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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
T950(E)(A2)T
APRIL EXAMINATION
NATIONAL CERTIFICATE |
MATHEMATICS N3
(16030143) -
- -13200-16:00
n paper consists of 7 pages and a formula sheet of 2 pages.
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(16030143) 2 T9SQ(E)A2)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N3
TIME: 3 HOURS
MARKS: 100
NOTE: Diagrams are NOT drawn to scale.
INSTRUCTIONS AND INFORMATION
1, Answer ALL the questions.
2. Read ALL the questions carefully.
3. Number the answers according to the numbering system used in this question paper.
i 4. Show ALL the calculations and intermediary Stops...
5. Questions may be answered in any order but sibbsections of questions must NOT be
| separated. soy AN
, 6. ALL final answers nitust be accurately approximated to THREE decimal places.
7. ALL graph siork «must be done in the ANSWER BOOK. Graph paper is NOT
supplied. ce
8. A forinula sl ect is attached to this question paper. The list is NOT necessarily
ete, Any other applicable formula may be used,
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3 T9S0(E)(A2)T
(16030143) ; 3
QUESTION 1
Ll Simplify the following expression without using a calculator.
i od 1
xr+x? {# -x |
(4)
12 Determine the value of p if x—2 isa factor of the following function:
f(s) =2x° + px? —4x4+5 } (3)
1.3 Factorise as far as possible in prime factors:
1.3.1 ay -2x-2y @)
1.3.2 5 2
dx? -3-—
ft lig «)
1.4 Simplify:
p-l2l, p-2_, 2p-22
pa—4 <Qpt22 pr+2p+i(pt2). - ©)
[20]
QUESTION 2
2A Solve for x:
21d. AP x9? 51 (3)
12 oF RFF =12 (4)
So Jog ¥ + log 3—log(3x—1) = log 4 (4)
‘; . \ *s -
‘The aréa of a rectangular swimming pool, 10 m long and 3 m wide, doubles if its
length and width are increased by the same amount. Calculate the dimensions. (4)
2.3 Make m the subject of the following formula:
W- 2mgh
\ mr? +1 (4)
2.4 Make T the subject of the following formula:
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(16030143) 4 T9SO(EMA2)T
2.5 Determine x by completing the square:
3x =10~x°
(4)
[27]
QUESTION 3
3.1 In FIGURE 1 below L(-5;-2), M(-1;-6) and K(5;4) are vertices of trianglé KLM in a
Cartesian plane. “
Y¥
K(5;4)
a x
“ 7
L(-5:-2)
M(-1:-6)
y
FIGURE 1
Calculate the coordinates of the midpoint of MK. (2)
Determine the gradient of LM. (2)
3.1.3 Determine the length of LM. Leave the answer in surd form, Q)
3.1.4 Determine the equation of the line parallel to LM passing through the
point K. Leave your answer in general form. GB)
3.2 Determine the equation of the straight line that passes through the points A(2;4) and
B(-1;-3). Leave your answer in gradient-intercept form. GB)
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3.3 Calculate the angle of inclination of the line in QUE: (3)
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(16030143) -S- T9SO(EWA2)T
3.4 Determine the equation of the line AB that is perpendicular to the line f(z) = ne in
FIGURE 2 below. AB cuts the x-axis atx = 6. Leave the answer in surd form and in
gradient-intercept form.
Y i
’ (3) |
FIGURE 2
fda}
indices and in surd form.
4.1.1 yeni a Wx (3) |
+x?
4.1.2 = 5 @)
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(16030143) -6- T9SOKE)(A2)T
42 Consider FIGURE 3 below. Determine the x-coordinates of the turning points P and
Q of foyae 9x
Q
FIGURE 3, 9]
QUESTION 5 :
5.1 The design of a bri ein FIGURE 4’shows an arch represented by a parabola, which
is defined by the equation y =2x - 0,252". ;
ww.dragoart.com
FIGURE 4
5.1.1 Draw the graph of the parabola. QB)
5.1.2 Use the graph to determine the height of the bridge (in meters) if the line
y = 0 represents the water level of the river. qa
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(16030143) -7- TISOCE AQT
5.2 Draw the graphs of the following trigonometric functions on the same system of axes
for 0° <x <180°:
y=3sin2x and py =2cos3x
ALL values at the point of intersection with the system of axes and the co-ordinates
of the turning points must be shown. net
QUESTION 6 ;
6.1 Simplify the following without using a calculator:
tan(180° ~ 4)Vi—sin? A
cos’ (180° + A) +sin? (360° ~ A) ; ; (4)
6.2 Make use of basic trigonometric identities to prove thai:
bcos f sin B =2e0secB o
sinfi i+cosf . : (4)
6.3 Calculate the values of a that will Satisfy the following equation for 0° < @ <360°:
sina =—2cos a GQ)
64 Consider FIGURE 5. From the top of a cliff 60 m above sea level the angles of
depression of two ships ori'the same vertical plane as the observation point are 20°
and 25° respectively: Calculate the distance between the two ships.
a
FIGURE 5 4)
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(16030143)
MATHEMATICS N3
FORMULA SHEET
Any applicable formula may also be used.
1. Factors/Faktore
“le T9SOCEKADT
2. Logarithms/Logaritmes
ab =(a—-bYe’ +ab +5?)
ae +b? =(a+ 6a’ ab +b?)
3. Quadratic formula/
Kwadratiese formule
pa lbt vb? = 4ae
2a
4. Parabola/Parabool
year tbx+c
_ 4ac~b?
, da
-b
xo
2a
5. Circle/Sirkel
logab=logatlogb
a
log—=loga—logb
og oga-log
loga” =mloga
log, a=
log, 6
log, a@=1..Ine=l
qlee! =te.e"™ =m
6. Straight Line/Reguitlyn
ey ar?
x
D=—+h
4h
x= 4Dh- 4h?
yoy smlx- x)
Perpendicular:
Loodreg:
m,®m, =-1
Parallel Lines: °
Ewewydige lyne: m, =m,
Distance:
Afstand: D= (x, -x)/ +(y,-»,)
Midpoint:
Middelpunt: P= Gemeaee
Angle of inclination:
Hellingshoek:, @= tan” m
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ta
(16030143)
7. Differentiation/Differensiasie
dy _™ f{e+h)- se)
dx 30 h
Sle" )= nx”
Max/ Min
Maks/ Min
For turning points:
Vir draaipunte: f'(x)=0
8, Trigonometry/Trigonometrie
sin@=2 = u
r cosec?
cos@=—= !
r secO
tno=2-—1
x cote.
sin? 6 +cos’ @=1
1+ tan’? O=sec’ 0
1+cot? @= cosec’@
sin 0
cos@
cote = cos@
sind
sind sinB_ sinC
a 5 ce
a? <b? +c? 2d wnloaded from
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hlayiso.com
TOSOCEXKA2)T
higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
MARKIN G GUIDELINE —
Ces:
NATIONAL CERTIFICATE
APRIL EXAMINATION
MATHEMATICS N3
2 APRIL 2015
This marking guideline consists of 9 pages.
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MARKING GUIDELINE -2. T930(E)(A2)T
MATHEMATICS N3
QUESTION 1
it tt toa foot es
waxy 2b x2-x? x24x2-) x2? -—x 2?
1 1 a
aVx4—~| vx-7= Hxttuet~x? 4px?
Te
oy?
