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QUESTIONS AND SOLUTIONS
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Spot On Mathematics study guide, 254 pages. Read online or get the free PDF download.
Subject: MathematicsMultiple grades254 pages
Mathematics. Study guide, 254 pages. Read online or download the PDF.
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
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(16030175) -2- T10S0(E)(M30)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N5
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1. Answer ALL the questions.
2. Read ALL the questions car
3. Number the answers accordii
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(16030175) -3- TLOSO¢E)(M30)T
QUESTION 1
11 11.1 dx-3
Determine In y if In y=lim
ny if iny=lm— (2)
1.1.2 Now, determine the value of y qd)
: 1x)?
1.2 Determine lim (=*)
xt x* —x—xInx 4)
[7]
QUESTION 2
2.1 Given: y -_!
4x?
: so. ay
Determine the derivative —, (4)
2.2
2)
(4)
(2)
| 2
2.3 Determine ed with the aid lic differentiation if y= yx ~ltandx
dx Inx (4)
2.4 Given: y = arctan x
24.1 Make a neai, rough sketch of the graph of y for the range i 5 4 @)
2.4.2 Derive a formula to determine » if y = arctan x
dx Q)
2.5 Given: the implicit function e**” -— tL 2y?
x
Calculate the derivative a F
dx (4)
[25}
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(16030175) ~4- T1050(E})(M30)T
QUESTION 3
3.1 Given: f(x)=x° +x? -2x-1
3.1.1 Determine the coordinates of the point of inflection of f(x). (2)
3.1.2... Draw up a table of values for x and f(x) with x ranging from x = 1,1 to
x = 1,4 using intervals of 0,1. Round answers off to ONE decimal.
Draw a neat graph of ffx) between these values. (4)
x=1,23.
Use Taylor/Newtdii's method to determine a better approximation of this
(4)
3.2
4 should be as
6)
3.3 from a fixed
Obtain an exp ‘or the velocity of the o! after ¢ seconds and then calculate
the values of ¢ Scbject is at rest, 4)
[19]
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(16030175) -5- T1050(E)(M30)T
QUESTION 4
4.1 Determine [Se by resolving the integral into partial fractions. (5)
x° -2x-3
42____ De Determine [ _ydx in each of the following cases:
42.1 See Ses,
(3)
4.2.2 howe dx (3)
42.3. [sin 3nx.cos 8m )
(5)
43
(4)
[22]
QUESTION 5
5-1 Given:
5.1.1 Za enclosed between the two curves, the
show the representative strip and the
(3)
5.1.2 itlide of the enclosed area. A)
5.1.3 Calculate the volume generated when this area rotates about the X-axis. (4)
1
1
5.2 Calculate the value of
J V6 —9x? (4)
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(16030175) -6- T1050(E)(M30)T
5.3 Determine the second moment of mass of a rectangular lamina of mass (7m) about the
axis parallel to one side of the lamina. The distance from the representative strip to
the axis is x. The lamina’s length and breadth is a and 6 respectively.
Y a
(a;b)
QUESTION 6
6.1
TOTAL:
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(4)
[19]
GB)
(3)
[8]
100
(16030175) “1
MATHEMATICS NS5
FORMULA SHEET
Any applicable formula may also be used.
TRIGONOMETRY
cos*x = 1 — sin’x
sin? x + cos? x = <<
sin’x = 1 —cos’x
1 + tan’ x = sec” x ————>» tan?x = sec?x - 1
1 + cot” x = cosec” x ——-» cot?x = ¢
sin 24 =2 sin A cos A
cos 24 = cos’ - sin’A
cos? A
cos (4 + B) }
tan A+ tanB
+ B) = ———_—.
tan (42 B)= Tan Atan B
dv
cos 4 cos B=% [cos (A + B) + cos (4 - BY]
sin A sin B= % [cos (4 - B) - cos (A + B)]
sinx , 1
tanx = 3 Sina = : 5; COSx =
cosx
COSEC X secx
cos x 1
cot x =— 5, cos ecx = — ;secx =
sin x sin x cosSx
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BINOMIAL THEOREM
non - -) n(n-N(n—-2)_ n3,3
eth)" =x" txt 14 2 Wt +
(eth) = 2" +m a Taxa A
DIFFERENTIATION ___
F(@)
vrate
PRODUCT RULE
y = ux). v(x)
dy dy du
S=u,—t+v.—
dx dx dk
=u-v +eeH
CHAIN RULE
y =fue))
wv _ ey du
de du dk
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(16030175) 3+ T1050(E)(M30)T
d
Se) Zl [ f@ac
x” nx”? a +e (n#-1)
n+1
axtb
dete
a
In(ax)
Se)
al?
In fix)
sin ax
cos ax
tan ax
cot ax
Sec ax
cosec ax
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a sec’ ax _intseo(axy}+C
a
datsin(ax)]+C
a
- a cosec” ax
«tn [sec ax + tan ax) +C
a
e(s)
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a sec ax:tan ax
= a cosec ax:cot ax
(16030175) 4. T1050(E)(M30)T
fe) ai) fo
sin f(x) cos fixdef"(x) _
cos f(x) - sin fx)of Gx) _
tan fx) sec” fix)of'" (x) _
cot f(x) - cosec” f(a
sec fix)
cosec f(x)
sin” f(x)
tan" (G0)
cot? f(x) _
sec” f(x) _
cosec? f(x) _
fOWVLF@P -1
sin’(ax) a x BnGer) + ¢
cos*(ax) _ 7 + see +c
tan’(ax) __ 5 tan (ax) — x +€
cot?(ax) __ - cot{ax}—x+C
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(16030175) 5
INTEGRATION
J 00) g'(@) dx = fo) - J £1) g(a) dx
[U/@O"* p(a)dx = Yor". C (#-1)
J Fey Ba BIE+C
APPLICATIONS OF INTEGRATION
AREAS
4A,=[) yar; 4,=[) 0, -y2)de
SECOND MOMEN
=f? 2a-7 =f? 2
=f dA: 1y=[) rdA
MOMENTS OF INERTIA
Mass = density x volume
M= pv
DEFINITION: [=m
GENERAL: J=f r’dm= pf’ ray
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA... .
MARKING GUIDELINE [
NATIONAL CERTIFICATE
MATHEMATICS N5
APRIL 2017
This marking guideline consists of 10 pages.
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MARKING GUIDELINE -2- T1050(E)(A)T
MATHEMATICS NS
INSTRUCTIONS AND INFORMATION
1. ¥ equals 1 mark.
2. Half marks are not allocated, unless indicated otherwise.
3. Where formula is required, using the wrong formula is a principle error and NO
marks are allocated.
4. Candidates should show ALL formula, intermediate steps and simplify where
possible,
5, ALL final answers must be rounded off to THREE decimal places (unless indicated
otherwise).
6. Questions may be answered in any order, but subsections of questions must be kept
together. If subsections are separated, the candidate can be penalised by ONE mark.
7. Where a candidate copied incorrectly from the question paper, and the standard of the
question is still the same, the candidate will be penalised with ONE mark. If the
copying error simplifies the question and makes it easier, the candidate forfeits the
marks,
8. Questions must be answered in blue or black ink. Answers in PENCIL are not
marked as it is regarded as rough work.
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MARKING GUIDELINE 3
3.
MATHEMATICS N5
QUESTION 1
1.1 111 -
iny=timY 3
x9 x—9
=
jiim2¥2
9]
aly
6
LL2 yy =1,181
1.2 —x)?
tin =
xl x —x-xInx
: —2+2x 0
= lim —————_. —
x1 2x-1-Inx-1 vv ry
“aT
x
=2 v
QUESTION 2
21 ye 1
V4 x?
1 1
Worthy drt
P0)= fi +H 2x
vx -2xh-W 1 yy
= lim .,—
b0 Dx" (x+h)? h
h(-2x-h) 1
slim ty
40 2x"(xt+hy kh
x
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T1OSOCEMA)T
Q)
qd)
(4)
[7]
4)
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T1050(E)(A)T
MARKING GUIDELINE 4.
MATHEMATICS N35
2.2 2.2.1 1,
y=arcsin(e? +3)
1d
2 v
wy 3
dx loo .
1-2? y x
2a (2)
22 2.2.2 3x °
= a
tan’ (--5
G ) Vo
4 4-4
ay 3.tan’(—5)—3x3tan’(—5).— >
de (tan(A-5)? ¥ :
x (4)
2.2.3 ys gieos’ x)
y=Cos' x y
De 2cosx.-sinx ~
dx 2)
23 _ vx? -1tan4x
» Inx
Iny = | In(x? 1) + Intan 4x—In(n x)
ld 2& 4 Asec? 4x _
1
ydz 2G?) tan dx a v¥v¥ ONE mark per term
Wy *_,Asectde 1)
dx x1 tandx xinx (4)
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MARKING GUIDELINE -5- T1050(E)(A)T
MATHEMATICS N5
2.4 2.4.1
¥- coordinates
~~ asymtotes
v- shape.
(3)
2.4.2 y =arctan x
tan y=x
sec? yal v
dx
dy 1
dx 1+tan? y v
dy 1
de 1+x
(2)
2:5
sec y
e ~1uny
x
v v v
sec dl dy
e” sec y tan y+ = 4 —
y Vox 2 Ve
dy _& secytany—4y
ax 1
x (4)
[25]
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MARKING GUIDELINE 6: TIOSO(EYAYT
MATHEMATICS N5
QUESTION 3
310 BA ptgyex? ex? 2x]
yl=3x? +2x-2
6x+2=0 %
x=-0,333
~ for both x and y
=-0,26 Q)
¥- values
v- shape
(4)
3.1.3 xi +x7-2x-1=0
-— 0,086
=-——_ = 0,0172 vvv .
