# Mathematics Grade 12 Paper 1 NSC June 2013 Memo

> Grade 12 Mathematics Paper 1 memorandum from the NSC June 2013, 21 pages. Read online or get the free PDF download, question paper linked when available.

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- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2013
- Exam period: NSC June
- Paper: 1
- Curriculum: CAPS
- Pages: 21

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Mathematics · Grade 12 · NSC June · 2013. Memorandum, 21 pages. Read online or download the PDF.

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basic education

Department:
Basic Education
REPUBLIC OF SOUTH AFRICA

SENIOR CERTIFICATE EXAMINATION

MATHEMATICS Pi
HIGHER GRADE

2013

POSSIBLE ANSWERS

MARKS: 200

This memorandum consists of 21 pages.

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Mathematics/HG/P] 2 DBE/2013
SCE — Memorandum
QUESTION 1
il 7a 3x(2—x)=7x-2
—3x? +6x-7x+2=0
3x* +x-2=0 Y standard form
Gx-2\x+1)=0 Y factors /substitution
25 or x=—1] (ca)
OR ¥ both values
3x(2—x)= 7x-2
—3x? + 6x-7x+2=0
pe (y+? 4-3)
_-I+v@?-4GX2) ,  ORx a=) a
2G)
_-1tv25 _-145
6 6
=tor-1 Vv @)
112 | x43 .¥x-5 =3
-—5)= Wrong factors,
RE) Coat aks ® 2 ¥ square both sides
x? -2x-15=9 breakdown: 5
x? -2x-24=0 Y standard form
(x-6Xx+4)=0 7 factors
60 on fe 4 NIA both values
“.x=6 (since x> 5) (5) v answer
1.13 3-x? =2x+1
x? +2x-2=0
—b+ Vb? ~4ac Y standard form
oe ) ¥ substitution in formula
_—@)+ Vey? - 4-2)
2()
[p_>y2 v Vi2
-2)tyC2)° -4-DQ)
OR x=
2(-1) YY values
a= 242 -1 incorrect rounding off
2
=0,73 or —2,73 0
Wrong formula: 5
Check the division line if
used correctly in subsequent
steps
G)
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Mathematics/HG/P 1 3