-22 (=n) v v = 2x
vx (vx 2 We
_x+l-x+1 v x x
dx
2 wr y
vx x
1.2 wx=2Q
f(x) = 2x3 + pr? 4045
f (2) = 22) + p(2y —4(2) +5 ¥
=16+4p-8+5
=4p4+13
4p+13=0
13
4 v
p=
13 13.1 vo -y-2x—2y
= (x- yet y)-2(et y)
= (xt y)(x-y-2) v
13.2 ; 2 v
2x* 3x? -2 v
“Vr
_Qe+DG'=2) v
x2
14 peri2l po? 2p-22
p-4 2p+22 p> +2p+1l(p+2)
_(P-1)(p+1) | p-2 Pet 2+ Mp +2)
(p-2)(p+2) Up+1) 4 p-11) y
_(p-U)(p41D_ (p-2)_ + ZT v
~i) Rewaloaded- trem hlayiso.com
_ptll v
4
MARKING GUIDELINE
eae
MATHEMATICS N3
QUESTION 2
21 14 33 32-4 = 3° P,
3x+2x-4=0
5x=4 ¥
4 v
x==
5
Dl fe en ZI
a se
27427 =12
2742?2=12 Vv
2? (1+2)=12
¥ v
2? =2?
x=4 v
2.1.3 log x + log 3—log(3x—1) = log 4
log x + log 3—log(3x—1) = log 4
v
log Ba =log4
(3x-1 Y
3x=12x-4 y
wat v
9
2.2 (10+ x) +x) =30x2
30 +13x-+ x? =60 v
2 =
x +13x-30=0 E
x=-15 orx=2
New length = 12 m y
New width = 5 m v
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T930(E)(A2)T
MARKING GUIDELINE
-4.
MATHEMATICS N3
23 we 2mgh V
mr +I
wmr? +I = 2mgh V
wir’ —2mgh =-w I
mwr? —2gh)=-wl v
as wil
aw 9 ="
” wr? —2gh mm 2gh-wr?
24 En
i=0,le?
t 900,
t 309
n= Ine?
0, v
int = 900
01 T
p 200. ___ 900 v v
nf. Int—In01
0,1
2.5 2 2
vs3c(3) -0-(3)
2 2
2
(=+3) =1042 v
2
x42at 10+2
2 4
xo~2+ 40 v
2 4
-347
2 2
x=2 or x=-5
v v
QUESTION 3
3.1 3.1.1 midpoint of MK.
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=p aade viv
= (2;-1)
T930(E)(A2)T
[
MARKING GUIDELINE -5-
3.2
3.3
3.4
MATHEMATICS N3
3.1.2 ~64+2
3.13 IM =f145y 4(-642)°
=Vi6+16 v
=V2x16
=4/2
3.1.4 yoy, =m(x—x,)
y-4=-la@-5)
prd=-x+5 Vv
prx-9=0
7
y-4=2 (2-2) Vv
or y= 2,333x - 0,667
Vv
@ = tan"! (7) v
Vr-Yy=m(x-»%)
y-02-V3(x-6) v
y=—v3x+6/3 v
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T930(E)(A2)T
MARKING GUIDELINE 6
a T930(EY(A2T |
MATHEMATICS N3 |
QUESTION 4
4.1 4.1.1
pense
x
=x! oy
D a ax
ax v
=2 224
Ye
wv
if
4.1.2 etx)
Sx” v
=tvy4,
» 5
B® 3a,t v v
ax 5 !
4.2 pax ~9x
® a3 ~9
de
13? -J=0 Y
x? =3
xaty3
”.X-coordinate of P =—/3 v v
X coordinate of Q=/3
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T930(E)(A2)T
7.
MATHEMATICS N3
MARKING GUIDELINE
QUESTION 5
4 meters
5.1.2
8 meters
5.1.3
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MARKING GUIDELINE
5.2
-3-
MATHEMATICS N3
7 7
T930(E)(A2)T
1
fi
i
'
fl
oe iaieiatiaiaianel att
120°
y=3sin2x
x-intercepts: 0°,90°,180°
y-intercept:0°
maxpoint:(45°, 3)
minpoint : (135°, ~3),
QUESTION 6
6.1
tan(180° — 4)VJ1—sin? A
cos* (180° + A)-+sin?(180° — A)
~tan Avcos* A
~ (-cos Ay +(sin Ay
__atan Acos A
cos’ A+sin? A
=—sinA v
v
y=2c083x
x-intercept : 30°,90°, 150°
y-intercept : 2
maxpoint:(0° 2), (120°, 2)
minpoint : (60° 2), (180°, -2)
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MARKING GUIDELINE
-9-
MATHEMATICS N3
62 (1+ cos B)(1 + cos 8) + (sin £)(sin f)
sin B(1 + cos f)
_1+2cos B+ cos” # +sin* B v
sin B(1+ cos £)
_ 20+ cos B) v
sin B(1 + cos 8) v
2
sin J v
=2cosecf
. LAS = RAS
6.3 sina _—2cosa@
cosa cosa
tana =-2 v
a = tan” (-2)
reference o = 63,4°
a =116,6°
or
a = 296,6°
64 In A ABD tan 20° = 22
BD v
BD = 164,849 m
IA ACD tan25°= 20,
cD
CD =128,67 m
a v
BC=BD - CD=36,179 m
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T930(E(ADT
TOTAL:
{
U
100
Copyright reserved
higher education
& training
Depaniment:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
T940(E)(NI8)T
NOVEMBER EXAMIN.
on paper consists of 6 pages and 1 formula sheet of 2 pages.
<
is
a
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(16030143) TOOCE)NIS)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N3
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1. Answer ALL the questions.
2. Show ALL the calculations and intermediary ste
7. A formula, sheet i
complete. Any othe
ached thisgquestion paper. The list is NOT necessarily
x applicable: ormula may be used.
ue
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(16030143) 3.
QUESTION 1
1.1 Simplify the following WITHOUT using a calculator:
Ld (log, 2) (og, b)
1.1.2
ve
x Vx
113 aig -3/35
J2+¥8
1.2 Make use of the factor theorem to prove that xe afactor of the function
f(x)=x? -39x +70
1.3 Factorise the following two expressions
pi 4m? +4in—1
14
QUESTION 2... |
2.1 “ Solve for x in the following equations :
2A See 5 =x <3
2.4.2 log, x +log, x’ tlog, x° =18
2.13 1
tae (t) on
3
2.2 Make + the subject of the following formula:
P=0,4e""
23 Make x the subject of the formula:
_ [xrl
x+2
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T940(E)(N18)T
G3)
(3)
(3)
@)
@)
[20]
GB)
GB)
™)
(4)
4
(16030143) 4. T940(E)(N18)T
2.4 — Solve for x by completing the square:
x? =12x+10
2.5 Calculate the coordinates of the points of intersection of the graphs defined by the
following two equations:
x
=+2y=17
rane
Jara
4
2.6 The length of the rectangular field is 6m longer than snes
QUESTION 3
exzaxis at E and P and the
3.1 Consider FIGURE 1. The parabola 7 ips it Re skelebgouts ¢
y-axis at M. The straight line ¢ pr x ya a E and M respectively. The
functions fand g are defined “a yes “> rand g(x)smxte.
Shee
N |
x
FIGURE 1
Determine the following:
3.1.1 The co-ordinates of M, E and P which are the intercepts of the graphs
with the system of axes .
3.1.2 ‘The co-ordinates of the turning point N of the graph of f
3.1.3 The values of mand cin g(x)=mx+e
314 Renmilonsagiiaempdalayiso.com
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@)
(16030143) -5- T940(E)(N18)T
3.2 Determine the equation of a straight line that passes through the point (-1,2) and
which is perpendicular to the line y= 4x+ 2..Write your answer in general form.