4,9987 for approximate value
v
r=1,23 + 0,0172 =1,247 4
3.2 Vex2h
na? vy
x
A=x’+4xh V
A=x? +4(205)x7
27-820 _ 9 v
x
x= 410
x=7,429 ¥
h=3,714 ¥ G)
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MARKING GUIDELINE -7-
MATHEMATICS N5
33 xeit a1 8-1
vals? —24-8 ¥
(@-4)(t+2)=0 v
t=4¢=-2 V
NA ¥
QUESTION 4
4.1 j x-8 ,
x? -2x-3
x-8 _ A + B
(x-3)(@+1) x-3 x41
x-8= A(x+1)+ B(x-3)
x=-tn2=B v
4
eg ed ¥
4
oe
4,4 4
x-3 x41
v ov
5 9
=——In(x~3)+—In(x + D+e
4 (x ~ 3) 4 (x+)
42 42.1 joa
us 8
2
3
dea dy V
1
footu.3du v
=3In(sinvx)+e ¥
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(4)
[19]
6)
@)
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MARKING GUIDELINE -8-
MATHEMATICS NS
4.2.2 fe 62" dx
u=-2x'
de =: di
—8x
je ilies ae] ”
= Jota v
-1 6"
8° Iné
4.2.3 f sin3zx.cos8ax dx
+¢ V
: f sin] 1+ sin(—Sax)dx
v v
1,—cosllax cos(—Sax)
= —[——___ - te
=e 5x
4.2.4 2
vv quotient/remainder
4
foe 424
9 — 2)
yov ~ ONE mark per term
wo2 4
=—+=x+—InGx-2)+e
3.2 9 27
430 f£@8' = f@a@ - [fed
x" Ina’dx
v v
roe —f 78 ee
=5x fasta v
6
35x cv
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3)
Q)
6)
(4)
[22]
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MARKING GUIDELINE -9-
MATHEMATICS N5
QUESTION 5
5 5A 1
T10S0(E)(A)T
v - shading and strip
(3)
5.1.2 1s
Je —x'dx Vv-ONE mark for limits/ONE mark for equations
0
4
= le" a .s v
=3,216-1
=2,216 ¥ (4)
513 Ls
V=nfe*—x%dx ¥
0
e x1,
v =a =v
al > 7 a
V = a{7,602—-0,5] ¥
= 12v
V = 22,312 (4)
5.2 1 1
9 ¥16 - 9x?
If 1
=f dk v
34 oy _
3
al +t 3X. 1 vo
= 3 sin ¢ 4 Mo
=; [0.848 -0]) ¥
= v
= 0,283 (4)
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~-10-
MARKING GUIDELINE
MATHEMATICS N5
4.3
QUESTION 6
6.1 a _ 2x? -1
de 10”
10” dy = 2x? ~
62 oa —2e* +
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TOTAL:
T1050(E)(A)T
(4)
[19]
QB)
qd ee 4
o£ "2 7p tansx—L2x+c¢ v~ penalise with ONE mark for every wrong term
2 YY penalise with ONE mark for every wrong term
3)
18}
100
7 her education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
pana
22, November 2016 (X-Papéx)
ys 09:00—12:00 ay
/ Scientific caleutators may be used.
This question: paper consists Gh6 pages und a formula sheet of 5 pages.
ane
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(16030175) -2- T890(E)(N22)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS NS
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND roy 108 5
1. Answer ALL the question
2. Read ALL the questions Psa
3, Number.thSainswers accohigg to the num
"Re,
4. Show
ed in this question paper.
5. ot final answers a rounded off to THREE Seong laces
on
6. Questibas may beslperel in any a but subset of questions must be kept
together> .N y al %
a “s at
7. Questions _* answered in blu or black ink.
i)
8. Write neatly and ieeibly.
“Sh. |
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(16030175) -3-
QUESTION 1
Ll
so Inx Inx GB)
112 jim * 72"-5
4300 x2 Oy (2)
ree =,
1.2 Determine whether f(x) =4% rs is continuous at x =—
y rl
«
QUESTION 2 |
2.1 Deteffine the derivative of Ke (x) =¥/2x ° fiom first rl (6)
é -
2.2
(4)
(3)
(3)
2.3 Given: A particle moves along a line such that displacement (in meters) in time
(t seconds) is given by
Determine the velocity at ¢ = * seconds with the aid of logarithmic differentiation. ‘6
2.4 Given: cos(x+ y)+sin x+sin y =2
Make use of implicit differentiation to calculate ——
dx (4)
[24]
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Determine the following limits:
1.11 lim-—-- +
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(16030175) -4- T890(E)(N22)T
QUESTION 3
3.1 Given: f(x) =3x° —4x? —2x+2
3.1.1 Determine the coordinates of the turning points of f(x). QB)
3.1.2 ..... Verify. by.using.a table that the equation 0 = 3x3 = 4x? ~2x+.2. has a root
between the points x =1 and x =2.
Use values on the table: ~1Sx52 @)
3.1.3 Hence, make an sketch of the graph of the function f(x). (3)
-_-
3.1.4 If the i of f(x)is estimated as 1,7, use Taylor's/Newton's
t
method to detégiine a better approximation of this root. (2)
3.2 A farmer Wnts to enclége a field rinse mat x meter and y meter
Th esst of fencing is R36,00/m for the length and R45, ni for the breadth. He has
ait atfount of R56 000,gppevailable. ‘
f os
BET Give the fons of the area and the cost i ’ of x and y. (2)
3.2.2 “tiga (5)
3.3
HINT: Y=ar'h (4)
[21]
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(16030175) -5- T890(E)(N22)T
QUESTION 4
41 Determi ef 1= cos2x
~2x+sin 2x
4.2 Determine J ydx in each of the following cases:
42.1 J 2x. dx
4.2.2 P= 4
4.23
424
\ a yA
dott” a?
43 Delerfnine f ydx by resolving the integrand into partial fractfons:
8 ‘ x
roy
44 Determine
QUESTION 5
5.1 Given: The curves X=—y’ +4 and x=4-2y
S11 Sketch the TWO graphs. Shade the area enclosed by the two graphs and
show the representative strip with dimensions x and dy.
5.1.2 Calculate the magnitude of the enclosed area towards the Y-axis.
5.1.3 Calculate the volume generated when this area rotates about the Y-axis.
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@)
3)
G)
G)
(3)
6)
(2)
[23]
(3)
4)
(4)
(16030175) -6- T890(E)(N22)T
5.2 Determine the second moment of mass of a rectangular lamina of mass (m) about the
axis parallel to one side of the lamina, which bisects the lamina.
The distance from the representative strip to the axis is x, The lamina’s length and
breadth is a and b respectively.
AY
Pe a »X
PY) (4)
Wy [15]
QUESTION 6
oA
(:
i= a&
6.1 Solvg the differential equation e“*” a
_ Po ly (4)
“ard ee
6.2 Detetmine the genefal Solution of the differential ¢
Se , =
4 Do,
d’y \ “s
ae % (6)
i [10]
¢ id
TOTAL: 100
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(16030175) -L
MATHEMATICS N5
FORMULA SHEET
Any applicable formula may also be used.
TRIGONOMETRY
sin? x + cos’?x=1
1 + tan” x= sec" x
1+ cot’ x = cosec” x
sin 2A =2 sin A cos A
cos 24 = cos"A - sind
na
iy é J
on ad” « i (3 .
sin (A+ B)=sin Aviad Zin Bcos A ~~
cos (4+ B)=cos A cosBsinAsinB |
tanA+tanB :
tan (4B) = 7 tan AtanB
sin A cos B= 4 [sin (4 + B) + sin (4 - BD]
cos A sin B= % {sin (A + B) - sin (4 - BY]
cos 4 cos B= [cos (A + B) + cos (4 - B)]
sin A sin B= % [cos (4 - B) - cos (A + B)]
sinx 1 1
= 3 . cos x =
cosec x secx
tan x = 3 sin
cosx
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(16030175) 20 T890(E)(N22)T
BINOMIAL THEOREM
(xt hy)" =x" + nx"! i MD We...
DIFFERENTIATION
S@
r=ate
PRODUCT RULE
y=ulx). v(x)
a
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(16030175) -3- T890(E)(N22)T
fs) =f) f(ayax
n+
n+l
0 axte
& & e+e
sinx
cos x
tanx
cotx
4 +tanx] +¢
In (cosecx - cotx) +c
sec x oN
cosec x
sin” x
cos" x
tan™ x
cot? x
sec’ x —— _
xy? -!
cosec" x —_ _
ax? -l
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(16030175)
» P 4
INTEGRATIO@ P 4
4 % mn
J f(x) 2) de = ABs | £°G) gta) dx |
FO ~~
Fray =n fate 4
Lor f) e- OT +6
f(x) __A4 B
(ax+byext+d) axtb cxt+d
fi) A B c
\
Z
(@+ay” Gta) Gta Gray Gray
APPLICATIONS OF INTEGRATION
AREAS
b b
4A,= fxd A= for — y)dx
4=f. xdy A, =f (4 — 4)dy
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(16030175) -5- T890(E)(N22)T
VOLUMES
Ve=a[ Pde :Ve=nf? bp -yF)ae
Vy =a xdy iVy =a (2 - x3 ay
SECOND MOMENT OF AREA
t.=[) Pda; t,=f" Pad
MOMENTS OF INERTIA
Mass = density x volume
M= pv
DEFINITION#// = m 7?
GENERAL”
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higher education
& training
Department:
Higher Education and Training
‘REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE
NATIONAL CERTIFICATE
NOVEMBER EXAMINATION
MATHEMATICS N5
22 NOVEMBER 2016
This marking guideline consists of 9 pages.
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MARKING GUIDELINE -2- T890(E)(N22)T
MATHEMATICS NS
INSTRUCTIONS AND INFORMATION
1. equals mark.
2. Haif marks are not allocated, unless indicated otherwise.
3. Using the wrong formula when one.is required is a principle error and NO marks
should be allocated.
4. Candidates should show ALL formula and intermediate steps and simplify where
possible.
5. ALL final answers must be rounded off to THREE decimal places (unless indicated
otherwise).
6. Questions may be answered in any order, but subsections of questions must be kept
together otherwise candidates can be penalised with ONE mark.
7, When something has been copied incorrectly from the question paper and the
standard of the question is still the same, the candidate will be penalised by ONE
mark. If the copying error simplifies the question the mark will be forfeited.
8. Questions must be answered in blue or black ink. Answers in PENCIL will not be
matked as it is regarded as rough work.
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MARKING GUIDELINE -3-
MATHEMATICS N5
QUESTION 1
1.1 1.1 li x 1
xe Inx Inx
1.2 _ [+9
rey= 2 v
undefined
discontinuous ¥
QUESTION 2
21 F(x) =V2x
“¢ binomial theorem
£42 9 S37 1
: 203 sina A-px" Ft) Aa
f'@)=lim -
= ss
WaW2x 3 ~2y9¢3 Ay v
=lim—23 9 2
20 h
i
3x3
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T890(E)(N22)T
(3)
(2)
2)
|
6)
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MARKING GUIDELINE -A-
MATHEMATICS N5
= 2.2.1 y =n? 32 1332
ev v v v
dy _ 3x. 1 3e*-3xe* 6x
a an 3x° ge 3x”
e
2.2.2 y = Sarccos ec(x sin 2x)
._ S0sin2x+2xcos2x) “~~ product rule
xsin2x(xsin2x)?-1 ¥
2.2.3 y =4ycos'(x?—x-1)
v v v
I
y= 40s! (x? ~x—1).-sin(x? —x-DQx-1)
2.3 s(t) = pos!
