# Mathematics Grade 12 P1 KwaZulu Natal November 2015 Memo

> Grade 12 Mathematics Paper 1 memorandum from the KwaZulu Natal November 2015, 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2015-grade-12-math-trial-exam-paper-1-kzn-memo/29ab746e-1936-45ac-8ac2-61f4f5c9fdfa
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2015
- Exam period: KwaZulu Natal November
- Paper: 1
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 12 · KwaZulu Natal November · 2015. Memorandum, 11 pages. Read online or download the PDF.

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## Document text

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### Page 1

Basic Education

Pd : **
Fo MATHEMATICS P1 ;
* _ PREPARATORY EXAMINATION
woe SEPTEMBER 2015
* MEMORANDUM :
Merrie terre er ere eer
NATIONAL
SENIOR CERTIFICATE
MARKS: 150

This memorandum consists of 11 pages.

KwaZulu-Natal Department of Education
REPUBLIC OF SOUTH AFRICA

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Mathematics P1 2 September 2015 Preparatory Examination
NSC-MEMORANDUM(KZN)
QUESTION 1
L111) | x(x-5)=0 2
x=0 or x= ¥v Answers
1.12 | 4x? ~5x=3
4x? — $x —3=0 ¥ equation in std. form
yal (-5) £ J (-5)* - 4(4)(-3) Y substitution
2(4)
x=1,69 or x=-044 vv answers 4
113 | 2° @x+1)<0
2 >Oforalixe R ¥2" >Oandxek
“3x+1<0 ¥ 3x+1<0
1
x<~— ~ answer 3
1.1.4 a
x-3x? =4 1
x-4=3Vx ¥ isolating 3Vx or 3x?
9x =x? ~ 8x +16 ¥ squaring both sides
x’ ~17x +16=0 Y trinomial
(1X - 16)=0 v factors
x=1 or x=16 ~ both answers
¥ selection 6
nia !
OR i
|
L
x-3x? —-4=0 ¥ standard form
1 1 ¥ factors
[#-4]e-1 =< '
a a Varad
= = 1
x?=4 or x? =-1 ¥ va-]
x=16 no solution Y answer
¥ rejecting 6
1.2 fg 2028
fg7030 _ 7s
/g2028
yor (yor -1 ¥ factorising denominator
vs v simplifying
80
2 Y answer 3
80
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Mathematics Pt 3 September 2015 Preparatory Examination
NSC-MEMORANDUM(KZN)
13 24” and
x 4+2y=29
3xtl=2y > Y for 2y as subject
w4+2y=29 > (2)
Subst. (1) into (2)
x? 43x41=29 ¥ substitution of 2y
x? +3x—-28=0 ¥ std. form
(x+7)(—4)=0
ve? or x=4 vx values
y=-10 of yee vy values 6
[24
QUESTION 2
211]3:;2;6:2;9.:,2.,
12; 2 vw answers 2
2.1.2 | a=3,d=3,n=50 ¥ for a, d and n values
50 50
So = ray +150] or Ze) 3(50-1)] Y for substitution into formula
= 3825
a=2 and is a constant sequence
“Ss = 50x 2
= 100
Swoy = 3825 +100
= 3925 5
2.2 §,=at+artart+..tar* > () ¥ for equation (1)
rS, =artar+.. tar" — (2) ¥ for equation (2)
(2)- a):
7S, -S, = ar" -a ¥ subtraction on LHS and RHS
S,¢-= alr’ . 1) | ~ factorising
alr” = 1)
1S St
in rl 4
[
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Mathematics P1 4 September 2015 Preparatory Examination
NSC-MEMORANDUM(KZN)
QUESTION 3
3.1 24
Sy 1124 6x4 3x? tee
a=6r=2 Ya value
1s T value
S= of2 ~1) “substitution inte formula
2-1
= 196602 answer 4
3.2 -l<r<l ¥ for condition
1
-teltyey Y for r=—x
2 2
-2<x<2 ¥ answer 3
3.3 r>l1
test vr>l
2
2 v4
x> answer 4
19]
QUESTION 4
4.1 2a=4
a=2 ¥ avalue
3a+b=2
b=-4
vb value 4
atbht+e=5
c=7 ¥c value
T,=2n? -4n+7
¥ general term
OR
2a=4
a=2 ¥ a value
3at+b=2
hea-4 %b value
T) =e=7 ¥c value
LT, =2n? -4n+7 4
¥ general term
OR
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Mathematics P1