3.3 Determine the value of p if:
3.3.1 p isthe x-coordinate of the nsidpoint of the line segment joining B,-D
and (1,-3)
3.3.2 p is the angle of inclination of the line y= 3x4
3.3.3
3.3.4
3.3.5
(5x2)
QUESTION 4
41 ar
Determine lim
hod
. Write your answer with
QUESTION 5
5.1 Solve the following equation if 0° <9 <360°::
3cotO+2=9,5
9.2 Use trigonometric identities to prove that tan? - sin? @ = tan? O.sin’ @
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(3)
(10)
[22]
@)
@)
@)
[ty
@G)
4)
(16030143) 6 | T940(E)(N18)T
5:3
5.4
5.5
Consider FIGURE 2. A window washer on a ladder looks at a nearby building 100 m
away, noting that the angle of elevation of the top of the building is 18,7° and the
angle of depression of the foot of the building is 6, 5°. How tall is the nearby building?
18,7°
@ Fig 4)
Consider FIGURE 3. Calculate the length Fusing the sine rule.
ae
A
58° 72°
B 17m : c
FIGURE 3 (2)
Draw the graphs which are represented by the following trigonometric equations on
s the same system of axes for 0° < x <180°: s
f(x) =cosx
g(x) =sin(x—30°) (4)
[17]
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(16030143)
7. Differentiation/Differensiasie
-2e
dy tim flx+h)— f(x)
ax hs h
("=n
Max/Min
Maks/Min
For turning points:
Vir draaipunte: f! (x) =0
8. Trigonometry/Trigonometrie
sin? -2. |
r cosecP
cos@= aa |
r secO
tan 0 = ca u
x coté
sin? + cos? @=1
1+tan’ @=sec? 0
1+cot? O=cosec?@
sind
tanO=
cos0
cotd= £08 é
sin 8
sind sinB sinc
a b ec
a =b? +c? —2becos A
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T930(E)N18)T
Downloaded from hlayiso.com
16030143
FORMULA SHEET
Any other applicable formula may also be used.
1. Factors/Faktore
-1-
T930(E)(N18)T
2. Logarithms/Logaritmes
a-b =(a~ba? +ab +b?)
a +6? =(a+ Va --ab+b?)
3. Quadratic formula/
Kwadratiese formule
—btvb? -4ac
2a
x=
4, Parabola/Parabool ae
year tbxt+e
_ dac~b?
y 4a 2
logab = loga + logb
a
log—=loga ~logh
oF g 08
log, a
log, b
log, @=
¢
loga” =mloga
log, a= = b
log,
log, a=1..Ine=1
e Inve
all ape, =m
6. Straight line/Reguitlyn
Downloaded fr
Copyright reserved
YrW= mlz =x)
Perpendicular:
Loodreg: m, em, =-L
Parallel lines:
Ewewydige lyne: m =m,
Distance:
Afstand: D= VG, — x y + (y, “yy y
Midpoint:
Middelpunt: P= St %e a he
2 2
Angle of inclination:
Heil G2 Om
&
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higher education
&training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE
NATIONAL CERTIFICATE
NOVEMBER EXAMINATION
MATHEMATICS N3
18 NOVEMBER 2014
This marking guideline consists of 9 pages.
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MARKING GUIDELINE -2- T940(E)(N18)T
MATHEMATICS N3
QUESTION 1
Ll LAE (log, a) (log, &) log, b ¥ apply law
1 loeb orlog, ax log,a ¥ ¥ simplify and
= 084, 1080 v * write it as ONE
logx loga = log. log
_ logb v
logx v
=log, 5
1.1.2
ve +
* * “common
_ axle tay ¥ denominator
nfs ¥ factorisation
x x+y) ¥ divide and
at ed v rationalized
xx wx denominator
_fl+y)
vx
_ved+y) ”
x
113 2x3V2 —3x 42 , Vorime factors
nn aay Time tractors 10
v2 + av2 roots
_ V2(6-12) v Y factorisation
J2(1+2) ¥ value
=-2 ,
¥
1.2 Let x=-7
= “value of x
S(-ND=-7 -3%-7) +70 v
v . :
=—343 + 273+70 ¥ substitution
=0 eduction
“x+7 isa factor of f(x) v v
13 13.1 p-4m +4m-1
¥ factorise
=p -Qm-)Qm-1) v , ¥ factorise
=p? ~(2m-ly ¥ simplify
¥Y s[p-2m+l][pt+2m—-i] v
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13.2 3|(4a? — 9p? 1] v ¥ factorise
=3[(20 + 3b\(2a — 36)| “factorise
MARKING GUIDELINE 23.
MATHEMATICS N3
v
14 x? 42x x3 x3
+ x
+4 x +dy-4 x42
_ x42), ee +4@-D) x-3
x +4 x~3 x12
v =X +2) +4 G@-D x3 v
+4 x3 x+2
=x(x~1) v
QUESTION 2
x+5=(x43) v
X+5 55° +6x4+9
x +5x+4=0 v
(x+(x44)=0
x=-lorx#-4 4
2.1.2 log, xt+log, x’ +log; x° =18
log, x” =18
Glog, x =18
log, x =2 v
x= v
x 25 v
2.1.3 ye
gag? fo] an]
()
gel 4+ 3%? 3° =11
¥#(34+9-1I)s1l vv
F=l
3 23° v
x= v
T940(E)(N18)T
¥ division sign
V factorise first
term
¥ grouping
Y answer
¥ square both sides
¥ factorise
solve
(penalise with 4%
mark if -4 is not
shown as invalid)
v write it as one log
by applying law
Y applying
definition of log
“value
Y applying
exponent law
¥ factorising
¥use exporient rule
to find value
¥ value
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MARKING GUIDELINE 4.
' MATHEMATICS N3
2.2 P=0,4e"
Ppt y
0,4
0,4 v
23 xl
x+2
H? (x4+2)=x~1
HPxt2H? =x-1
HW?’ x—x=-2H? ~1
x(H? —1l) =-2H? -1 v
_ ~(2H? +1)
Pl
24 x? -12x-10=0 v
2 2
x? -120-(-2) = iox[-2)
2 2
(x-6) =46
x-6= 4/46
x= 64/46
x=12,782 or x=—0,782
v v
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T940(E)(N18)T
¥ manipulate
“using logs
“simplify
“make r the subject
v square both sides
¥ cross multiply
V take out t as
common factor
¥ simplify
VY write LHS as
perfect square
¥ x subject of the
formula
¥ both values
¥ correct rounded
off-penalize only
with one here
MARKING GUIDELINE -5-
. MATHEMATICS N3
2.5
S+2y=17 ._(l) V
taye7 (2)
4 v
x+l0y=85 2.x =85—-l0y v
Replace x by 85-10y in equation 2:
y44x=28
y+4(85-10y) = 28
v
y+340—40y = 28
—39y=-312
y=8
Replace y by 8 in equation 2: Y
8 ee7
4
x=§
2.6 Let breath be equal to x
Then length (3x)+6
Perimeter = 2(1+b)
188 = Ax+3x+6] y
8x +12 =188
x=22m v
3x+6= 72m
v
QUESTION 3
3.1 3.1.1 let y=0
letx =0 >
Osx —4x4+5
y=0-0+5 gn 5
yess ee 4 v
orx=1
“4M =(0;5) Vv
«FE =(-5;0) and P = (1;0)
3.1.2 -(-
ee)
2(-1)
TO40CE)(N18)T
¥ simplify
¥ substitutions
“simplifying
¥ value of x
¥ value of y
¥v Any steps
v answer
y¥M
YE
¥x-coordinate
¥y-coordinate
x=-2
Dopnletdpd from hlayiso.com
“y=
oN =(-2:9)
MARKING GUILIELINE 6
; MATHEMATICS N3
3.1.3 _ 5-0 yrmrte
ee ors S=1@)+o v
smal v .c=5
32
3.3.1
3.3.2
3.3.3
3.3.4
3.1.4 EM =J(%,-4)Y +0. -"y
= (-5-0)? +(0-5)”
=V254+25
= 50. 5V2
v
1
=4 n, =—-—
m, m, r
yry, =m@—x,)
y-2=-Vi(x+l) vy Vv
. po-5 Downloaded from hlayiso.com .