Ins=cost.Int ~
cost
148 | snp ne + ¥
s dt
2 = s(-sintInt + 2088)
at t v
for t=4:5=08%3
velocity = 0,843(0,17140,9) “
=0903
2.4 cos(x+y)+sin x+sin y=2
v v
—sin(x+y).(1 +2) +cosx+cos ye =0
& (sin(x-+y)-+e08y) =-cosyt+sinixt+y) vv
dy _~—cosy+sin(x+ y)
dx —sin(xt+ y)+cosx
Diiv
d&
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T890(E)(N22)T
(4)
(3)
(3)
(5)
A)
[24]
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MARKING GUIDELINE 5-
MATHEMATICS NS
QUESTION 3
3.1 3.1.10 f(x) =3x?-4x? -2x42
3.1.2
3.1.4
3.2 3.2.1
3.2.2
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9x?-8x-2=0 VY
x=1,092 of x=-0,203 ¥ for both x-values
y=-L047 of -y=2;216 for both y-values
0 =3x°—4x7-2x+2 : -lSx<2
x-l| 0 1 2v
y f 2-1 «Ov
a
Y y-intercept
Y turning points
v shape
-1 fy XC) 2
e=— 7? _ i709 v
10,41
r=1,7—0,709=1,5291 ¥
A=xy v
C= 26x) +2(45y)
C=72x+90yp ¥
56000 = 72x+90y
56000—72x _
9
A=622,22x-0,8x" y
622,22~16x=0 ¥
388,89 =x
B1Lll=y y
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TR90(E)(N22)T
@)
2)
(3)
2)
(2)
6)
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MARKING GUIDELINE -6- T890(E)(N22)T
MATHEMATICS N5
3.3 V=arh
r=1,5m
dV > dh
= = ary. Y derivati
dt (7) dt derivative
3% mrs) ~ substitution
dt
0,424m = “ v
(4)
[21]
QUESTION 4
4] { 1-cos2x d a
—2x+4sin 2x apg oak sin' 2s) ==24 2 cos 2x
v
~20- cos 2x) &
— | -2x+sin2x
-2
. v
= Sineax +sin 2x)+¢
; @)
42 421 faxe™de
20% v
2x —— [26 dx
2
=xe*-e* 40 0 B)
4.2.2 3
[= 4 de
xtl
Long division:
¥¥ quotient remainderY
J 2x? -2x4.24— de
x+l
2x,
= 3% +2x-6In(x+1)+e¢ ~~ ONE mark for two correct terms
(6)
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MARKING GUIDELINE 7. T890(E)(N22)T
MATHEMATICS NS5
4.2.3 sin’ 3x
Je a
sin3x
fsiv’ 3xdx
11”
= J5=s00s 6xdx
2.2
_ lv sin 6x
2 12
(G3)
4.2.4 for [58 + 1d
u=2x' +1
dx= ae v
6x?
[oxta de
6x?
ov
i}
Alu
vos
v
5 3
==(2x7 +1) +e
9 (G3)
43 pe ae
4x? -1
4° -3x+l _ A +2
(2x-1(2xt1) 2x-1 2x41
4x? -3x4+1= A(2x+1) + B(2x-1) ¥
x=-05:-L28
4
: Z
J --he
(2x-1) 2x41
= diag) -2
= inex )) g inax+ +e (5)
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MARKING GUIDELINE -8- T890(E)(N22)T
MATHEMATICS NS
44 [ sin(aresin x de
J xd ¥
slaxtev Q)
[23]
QUESTION 5
5.1 SAL x=-y'+4 and x=4-2y
v - parabola graph
Fy Y - straight line
Vv - shading and strip
| ——~ |
(3)
512 2,
Jo’ +4)-(4-2y)dy - equations
0
-y
=F3-+9"lo v ¥ - integration
=1,333 ¥ (4)
5.13
2
V =nf-" +4) —(4-2y)'dy V - equations
0
2
=a] y‘-12y?+16ydy_- simplification
QO
5
=a —4y° +8y’2 “> integration
= 7(6,4)
=20,106 (4)
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MARKING GUIDELINE
5.2 4
2fx be Y
! ab
z
=27 fxtde
4%
a 24°
QUESTION 6
6.1 2xt3y a =e S8y
é€
dy
62 2 =210%4
v v
dy 2.10 cot2x
=—— -—— -4x4+
dx 41n10 2
v v
10 In(sin2x)
2.4In?10 4
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-9-
MATHEMATICS NS
v
e
v
2x? +ex+d
T890(E)(N22)T
@)
[15]
(4)
(6)
[10]
TOTAL: 100
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
TS90(EA2)T
AUGUSEEXAMINATION
NAGPIONAL CERTIFICATE
is
MATHEMAT CSNS
J
2,August 2016 (X-Paper}
This question paper.gonsists 6f'6 pages and 1 formula sheet of 5 pages.
“OP a
a
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(16030175) -2- T890(E)(A2)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS NS
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMAT OR,
1 Answer ALL the questiong.
2. Read ALL the questions
3 Number thegnswers
4. Show/AKL intermediate steps \
5. final answers must be rounded off to THREE decimals.
6. ions may be wife in any order, but subse:
togethe. hd }
7. answered in uy or biaeietne
A
Fe
8. Am
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(16030175) 3.
QUESTION 1
1.1 Determine In y if
Ll Iny=lim
x3 Insin 3x
1.1.2 Now, determine the value of y
1.2 Determine limx’.e*
a
QUESTION 2 |
2-; @
2.1 Given: = et ee
Desiegpine the derivative f'(x) using first principles.
j
op ee)
T890(E)(A2)T
2.2 Determine 2 in eagif6f the following cases (simplificatijin is NOT required):
aN y- 4 —
‘ PS 4 ~ A
- S ‘ . A “
221 aie \ a
222 yy =[In(Cdsec2x))
2.2.3 y=2et 3
23 Determine 2 with the aid of logarithmic differentiation if y = (9—.x*)"*
2.4 Given: y = arcsin x
dy
2.4.1 Derive a formula to determine G, if y = arcsin x
2.4.2 Hence, determine 2 if y =aresin (5.107)
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GB)
(1)
(3)
7}
(4)
(4)
(3)
@)
(4)
(2)
(2)
(16030175) -4-
2.5 Given: the implicit function x? +2= y?.cot3x
Calculate the derivative 2
QUESTION 3
3.1 Given: f(x) = x(x? -3x-9)
3.1.1 Determine the coordinates of the point of inflection of f(x)
T890(EMA2)T
(4)
[25]
Q)
Q)
3.1.3 Draw a neat gtaph of JSG) between the values in QUESTION 3.1.2 if the
turing points bal given as (3;-27) and (-1;5)
3.2
HINT: V=mh
3.3 The sum of the diameter and height of a cylinder is 6 cm.
Calculate the dimensions of the cylinder to ensure a maximum volume.
QUESTION 4
4.1 Given: _ A + B
i x? ee) (10+ x)
4.1.1 Determine the numerical values of A and B.
3
4.1.2 Hence, calculate In 4x
1
~“x
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@)
@)
@
6)
6)
[21]
(2)
(3)
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(16030175) -5- T890(E)(A2)T
4.2 Determine f ydx in each of the following cases:
4.2.1 yex? sine
x
4.2.2 y=cos? 5x
4.2.3 y =cos12x.sin3x
4.2.4 .
ay
(4x3) (12)
43
\ (4)
[21]
‘
4
a, 4
fe mca
9
y
ad
« > XxX
4= y
vy
3.1.1 Prove that the points of intersection are:
(0;0), (2,828;1,414) and (—2,828;-1,414) (3)
5.1.2 Calculate the area of the bounded region in the first quadrant. (4)
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(16030175) -6- T890(E)(A2)T
5.1.3 Calculate the volume generated when the area in QUESTION 5.1.1
rotates about the X-axis.
52 $2x?
: Calculate the value of Jie
x
1
5.3 Show that the volume of the figure formed when the line y = Gor rrotates about
: h
the X-axis equals 7 =
sry
x
QUESTION 6 Te
——— |
6.1 Solve the differential equation ede—y'dy =e" dx
6.2 Determine the general solution of the following differential equation:
(1+ x)sin y.dy = cos y.dx
TOTAL:
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(4)
(3)
(4)
[18]
@)
(4)
[8]
100
(16030175)
3. T890(E)(A2)T
fa)
= 70) {player
sin x
cos x
tan x
cot x
sec x
cosec x
sin? x
cot” x
sec’ x
cosec' x
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n+]
nax”!
n+1
axtc
& ete
Ina
i
aafx? -t -
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(16030175) 4. T890(E)(A2)T
fo = 40) Sf (dx
INTEGRATION™.) #
[ f() 8) dx = £9) eo £1) g(x) dx |
oN
SO “a
Fag @ fe) te Des,
atl
Ler se ar-YET
fe) A
(axtb\ex+d) axtb oxtd
f@) A B c Z
(xta)" (eta) (xt+a)? Gra) Otay
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(16030175) De
BINOMIAL THEOREM
(xt hy =x" 4 nx" h MaDe Reo.
DIFFERENTIATION
-_£@
£@
roate
PRODUCT RULE
=u) . vO)
dy dv
CHAIN RULE
y=fu) -
w_w ha
dx du ax
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(16030175) -1- T890(E)(A2)T
MATHEMATICS N5
FORMULA SHEET
Any other applicable formula may also be used.
TRIGONOMETRY
sin’ x + cos’ x= 1
1 + tan? x = sec? x
1 + cot” x =cosec” x ,
sin 24 =2 sin A cos A
cos 24 = cos" - sin
Wa
cos" A= 4+ A coe2d i , A y
sin (A+ B)=sin A coi
tan A+ tanB
1+ tan Atan B
tan (A + B) =
sin A cos B = ¥4 [sin (A + B) + sin (4 - BY]
cos A sin B= [sin (4 + B) - sin (4 - B)]
cos A cos B = 4 [cos (4 + B) + cos (A - BY]
sin A sin B= [cos (4 - B) - cos (4 + B)]
sinx | 1 1
tanx = 5 sinx = 5 COSX =
cosx cosec x secx
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(16030173) 5
APPLICATIONS OF INTEGRATION
AREAS
b b
A, = foe ae J Oirydde
4=f. xdy ; A, =f. (% — x,)dy
VOLUMES
~ OF AREA
haf r thy iis AM
Mass = density x volumé™)
wee
M= pv
DEFINITION: J=m??
b
GENERAL: 1=[” rdm= pf” Par
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T890(E)(A2)T
higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE
NATIONAL CERTIFICATE
AUGUST EXAMINATION
MATHEMATICS N5
2 AUGUST 2016
This marking guideline consists of 10 pages.
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MARKING GUIDELINE 2 T890(E)(A2)T
MATHEMATICS NS
INSTRUCTIONS AND INFORMATION
1, Y equals 1 mark.
2. Half marks are not allocated, unless indicated otherwise.
3. Using the wrong formula is a principle error and NO marks will be allocated.
4, Candidates should show ALL formulae and intermediate steps and simplify where
possible.