SCE — Memorandum

DBE/2013

OR
3x? =2x41
x? 42x=2
2 2
2 ie 1 ¥ complete square
of ye (23) = +(24) Y write as perfect square
2 on LHS
(x+1) =3
x+1=4V3
Y find square roots
x=-14 3 sq
=0,73 or —2,73 ¥Y values
7 -1 incorrect rounding off
4
114} x-35—
x .
multiply by x (since it is positive) to get v oa by a non-neg
2 factors
Nae V~15x54
(e+1)x-4)< 0 Gee
-ls<x<4 and x>0
“ O<xs4
4 (4)
OR x-3s-
x
x? -3x—4 vB a3x-4
——_——-s0 5
x ;
(&+1)-4)<0 since x >0 ain
-l<x<4 and x>0 OF *s
1 O<x<4 dee
Wrong inequality sign
or using equal sign:
-1 4 : marks
OR
x° —3x-4 <0
x
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Mathematics/HG/P 1 4 DBE/2013
SCE - Memorandum
ne |-2-x|>5 v¥ removing | | sign
—2-x<-5 or —2-x>5
x>-24+5 or x<-2-5 V¥x<-7
x<-7 or x>3 VY x>3
OR the solution set
C;-7)UG;%)
OR ;
(-2- Oe >25Vv Incorrect squaring: v square both sides
4+4x+x°>25 1 marks ¥ factors
x +4x-21>0 4 Yx<-7
(x—3)xt 7) >0VY Yx>3
x<-7 Yor x>3 ¥
OR
|2+x|>5
v v
—~<¢ >
7 2 sj
x<-7 Yor x>3 Vv (4)
12 2 =a @
Y=V20— x7 eee (2)
From (1); 2” =227-6 ¥ 2% =2?*6
_ y=2x-6 ... 8) ¥ equation (3)
Equating (2 and (3) or substituting (3) in (2):
2x-6 = V20-x? Y substitution
4x? -24x +36 =20-x? Y standard form
2 -
5x" —24x+16=0 ¥ factors/subst in formula
Gx-4Yx-4)=0 Y both values of x
x= 4 orx=4 Vv y= 2(CA)
y =-2 or y=2 Y y=-2 (CA)
x=4 and y=2 (since y> 0) (9) v answer
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Mathematics/HG/P | J DBE/2013
SCE — Memorandum
OR
= 92x-6
12 | From GQ) 2” =2?*6 ee
até (3) ¥ equation (3)
2
Substitute (3) in (2):
i +6)"
ye y-(225} Y substitution (CA)
2
y? =20 Yes
v
4y? =80—y? -12y-36 standard form
Sy? +12y-44=0
(5y +22Xy—2)=0 Y factors/subst in formula
) n Y both values of y
Very 2
Substituting y Mee in (3) gives: Yx= 4 (CA)
x=forx=4 Vv x= 4 (CA)
x=4 and y=2 ¥ answer
13 x +mxt+1=0
y
yaxrtmetl
a t+ 1) 5
+H x
fe)
(4)
*. Roots are non-real. “
OR
Roots are equal, therefore, m’—4 = 0, m’=4v
Forx?+mx+6=0,Y A=m?-24=-20V
.. roots are non-real / imaginary . ¥
OR Y substitution into formula
x +mxt+1=0
Equal roots: -. A=b? —4ac =0 Y both values of m
m? — 4(11) =0
(m+2\m-2)=0 YA=-20
m=-2 or 2 Y conclusion (non-real)
When m=-2: A=(- 2) —4(1X6)=-20
the roots of x” + mx + 6 = 0 are non-real
When m= 2: A=(2) -4(1)6)=-20
the roots of x” + mx + 6 = 0 are non-real
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Mathematies/HG/P 1 6 DBE/2013
SCE - Memorandum
14 | x?_(k45)e4+2k=0
A=(ck~5)? —4()(2k) Y substitution in A
=k? +10k+25—-8k
=k? 4 2k4+25 Y simplify to # + 2k +25
. ; v completing the square:
ee aah R42k+1424
=(k+1P +24 v (k+ly +24
 A>O forALL keR Y conclusion
~. Roots are real for allk eR (5)
OR
From  +2k+25
dac—b? VV +2k+25
Minimum value of A = ————
4a > ¥ calculate the min value
_ 40)(25)-(2) _ 96 _
a 4() _ ra! Y value of A
y :
<A>0 forallkeR mate
OR
If y= +2k+25, VY
the discriminant is 4 — 100 =-96 ¥
So the graph of y does not cut the k-axis. V
Since a=1>0,  +2k+25 >0 for all values of k. ¥
(39}
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1 PUBLIC EXAMINATIONS

Mathematics/HG/P 1 7 DBE/2013
SCE — Memorandum
QUESTION 2
2.1 | Given: f(x)=-x?-2x+3 and g(x)=|x+3|
211 Ro f'(x)=0
-2x-2=0
x=-]
FCI=-C1P -26-1)+3 gr y= sagt
(-1)(3
=-1+2+3 Al
=4 =4
Turning point is c 1 34)
y-intercept: (0 ; 3)
x-intercepts: —(x + 3x—-1)=0
x=-30rx=1
. (330) or (1; 0)
For g: (0; 3) and F3:9) For ft
k vv turning point
. vv x-intercepts
(- 1 4) v shape
0; 3) For g:
g Y 330) & (0; 3) arm
Y other arm
7 x If graph is not drawn:
~ 3530) (150) (033) ¥
(330) v
C4) vv
If not in coordinate form:
Maximum of 2 marks
()
2.1.2/@) | e(x)>3
.. graph of g is above the horizontal line ¥ statement/reasoning
y=3. y =3 intersect y = x+3 at (0; vix>0
3)and—x—3 at (-6;3) v x<-6 (CA
5-0 of 7-6 (3) | applies from graph)
) | #&)> e&) vv answer (CA
-3<x<0 (2) | applies from graph)
© -x? -2x+k=0
ie. —¥? -2x+3=3-k
has non-real roots when the horizontal line
y=3~-k and the parabola
ye =-x* —2x+43 do not intersect.
That happens. when 3-4 > 4
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8
SCE — Memorandum