T940(E)(N18)T
¥ Vanswer in
surd form
¥ gradient
¥ substitution
¥ simplify
vv value of p
¥v value of p
vv value of p
vv value of p
MARKING GUIDELINE
QUESTION 5
5.1 3cotd+2=9,5
cotO =2,5
tan@ = 0,4
-8-
MATHEMATICS N3
0=21,801° or @=201,801°
v
v
5.2 tan” @—sin? @ = tan? @.sin’ @
LHS = tan’ @ —sin’ 8
_ sin’ @
z= ~sin’ 6
cos’ @
_ sin? @—sin? Ocos’ 0
cos’ @
_ sin? 6(1— cos’ 0)
7 cos’ 0
= tan? @.sin? 6
=RHS
wv
5.3
tan18,7° = BC
100
BC = 33,848
tan6,5° = ep
100
CD =11,394
Height = 8C+CD
= 45,242m
5.4 in 72°
ap alsin?
sin 50°
=21,106m
v
4
v
v
T940CE)(N18)T
¥ manipulate for 0
¥ ¥ two answers
¥ applying quotient
identity
¥ finding die LCD
¥ factorise
¥ proof
¥ v determining
BC
“determining CD
VYadding BC and
CD to determine
the height
“replacing
Y answer
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7 foe]
og i ne 2g
eB RIN 2 Bo R a > Bo
= gg ¢ 28 £ & Ff 22
z 5 g Ca 3 8 B os
& 1 8 8 Paes) if 2 DS
5 - 8 ga 41 8 2S
= ~ S#8 EES BS £A255 88
z S SSE 62868 wv Soe 6&8
7
:
:
£
— t
:
:
- :
:
:
:
_
— ieee n define
:
2 H
S fem eee ok tee y
= »
'
as :
ma 1
EE piateteteted Tr
3
a
On
aQ
i
tees wow ee ew He YUL
4
a
th
a
1
an —
23 :
za i
a 1m
Dp -
5 my
ie) '
6 :
: i
5.5
graph
TOTAL:
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MARKING GUIDELINE 7
MATHEMATICS N3
3.3.5 Loy ag or y-intercept
4 Oy? =81
anes oy? =81
x vo
3 ac =! y =9
p=B vv y=s3
p=
QUESTION 4
4.1 _
jim ft Fe)
A h
~4x? —8xh—4h? —(4x7)
= lim
0 h
= vw
tim AHOX +A) v
30 h
=-4(2x+0)
=—8x v
4.2.1 1
yaz-4¥x
x
1
pox —4xt v
1
a4 Ay ayt!
* v
=—-4x5-x
. v
41
~ 33 fe v
4.2.2 _ x(x? -D
x-1 v
x{x+1(x-1
yo2le DG =D
x-l
per +x
. ® oy st
T940(E)(N E8)T
vw value of p
¥ multiplying of
2¢c+h)? and
substituting
¥ factorising
¥ simplifying
“answer
¥ rewriting by
applying
exponential rules
y ¥ applying
differentiation
rule
¥ writing it as
positive exponent
and in surd form
¥ factorise
vv differentiate
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1
i
1
higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
T930(E)(M31)T
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(16030143) je
QUESTION 1
Ll Simplify the following WITHOUT using a calculator.
Lil Ly
7 92 _ 3H
ae
1.1.2 log V27 +log V8 —log Vi25
log6-log5
1.2 Factorise as far as possible in prime factors:
1.2.1 22-2) £9(x—2)-5
122 (m-=ny36x" + 49ny? — 49m”
13 Given that x+1 isa factor of f(x),
Determine the other factors if /(
14 Simplify:
V3x410-x=2
a2 axes 23
x x
2.2 Make 'x' the subject of the formula by completing the square:
x? +8x—S5a=0
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T930(H)(M31)T
GQ)
G)
@)
@)
GB)
(3)
6)
B35]
@)
(3)
Q)
wa
(16030143) 4 T930(E)OM31)T
23 Make ‘t’ the subject of the formula:
” ¥, = Vie : (4)
24 Thomas is 4 times as old as John. In 12 years’ time Thomas will be twice as old as
John. What are their present ages? QB)
2.5 Determine the points of intersection of the graphs represented by @ following
equations algebraically:
xy? =25 .
i yext5 (5)
| (21)
QUESTION 3
L, 3. Complete the following sentences by fii Sissine values, expressions of
| equations. Write only the missing ans ot fe question’ number (3.1.1-3.1.3) in
the ANSWER BOOK. “ty
! 311 P(-};-1) and QC3, foordinates of the midpoint of
o PQare... (2y
412 The acute anglBg
; equal t (2)
|
3.13 The & ()
3.2 In the dj pram be e
‘ is atangtint to thyrtite al'point P.
Y-axis
em
P(-3,2) oN
X-axis
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(16030143) 5 ( T930CE)(M31)T
3.21 Calculate the length of OP. - ; (2)
3.2.2 Determine the equation of the circle. (2)
3.2.3 Determine the-equation’of the taagent AB.. 4
3.3 Determine the equation of the straight line that passes through point O (0;0) and
which is parallel to the line 18x —3y +9=0 . GB)
[16]
QUESTION 4
4.1 Consider the function y =1~-4x"
. 4. yp
4. Determine the derivative, =,
che @)
112 Detormins the gra Gy
4.2 Determine a if
ak
A)
mown system of axes. Calculations need NOT be shown. ALL
oints @¥ intersection with axes and the coordinates of the turning points
Cauiyemiust be shown,
i 2
ane ae
256 dd Q)
yo3x? 2x3 (4)
433 x yy
4°43 (2)
(17)
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(16030143) -6- T930(E)(M3 1I)T
QUESTION 5
| 5.1
5.3
5.4
5.5
Calculate the exact value WITHOUT using a calculator:
sin? 120°x sec? 150° + sin 150°
* cosec 30°x cos 60° G)
Calculate the value(s) of 8 which will satisfy the equation if 0° <@<360°:
2sin@cos@—sin@ =0
é 6)
From the top of a tower, 100 m high, the angles of depression 0 5 ghia gon the
ground, due west of the tower, are 41,45° and 21,68° x 5
Calculate the distance between the TWO object (5)
Given : AABC with, A= 60°, AB= V2 ag
Calculate: sy
A
60°
Va V8
B A c
The value of ‘x’ 63)
5.4.2 The area of the triangle. Q)
Sketch the graph of 3y =sin3@ for 0< 8 <120° (3)
[21]
TOTAL: 100
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MARKING GUIDELINE “2- T930(E)(M31)T
MATHEMATICS N3
QUESTION 1
Ll Ltd Lo 9X Q
9g? 3"! oa _ ee
xe? — 4 sa .
ee TG > LL
(ey 3 j ° 5 tO on
= ye 4 a uae 2
~. > - >
38 343°!