5. ALL final answers must be rounded off to THREE decimals unless indicated
otherwise.
6. Questions may be answered in any order, but subsections of questions must be kept
together. If subsections are separated, the candidate can be penalised with ONE
mark.
7. Where something has been copied incorrectly from the question paper, and the
standard of the question is still the same, the candidate will be penalised with ONE
mark. If the copying error simplifies the question and makes it easier, the student
forfeits the marks,
8. Questions must be answered in blue or black ink. Answers in PENCIL are not
marked as it is regarded as rough work.
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T890(E)(A2)P
MARKING GUIDELINE 3-
MATHEMATICS N5
QUESTION 1
Ll Ld Iny <li v
ay ti osx
sin3x
1
Iny = lim tan 3x
Iny=0
¥ (3)
112 y=l¥ (1)
: x?
120 Ne
ree
=lim v
2
=0 v
@)
[7]
QUESTION 2
21 2-x
IQ) = 3
flet+hy 222th _ 2-4-h ,
3-2(x+h) 3—2x—2h
. , 2-x-h 2-x_1
‘(Od = lim( A" —
LO) nae 2h 32x
Fe) = Him 222 AMG = 20) B= 2x-2wYO—z) 1 ;
am (3—2x—2h)(3—22) h
h 1
\(x) = lim ————2____. =
PO)=|ia Gy -tHG—2In hk
1
I'(x) = v
(3—2x)" (4)
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MARKING GUIDELINE 4. T890(E)(A2)T
MATHEMATICS N5
_ l-x
2.2 2.2.1 aa
an x
y YY a
dy ~1(vtanx)— (2), (tanx)? sec’ x
a (vianxy ” (4)
2.2.2 y =[In(cosec2x)]°
& ¥ - cot 2x
& = S[In(cosec2x)]* = OOS ee x 00k 2x
cosec2x ” @)
2.2.3 ys Demsinsx Me
v v
. 5 ~ 1 2
y= dees acme Ma + 20S?
1-(Sx) (2)
2.3 y=(9-x°)™*
w
In y = (In x) In(9-x°)
—3x? vw
1h 21 nox) +Inx. 3x
ydx x (9-x’)
. 3 4,2
& _ Om) ine. 3x v
& (9-x') (4)
2.4 2.4.1 y = aresin x
sin y=x
ay
cos y—=1
y ax
dy 1
dx cosy
wt v
dx -sin? y
w__s v
dy Vi-x
(2)
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MARKING GUIDELINE 5-
MATHEMATICS N5
24.2 y = aresin(5.10*)
dy _ 5.10% Inl0 v
dx ft-(5.10")? Y
2.5 x? +2=y? cot3x
v ¥
2x =3y? D cbt3xty’.—Beosee?3x
2x+3y’ cosee?3x =3y" cote ¥
2x+3y* cosec?3x _y
3y’ cot3x dx
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(2)
(4)
[25]
MARKING GUIDELINE -6-
MATHEMATICS NS
QUESTION 3
3.1 3.1.1 f@=xe —3x-9)
3.1.4
3.2 3.2.1
3.2.2
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S(@) =x -3x? -9x
S'(x) =3x? -6x-9
f'"(®) =6x-6
6x-6=0 |
x=l
y=-ll v
X}-1 0 1 2 3
Y|5 0 -11 -22 -27
v — y-intercept
: 2 4 ¥ — shape
a)
ex - LOO) _ 0869 _ ying vy
FCO) | -837
a=-0,1+0,104 = 0,004
T890(E)(A2)T
v¥¥ (Any 2 correct=1 mark)
All these units indicate rates of change“ of the volume, radius and height
tespectively. “
a ra
DY age Ky? th
dt dt dt
aVv ” A
a 27(11)(0,5)(30) + #11)? (-1)
cla = 656,593
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@)
Q)
G)
Q)
6)
MARKING GUIDELINE -7- T890(E)(A2)T
MATHEMATICS NS
33 D+h=6
D=6-h
2 wv
V=mh
yan3—")h ¥
3
3
¥ =9nh—3ah +
2
V'= 90-642
0=2(36—24h+3h7)
0=(4-6)(h-2) ¥
h=6;h=2
r=0;r=2
[21]
QUESTION 4
41 4.1.1 j a 4,38
100-—x? (10-x) (10+x).
4x= A(10+x)+B(10-x) ¥
x=-10:-2=B
-iIn-4— w
x=10:A=2 (2)
4.1.2 3 ay
S07
+ 2 -2
aos v
4(l0-x) (0+x)
=[-2InQ0-x)-2n0+x)] |,
= (-9,022)-(—9,19)
=0,168 » (3)
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MARKING GUIDELINE
4.2 4.2.1
4.2.2
4.2.3
4.2.4
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-8-
MATHEMATICS N5
2. 1
y=x? sin—
x
1
u=—sde=—x' duw
x
Jez sin ux? du ¥
=~ fsin udu
=cosut+ec
I
=cos—+cr
x
y =C0s" 5x
11
Jz + zoosl0xdx v
1” sinlOx ¥
=ix+ +
2 20
y =cosl2y.sin3x
J + (sin 5x) sin 9x) v
= 1 eoslsx + hs 9x
2 15
J+e
3 fox? Vv
tL.
=csin'xt+e Vv
5
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(4*3) 12)
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MARKING GUIDELINE -9-
MATHEMATICS N5
43 fx” In2x dx
F(x) = In 2x gia?
, 1 xt
x)=— =—
f'@) x 80) =
14 14
in2x2—— [1 ge vy
14 4x 14
=In2x 2 — cae
‘14-414
=In2x 2 = 4¢
“14 196° ¥%
QUESTION 5
5.1 5.1.1 Vea
2
x -8x=0
x=0;x? =8 ve P ;
y =0;x = 42,828 one mark per x-value
y=tl4l4 -
5.1.2 2,828 1
J * ary
0 2
=? 5 x 2908 v
= 4 re
=0)-@ y
=l y
5.1.3 288 2 02 ¥
au J x a
ry 4
32 x
=x} — pes oy
Be ygh
= A{(1,508)-(0)]_ v
=4,738 y
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T890(E)(A2)T
(4)
[21]
@G)
4
(4)
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MARKING GUIDELINE -10-
MATHEMATICS N5
5.3
QUESTION 6
6.1 e*dx— y dy =e"dx
¥ w
dxfe* —e*) = y'dy
v sv
e+e* +c0=—
6.2 (i+ x)sin y.dy = cos y.dx
v 1 i
tan =—
dy 1+x
Inst y = Ind #%) +c
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T890(E)(A2)T
GB)
(4)
[18]
(4)
(4)
[8]
TOTAL: 100
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
(ForoGiyadyr
APRIL EXAMINATION
CERTIFICATE
(16030175)
4 April 2p 16 (X-Pap
09:00-12:00
This question paper cons! f 6 pages and 1 formula sheet of 5 pages.
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(16030175) 2- T970(E(A4)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS NS
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1. Answer ALL the questions.
Read ALL the questions care:
js question paper.
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(16030175) 3
QUESTION 1
1.1 Determine the following limits:
111 lim—*
x30 xe* ~ 2x
lim (sec x — tan x)
xr
2
1.2 Determine whether f(x) =
QUESTION 2
2.1 Determine th i first principles.
2.2 Determine —in & i is not required):
2.2.2
~ 2Incosec4x
2.2.3 y =arcsec(2.4™)
2.3 Determine ke with the aid of logarithmic differentiation if:
aresin(sin y) =(x* )
2.4 Given: 3x‘ — xy =2**
dj
24.1 Determine the slope = of the tangent at the point: (1;-5).
2.4.2 Hence, determine the equation of the tangent al this point.
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TOTO(E\(A4)T
(2)
3)
2)
7]
(4)
(4)
6)
2)
(4)
@)
Q)
[24]
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(16030175) -4- T9T0(EMA4)T
QUESTION 3
3.1 Given: f(x) = x(x’ -5)-4
3.1.1 Determine the co-ordinates of the turning points of f(x). @)
3.1.2 Verify, using a table, that the equation 0=x(x°-5)-4 has a root
between the points x=2 and x=3.
Use values on the table; O<x<4 (4)
3.1.3 Hence, make a neuit sketch of the graph of the function f(x). (2)
3.1.4 If the positive raokof svi estimated as 2,7, use Taylor's/Newton's
method to determitie Qhetter approximation of this root. (4)
3.2 Two sides of alivctingle are lengthened at a sate vi'S cm/s wiitethe other two sides
are being shotiened in)such a way that the figure remains a rectangle with a constant
area of 50 ci’.
3.2.1 Calculate the rate of change of the perimeter of tie rectanjle when the
length of an increasing side is 7 cm. (5)
3.2.2 Prove that when the rate of change of the perimeter is zero, the figure
must be a square. > (2)
3.3 A particle moves in 4 )§iraight line accofting to the distance formula
s(t) =V#G-3t-2).
3.3.1 Calculate the velocity of th2\particle after,2,7 seconds. 4)
3.3.2 Calculate the acceleration after 2 seqiuids. (3)
[27]
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(16030175) 5- T9TO(E\(A4)T
QUESTION 4
44 Determine: fe +e)‘ (e? -e")dx Q)
4.2 Determine j ydkx in each of the following cases:
42.1 y=
~~ 1+ cosx (3)
4.2.2 G3)
4.2.3 (2)
4.2.4
(3)
43 Determine
= 9)
44 Determine: J 4)
[22]
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(16030175) -6- ‘1970(EW AAT
QUESTION 5
5.1 Given: The curves f(x)=VI16—x° and g(x)=4-x
No
es about the x-axis.
5.2 Prove that f-se"
0
QUESTION 6
6.1 Solve the differential equation:
dy _ tan’ y
dx cosec?x
2
6.2 Determine the particular solution of the differential equation = aes + ; x40
for which y=2 and y'=—3 when x=1.
TOTAL:
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GB)
4)
(4)
[11
(4)
(5)
[9]
100
T970(E)(A4)T
(16030175)
DIFFERENTIATION
__ f@
f (a)
r=ate
PRODUCT RULE
y= ulx).v@)
A
dx ax dx
=u +vu'
QUOTIENT RULE
- u(x)
v(x)
yt _ &
Dds ds
dx y
_ vu! —av'
y
CHAIN RULE
¥ =flu@))
ay _ dy du
dx du dx
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(16030175) -1- TS70(E)(A4)T
FORMULA SHEET
Any other applicable formulas may also be used.