DBE/2013

OR
No roots when the graph is shifted
vertically downwards through more than 4
units. Y The y-intercept is then <—1 ¥
ie k<-1V¥
IF
- (9% a
For non-real roots: (—2) - 4( it) < a Maximum: 1 mark
<
k<-1
(d) | For -3 <x <0; g(x)=x+3
Hx) = f()- g(x) v identify equation of g
=x? -2x+3-x-3
=x? —3x Y equation
-{21a03}e3 ha)=-x? ~3x
3/7 9 ~ completing a square
: {x + 3) Be
2 4
: Pop ee Y answer
Therefore maximum value is: — 4
4 (4)
OR
x)= f(x)- g(x) Y identify equation of g
=-y?_ oo
_ i Sit ae Y equation
a 3 A(x) =-x? -3x
Max value = —— au
4a
—~1)(0) — (~3)?
= 490) = (3) substitution
4-1)
Therefore maximum value is: 2
4 v answer
OR
A(x) = F(&)—a() noe ;
= oy? 2x 43—-x-3 Y identify equation of g
=-x? -3x ¥ equation
2
Maximum when A'(x)=0 AG)=—x" —3x
That is: T 2x3 = 0 Y h(x)=0
x=->
2
~3)\.(-3) _of_3).9
2) \ 2) V2) 4 Y answer
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Mathematics/HG/P 1 9 DBE/2013
SCE - Memorandum
OR
Roots of ~ x” —3x =—x{x+3)=04%
are Oand—3. “
0+(-3 3
Maximum value is at “3 ) “3 v
~37_343)29v
Maximum value = 2 2 4
22 (x)= b* and a(-1d]
a_l
22.1 | ¢-N=b ar; 1
Y substitute a(-13]
= ay 4
b=4 Q)
2.2.2 | t(x)=-q(x)=-4" vv answer
OR #x)= —5F Q)
=a
2.2.3 q , v interchanging x and y
oy = loge x v answer
q2'@) = loge @)
2.2.4 J Y shape of q
Y shape of t
q
Y shape of g™
“1 Y point e.g. (1;-1
; | ¥ point e.g. (t;-1)
(43) Y point e.g. (1;-2)
x
-1
G)
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Mathematics/HG/P 1 10

SCE — Memorandum

DBE/2013

2.3

B(2;9)

2.3.1 | Substitute into _k=(29)= 18

(1) _| ¥ answer

2 24
23.2 yotvr’—-x

7? = (2)? +9 =85

¥ formula (positive)

—— v 85
a y=4v85—x? v answer
OR Y formula (7 +7 =r’)
v 85
x’+y=85 and y>0 3) |“ y>0
C032
233|°C? ¥ CQ ;2)
-c-Yo
Moc =
xc —Xo
2-0
m=
9-0
2
_— Y gradi
9 Q) gradient
[35]

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Mathematics/HG/P 1 i DBE/2013
SCE — Memorandum
QUESTION 3
3.1 | Ifa polynomial f(x) is divided by (ax +5) until the remainder Y state theorem

Proof: Any polynomial

can be expressed as
S(e)=(@% +b).O(%) +R

where Q(x) is a polynomial in x and R is independent of x.

contains no more x terms, then the remainder is given by /| t 4)

F(x) =agx" +. ax") +ayx"? +..44, x +a,, a;€R and neN

v express f(x) as
product of factor
& quotient +R

F)=(+8). +R ¥ calle
2 fh 2)=(0.-8+b).0C-2)+R s4)
=0+R=R oy | teers
3.2 | f(x) = 4x? + ax? +x -3
Since f(x) is divisible by 4x? -1
fG¢)= (4x? -1)x43) vv
-. other factor is (x +3) ¥
and f(x)=4x° +12x? -x-3V
a=12 Y and
b=-1v ©
OR
4x? —1 = (2x—1f2x+1)
SF (x)= 4x3 +ax? +bx -3, v
1 1 remainder/factor
3] =0 and f- 4 =0 theorem
y as : +2)-3 _ Y substitution
2 2 2
Ltt e-3=0
4
FRG) (1)
f- i] =0
2 Y substitution
3 2
2 2 2
1414-15320
24° 2
Py) a (2)
()+@Q): 2a=24 ¥ value of a
; = 12 v value of 5
and b=-1
f()=4x3 +12x? —x-3
= (2x41)\2x-1)x+3) oa
«. other factor is x +3
{10}
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Mathematics/HG/P 1