- L$
ay 2 tr 8 2
1 :
fs) Y 3 —) hee 4
Ss 4 ut x Quit,
39 ence
2 gov €
oa a Moe obY
Saye, ye
9 3 9 27 (
3)
1.1.2 log V27 + tog V8 —log S125 log (Gy + log 4/2) —log Jy’
log 6—log5 ioe6 “ee? ; .
log ee, joo VO y (le ao}
___V125 °8 ia ns eee
6 ~ 6 loc) br
log 5 v a :
log 105 loo ay ay ae
= 5 (5) | oe a Ic
log 6 = 6 J
PLS log} ~ fp
“LS 4 lose
5 bee
6y ov ;
log 6
_ 5 log z
5 oe( |
v 3 6
5 gl 3 6
2S aS 285) 3
6) *2 ef 24 af .
log 5 lo (§ 2
*\s (5)
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°
MARKING GUIDELINE -3-
YO30(EKM3 IIT
|
MATHEMATICS N3
or 2(e—2)? +9(x-2)-5
12 Lat Mx 2F + 9(x-2)-5
lei x-2=k = Uxt dv +4) 9x 18-5 %
ode +9k-5 = 2x? 8x48 49x-18-5
=(2k-IMk+5) ¥ & =I ¢x-td
Factors of 2(x—2) +9(x-2)-5 = (2x -5\(x +3)
=[2(x—2)~H[(r 2) +5] 7
| =(2x-S)(x4+3) v
, B)
| 1.2.2 Gn—n)36x" +49nyr —49my"
= (mn —ny36x7 — 49)" (m1 -n) ‘x
=(m—n)(36x" 499") a
: =(m—1)(6x+ Ty)(6x-7¥) v
@)
13
: xo-x-12 v
yeljx? -13x-12
i - x?-13x
-xi- x
-12x-12
-12x - 12
2 f(s) = (et DOP =x -12)= (x4 De 4043)
14d ( 1 1 [: ‘|
soy Flat
xy yx
yx
vy
x+y
xy
OF NO),
xy
yr-x
xy
r =
xy
(x+y)
v
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MARKING GUIDELINE ade
MATHEMATICS N3
12 4x 2x SX
3x¢1 1-3x 9x’ -1
4x 2x 5x
= ! + v ¥
“3x4l 3x-1. Gx+)Gx-1)
— 4xGe-1)-2xGx+D45x7 0 y
Gx+DGx-D
_ Lax? ~4x— 6x" - 2x +5x° Y
Gx+NGx-1
Lx? - 6x
* Gx4DGx—-1)
QUESTION 2
2.1 2.1.1 Px+410-x=2
v3x+10=(«42)) Y
o3x41l0sx 44x44
ow +x-6=0 v
T930(H)(M3 7
(x+3}(x-2)=0
exb3=0 of 6-2=0 | ; ao 0”
x=-3 or x=2 g wo?
ae)
X #-3 (1.a) extraneous solution “
x=2 is the only solution
any clear rejection of the root is acceptable.
2.1.2 3 x43 . 2
ataet5= : mc ae xt} _ =
A Btx 4 Sxex43 =e -
wx? +4x=0 y a ate
wex(xt4)=0 v a
x=0 or xt4=0 e mm ~ on
an x#0 7 x=4 VY oe . .
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Ns
nin
(3)
MARKING GUIDELINE “5 T930(EMM31)T
MATHEMATICS N3
Ae
2.2 vy +8x-Sa=0 ~ my
wx 48x = Sa CX “OS Wa iS
Lx £8x416=Sa+16 % Ne ean
J (xt4y =5a416 \ wv va
- a, ~
wxtdat/Satl6 ’ fu aye ) C
oxe-diJ5a+16 4% < oe" Ss (3)
Vv cf
2.3 Babe ay
Jub se v
ort o Inf, -In¥,
a r
a a (4
2.4
AGE NOW AGE IN 12 YEARS
John x (x+12)
Thomas 4x (4x+12)
age of Thomas=2 x age of John
4x+12 =2(x +12) v
4x+12=2x+24
2x =12 v
x=6
present ages = 6 & 24 yeas VW . : GB
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T930(E)(M31)T
MARKING GUIDELINE -6-
MATHEMATICS N3
2.5 Vey sie (1)
YH OES coerce (2)
replace y by x+5in (1)
x4 (x45) =25
vex 410xt25=25 Vv
2x? +10x=0
2x(x+5) =0
r=0% or x=-5 i. ¥
If x=0 then y=5
orifx=-5 theny=0 ”
QUESTION 3
cm 3.1.1 ae ve
(-2;1)
3.1.2 ; aA
45°
3b m= 2 v
32 Bed oP =V(-3" +P wee . aa We
=/9+44 =J13 units due er set \
pw
ag et ee
3.2.2 e4yer’ twee o
attr s(/By 7 vi
arey sl 2
3.2.3 Mop *Myy =—) (OP 1 AB)
2
ay mae =-] a
©. May = ¥ - uel
AB 2 oe . _—_
yr y, =m(x-%) x td
3
wyr-2=—(x+3) Vv
y 5 )
a 3? .
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13
. 3 13
. “Rosa 5 v K
G)
[21]
MARKING GUIDELINE “1 T930(B)(M31)T
MATHEMATICS N3
3.3 -
[8x -3y4+9=0 v . -
y ue ft (ec 0)
p= 6x43 “4 -
“Required line: p=6x 9) / . cee .
: f
G
[16
QUESTION 4
41 4.1.1 pelea?
oath hh
40 A
— d(x? +2xh +h ?)-1 44x?
= lig ——
had A
_ 14x? ~8xh—4h? 14402
= lim
hod hh
_ -8xh— 4h?
= lin, ——- v
ho0 fh
hat h
=-8x a Vv (4)
41.2 .
if x= then ® gq) = 8 Y
he a
4.2 2 oy
fi -Vx-=— >
x
1 v \y -
of) sx? -14+2x! v ce ay
wf) aan? 2x7 - oa . .
2 a
: 1 2 vy
oe X)= Re - T v Vv .
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TO3O(E)(MB INT
fe!
MARKING GUIDELINE 8.
MATHEMATICS N3
43
ms “
— flare i
i . : ' H
if /
| a
aN yy 16“ x
a
-12 _.
MARK ALLOCATION ___ | MARKS
x iniercents at (16;0) and (- 16.0 50) 1 ee
y intercepts at (0:12) and (0:-12) i rn
43.2
or j-x#90
or vel N
a A
0,0) or Gt} i
Iw I = 0
-x?(3-2x) =0
- y=Qorv=Oor3-2x=0
x =0 ~ ye
y-intercept y=0
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MARKING GUIDELINE ~9-
MATHEMATICS N3
Y
a
3
v
v
2
1)
1
a
NN “ ‘
2 \
0]
— oy 1 — : a , —e
-2 15 1 -0.5 0 05 { uy
(0:5) \
: v
:
2
& points)
“PO3OCEM3 LYE
1 mark for tursing point(0;0)/ I
| y-imtercept al O/x-interceptatQ oe
. 3 \
x-intercepts at 5 and 0
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(4)
MARKING GUIDELINE -10-
MATHEMATICS N3
T930(E)(M31)T
4.33
¥ 7
3
v
“ed 6 a
| ALLOCATION OF MARKS MARKS ee
L mark for x-intercept (439) i
Limark for y-iniercept ai (0,3) en se ee
QUESTION 3
3d sin® 120°xsec? 150° + sin 150°
20» eos BAP
| v
2
v
5.2 2s8in @ cos@ —sin @ =0
v.sin@Qcos@-l=0
“sind =OQor v - 2cos@-1=0 “4
8 =0° or 180" or 360° 0=60° of300 YY
v vv
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(2)
le
[7]
3)
MARKING GUIDELINE “lie T930(E\(M31)T
MATHEMATICS N3
5.3 C
Tan6832=2- Tan48,55=—~~
100 100 .
