# Mathematics Grade 12 Paper 1 Gauteng Mock 2015 Memo

> Grade 12 Mathematics Paper 1 memorandum from the Gauteng Mock Exam 2015, 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2015-grade-12-math-trial-exam-paper-1-gp-memo/0b8bcf17-ed8a-401b-880a-245aeb5b3e53
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2015
- Exam period: Gauteng Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 12

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## About this document

Mathematics · Grade 12 · Gauteng Mock Exam · 2015. Memorandum, 12 pages. Read online or download the PDF.

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## Document text

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### Page 1

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INCE

PREPARATORY EXAMINATION
2015
MEMORANDUM

SUBJECT: MATHEMATICS P1 (10611)

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10611/15
GAUTENG DEPARTMENT OF EDUCATION
PREPARATORY EXAMINATION -~ 2015
MATHEMATICS
(First Paper)
MEMORANDUM
QUES TION 1 [21 MARKS]
Ll 111 x=-2 or x=V2 or x=-J2
Le [3]
[| 2-12 [| x*>9 of x?-950 1A
x23 or x<-3 1A for critical value of 3
2A for each inequality
Max 2/3 if ‘and’ is used
Full marks if ‘or’ is omitted |
[4] |
1.13 6"*_2?* — 8 6 1A for factorizing bases
or gfe 32x = 8.32%,22*
or gt 322 = Q3+2x 32x
— ar ICA =
22-23 ICA ||
aS 3
25 ICA
are
9
12 (+)? => 1A
————} = —- |
9
+1=4,/—
ane fa
[x=112 or _x=-3,12 2CA for each solution i
| 2x? +4x-7=0 1A standard form — —
| ~44f4? —4(2)(-7
x= elf 42) ICA substitution into correct
2(2) formula —
| x=112 or x=-3]2 2CA for each solution
| =e]

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10611/15
13 C2xtT+5 _ 1
aa] _ 1A correct substitution.
2(-2x +12) =x—-1 1CA simplification
x=5 i ICA
Substitutex=5: y=—2(5)+7
a y=-3 ICA
OR
Jo LE
vi 2
aS 1A correct substitution
5
Ce are 2 1CA simplification
Lode 1CA
substitute y=-3: x= pati
—
ae ICA
{4]
1.4 1-p>0o 1A
Ded ICA [2]
QUESTION 2 [12 MARKS]
|
21 (
sr 1A for —=
2 2
pa
7 5 1CA for multiplication with x
(2]
2 | xt+xt+2 TA for x+2
= 2x+2 1CA for adding to x
(21.
22 ¢ IM
Seda eery iCA
x? +6x=0 1CA for simplified form
x(x+6)=0 1CA factorizing
x=-6 1CA choosing correct x-value
[5]
23 The series does NOT converge. ICA

2M _ for any logical explanatior.

r=3. toconverge, —l<r<l

[3]

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10611/15 *
QUESTION3 [8 MARKS]
3.1 | (250)? + 4(250) 1A substitution
= 63500 ICA
[2]
3.2 Foo = Sica ~ Soo 1M
= 100° + 4(100) —[99? + 4(99)} 1A correct substitution
= 203 ICA
OR
T=5;T,=7; T, =9 1A for first three terms
Dooo = 5+ 99(2) 1M for correct formula
= 203 1CA
[3]
3.3 n? +4n =1440 1A

(n+ 40)\(n—36) =0

ICA correct factors or correct
substitution into correct formula.

n= 36

1CA choice of correct 7.
Full marks for answer only

{3}

QUESTION 4 [6 MARKS]

| 4.1 2; 6; 12; 20 1A for first three terms
1A for the 4" term
[2] —
4.2 a=], b=1; c=0 3A finding a, b, c using any

correct method.

T =n? +n or T, =na+l

”

Trop = 100(100+1)

=10100cm?

ICA

[4]

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10611/15
QUESTION 5 [15 MARKS]
5.1 |5.1.1 | 4=6500000-0,3)* 1A substitution into correct
formula
=R156 065 1CA
{2]
5.1.2 | 4 =650000(1+0,15)* 1A substitution into correct
formula
=R1 136 854,06 ICA
| [2]
Bye) 1136854, 06 ~ 156065 = 980789, 06 ICA
0,095)" 1M for use of correct formula
x (: + oe} -1 1CA us of correct formula
980789, 06 = ———_—____—___+
j 780789 0,095
12
= R16 875,92 1CA
[4]
52 | 521 0,092)" 1M
wx 11636,02 (: 102) 1] 1A correct substitution
oB =1275000/ 1422092) _ ®
12 0,092
12
= R1056 675,39 ICA
[3]
REELS) 5
0,092
1056675 39(1 +o) 1CA substitution into correct
formula
= R1097807,15 ICA
0,092) eS
x{1-} 1+ Te 1CA substitution into correct
formula
1097807,15 =—-—______*_/
97807,15 0,052
12
= R12297,82 ICA
[4]
a
i
Ea
|
| QUESTION 6

[3 MARKS]

oN

1M for correct shape
1A for horizontal asymptote
1A for negative y-value

3]