TRIGONOMETRY
sin’ x + cos’x =1
I+ tan’ x =sec’ x
1+ cot’ x =cosec’ x
sin 2A = 2 sin A.cos A
cos 2A = cos’ A - sin? A
2 tan A
tan 2A = Z
1-tan* A
sin? A at = De
2
cos” azi + —cos2
2 2
sin (A + B) =sin A.cos B + sin
cos (A+ B) = cos A.cos B F sin A.sin
tan (4+ B)= tan A+ tan B
1-tan A.tan B
sin A.cos B = % [sin(4 + B) + sin (4 - B)}
cos A.sin B = 4 [sin(4 +.B) —sin(A - BY]
cos A.cos B= [cos(A + B) + cos(A - B)]
sin A.sin B= % [cos(4 - B) - cos(4 + B)]
sinx . 1 1
tanx = jsinx = 5 cosx =
cosx cosec x secx
BINOMIAL THEOREM
(xt A)" =x" 4+ nx" th + MOD ray? +...
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(16030175) -3- T T970(E)(A4)T
d
fe) af [fea
atl
ax’ nax™! =
n+1
a 0 ax te
sin x
cos x
tan x
cot x
sec x
cosec x
sin’ x
cot” x
1+x’
sec” x —— _
ax? -1
cosec™ x > _
xa)x -I
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(16030175) 4 T97O(E)(A4)T
(axtbext+d) axtb cxt+d
fe) _ 4, 8B Cc Zz
(xt+a)’ (x+a) (x+y “Ga+ap "Gray
APPLICATIONS OF INTEGRATION
AREAS
4, =f) yar; A, =) (v2 -¥,)4e
A, =" xdy; A, =. (a —a)ey
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(16030175) -5- T970(E)(A4)T
VOLUMES
V,=af yids s¥,= Af! (vy -») dx
Vv, =n xdy iV, =a (x, x, )ay
SECOND MOMENT OF AREA
I= { dA Ty = { dd
MOMENTS OF INERTIA
Mass = density x volume
M=pV
DEFINITION: J=
GENERAL:
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE |
N
NATIONAL CERTIFICATE
APRIL EXAMINATION
MATHEMATICS N5
4 APRIL 2016
This marking guideline consists of 10 pages.
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MARKING GUIDELINE -2- T970(E)(A4)T
MATHEMATICS N5
INSTRUCTIONS AND INFORMATION
1. Half marks are not allocated, unless indicated otherwise.
2. Where formula is required, using the wrong formula is a principle error and NO
marks are allocated.
3. Students should show ALL formulae and intermediate steps and simplify where
possible.
4. ALL final answers must be rounded off to THREE decimal places (unless indicated
otherwise).
3. Questions may be answered in any order, but subsections of questions must be kept
together. If subsections are separated, the student can be penalised by ONE mark.
6. Where a student copied wrong from the question paper, and the standard of the
question is still the same, the student will be penalised by ONE mark.
If the copying error simplifies the question and makes it easier, the student forfeits the
marks.
7. Questions must be answered in blue or black ink. Answers in PENCIL will not be
marked as it is regarded as rough work.
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MARKING GUIDELINE 3 T970(E)(A4)T
MATHEMATICS NS
QUESTION 1
Ll 1.1.1 e
lim.
x90 xe" — 2x
ll
8 ol-
Q)
1.1.2 lim (sec x — tan x)
BOD
2
= tim =8inx Ov
ee cosx 0
_ COS x
= lim — v
sof —sinx
=0 ¥ @)
1.2 7 _ 3-27 _ -54_
F(@)is continuous aix=-3 Vv (2)
7)
QUESTION 2
24 -
WENK: [imeem tig stim SM ay
io 430 ft
lim cos(x+h)-—cosx
40
.__ cos x.cosh—sin x.sinh—cos x
=lim
h0 h v
sh-l . — sinh
—sin a v
. co!
=limcos x.
oat)
=cos x(0)—sin x(1)
=-sinx y (4)
2.2 2.2.1 y =cos* (x? —4)+cos(x? —4)*
= 4cos* (x? —4).—sin(x? —4).2x — sin(x? -—4)4.4(x? -4)°.2x
¢ D (x 74) Ae ) @)
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T970(E)(A4)T
MARKING GUIDELINE 4.
MATHEMATICS N5
v v v v
2.2.2 _ -
6tan¥1—x.sec” V1-x. | aIncosecdx —3tan? y1— x speed cot Ax
_ 2v1-x cosec4x
(2incosec4xYy
6)
2.2.3 y=arcsec(2.4**)
3x
dy _ 6.47 In4 Vv
de 2.4/2.4")? -1 Q)
2.3
yas
Iny=e*.Inx
1® eine
yas x
® oye (ef Inx+)
ax x
(4)
2.4 2.4.1 d
12x° -1y-x =3.2* In2
x —ly-x hk v
12x7-y-3.2"In2_ dy
: x dx
at(,-5)
ay _ 12 -—(-5)-3.81n2
dx 1
» = 0,364 v
a 8)
2.4.2 y=mete
-—5=0,364(l) +c v
-5,364=¢ v
equation : y = 0,364x — 5,364 (2)
[24|
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MARKING GUIDELINE “5 T970(E)(A4)T
MATHEMATICS N5
QUESTION 3
3.1 3.1.1 f(a=x? -5x-4
3x? -5=0 ¥
x=1,291 or x»=-1,291 mark for both x-values
yp =-8,303 y= 0,303
» mark for both y-values (3)
3.1.2 Use values on the table: OS x <4
xX | 0 1 2 3 4 we'v marks for any 4 correct values
y | 4 8 6 8 40
y-value changes from negative to positive indicating a root” (4)
3.1.3 40
8 # shape of the graph
y-intercept
< ZN >
-2 / -1 9 1 2/3 4
-6)
8
’ (2)
3.1.4 w
we
a=2,7—0,129= 2,571 » (4)
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MARKING GUIDELINE -6-
MATHEMATICS N5
32 32 Azay
50=14y +
50x7 =y
P=2x+100x"! ¥
dP 9% 100 dx
ada ea
ld
o =-0522 v
32.2 &=0 only if <.% v
Therefore: x=y
Meaning the figure is a square
3.3 3.3.1 i 3 5
s(t) =32? -31? -2?
1 3
yop 27 vw
wi 2 2
V= 23,987 “we
3.3.2
a=3- 4B we
4t? 4¢?
a= -7,159 v
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T970(E)(A4)T
(5)
(2)
4
(3)
27
MARKING GUIDELINE PV
MATHEMATICS N5
QUESTION 4
4.1 eo cass
fete Le
5 ww
4.2 4.2.1 sinx
ys
7 vV1+cosx
u=l+cosx
du
-sinx ¥
sinx du
Saas
v
dx=
a
= fu? du
i
—2u? +e
—2Vl+cosxt+e ”
4.2.2 y=x.sec’ x
S(@=x g'(x) =sec’ x
f'@)=l a(x) =tanx
x.tanx— f tan xde
x.tan x—In(secx)+e
we a
4.2.3 y =cos6x.cos2x
Joos(8x) +cos(4x)dx
sin8x sin4x
= 3 +
” ¥
4.2.4 2
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T9T0(E)(A4)T
2
(3)
(3)
(2)
(3)
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MARKING GUIDELINE -8-
43
44
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MATHEMATICS NS
xe-2
OP ae
x-l _A B
(Oirt+2 x x42
xP -1=A(x+2)+ Bx v
9
x=2:—=B
3 v
x=0:-L=4 a
2
1 9
x (x+2)
al iny-2ines 2 +e
2 2
we “
~, 25
= |x+5+——d&k
fx x—5 ad
x?
ay t5xt25lnG@—5) +e
a w
T970(E)(A4)T
@)
4)
[22]
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MARKING GUIDELINE
QUESTION 5
5.1 5.1.1
§.1.2
5.2 : en
=-5 a
—-s
5.1...
=-Lylo ¥
e
=0-1) ”
Ss
_75
=—
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-9-
MATHEMATICS NS
[vies -(4-xde ov
0
2
=(ssin“ 3 +5 16—x? ~4r4]}
=( 4,566-0)
= 4,566 ¥
4
n[Vi6—x? - (4-2) ax v
0
; 3 8 x
= n{16x-7 16x + - =
= A[(21,333) - (0)]
=67,02 ¥
4”
T970CE)(A4)T
(3)
4)
4)
a)
Please turn over
MARKING GUIDELINE -10-
MATHEMATICS NS
QUESTION 6
6.1 a _ tan? y
dx cosec*x
1 dy= ! dx
tan? y Cosecx
cot? ydy = sin? xdx
ca 2 liv
(cos ec“ y—1)dy = G08 Hae
“1
sin 2x
Toby Ysa 4
Cw
6.2 w
13 2
7 ede ee “
6 4
-x* 3x3 am?
++
24 12
w
ax
+ext+d
y=
wv
-~3=3,725+¢
-6,725=¢
2=-4,946 +d
6946=d v
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T970(E)(A4)T
4)
(5)
19]
TOTAL: 100
higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
T960(E)(N25)T
NOVEMBER EXAMINATION
NATIONAL CERTIFICATE
MATHEMATICS N5
(16030175)
25 November 2015 (X-Paper)
09:00-12:00
Scientific calculators may be used.
This question paper consists of 6 pages and 1 formula sheet of 5 pages.
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(16030175) 2- T960(E)(N25)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N5
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1,
2.
bed
Answer ALL the questions.
Read ALL the questions carefully.
Number the answers according io the numbering system used in this question paper.
Show ALL intermediate steps and simplify where possible.
ALL final answers must be rounded off to THREE decimal places
Questions may be answered in any order, but subsections of questions must be kept
together.
Use ONLY blue or black ink.
Write neatly and legibly
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(16030175) 3 T960(E)(N25)T
QUESTION 1
1.1 Determine:
2
LL iny = lim
x90 tanx
1.1.2 The value of y
1.2 Determine lim Insecx
x90 xe *
QUESTION 2
2.1 Given: —- = S08 ecx
Determine the derivative oy making use of the derivatives of sinx and cos x
2.2 Determine o in each of the following cases (simplification is NOT required):
se
2.2.1 y <0
2.2.2 y=e™ In(ax’ — bx +c)
2.2.3 y =aresin(10**)
23 Determine ”. with the aid of logarithmic differentiation if y= (cos3x)™"
2.4 Given: y = arecos x
2.4.1 Make a neat, rough sketch of the graph of y for the range [0; 7]
dy
2.4.2 Derive a formula to determine 7, if: y= arecos x
2.5 Given the implicit function: sin(5x—2y)=4
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Calculate the derivative 2
Ie
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3)
¢3)
GB)
[7]
3)
(4)
(3)
2)
(4)
G)
(2)
(4)
[25]
Please tum over
4 T960(E)(N25)T
(16030175)
QUESTION 3
3.1 Given: f(x) =2x° —4x? —8x+8
3.1.1 Determine the coordinates of the turning points of f(x)
3.1.2 Draw up a table of values for x and /fx) with x ranging from x = -2 to
x= 4.