x= 251 544m y=113,228m v
.. Distance between objects = 251,544 m— 113,228 m
; = 138,316m v
i 1 100 a
| ortan 41,45° = 100 y=
x ian 21,65
i . xa 190 _ = 251,929 0
} fan 41, 45° we “Distance= yx
, = 113,228 = 133,701 (5)
5.4
aA veer + O2y —2V8N2Cos0
. 7
, «ff +2 wf
' v
= V6 Ol 2,449 we Q)
542 a
: Area =~ beSind
2
bog Se .
see V2SOUS ¥
=1,732 mi? v ()
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MARKING GUIDELINE -12- T930(E)(M3 1)T
MATHEMATICS N3
$5 3y=sin30
_ sin3@
3
|
nineteen
aed ne ca. 9ae | { i ra
| mark for y-intercept at OP i oo T
TOTAL: 100
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aa% yp higher education
& training |
_. Department ——-——. - --------—-
Higher Education and Training:
REPUBLIC OF SOUTH AFRICA
T930(E)(I28)T
AUGUST EXAMINATIO
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(16030143) 2 T930(E)I28)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N3
- 2 cee ve. TIMED. 3-HOURS -—--- - --
MARKS: 100
INSTRUCTIONS AND INFORMATION
1. Answer ALL the questions.
NO
Show ALL the calculations and intermediary stey
separated,
4. ALL final answers must be accuratel,
5. All graph work must be done i
‘question paper. The list is NOT necessarily
a may’ be used.
6. A formula sheet is attaclied tot
complete. Any other applicabie for
Write neatly and-Jegi
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(16036143) “3. ‘7930(2)028)T
QUESTION 1
ll Simplify the following WITHOUT using a calculator: (5)
Qt x 3 gt
1.2 Prove that (log, a) (Jog, c)(log, b) =! 6)
1.3 Determine the factors of the following fimetion if x ~1 is one of the factor:
fsx) 43x -x-3 (4)
14 Factorise the following expressions as far as possible in prj
14.1 x (x-1)+(-x) GB)
14.2 @ +2a—3+ab—b G3)
1.5 Simplify the following:
ab @ ~2ab+
Qa-b 4a? -B2 4
[22]
QHESTION 2
G)
3)
“x4 “Y -1 (4)
2.3 Make ‘1‘ the subject of the formula :
P=100e°" i)
24 Make ‘r? the subject of the formula:
a=nr'+,420Wnloaded from hlayiso.com 0)
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NN
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WS
(16030143) “4.
T930(CE\U28)T
2.5 Calculate the coordinates of the points of intersection of the graphs defined by the
following two equations:
y= 2x? -8x-10 (4)
p= 9x
2.6 The product of two consecutive odd numbers is 143. Calculate the numbers. (3)
[26]
QUESTION 3
3.1 AdBC hias vertices A(2;3), B(-2;-1) and C(4;)).
3.1.1 Draw AABC ona set of axes. qd)
3.1.2 Determine the gradient of AB. ()
()
2)
@)
@)
G)
3.2 iH COLUMN B that matches an item in COLUMN A. Write
xt to the question tumber(3.2.1-3.2.4) in the ANSWER
COLUMN B
y A x+y =10
B =+%=100
1 10 Q)
lo 7 *
2 2
~ 47 2100
1 10
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‘F930(E)28)T
Lo
Ne
bo
D x=y
y
Copyright reserved
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(16030143) 6- T930(E)(J28)T
QUESTION 4
Determine o by making use of the rules of differentiation. Write your answer with
positive exponents and in surd form when applicable.
4LL 2 :
. = ae Lt
* x @)
4.1.2 x8 —1
arn Q)
42 Determine the gradient of the tangent to the following 4 Ren neu :
ya (x 341) @)
» ) AS :
$.1 Make use of basic trigonometric identifiés to prave thas,
I+sin@ + cos? é :
QUESTION 5
=2sec0
cos@ 1 +sin8 ae (5)
5.2
Q)
A
259
80
46
B oo Cc .
5.3 Solve the following equation if 0° < B <360°:
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MARKING GUIDELINE -2- T1020(E)(528)T
MATHEMATICS N3
QUESTION 1
; MARKING
: INSTRUCTION &NOTES
i Ll gett y agai ¥ writing 8 as power with
| — v base 2 .
a wo ¥ writing 4 as power wit
28x22? 2 base 2
! ape v ¥ simplifying root
2x2 Y applying multiplication
| v rule of powers by adding
' aq DoHGH-2at2 74 16 exponents
¥ ¥ simplifying ()
i
| 12 LHS =(log, a)(log,c)(log,b) RUS =1 ¥ ¥ applying log rule where
\ 1 logb bases changed
= OBE OBEY 108”, ¥ ¥v ¥ simplifying
loge logb loga
=1=RHS v
i G3)
i
; 1.3 AK 43 v v¥ for correct quotient
x= xP 43x? - 2-3 ¥~ two factors-
2 v (x4+3)(x+1)
3 1 2
wie no credit to (x-1) it
; t4x?-x was given
| 4x? - 4x
+3x-3
3x-3
0-0 v v
Factors : (x-1)(x+3)(x+1) i)
L4 LAA 2 (e-D+(1-x) ¥ taking out -1 as common
2 v factor
ax(x-1)-(x-D v Y factorise
=(x-DOr-1) Y applying factorisation of
= (x-D(e-D(e+1) v difference between two
squares (3)
1.4.2 a +-2a—3+ab—b . Y factorising of quadratic
° . trinomial
_ 2
=(@ +2a—3)+ab—b “taking cut of common
=(a+3)(a—1)+b(a-)) v factor
=(a-Dla+3+d] Yoev ¥ taking out of common
factor Q)
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MARKING GUIDELINE -3-
MATHEMATICS N3
15 a’? a? —2ab+b' a-b
Qa-~b da -B a tb
v v ¥
_ (a+bXa~b) | Qatb\Qa-b) a~b
2a~b {a-bya-—b) atb
=2at+b FT
v
QUESTION 2
21 da’ ~12a-7=0
4a -(2a=7%
a—3a2e
4
3V. 7,9
a ~3a+|-=|)=—4+2
( [-3)) a4 Sf
(«-2) ~16_4 /
2) 4
3
aq~—~=42
5 f
a=242
2
7 1
Qa=—- Of a=~ >
2 2
2.2.1 yt
araars( 2 =42
5 v
22 +2727 +27 =42
B[P+reil=42 v
2* [5,25] =42 v
3 v
2% =2
x=3
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TLO20(E)323)T
¥ change of division sign to
multiplication sign
Y (Vx2) factorisation using
difference between two
squares
Y factorisation of trinomial
simplifying
4)
[22]
“taking out 4 as common
factor
¥adding the
. 2
oe: of :) both sides
v writing [hs as a square and
simplification of RHS
¥ get rid of squares by drawing
square roots both sides-
Mark will not be awarded if
tis left out.
¥ (x2) two answers
(5)
Y simplifying of negative
power
¥ taking out common factor
¥ simplification of power
¥ solving by equating like
powers
4)
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(16030143)
MATHEMATICS N3
FORMULA SHEET
Any applicable formula may also be used.