QUESTION 4

12
SCE — Memorandum

DBE/20}3

ay | ovenx_

=
x* —yyx”
vxly- vx)

“ahs
1

vv

G)

Y factorising numerator
v factorising denominator

Vv dividing

Y factorising numerator
Y factorising denominator

Y dividing

17 vY
ee
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Mathematics/HG/P] 13 DBE/2013
SCE ~ Memorandum
4.2 p+q
pitq
11
=(p+q)+|—+—
Pq Y positive exponents (b+2)
+
=(p+4) (22)
Pq
=(p+q)x ( = } Y simplification
pt+q
i'l G)__ |“ answer
2 1
43 143.1 x3 5x3 +4=0
1 1
[? = ‘| = J =0 Y factorising
1 i :
PO ee ee (1) ~ both values of x?
x=4 or x= aN
x=64 or x=1
(4) | ~~ answer (each value)
OR
1
Let x3 =k
k? -5k+4=0
(e-4Xk-1)=0
k=4 or k=lv
1 J
w=4 o x=]
x=(4? or x=()
— i=.
43.2]  log,(x+7)+2log,5=4
log, (x +7)+log, 25=4
logs 25(x+7)=4 Y write as single log
25x +175 =3' =81 ¥ write without logs
25x =-94
94
jae v
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Mathematics/HG/P 1 14

SCE — Memorandum

DBE/2013

OR
log; (x+7)+2log, 5=4
log; (x +7)+log; 57 = 4log; 3
log, 25(x+7)=log,3*  v
2 25(c+7)=3* v
25x =-94
94 v

x=-S

25

4.3.3

logi9(2x—5)< logo, (x-3)
x>3 and 2x>5 so x>3
logio(2x—5)< —log;o(x—3)
log19(2x—5)+ logy9(x—3)<0

log g(2x~-5x-3)< 0

2x? 1x +151
2. 2x? -11x+14<0
. x—7x-2)<0

x>3 and 2exst

ae oe
2

OR

logio(2x-5)< logo, (x-3)
x>3 and 2x>5V sox>3Vv
logyq(2x—5) < —log;o(x-3) ¥

1
logi9(2x—5) < logyg —-
x-3
gx-s)s tv
x-3

2x? -11x+15<1 (ince x—3>0)%

2x* -11x +1450
v. (2x-7)x-2)< 0%

2. x>3 and asxsiv

sexs vv
2

(9)

¥x>3 and 2x>5
vv x>3

V —logio(x—3)
v inequality (1)

V 2x? -1x+15<1,
Y factors
Y solutions (2)

Y end points v inequalities

IF x>3 is missing and
answer is given as 3 <x < s

award max 5/9 marks

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Mathematics/HG/P 1

15

SCE — Memorandum

DBE/2013

44 Let log,x=M «.a°
and log, y=N -.a ¥Y writing in exponential form
coxypsa” aN =qtt¥ Y exp. law
- log, xy =M+N Y log, y=M+N
=log, x+log, y (4)
[26]
QUESTION 5
5.1 |1343;7;10;...
a=1 and d=3
5.1.1 Tr=a + (n-1)d v value of a
Tos = 1 + (66-1)3 Y value of d
= 196 v answer
(3)_| Answer only: 2/3 marks
5.1.2 | Even sequence: 4; 10; 16;22;... ¥n=33
a= 4; d= 6 and n=33 Y value of new a
Sy ="ha +(n-1)d] Y value of new d
2 3 Y formula
53 =F P@)+ 63-6] (6) | V substitution
= 3 (8 +192)
=3300 v answer
OR
Tas =4+32(6)=196 VV %4
n
S,=—la+T,] ¥
. 2 7 Y formula
33
S33 =—|4+196} Vv (33 and 196
Dery @ Aco) ¥ “substitution of 33 &196
=3300 Vv ¥ answer
©
2 D4... :
5.2 {5.2.1 x53 *=15+5+34+3+ Iino ¥ expansion
. a=15; r=t expansion,
' : give 1 mark v value of r
Se = 2 for a=15
I-r
15 ¥ formula & substitution
fe 1
3
a 22,5 v answer
2 @)
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Mathematics/HG/P | 16 DBE/2013
SCE — Memorandum
5.2.2 55.327 =15+54+54+84---45.9?" = zeae
r=l 81
S, = _ali-r”) Y formula
l-r
fF]
1820 ¥ substitution LHS
1- ao “8, substitution RHS
n
1-(3)- 1820 2 GF 0,998628..,
we -(tY' = 0,998028...
(LV 28
3 729
Wee 728 |
=| =1-——=— =0,001371...
(3) 729 729 v Qf =.= 0001371...
n 6
(2) - 4 Aine 1og0,001371..
3 : log 3 v answer
n=6
Answer only: 2/6
(6)
5.3.1 | Salary: a = R108 000, d = R8 640
Tr=a + (@-—l)d Y formula
Ts = 108 000 + (8-—1)8 640) V substitution
= R168 480 v R168 480
-. gross monthly salary = nae
=R14 040 (4) | ¥ answer
OR
Present monthly salary R9000 ¥
Monthly increase R720 “
Gross monthly salary = 9 000 + (8 — 1)720 ¥
=R14 040 ¥
5.3.2 | Expenditure: a= R72 000, r -(1 +3] =11
. ¥ value of r
Ss —a-r Y formula
0
1-r
ae 72 ooo(t-f1,1F) Y substitution into formula
=
1-11
° %
= R823 383,94 ~ R823 384 (4) | ~ nswer
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SCE - Memorandum