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4061115
‘QUESTION7 [19 MARKS] —
a -2x? ~4x+30=0 1M equating to 0 a)
(x+5)(x-3) =0 1A correct factors oo
A(-5; 0) B(3;0) 2CA for both intercepts
-1 if not in coordinate form |
[4] oo
7.2 y =-2(x? +2x-15) 1A for taking out -2 |

y=-2(x? +2x4+1-15-1 2
, ICA (2) added and deducted

y =~2[(x+1)? ~16] ICA incomplete square form |
y=-2(x+1)? +32 1CA |
Tpt 1:32) ica 2
OR

= ooo 1A substitute into correct formula

ae)
edie ICA
y=32 1CA

y =-2(x+1)? +32
1CA in correct form

Tpt (-1,32) ICA
5]
aa y =-2(1)? - 4() +30 1M substitution into equation
y=24 - No ICA value and conclusion
OR p
2x? —4x4+30=2x+10 IM finding point of intersection
x’ +3x-10=0
x=2 or x=-5 -No 1CA for values with conclusion |
— [2] a
74 2x? —4x430=mx430 IM for equating |
—2x* +(-4—m)x-2=0 1CA for standard form
A=0 iM = 7
(-4~m)? —4(-2)(-2) =0 1CA correct substitution
m=0 or m=-8 2.CA for each value of m ad
[6] ——_—_
TEs) (-29;3)—{-5} or x<3; x¥-5 2A
[2]

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10611/15

| QUESTIONS [il MARKS]

| [
 x=2 1A

8.1 Vas
HA: y=-] 1A

- [2]

82 | i
y-intercept: ae. IM

3 ICA

x-intercept: ye 1=0 1M

x=3 ICA

[4]

iS)

1A intercepts with axes
1A. asymptotes

: 1A Shape

[3]

8.4

1)
i
I
1

_

1A correct substitution

nN

Pa

i
09 | Ba) a
|

1CA

| [2]

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0611/15
QUESTION 9 [6 MARKS] :
9.1 9 1A correct substitution a
—2=log,| —
4
ae
sg 1CA exponential form
ea 2 '
5 ICA
E [3]
9.2 y=~log, x or y=log, x
3 2 1A
{4
9.3 g(x) >-2 1A
{i]
9.4 (—©0} 00) 1A
[4
QUESTION 10 [14 MARKS]
10.1 [1011 | fG+A)-f3)
h 1M correct formula
-23+h) +1-[26" +1]
= h 1A correct substitution
-12h—2h?
it 7 1CA correct simplification
_ h(-12—2h)
2 ICA
=-12-—2h Max 1/3 if ‘limit’ is used
oo OR
fath-f@
h 1M. correct formula
—2(x+ hy +1-[-2x? +1]
=  ™ 1A correct substitution
_ -4xh—2h?
= h : 1CA correct simplification
_ h(-4G)—- 2h)
_ h 1CA substitution of x=3
=-12-2h Max 1/3 if ‘limit’ is used
[4]
10.1.2 | limtl2-2h) 1M using limit
h-0
=12 ICA
[2]
10.1.3 m=0 1A
[1]

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10611/15
10.2 | 10.2.1 | y=16-16x+4x" 1A for simplification
7 a
Ve Ok Fa oe 2CA
(-1 for incorrect notation)
[3]
10.2.2 as
yax~ax? ta 1A for simplification
reels
yi=l-Sax? 2CA
oe
7] a 1CA positive exponents
2x2
[4]
QUESTION 11 [11 MARKS]
ILA ‘The graph passes through the origin.
Or
f(x) = (~-0)? (@-a)
1M_any logical explanation
{1}
11.2 7112.1 | f(x) =2(x-2)? 1A factorization
a=2 ICA
[2]
1122 | £0) 0 1M
3x? -8xt+4=0 1A correct differentiation
(3x —2)(x-2)=0 1CA factors
2
le ICA
[4]
11.3 pO 3(0)? —8(0) +4 1CA substitution into derivative
m=4 1CA
{2]
114 4x2 44x = p-2 1M taking 2 to RHS
p<2 1CA (full marks for answer only)
[2]

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10611/15
QUESTION 12 [9 MARKS]
|
| : x
- [+9 -2] <0 1M for inequality
x > 120km/ hour 1A for inequality sign
ICA for 120km/hour
Full marks if units are omitted.
[3] |
| 3
P=40x? ~—
| 3
P(x)=0 1M
80x—x? =0
1CA. ic
x(80—x)=0
1CA_ correct factorization :
x = 80km/ hour ICA
40 303
Ora ICA
ICA

Rae ene (aay
Re533 el

Full marks if units are omitted.

[6]

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10611/15
| QUESTION 13. [15 MARKS]
ie i 7
6 IG 2A for both outcome
L Ur IM for addition |
l9
[73 1CA for simplification
— [4]
13.2 | 13.2.1 Ss
3A for each section
[3] .
1322) 1,541.)
6 12.3 1M
ol
w= 75 1A
1 1
P(A) ==4+—==
‘) 6 12°54 1CA
[3]
13.3 36 000 1A
E y
134 [13.4.1 [291 2A a
[2]
1342 | 33) 2A
a | [2]

10

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