by HUERATOR
; i (xt) only — 0/4
3X + wen . pe .
20 Tt CAY simplification
x+1 (4)
2 CAYCA VYanswers
x<-lL oor x2- =
3
OR OR j
I
41 +3=0/, AY setting up equation to calculate
| xX — intercept
3=
x+1
3x4+3=1
3x S2
2
x= a CAYx — value
2
x<-l oor V2 op CAY CAY answers
3 (4)
X=a or X=C Av AY answers (2)
If Candidates solved simultaneously - i ;
have irrational roots - full marks
45
GSxSC AAY Veritical values, inequality (2)
( If Candidates solved simultaneously .- f- “A
416 have irrational foots - full:marks

Ay

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Mathematics _ March Common Test 2018
4.7 1 2
AQ) = ~z(«+2) +1 AAY Yanswer
(2)
4.8 Maximum value of A(x) =1
CAY 1
Therefore maximum value of 3!"
CAY 81
=31=81 (2)

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Mathematics

March Common Test 2018

49 Ji, a
Be +5) =k Answer only
lo +5 =k full marks AY substituting #= 0
3
25 i AY answer (2)
—=k
3
: [19]
QUESTION 5
5.1 : yo Ff AY shape
i AY y= intercept = 1
1
* 4
y
Ko
,
’
4
F
y
4
‘a
’ @)
5.2 / 1
yo log, x =—log, x = log, ~ AAYY answer
3 ~ (2)
5.3 see graph in 5.1 to allocate marks f':CAY shape
AY x—intercept at 1/(1;0)
AY y=x line G3)
5.4 g(x) =log,(-x) ;.x<0 OR
3
g(x) =—log,(—x); x<0 OR
g(axy= toe (—} x<0
me, ¥VCACA answer [any form] (2)

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Mathematics 9 March Common Test 2018
5.5 f'@)2-1
log, x2—-1 or log, aes al CAV setting up incquality/equation
3 3
x<3 x=3 CAV x-— value =3 (4)
O<xs3 CACAY V critical values and
inequality
: [13]
QUESTION 6
6.1 A=P(l-i)"
oP = P(1—0,1235)" If P=tPU — 01235)" AY substitution into correct formula
1 Breakdown ~ 0/3
— = (0,8765)”
2
N = 1085 9765 > MY use of logs
n = 5,258342252 years
Time period will be Syears and 4 months CAY answer (3)
OR OR
A=P(l-iy”
3? = P(1—0,1235)" AY substitution into correct formula
_ (0,8765)"
5 ,
logs = log(0,8765)" MY use of logs
, = 10808765
7 1
log—
1 = §,258342252 years
Time period will be Syears and 3 months CAY answer (3)
6.21 | 4=P(+iy" _ -_
=900 0001 + 0,1025 } AY substitution into correct formula
712
= R915440,66 CAY answer
Q)

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Mathematics «ds ee March Common Test 2018
6.2.2 pe a= (4+i9")
t AY substitution of i and value
[ ( aes)" into correct formula
xj 1-] 1+
12 CAY substitution of P into correct
15440,66 = .
aus 0,1025 formula
12 y (3)
x = R9009,43 CAY answer
6.2.3 | Balance outstanding after the 99" payment :
— afl-(L+"]
‘ 8 AY substitution of 1 =139 into
9009,43] 1—| 1+ 0,1025 : correct formula ;
12 CA” substitution of x into correct
= 0,1025 formula
12 CAY R731384,7351
= R731384,7351
Value of the loan afier 3 months of non - payment :
0.1025\ CAY substitution into correct
A =73138,735 (1 + | = R750287,0104 formula
. CAY R750287,0104
The new monthly instalment to settle the loan :
7 136
0,1025
750287,0104 = ———_______= AY substitution 1 =136 into correct
0,1025 —
: formula
12
x = R9349,11 We ,
CAY answer (2)
[15]

Total Marks : 75

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623. Out-Bal = A - fy
= pAyT — x Canard

——__-

GY
= 4iSy 40, bh (i+ id, 2k) 100443 [C4 9a) i|

ig 2%
[a

= 2124513, %0s — 1393 428,418

=R7 31384 585 46

oad

Value of Loan fue Zmuth = TBI3EY 5 y eu (1+ lyase)
of Non payment

= 78026 7, 16ds
Sa

New mom Unk payment x = TSO287, 166 X ib,25%
12
if | - + Ions me]
= 93-49 108827

R43451
Oo

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