3.1.3 Draw a neat graph of f(x) between the values determined in
QUESTION 3.1.2
3.1.4 An estimated root of the graph is 0,7.
Use Taylor's/N ewton's method to determine a better approximation of this
root.
3.2 An open cardboard box with a square base is required wo hold 163 m.
Determine the dimensions that require the least amount of material, and neglect the
thickness of the material and waste in construction.
Veh /
A=4xh+x? |
A
Po
x
3.3 An electric current is varying through a resistor of 1692. The rate of change of the
current is 0,6 A/s when the current is 12A.
Find the rate of change of power developed with the current at 12A.
HINT: P=I?R
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Q)
(2)
(2)
(4)
(3)
(4)
[19]
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(16030175) -5- T960(E)(N25)T
QUESTION 4
4 3 by 2
4.1 Determine f 3x0 43x" Sx tanh, by resolving the integral into partial fractions.
x°+x-2 (5)
42 Determine j ydx in each of the following cases:
. 3
421 _ sin® x
secx (3)
x
422 ya
2x? +3 (3)
arctanx
42.3 yoo —
l+x (2)
42.4 yo
3x-2 (4)
43 Given f /@)8'@) = S@).8(2) - ff C.g(dde
Determine. f2xe*dx (4)
[24]
QUESTION 5
1
5.1 The sketch shows the area bounded by the curve x= ae x=y’ and theline y= ry
y
ay
a
v
<
‘
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(16030175) -6- T960(E)(N25)T
5.1.1 Prove that the point of intersection of the two curves equals (2,52; 1,59). (3)
5.1.2 Calculate the area of the bounded region. (4)
5.1.3 Calculate the volume generated when the area in QUESTION 5.1.1 rotates
about the Y-axis. (4)
é 1
5.2 Calculate the value of : |——«
1vi-x (4)
5.3 Determine the second moment of area of a rectangular lamina with coordinates (a;5)
about the axis parallel to one side of the lamina. The distance from the representative
strip to the axis is ‘x’. The lamina's length is 'a’ and breadth is 'b’.
|
- (a;b)
X »
xX_y,
“y (4)
{19]
QUESTION 6
6.1 6.1.1 Solve the differential equation Wz e™
dx (3)
6.1.2 Determine the equation of the curve through the point (1;-2). (2)
6.2 Determine the general solution of the differential equation (1 — y*)dx = —2xydy (4)
19}
TOTAL: 100
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(16030175) -1- T96O(E)(N25)T
FORMULA SHEET
Any other applicable formulas may also be used.
TRIGONOMETRY
sin’ x +cos’x =1
1 + tan’ x = sec’ x
1 + cot’ x = cosec’ x
sin 2A = 2 sin A.cos A
cos 2A = coy’ A-sin’ A
2 tan A
tan 2A = — 72,
t—tan* A
sin? 421A cos24
2 3
cos? A= i, | cos2d
2 2
sin (A + B) = sin A.cos B+ sin B cos A
cos (A+ B) = cos A.cos B F sin Asin B
tan A+tanB
1-tan A.tan B
tan (4 + B)=
sin 4.cos B= % [sin(A + B) + sin (4 - B)]
cos 4.sin B= % [sin(4 + B)— sin(A - B)]
cos A.cos B= % [cos(A + B) + cos(A - BY]
sin A.sin B= % [cos(A - B) - cos(4 + B)]
sinx , 1 l
tanx = 3 sinx = 3 cosx =
cos.x cosec x secx
BINOMIAL THEOREM
(x thy ax" tnx h nln 7 1) vn?
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(16030175) 2
DIFFERENTIATION
f(a)
f@
r=ate
PRODUCT RULE
y =ula).v@)
ak dx dk
=uv't+vu'
QUOTIENT RULE
= UO)
v(x)
CHAIN RULE
v=flue))
dy _ ay du
dx du dx
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T960(E)(N25)T
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T960(E)(N25)T
(16030175)
f@) £ Fx [¢@)ar
dx
1 n+1
ax" nax” ——+e
ntl
a 0 arte
& é ete
ao
a a lna te
Ina
log. x =
x
1 1
OB x
Ba xlna —
sin x cos X -COSX re
cos x -sin x sinx +¢
tan x sec” ¥ In (sec x) +€
cot x -cosec” x In (sin x) +
secx sec x.tan In [seo x + tan x] +
cosec x -cosec x.cot x In [cosee x - cot x] + ¢
ca 1
sin” x a cara _
yl-x#
ot 21
cos" x _
1x?
4 1
tan” x z _
1l+x
-l
cot” x - _
d+x*
sec’ x !
xy eal a
a ~i
cosec” x =
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T96O(E)(N25)T
(16030175) 4.
d
fe) gf [7a
| _ sin” 6 te
ax a
i 1 af Xx
pan _ zi [2+
: ‘()
—see”|—J+e
xfer -a - a a
eax? @ on! (2)+28 -x +e
a
! 1 {x-a)\
it 1 atx
ax ~~ puss] [
INTEGRATION
J/@)e'@)= SO) e@) - J HO) ee
jE a= in sles c
(LrCol F ear = £0" 5.
ntti
fe) AB
(ax+b)extd) axt+b oxt+d
f(x) __ A B c Zz
(x+a)" 7. @+a)- (x+a)’ * (x+a) Feet («+ a)’
APPLICATIONS OF INTEGRATION
AREAS
A.=f yar; A= [ . — yee
A, = [ xdy; A, = [ (x, — x,
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(16030175) -5-
VOLUMES
v, sap ytd 5V,=2f'(y.? — 9?) ax
V, =a x*dy; vy, = af x," = x" )dy
SECOND MOMENT OF AREA
t=! Pda st, =f? Pa
MOMENTS OF INERTIA
Mass = density x volume
M=pV
DEFINITION: J=mr
Ab
GENERAL: J ai dm = rl, rdV
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T960(E)(N25)T
higher education
& training
clarks Depariment:
ey Higher Education and Training
ae” REPUBLIC OF SOUTH AFRICA
—
MARKING GUIDELINE
an
NATIONAL CERTIFICATE
NOVEMBER EXAMINATION
MATHEMATICS N5
25 NOVEMBER 2015
This marking guideline consists of 9 pages.
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MARKING GUIDELINE -2- T960(E)(N25)T
MATHEMATICS N5
INSTRUCTIONS AND INFORMATION
1. equals lmark.
2. Half marks are not allocated, unless indicated otherwise.
3. Where a formula is required, using the wrong formula is a principle error and NO
marks are allocated.
4. Students should show ALL formulae and intermediate steps, and simplify where
possible.
5. ALL final answers must be rounded off to THREE decimal places (unless indicated
otherwise).
6. Questions may be auswered in any order, but subsections of quesiions must be kepl
together. If subsections are separated, the student can be penalised by ONE mark.
7. Where a studeitt copied incorrectly from the question paper and the standard of the
question is still the same, the student will be penalised by ONE mark.
If the copying error simplifies the question and makes it easier, the student forfeits the
marks.
8. Questions must be answered in blue or black ink. Answers in PENCIL are not
marked as it is regarded as rough work.
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MARKING GUIDELINE -3-
T96O(E)(N25)T
MATHEMATICS NS
QUESTION 1
11 1.1.1 e
lim—
x90 xe* — 2x
u
8 ole
AON
2)
lim (sec x — tan x)
xs
2
=lim
Insinx (0. a
x95 COSK 0
_ ~COSX
= lim ¥
xot —Sinx
2
0
a
=0 @)
L2 ay _ (3) -27_ -54_
f= Ty 6
f(x) is continuous at x =3 “
ow
(2)
[7]
QUESTION 2
24 _ F
pint: tim 20S! 9 stim SIMA Ly
td i0 ft
._ cos(x+h)-cosx
tim ———+__+>———
hoo h
cos x.cosh— sin x.sinh—cos x
=lm Oa
4290 h
: cosh-l . — sinh y
=limcos x. — sin x.—
00 h h
= cos .x(0)-sin x(1)
=~sinx
4)
2.2 2.2.1 y = cos* (x? ~4)+cos(x? —4)*
= 4cos*(x* — 4).-sin(a°—4).2x - sin(x? —4)*.4(x? -4)'.2
( 5 ( ) LD ( ) & y 2x
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MARKING GUIDELINE 4 T960(E)(N25)T
MATHEMATICS NS
2.2.2 _3tan? VIl-x
* 2Incosec4x
ee ae “
6tan-¥1—x.sec’ V1—x. =) a Incos ec4x —3tan? v1- x. Boosecde. cot 4x
2Vl-x cosec4x
(2Incos ec4x)?
w
(5)
2.2.3 y =aresec(2.4*")
as
wy 6A ns
dx 2.4" (2.4) -1 (2)
2.3 arcsin(sin y) = x")
pax”
Inyee™.Inx
! 1D _ ys nxt we
ydx x
dy . ew
Sax (lnx+—
= x ( x z ) (4)
2.4 2.4.1
; 12x° -ly-x% 23.2" In2
yo dx 7
12x°-y-3.2¥n2_ dy
x dx
at -5)
dy _12~-(-5)-3.8In2
dx 1
® 0364 ¥
ax G3)
2.4.2 yrmxtc
—5=0,364(1) +e
—5,364=¢ ”
equation: y = 0,364x -5,364 Q)
[24]
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MARKING GUIDELINE
-5-
MATHEMATICS NS
QUESTION 3
3.1 3.1.1 f()ax? -5x-4
¥
3x? -5=0
x=1,291 or
y =-8,303 y = 0,303
“mark for both y values
T960(E)(N25)T
x =—1,291 » mark for both x values
4) wv’ marks for any 4 correct values
312 X,0 1 2 3
y |
\4 8 6 8
y-value changes from negative to positive indicating a root ”
40
3.13 40
8
< .
6
8
NT
3.1.4 ”
_ fQN 2183 _
F'Q27) 1687,
0,129
a=2,7-0,129=2,571 y
(3)
(4)
shape of the graph
y-intercept
4
(2)
(4)
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MARKING GUIDELINE
3.2
3.2.1
3.2.2
3.3 3.3.1
3.3.2
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-6-
MATHEMATICS N5
x
A=xy
50=xy +
50x71 =y
P=2x+100x'! ¥
aP _ dx 100 de
dt dt x* dt
00
aP = 33)-4 —3
dt: T v
aP __o192
dt
dy
—=0 only if -——=->
y ae
Therefore: x=y
Meaning the figure is a square “
1 305
(0) =302 -322 -2
3
ye 9858 we
2/2 2 2
v= .23,987 vw
-3 9 185 we
a=—y-— yt
402 49?
a= -7,159 ¥
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T960(E)(N25)T
6)
(2)
@
(3)
[27]
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MARKING GUIDELINE
QUESTION 4
4.1
-J=
MATHEMATICS N5
(er +e") te OH
5
4.2 4.2.1 sinx
v1+cosx
—2Vl+cosx+e~
4.2.2 y=xsec? x
f(x)ax
J'(@)=1
g'(x) = sec” x
g(x) =tanx Y
x.tan x f tan xd
x. tan x — In(
“
4.2.3 y=cos6x.c
secx)+c
¥
OS 2x
f cos(8x) + cos(4x)dx
_ sin8x
8
4.2.4 2)
2 an”!