1, Facters/Faktore
2, Logarithis/ Logaritmes
T9300E)(J28)T
a -8 =(a-D(@+ab+b’)
a+b = (at bya -ab+ 8?)
log ab=log a+ log b
log slog a ~ log b,
3. Quadratic formula/
Kwadratiese formule
log, a
log, a=
log
r ~bt yb? ~ 4ac
| 2a
4. Parabola/ Parabool
| 2
! voar* +hxr+e
_ dae — b°
da
5. Circle/ Sirkel
ght line/ Reguitlyn
2 2 2
[ x + po =P gs,
Perpendicular:
Loodreg: m,-imy= ~]
Parallel lines:
Ewewydige lyne: my =m
Distance:
Afstand: = D= (ez ~ xy)? +@ - y,)?
Midpoint:
Middelpunt: P = (72 : arm
2 2
Angle of inclination:
Hellingshoek: 6=tan/m -
°
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(16030143) 2 T930(E)(I28)T
7, Differentiation/ Differensiasie
dh _ lim fern) - se)
dx hoo h
Max/Min
Maks/Min
For turning points:
Vir draaipunte: f ‘@)= 0
8. Trigonometry/ Trigonometrie
el
sing=%=—+ 7
rk cosecO i
COSO = = a :
vr sec
1 7
tan@ == = —~— i
x coid i
sin?@ + cos’6 =1
D+tan?6 = see?
ind _sinB _ sinC
a .b° ¢
Pap pe? 2be cosA
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' MARKING GUIDELINE “4.
MATHEMATICS N3
2.2.2 2x _ 4 4-x
x-1 xt) x1
| 2xx+)—-4(x-1) 4x
i (z-DY@t) — («-Darh
; 2x? 42x—-4e+4a4~x
2x*-x=0
x(2x-D=0 y
X=U or ral
‘ 2
; yf v
; 2.3 P=[00e°°"
i Pam Vv
100
In a =In(e*™)
| 100
In P—1In 100 = —0, 32Ine v
i In P—1n.100
i oo = OF
, ~0,3 v
In f00—1n P
on SI OF
0,3
| (2)
' AP lL, v
0,3
24 az ar +7rs
2 v
ar’ +ars-a=0
pe estas’ +40 v
20
25 y= 2x? -~8x-10
y=-9x
> 2x" -8x—10=—9x
2x? +x-10=0
(2x4+5)(x-2) $0
xa or xa]
2
v ¥
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T1020(E)28)T
¥ correct LCD(denominator)
Y rewrite fractions to
LCM(numerator)
¥ simplification
¥ (¥x2)-solution
Check the roots
@)
¥ manipulation
Yusing logarithms to manipulate
¥ / simplify
co)
¥ using correct values in the
place of a,b,c in the quadratic
formula
¥ manipulation and simplification
2)
¥ v finding the x-coordinates
vv finding the y-coordinates of
points of intersection
4)
Please turn over
MARKING GUIDELINE -5-
2.6
MATHEMATICS N3
5 nol
-2;22—Jand (2; -18) vo ov
(3322 )and (2; -18)
XX(x+2)=143 -
x’ +2x~-143=0 Vv
(«+ 19G-1)=0
x= -l3 x=I]l
vw vw
Two numbers: 1] and 13
Or -Ll and -13
QUESTION 3
34
3.14
T1020(E)(328)T
Yany steps to show how the
student determined the answers
vv answers
Students could determine answers
also using inspection.
3.1.2 ~
May = +s ts
X,-X,
_3-CD
2-(2)
4
= t=]
4 v
3.13 My, =
a=tan'(1
an” (1) y .
=45°
(3)
[26]
¥ sketch of
triangle
(hy
¥ substituting values and
simplifying
q)
“substituting value and
simplifying
(1)
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°
Please turn over
e >
MARKING GUIDELINE -6-
MATHEMATICS N3
3.1.4 M.= ¥,-¥,
eX, - Xe
MyXM,. =Mx(-D=~
Therefore the two lines are perpendicular.
3.1.5 Parallel lines have equal gradients.
The gradient of the line through C(451)
and parallel to AB with M,, =1 y
Y-Yo =M(X- Xe)
y-l=MGa-4) “
ysx-4t+l
yox-3
2 2
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TLO20(E)028)T
finding the gradient of
line AC
¥ multiplication of gradients
to show that lines are
perpendicular
(2)
Yusing identity on parallel
lines to find gradient of line
AB
¥ finding y-intercept by
substituting
equation
(3)
¥ substituting
“simplify
Y substituting
¥ simplify
4)
Please turn over
MARKING GUIDELINE -7-
MATHEMATICS N3
SAT Dag =p) +04 9)
i
v¥2-(-2)" +3-C by
= V¥16+16 v
= 32 Y
Dog = (%4~ Hn) + (Y4-e)
= ¥B-)?+@2-07
v
= V4+4 v
= B
AB=V32=J4x8=2V8 v
AB =2DE
3.2 3.2.1 G v v
3.2.2 B
wv vw
3.2.3 D v v
3.2.4 I
v v
QUESTION 4
41 4.1.1 2
ya2vx-2
x
t
=2x2-at ¥ v
j voev
YL x 242x7
° dx .
-L,2
vx x
v v
T1020(E)(528)T
Yusing the distance formula
and substitute
¥ simplify
“using the distance formula
and substitute
v simplify
“rewriting surds to prove
that AB=2DE
¥ ¥ Choosing the correct
equation to describe the
gtaph
¥ ¥ Choosing the correct
equation to describe the
graph
¥ “Choosing the correct
equation to describe the
graph
“¥ Choosing the correct
equation to describe the
graph
Vrewriting square root
rewriting a fraction
¥ differentiate function
V differentiate function
rewriting function
rewriting function
G)
(2)
@)
[25]
(4)
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MARKING GUIDELINE -8- T1020(E)(J28)T
MATHEMATICS N3
4.12 xt] Y (xv )simplify fraction
' eae ¥ differertiation
(xt -1y(x +1
i x +1
i ys x'-1 V
: dy 3
| —i=4y
dx v Q)
42 y= (x? —3x+D ¥ multiplication
=p4p—3e-3 V differentiation
POR EH Od v “substitution
i f ()=3x +2x-3 Y
FQ) =3(2Y +2(2)-3
=3(2) 44-3
| 213 ¥ (3)
[9]
|
i
|
| QUESTION 5
i
3.1 l+sin@ cos? ¥LCD
| cos + ising 2secd v 2 rting fractions to
LHS = (ysind)(l a A+ cos “cos RHS =2sec@ “applying square identity
cos @(1+sin A) y cos? x=1—sin? x
“simplifying
_L+2sin@+sin?@+(I-sin’?@) ~ Vindicating that ths = chs
cos(1-++sin 8)
_ _2+2sin@
cos@(l+sind) *
° _ 2(i+sind) . .
cos A(1+sin A)
_ 2
cos?
=2secO
LHS = RHS v . ° (5)
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MARKING GUIDELINE -9-
5.2
5.3
5.4
5.5
Copyright reserved
MATHEMATICS N3
“ad? = 46° +80? -2x46%80%Cos25
a’ =2116+ 6400-2 46x80x0,906
a? =1845,575 . v
a= 42,96 units Y
x
2SecH+4=0
sec B=~-2
reference angle = 60°
.B=120° and B=240°
v v
¥
cD
40
CD#=14,558 yy
tan 24° =f?