DBE/2013

OR
Expenditure: a= R72 000, r -(\ +3) =11
v value ofr
S, _a”-1) Y formula
r-l
7 72 ooo Lik - 1) Y substitution into formula
a oAd= I y
= R823 383,94 = R823 384 neil
{26}
QUESTION 6
61 f= lim ZEt4-20) Y formula
h0 4
= limi (c+? - Ql -x? Y substitution into formula
rev) h
2
= lim accu eLe -1 mark if Y simplification
moh lim = L&e#-/@)
a(- 2x- h) noo h
= lim ————
hoo h Answer only, 7 _
- lim (-2x- nh) no mark ¥ factorising & dividing by h
=-2x (5) | ¥ answer
OR
f(@)=1 -x?
S(x+h)=1-(x+AP ¥ 1 (eh?
=1-x? -2xh-7?
Sle+h)— f(e)=~2xh 1? v ~2xh-#?
Oy) = fim LEta-£)
f @) a) : Y formula
= lim h(—2x—h)
h>0 h gy an
= lim(-2x-A) factorising & dividing by h
hv
=-2x ¥ answer
2
62 }o21| y=*-3
2
1» 3 Y exponential form 3x7?
ype-x* -3x
Pr ; vv derivative (each term)
oe ol ara : ,
ee) 3{-32*) - ifNO 2 or y
=x+9x7 @G)
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Mathematics/HG/P I 18 DBE/2013
SCE — Memorandum
3 3
6.2.2 x)=| x+—= fx- =
an Aeron
x? ~— ‘pansion
ve Y ial fc
ay exponential form
=x? 9x3 ie
g(x) =2x+ 6x73 (4) YY derivative (each term)
63 Wx) = 2x3 + 7x? +4x-4
At point of contact h'(x)= 4 ¥ e)=4
(xe) =
6x? +14x4+4=4 v derivative
6x? +14x=0 Y factors
2x(3x+7)=0
=-2
x=0 or x=-} Vxs-F
. x= (x<0)
AL 2)= 2-2) +7-2f +4{-2)-4
( ) i. ( iy ( 3) ¥ substitution of x=-2
SOT Y y-coordinate
i i Tid
Therefore point of contact is C $-#
VF AFC) ceesreee (1)
. yu a at Ire Y substitution into eq. (1)
c= 2817
3 27
. = 235 =8,70 (8) Y value of c
[20]
PARTE) Fs
rat
EDUCATION.”
2013 -05- 20
PRIVATE BAG xX
PRETORIA gooT~
PUBLIC EXAMINATIONS
Copyright reserved Please turn over