5
NSC-MEMORANDUM(KZN)

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September 2015 Preparatory Examination

_ (nn - 2)

7. =T+(- Nd}, ¥ formula
n-1\n—2
= 5+ (n—1X2)+ {a- he 2) (4) ¥ substituting first and second
difference values
=5+2n-2+2n? —6n+4
=2n? -4n+7 ¥ simplifying
¥ general term 4
42 | First Difference n™ term = (4n-2)
4n-2= 2018 v4n—2= 2018
4n = 2020
n=505 ¥n=505
Toys = 2(506)’ — 4(506) + 7 = 510055
Fo, = 2(505)? - 4(505)+ 7 = 508037
Between the 505" and the $06" terms. “answer 3
OR
7, -T,) =2n* —4n+7-[a(n-1)? -4(n—-1)+7] |
2 2 ¥T,-T._, substitution
=2n? —4n+7-[2n* —4n+2-4n+4+7]
=4n-6
4n—-6=2018 v4n-6=2018
4n = 2024
n=506
Between the 505” and the S06" terms. Vanswer 3
[7]
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Mathematics PL 6 September 2015 Preparatory Examination
NSC-MEMORANDUM(KZN)
QUESTION 5
5.1 A=P(1+i)" ¥ ¥ correct
4 substitution into
2x = x{l + 0,085) correct formula
. log2 ¥ making 7 the
log(1 + 0,085) subject
=85 or years or 8 years and 6 months ¥ answer \
5.2.11 4 = P(+i)"
2 ¥ correct
= 350 ovo + 23) substitution into
correct formula
=R357919,30 answer e
3.2.2 ~ jy"
pal 497]
i
0135)"
xy 1-—|14+-4
( 12 } ¥ “correct
35791930 = —435 substitution into
—— correct formula
12
0,135

357919,30x 2"
12

238
(1-285) |
2

x=

v x as subject

x =R4328.60 v answer 4
5.2.3 | Balance on Loan after the 180” payment
178
in 43 2300 28) - 1
= 457919 a 285) _ ¥ ~ substitution
, 0,135 into formulae
12

= R188118,54 ¥ R188118,54

Total Payments = R4328,60« 60 = R259716 v R1030206,80
Savings = R259716 - R188118,54
= R71597,46 v answer 8

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Mathematics Pt 7 September 2015 Preparatory Examination
NSC-MEMORANDUM(KZN)
QUESTION 6
6.1 a
x*)=——+q. P(-2;4) isa point on
f= + © PC24) isa point onf
p=l v pvalue
gel ¥ qvalue
a
= 4]
» xt+t
Now subst. P(-2;4):
4=—4 4) ¥ sub d point P
24] subst. p, g and point
.a@=-3 ¥a value 4
62 3 iso
xt1 4 =o)
x+1=3 wl=3
x=2
T@;0) “answer 3
63 y=(x+p)+¢@
y=(e+D41 , “substitution of p and q values
into equation of line of symmetry
y=xt2
“c=2 ~ answer 2
OR
yexte
Point of intersection of asymptotes (—1 ; 1
l= ym ( ) ~ substitution of p and q values
~lelte into equation of line of symmetry
c-2 ¥ answer 2
im ka
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Mathematics P! 8 September 2015 Preparatory Examination
NSC-MEMORANDUM(KZN)
QUESTION 7
val a(x)=—x? +2x+3 and ACx)-axt+q.
CD = 6 units.
-x? 4+2e+3=0 vy=0
(x+1(x-3)=0 ¥ factors
x=-1 or x=3
B(3;0) ¥ coordinates of B
C(0;3) ¥ coordinates of C 4
72 DO; -3) v¥ answer 2
73 a=1 and g=-3 v¥ answers 2
74 ~x° 42x43-x-3 ~ equating both equations
x -~x¥-6=0
(x+2(x-3)=0 © factors
x=-2 or x=3 ¥ x values
y=-5 or y=0 vy values
E(-2;-5) answer 5
75 gyal +4 VAxis of symmetry value ;
_lex<l or x>3 0% ond points and inequality 4
[17]
QUESTION 8
8.1] p@)=log, x
x= log, y
Diiy=3* vv answer 2
8.2
¥¥ graph of p
¥ graph of p@
4
8.3 log, x=2
x=9
O<x<9 ¥vfor end points and inequality 2
8.4 x intercept of p is (1 ; 0)
x intercept of A is (-1 ; 0) vv answer 2
{10}