40
AD = 17,809
AD~-CD = AC(lenght of antenna)
17,809 -14,558= AC
tan 20° =
AC =3,250m
v
A=(45°:2) v
B=(225°:2)
v
P=(180°;-D v
T1020(E)(J28)T
“substituting using cosine
tule
¥ simplifying
¥ simplifying
QB)
¥ manipulating
Vv finding solutions
(3)
“ trigonometric equation
Vlength CD
Yilength AD
“length AC
(4)
¥(Vx2)
¥ (v¥x2)
¥ (4x2)
(3)
[18]
TOTAL: 100
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
TLOZO(E\AS)T
APRIL EXAMINATION
NATIONAL CERTIFICATE
MATHEMATICS N3
(16030143)
3 April 2013 (X-Paper)
09:00-12:00
Non-progranmuable and non-graphical calculators may be used
This question paper cousists of 6 pages and a 2-page formula sheet. cats
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“26 T1O7OCEMAS)T
QUESTION 1
Ll Simplify the following WITHOUT the use of a calculator:
Lil (7 +32) : (2)
Lt2 log, x x log, 8 (3)
113 24-28 +f54 3
oe (3)
2 Factorise as far as possible in prime factors:
124 5x*8 ~ 807 @)
1.2.2 2’ (p~q)4 3xlq~ p)~18(p~ y) By
SQx +1) -i7(e +46 @)
i Use the remainder theorem to determine the remainder when «+ is divided jnto the
followin
La
3)
phe 4)
(25]
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4
4
oO
QUESTION 2
ne
we
Solve for x in each of the following:
2 Poe 5
211 a
a -3
“3 =a
a
2.4.2
2.1.3
Make “d" the subject of the formula:
8 = Rasen D2]
Make "£" ithe subject of the formula:
The sum of thres consecutive uatoral numbers
TIO7OCE)\CAS)T
is 33. Detennine the smallest of the
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al
@)
@)
{#8]
~5- T1070(E)(A5)T
QUESTION 3
31 In the accompanying sketch, the graph of y=: -~x? + 3x +10 is represented,
DE is parallel to the x-axis
3.1.1 Calculate the length of AB Q)
3.1.2 Calculate the value of the x -coordinate of the turning point at C (2)
3.1.3 Calculate the length of DE (2)
iv i
oe oe ao 4B
f :
/ H
/ |
i \ !
32 Sketch the graphs of the following equations in the ANSWER BOOK.
h graph must be drawn on its own sysiem of axes. ALL vahies at the poinis of
. : te 2 |
intersection with the system of axes must be shown. Name the type of graph below
each sketch,
SA
6)
3.2.9
ae @)
3.2.3 ol
—_ @)
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PTO
~6- TLOTOQR) (AST
/ 3.3 In the diagram, the circle with the origin 0 as centce cuts the straight Hne I at the,
points A(0;5) aud C. The point M(2;4) Is the midpoint of AC and @ is the size of the
j acuie angle that J makes with the x-axis,
BL (2B
i
‘ 332 JJsterimins the coordinates of CO (2)
i Prove, by using analytical methods, that OM LAC (23
|
Saloulate the sie af O correct to ave decimal place (2)
ME SGG Glefgtata, ALNY i, A tg Ue Poult (324) end 1313 tae potat L2),
atermine the following:
3.41 The co-crd
The length of AD in surd form 1)
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PTO
-T- T1070(E)(A3)T
QUESTION 4
44 Determine /'(x) in each of the following by using the rules of differentiation. Leave
the answer(s) with positive exponents and in sud form.
Z a
401 f(s)=(- 2x2) Q)
4.1.2 i
f(xje-tve
$2 Deienmins the gradient of the tangent to the following curve at the point whers x= 3:
43 The diagram below represents the gfaph of:
yaa Gx? 49%
Determine ihe co-ordinates of the local turning points A and B.
B
oN 4
/ \
/ \ i
/ \ t
/
/
awa
a errr eran aS A WTR REE nope Nn soak saad Wh ew
0 A
| (4)
[10]
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Pra
TLOTO(EY(ASYT
QUESTION §
54
Calculate the exact value of the following WITHOUT the use of a calculator:
cot? 30° + cosec? ag? ~tan 4s?
(3)
5.2 Simplify the following:
i cos(180° ~ x)tan(360°~ x)
sin(goe.- x}tanflgg°+ 80° + (3)
| 5.3 Make use of basic trigonometric identities ta prove the following:
: I . i
/ Irian? x lec @)
i satisfy the following irigcnometrin equation for;
2M ys QB)
33 Two towers, CD and X ae separated by a distance DY = . The angle of
i elevation from Cte X is 60° and the angle of depression fom C: “ va ig 35°
: ~
{ Pont
i a
Determine the following:
5.5.4 Ths height of the tower CD (2)
55.2 The heigl ‘oad tow : (3)
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PrO
5.6
9. TLO7O(EYAS)T
The following graph represents the function f and g for xe[0°s190°] where
J (x) =acosx and g(x) =sinw
Find the values of a and ¢
TOTAL:
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(2)
(19
400
7
i
QUESTION i
Il Ld (432)
12)
1
2
. © 34 ne))
. 112 flog, aJ® + in ve
= ay? +5hn g vf
f
. me
/ aes i 3)
13
Ri)
j P
¢ (3)
\
; v
Y 3)
Copyright reserved Please turn over
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Fisuiyesil:
MATHEMATICS Fy
1.2 E21 5x8 ~ 80)?
=5(x% ~16y") y
=5(x" dye ~4y) v
12.2 x3 (p~q)+3x(g— p)-18(p ~ 9) J
=x"(p~ 9)- 3x(p 4) 18(p ~q)
=(p ~q){x? -3x~ 18) v
=(p-a)=- Oe +3) y
123 s(x 1) 17+ +6
let(x+i)=k
Sk -17E+6 V
(5k —2Xk ~3)
v
a
.
13
H4+i443
=? y
La idl
I
V
Copyright reserved Please turn over
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G)
(2)
G)
PIVOGS 8 Sab
1.4.2
45 ox a 13 ’
| x-3 x42 x? -~7-6
aif 5 _ 9x +13
i x~3 x42 (x-3hx+2)
; 14x +2) ~ 5(x-3)~ (02 413) v
= eS) (Pa 4 13)
(x-3)e+2)
| _ 14x + 28~Sx415—-9x 13
_ 30
130}
QUESTION 2
21 QA i
But =6
'
i whey
!
@
j 21.2
i
(
243
(3)
Copyright reserved Please turn over
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Vxe5=x-1
xe5e(e-i)?
x+55x?-2x4]
x? -3x-4=0
(x -4Xx+41)=0
wxed
xe-l
2.2 P=PR+VT
UR+Vi-P=0
. pa eye’ ~4ac
— 2a
2.3
Asta; Ped
Smallest no.=10
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PPogR as
@)
n=?
os
tO
(3)
(20)
Please turn over
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Ro ee Ce
VUES A SIH
QUESTION 3
31 3d
@)
3.4.2 ~b
+=—
2a
-3
yeot
-2
veSN esis
! 33
! Q}
| 32 3.24
j
Q
Copyright reserved Please turn over
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ve
=
iv
3.43
OULESTION 4
44 411
Copyright reserved
B=G3)
AD =(-3-1)' +42)
AD=4f(-4) +5?
AD = VAI Units
yy, =m(z~x,)
3-(-2)=mn{-3-1)
Elicit vie
ie)
@)
[28]
(2)
Please turn over
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QUESTION 3
* eee
Sneceme
=3+4-1
=6
5.2 00s x*— tan x
cos x4 tanx
=1
3.3 t I
=]
rt 2
see” x cosec
cos? x+sin?x=1
“LHS = RAS
54 tanx=-l
xetant~}
we 45°
Copyright reserved
Q)
(3)
13]
TOTAL: 106
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