=—= tan
v3
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sin4x
+
4
x
+e
V3
T96O(E)(N25)T
(2)
GB)
(3)
(3)
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T960(E\(N25)T
MARKING GUIDELINE -8-
MATHEMATICS N5
43 a)
x +2x
v-l 4, B
(x)(xt+2) x x42
xi -1=A(x+2)+ Be +
9
=-2:-=B
x 2 vw
x=0:-t24 “
2
1 9
more
x (+2)
=—dinx-2 ine +2) +e
2 2
a v 6)
4.4 25
= |x+5+——a&k
if x—5 we
= +5x+25In(—5) +e
“ (4)
[22]
QUESTION 5
5.1 5.1.1
[lie -a-ne y
=[Ssin '4 +2 Vi6—3? ~ 4x42; v
= (4,566-0)
=4,566
5.1.2 ‘
n{Vi6—x” -(4-x)'de ¥
CO)
3 2 3
= nfl6x ~*~ 16x 4 8 Eyre
3 2 3
(G3)
= a[(21,333) — (0)]
=67,02 4)
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-9-
MATHEMATICS NS
MARKING GUIDELINE
5.2 eo
=-S—]; ¥*
-s
5.1...
= sigue v
5
=-(@-l) v
s
=> ¥
S
QUESTION 6
6.1 ay _ tan” y
dx cosee*x
1 1
ee dx
tan’ y y coseex
cot? ydy = sin? xdx
“ v
(cosec’y —1)dy = 500s 2x)dx
“1
-coty-yerx-
2
sin 2x
cr
6.2 for which y =2 and y'=-3 when x=1
ad ~1x3 2
dy _-ix 43%
+iaxt+e #
ax 6
-x* 3x mm?
y= + +— +ex4d
24 12
~3=3,725+¢
-6725=¢ vy
2=-4946 +d
6,946=d _
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Ww
T960(E)(N25)T
(4)
11)
6)
(4)
19]
TOTAL: 100
y, higher education
mie & training
wi, Department:
4 i; § — Higher Education and Training
as aoe REPUBLIC OF SOUTH AFRICA
T970(EMA7)T
AUGUST EXAMINATION
NATIONAL CERTIFICATE
MATHEMATICS N5
(16030175)
7 August 2015 (Y-Paper)
13:00-16:00
This question paper consists of 7 pages and 1 formula sheet of 5 pages.
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(16030175) -2- T970(E)(A7)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N5
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1.
. 2
Answer ALL the questions.
Read ALL the questions carefully.
Number the answers according to the numbering system used in this question paper.
Show ALL intermediate steps and simplify where possible.
ALL final answers must be rounded off to THREE decimal places,
Questions may be answered in any order, but subsections of questions must be kept
together.
Questions must be answered in blue or black ink.
Write neatly and legibly.
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(16030175)
-3- T9TOE MATE
QUESTION 1
1.1
Determine the following:
2x
Ll Iny=lim2*-
nye tan x (3)
1.1.2 The value of y (1)
1.2 Determine lim Inseo x
x30 yer * (3)
7]
QUESTION 2
2.1 Given: y= cos ecx. Determine the derivative a making use of the derivatives
ofsinx andcos x. GB)
2.2 Determine 2 in each of the following cases: {simplification is NOT required)
xla
221 yo\gma 7
2.2.2 y=e™ In(ax? —bx +e) 3)
2.2.3 y= aresin(L0"*) 2)
2.3 Determine = with the aid of logarithmic differentiation if} y = (cos3x)"?*" (4)
2.4 Given: y = arceos x
ZAt Make a neat rough sketch of the graph of y for the range [0; 7] @)
dy
2.4.2 Derive a formula to determine GY if ¥ = 4reoos x (2)
25 Given the implicit function sin(5x — 2y) =4
Calculate the derivative &
dx (4)
[25]
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(16030175) -4- T970(EAT)T
QUESTION 3
3.1 Given: f(x) =2x* — 4x? -8x +8
3.1.1 Determine the coordinates of the turning points of f(x). (2)
3.1.2 Draw up a table of values for x and f(x) with x ranging from x= -2 to x=4. (2)
3.1.3 Draw a neat graph of f(x) between these values. (2)
3.1.4 An estimated root of the graph is 0,7.
Use Taylor's/Newton method to determine a better approximation of this
root. (4
3.2 An open cardboard box with a square base is required to hold 108 m’.
Find the dimensions that require the least amount of material
Neglect the thickness of the material and waste in construction.
V=x*h
A=4xh+x?
h
x
x
(5)
3.3 An electric current is varying through a resistor of 16€, The rate of change of the
current is 0,6 A/s when the current is 12A.
Find the rate of change of power developed with the current 12A.
Hint: P=1°R (4)
{19]
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(16030175) -5- T970(E)(A7)T
QUESTION 4
4 3_ ey? _
41 Determine j 3x0 t3x0 — Sx tan1 by resolving the integral into partial fractions.
xt 4x2 6)
42 Determine [vax in each of the following cases:
4.2.1 _ sin? x
secx (3)
x
4.2.2 y
“2x? 43 (3)
4.23 ee
1+ Q)
2
424 __*
o 3x2 (4)
43° Given: [ 7.2) = FO - fF @Oa@ax
Determine: f 2x.e* dx (4)
[21}
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(16030175) -6- T970(E)(A7)T
QUESTION 5
1
3.1 The sketch shows the area bounded by the curve x= 4 ;x=y’ and theline y *5
y
ay
1
ba F
« »X
v
~
5.1.1 Prove that the point of intersection of the two curves equal (2,52; 1,59). (3)
5.1.2 Calculate the area of the bounded region. (4)
5.1.3 Calculate the volume generated when the area in QUESTION 5.1.1 rotates
about the Y-axis. (4)
6.
5.2 Calculate the value of : —
1
vl-x7 (4)
5.3 Determine the second moment of area of a rectangular lamina with coordinate
(a;sbjabout the axis parallel to one side of the lamina. The distance from the
representative strip to the axis is "x". The lamina’s length and breadth is "a" and "b"
tespectively.
q (a;5)
x 4)
[19]
A
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(16030175) 1 T970(E)(A7)T
QUESTION 6
‘ . ees
6.1 6.1.1 Solve the differential equation m =e
x
6.1.2 Determine the equation of the curve through the point (1;,-2).
6.2 Determine the general solution of the differential equation
(l- y? dx = -2xydy
TOTAL:
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(3)
2)
(4)
19]
100
(16030175) -1-
MATHEMATICS N5
FORMULA SHEET
Any applicable formula may also be used.
TRIGONOMETRY
sin’ x + cos’ x = 1
1 + tan? x= sec? x
1 + cot” x = cosec” x
sin 24 =2 sin A cos A
cos 24 = cos"A - sin?
tan 24= 2 4n4
L—tan‘ A
sin? A =% -% cos 24
cos’ d =%4 + ’cos 24
sin (A+ B)=sin A cos B+ sin Bcos A
cos (4 + B)= cos A cos B F sin A sin B
an (4S) 15 tan dtan B
sin A cos B= % [sin (4 + B) + sin (4 - B)]
cos A sin B= [sin (4 + B) - sin (4 - B)]
cos 4 cos B='4 [cos (A + B) + cos (4 - B)]
sin 4 sin B= [cos (A - B)- cos (A + B)]
sinx , 1 1
tanx = ;sinx = COS x = :
cosx cosec x secx
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(16030175) 2
BINOMIAL THEOREM
n(n-1)
(xt hy =x" + nx" 1h aa Wt...
DIFFERENTIATION
_f@
f(@)
r=ate
PRODUCT RULE
y=ulx). vx)
dy dy du
S=u.—+7,—
dx dx dx
Suey teen
QUOTIENT RULE
- #Q)
v(x)
du dy
& ates
y dx. dx
CHAIN RULE
y = fu)
wb _y du
dx du dz
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T970E)(ATIT
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(16030175) -3- T970(E)(A7)T
d
fe) = fe) Up (ayae
atl
ax" nax") +e
n+1
a axte
é x & + c
x
x x
a @ Ina e
Ina
1
logex = —
x
1 1
lo;
Bax xlna 2
sin x cos x -cosxte
cos x ~sin x sinxt+e
tan x sec” x In (seox) +e
cot x -cosec” x In (sinx) +¢
sec x sec x‘tan x In [seex + tanx] +
cosec x -cosec x'cotx In (coseex - cotx) +c
oot 1
sin” x =
1— x"
cos” x 7
ie _
yi-x
4 1
tan” x 5
I+x
a 1
cot x Z
I+x
sec? x 1
ax? -1 _
4 1
cosce” x
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(16030175) TITOEMADT
d
Se) ke nic) Sf(dx
1 al (2)
_ sin” |—|+e
Ja? _ x a
1 1 a(x
—tan’|—jt+e
: ‘()
_ —sec|—|+e¢
xx? —a? a a
2
a? — x? © sin' (2 +2 P-P 4c
= 2 aj} 2
t L x-a
— In}: +e
x’ — a? _ 2u.\xt+a
f 1 at+x
—In +e
@—x _ 2a \a-x
INTEGRATION
$9) g(x) dx = £6) g(a) - J £'@) a) de
£2
F(x) dx=lIn f(a) te
att
Leon fe) d= LET +e
f@) __A B
= +
(axt+bjext+d) axt+b cxt+d
f@) A B c
(tay Gta) (tay (wta)®
- (x +a)"
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(16030175) -5-
APPLICATIONS OF INTEGRATION
AREAS
6 b
A, = J yds Ay = { (4 - ya)ax
b b
4, =f xdys4,=f (-m)ay
VOLUMES
boo b
Vy, =a] ¥ dx Ve =f (2 — y3 Jax
b 2 b (2 2a
Van], Vay Vy =n), ( — x3 Jdy
SECOND MOMENT OF AREA
Iy=[) Pda: 1y =? Pad
MOMENTS OF INERTIA
Mass = density volume
M= pv
DEFINITION: Z=m ?*
A b
GENERAL: 1 = [’ rdm = of. r'dv
T970(E)(A7)E
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higher education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE
NATIONAL CERTIFICATE
AUSUST EXAMINATION
MATHEMATICS N5
7 AUGUST 2015
This marking guideline consists of 10 pages.
CM RAL ati. .