### Page 19

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Mathematics/HG/P 1 19 DBE/2013
SCE ~ Memorandum
QUESTION 7
qa F(x)=a3 +bx and g(x)=x.
Turning point (—1 ; 2)
TA SED =aC1)P +6C1)=2
—A-b=2 eee i
a-b=2 a ¥ Eq (1)
f'd= Bax? +b Fi a ¥ derivative
7) — -1 marki Y f'-l)=0
fCy=0 derivative not rey)
equal to zero
fC) =3a(-1? +5 =0 V Fa. 2)
3atb=0 was. (2)
Q)+@: 242 oe) ee
“ a=i Dia)
Substitute in (1) or (2): =-3 (6) a2
qi2 F(x) =x3 -3x
I “(x) =3x?-3
"(x)= 0
J ®) Y derivative =0
3x°-3=0
3(x—-1Xx +1)=0
x=1 oF ee Y both values of x
AtC, f(l)= (1 -30)=-2 Y y-value (C)
-. C(1; -2) (3) | Answer only: full marks
OR 9
fEx)=al - x3)+a(-x)
= ax? + bx)
=-f(&)
.. f is symmetric about the origin
(~ 1 2) is a turning point ¥
-.(1_; 2) is the other turning point ¥
713 At A fx) =e)
 @-3x=x ¥ equating in terms of x
x -4x=0
x(x —2)(x + 2) =0 Y factors
x =Oorx=2o0rx=-2
At A, x=-2 Y values of x
oye? v y-value at A =y-value
atC
.. A & C have the same y-coordinates Y ya=Ye
.. AC is parallel to the x-axis (y = 0) or mac=0
OR
eee ae 2) 29 a
274 TT DEPARTMENT OF BASIC
:. AC is parallel tothe x-axis EnucaTion | (6)
2013 -05- 20
Copyright reserved . Please turn over
PRIVATE BAG X 110
PRETORIA 0001

i
s_ PUBLIC EXAMINATIONS

### Page 20

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Mathematics/HG/P I 20 DBE/2013
SCE ~ Memorandum
7.1.4 | x-intercept of f
f(x)=x7-3x=0 v f(e)=x° -3x=0
x(x- V3\x+ v3)= 0
x=0 or x=V3 or x=-V3 1 & BL43;0)
BL3;0) and _D{v3;0) 3) _|% 2b3;0)
TAS -~V3<x<0 VV -V3<x<0
or Yor “ x> 3
x>v3 (4)__| (CA from 7.1.4)
7.1.6 [(0; -3) (2) _| “each coordinate
72 7.21 VU =12—2x Y length
SV =xtan60°= 3x v breadth
Area = length x breadth
- Y substitution in formula
A= (12-2x)03 x)
= V3 x(12—2x) square units (3)
722 A=12W3x-23 x? oo. a) iad
Area will be maximum when: a =0 aa ay
IK dx Y derivative
ie. 12V3-4 V3x=0
4J3x=12V3 y
4 value of x
b
OR from Eq. (1): 2 oF v v answer
= 13, y
2-23
=3V
A= ¥3.3(12-2.3)
A=18y3 or 3118 sq.units | (4)
(30}
POP Sarees
- a MENT OF BASIC
~~ - Pe fos - e
ann EUCATION |
i
F
| 2013 -05- 20
i PRIVATE Bag x
| ipso OMG X 44
fomopremee PRETORIA D004
Lead SBLIC EXAMINATIONS
Copyright reserved Please turn over

### Page 21

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Mathematics/HG/P 1 21 DBE/2013
SCE — Memorandum
QUESTION 8
¥ x>1
8.1 v y<s
SEE GRAPH BELOW VY 8y <60—-4x
VV 2x<18-y
¥¥ shading of feasible
(8) __| region
8.2 | T=600x+400y (2)_| “answer
3 if oo Y correct gradient
83 | 8.3.1 oe mm (dotted line: SEE GRAPH) Y line at optimum position
(2) _| (CA from 8.2)
8.3.2 | 7 Comfort beds and 4 Deluxe beds (1) __| ¥ answer
{13]
TOTAL: | 200
20
7
15
n
a
kl
i}
3 D
8 wa wa
f 10 o
a be
fe o
io} 7
Fa = :
a n,
= 3
zs Z
3 a)
(050) 10 15
NUMBER OF co a BEDS
| SEPARTMENT GF
4 i. - JP B,
2013 -06- 20
Copyright reserved i PRIVAT Bac
i PR ORG ue

L_ PUBLIC E (AMINATIONS

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