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Mathematics PI 9 September 2015 Preparatory Examination
NSC-MEMORANDUM(KZN)
QUESTION 9
91 fQx)=x? +3x
1°) =tim flx+h)- 4)
h0 h
=i (x+AY +3(x+A)- (2 +3x) v f(x +h)= (+h) +3(4h)
am h ¥ substitution into correct formula
_ x +2xh+h? +3x4+3h—x? -3x an
= lim h ¥ simplifying
hao
xh +h? 43h
hoo h
_ AQQx+h43)
= tim a ¥ factors
hoo
=2x+3 answer 5
921 | a ity 35? Yaxt
dx
16 0
y=6x3 Y6% x3
7
* =20x3 or 20Vx7 ¥ answer 3
9.2.2 bons x 5x? 44x
S(t f= 4 ¥ factorising
f= xx? -5x+4) ¥ factors
(x4)
= x(e~1fx-4) % simplification of f
(x-4) “answer
=x -x
Sx) = 2x-1 4
| {12]
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Mathematics P1 10 September 2015 Preparatory Examination
NSC-MEMORANDUM(KZN)
QUESTION 10
1
10 y= fe the gradient of the tangent to the
x
curve is — 1
ya4yt ¥ exponential form
dy a _74 ¥ derivati
ok =-~4y? = 7 derivative
a4} Y derivative =—1
x
v=4
“x=-2 or x=2 ¥ x~ values
y=-2 or y=2
(-2;-2) (2;2) ~y —values 5
10.2.1) y= (x 1e- 4)’ ¥ (e—Dr-4y?
=(x- 1)? ~ 4x44) ¥ squaring binomial
=x ~9x? 424y—-16 2
10.2.2) y= x3 -9x? +24x-16
& = 3x? —18x+24=0 ¥ derivative and equal to 0
x” ~6x+8=0
(x—-2)x—4)=0 “factors
x=2 or x=4
ya4 or y=0 v y—values
B(; 4) ¥ coordinates of B 4
10.23 4&<-16 v inequality
| W16 2
10.2.4] f(x) =6x-18>0 ¥ 6x-18>0
x>3 answer 2
[15]
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Mathematics Pt 1 September 2015 Preparatory Examination
NSC-MEMORANDUM(KZN)
QUESTION 11
11.1 Velxbxh ¥ formula
= 5x(9 ~2x\(x) Y substitution
= 45x? —10x? 2
2 | 790% — 30x? v derivative
90x ~ 30x? =0 ¥ derivative equal to 0
vw
30x(3 — x)= 0 factors
vx values
x=0 or x=3
Therefore the box will have a maximum atx =3 | ¥ choosing x =3 §
[7]
QUESTION 12
12.1 | (x+0,2)x0,5= 0,2 ¥P(A) x P(B) = P(A and B)
x+0,2=0,4
x= 0,2 vx value... P(A only)
0,24+0,.24+03+y=1 v y=1-07
y= 9,3 vy value 4
12.2 15! 2t5!
P(X and Y together) = = ¥ 7
al
3 vi
3
1
P(X and Y not together) = 1 — 3
=2
3
Y answer 3
12.3.1 | 4 digit numbers(with repetitions) ¥ product
=9x10x10x10 ¥ answer 2
=9000
12.3.2 | 4 digit numbers(without repetitions)
=9x9x8x7 ¥ 9x9
=4536 v 8x7
v answer 3
12.3.3 | 4 digit numbers(with repetitions and last digit = 0)
=9x10x10x1 ¥ product
=900 ¥ answer 2
[14]
TOTAL MARKS: 150

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