- TS Mork4h—
CM) S wetticacke AK
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MARKING GUIDELINE 2- T970E)AT)T
MATHEMATICS NS
INSTRUCTIONS AND INFORMATION
1. Vv equals 1 mark.
2, Half marks are not allocated, unless indicated otherwise.
3, Where formula is required, using the wrong formula is a principle error and NO marks
are allocated,
4. Students should show ALL formula, intermediate steps and simplify where possible.
5. ALL final answers must be rounded off to THREE decimal places (unless indicated
otherwise)
6. Questions may be answered in any order, but subsections of questions must be kept
together. If subsections are separated, the student can be penalised by ONE mark.
7. Where a student copied wrong from the question paper, and the standard of the
question is still the same, the student will be penalised by ONE mark. If the copying
error simplifies the question and makes it easier, the student forfeits the marks.
8. Questions must be answered in blue or black ink. Answers in PENCIL are not marked
as it is regarded as rough work.
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MARKING GUIDELINE
-3.
MATHEMATICS N5
QUESTION 1
11 11.1
Iny = lim
x0 tan x
: 2 44
nye ee sec! x
112 -y=e?=7389%
1.2 sec x. tan x
; secx ¥
tim ie ee v
= lim —"*
~ x90 pF _ yg
__0
2-0
=0 v
QUESTION 2
21 y = cosecy = —
sin x
dy _0-l.cosx
dx sintx ~*
8 oe
dx = sinx.sinx
dy
2.2 2.2.1
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—=-—cotx.cosecx ¥
a&
xha
y= 1Q Sse
vv
a xIna Ina.cos ax+xInaasinax
= == 10cosax "19. costax
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G)
()
3)
[7]
(3)
(4)
MARKING GUIDELINE -4-
| MATHEMATICS NS
222 ye e™ In(ax? ~dxt+e)
v v
Bw be™ In(ax? — bx +2) + of 2b
de ax’-bxte y
2.2.3
y= arcsing 0")
v
4x
gy 4.10 InfO |,
ax
h — (10)?
2.3 Iny=~2x41 . Incos3x
v v vw
3
LD hi ayeni F-2 incos3x-+Vrana1 2s
ydx 2 Cos3x
dy ~Incos3x
== ~3y~2x+1.tan3x
dx WS v-2x41 1”
24 24.1
Y
vn
eed
i
2.4.2 COs y=x
v
—sin y =1
ax
& - ; BUT:sin y=./1-cos’ y
ad -siny
dy _ 1
a ’
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TITOCLEMKAT)T
@B)
2)
(4)
(3)
(2)
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MARKING GUIDELINE -5-
MATHEMATICS NS
2.5
dy
cos(Sx—Zy).[5 eae =0 v
dy ¥
Scos(Sx-—2y)- Ca -2y)=0
he
YL
~ Scos(Sx—2y)
dx —2cos(Sx-2y) v
dy 5
T970(EMATT
a2”
1)
[25]
QUESTION 3
3. 31d 6x? ~8x-8 =0
3x? -4x-4=0
(x+2)(x-2)=0
2
Fan gine2,
y=10,962;y =-8 %
[One mark per coordinate x AND y} (2)
3.1.2 x | -2]-1/ 04 1 2 3 4 v
foo [8 [10] 8 f 2] -8 [2 [40] |v Q)
3.1.3 K
Shape “
8 v¥
of V4 i
A 2)
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MARKING GUIDELINE -6-
MATHEMATICS NS
3.1.4 v
_ __ 1126 = 0106"
—10,66 y
r=0,7 + 0,106 = 0,806 “
3.2 x
A= dap)
A=432x1 4x2”
dA _—432
az x
~432+2x7 =0
x = 216
x=6 4
+2x¥
h=3
33 .P a
a2 RY
dt dt
dP vv
— =2(12).0,6(16
z (12).0,6(15)
aP
=230,4 %
at
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TI70(EMATIT
)
©)
4)
[19]
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T970(E)(A7)T
MARKING GUIDELINE -7-
MATHEMATICS NS
QUESTION 4
3x4 +3x> —5x? -i
“fs ic 4 x? Dow 1 A B
= 3x0 + Ft ——— et ee Be = +
(«+ 2)(x-1) @&+2)G@-1) (*+2) (x-D
v
= i
= [3x?-+ 3 dy fore =1:B=4 and forn=-2:Aars
arm a (x-D 3 3
vv
Tonle lin -pe vv
G)
Give the candidates two marks if long division is not used for partial fractions
4.2 4.2.1 sin? x
" seex
if sin’ x.cos xdv u=sinx
= fu’.cosx, au dy ay
cosx cos x
sa4
_sintx ov
GB)
4.2.2 _ x
© Ox? 43
alee 243% a
= Fin(ax? +3)te yy
OR w=2x?+3
=
u 4x
= In(ax? +3)
@)
.
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MARKING GUIDELINE -8-
MATHEMATICS NS
4.2.3 pica
acre
e” 2
f 5 du(l+x) u =arotan x
l+x
=e" te de=(14+x")du *
= ent tg
4.2.4 4
i xh 9 y _using longdivision
30 9 3x-2
2.
a 24 ingr-2) +0 vvv
6 9 27
43 22 2x
= 2x [2d vv
2 2
e*
2
dx vo
=xe~ —
+¢
TOTOCEKATIT
2)
@)
4
[24]
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“MARKING GUIDELINE ~9-
MATHEMATICS NS
QUESTION 5
5.1 5.1.1
4_2
—=y
y
4zy?
159=y ¥
=
1,59
x= 2,52
5.1.2 “i 4
=-y dy ¥
os ¥
3
=[4iny-“ he y
= (0,515) ~(~2,814) %
#3329”
5.1.3 439
a [(6y7)-O* ay ¥
6,5
~16 yas ¥
= a[— 78
= n[-12,095 -(-32,006)] “
= 62,552"
5.2 :
$4
|e
1 1-x?
=[aresin x]¢ v
= (0,551)—(1,57) VY
=-1,02
5.3 a vv
2
J bx'de fone for limits; one for integral}
b 3
“kv
ot
3
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TITOE(AT)T
2)
A)
4)
4)
4)
{19]
Please turn over
QUESTION 6
dl @
61 611 dy ogy
e”
[evay = | erax
o
y
apr te (3)
612 ~-e"CDeetgc
-e?elec
c =-10,107 ¥
—ey = e* - 10,107 ¥
2)
6.2 1 y
dye
-2x I-y? ay
1 1 vv
Zins Find-y') +e
(4)
1]
TOTAL: 100
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(16030175) -2- T970(E)(A1)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N5
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1. Answer ALL the questions.
2. Read ALL the questions carefully.
3. Number the answers correctly according to the numbering system used in this
question paper.
4A. Show ALL intermediate steps and simplify where possible.
5. ALL final answers must be rounded off to THREE decimal places (unless indicated
otherwise).
6. Questions may be answered in any order, but subsections of questions must be kept
together.
7. Questions must be answered in blue or black ink.
8. Write neatly and legibly.
Copyright reserved Please turn over
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Qe y higher education
sie & training
afo, | Department:
A) Higher Education and Training
Oy REPUBLIC OF SOUTH AFRICA
cooooacoco
T970(EA1)T
APRIL EXAMINATION
NATIONAL CERTIFICATE
MATHEMATICS N5
(16030175)
1 April 2015 (Y-Paper)
13:00-16:00
Scientific calculators may be used.
This question paper consists of 6 pages and a formula sheet of 5 pages.
Copyright reserved Please turn over
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(16030175) “3.
QUESTION 1
Lt
112 Prove that: lim 282° — 9
x30 Ke =
Show ALL calculations.
1.2 x
Consider the function f(x) = .
f v2x—3
Determine whether f(x) is continuous for x =1
QUESTION 2
2.1 . toes 1-x we
Determine the derivative of (x)= eae from first principles.
+x
2
2 Determine 2 in each of the following cases:
aaa yeeton® + tant (hy
x
2.2.2 y =cos ec2x.5™
2.2.3 _ Ssin 2x
In(cos x)
2.3 Given: y=arcsinx
2.3.1 Make a neat rough sketch of the graph of y for the range [0;7]
2.3.2 : .
3 Determine s if y = aresin(3x)
am Determine % if xsin x + ysin y = 76,29
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Determine the following limits:
11.1 : 3x
lim——;
xe x — +2
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T970(EMA1)T
Please turn over
@)
GB)
1)
8]
(4)
3)
Q)
3)
2)
Q)
G)
(16030175) 4s T970(E)(AL)T
25 Determine 2 with the aid of logarithmic differentiation if:
[23]
QUESTION 3
3.1 Given: f(x) =x? —5x-8
3.1.1 Determine the coordinates of the turning points of f(x) -
Show ALL calculations. (3)
3.1.2 Draw up a table with values 0 < x <3for the equation x° —-5x—8 =0.
Use the table and turning points to make a neat sketch of the function of
F(x). G)
3.1.3 Let x =3,7 be an estimated value of the rout. Use Taylor/Newton’s
method ONCE to determine a better approximation of THIS root to FOUR
decimal numbers. (4)
32 Prove that a rectangular field with a perimeter of 100 m will have the greatest area
when it is a square. (5)
3.3 The distance of a particle is given by f(x) =10tanx. f(x) gives the distance in
metres after x seconds
33.1 7 . +
Determine the velocity of the particle after 7 seconds, Q)
3.3.2 Determine the acceleration of the particle at this moment. (3)
3.3.3 Determine the distance travelled by the particle in 4 seconds. (2)
[22]
QUESTION 4
4.1 Determine [ ydx if
4.1.1 y= (x? -2x)? @B)
4.1.2 y=x.secx.tanx (4)
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(16030175) -6-
TITO(EYADT
3.2 Determine the moment of inertia of a circular lamina of radius r and mass m about
an axis through its centre and perpendicular to the plane of the lamina.
ry
v (4)
[16]
QUESTION 6
6.1 : : 2. ax
Solve the differential equation sin 1 +cos2yp=0
B)
6.2 a . @ y s 2 .
Determine the particular solution of ars = x° +2x° — x, given that
dy
—=Ly=2.x=1
hk Ly
d’y
Change the subject to ax? ; before solving (5)
(8)
100
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(16030175) -5- T9I7OCENAI)T
4.1.3 _ i
> Ty 6x (3)
4.1.4. _ +inx)
=" @)
45. __tanx
v= Incos x (2)
4.2 Given: 5s ~11
x? —2x-8
4.2.1 Using partial fractions, determine the numerical values of 4 and B
oxcll A,B
x? —2x-8 (a+b) (ex+d) (3)
4.2.2 : 5x-11
Hence, determine | —————
J x? -2x-8 63)
[21]
QUESTION 5
3.1 The sketch shows the area bounded by the curves xy =6 and y+x=5
v \
5.1.4 Show that the points of intersection of the graphs are (2;3) and (3;2). (3)
5.1.2 Calculate the area of the shaded region. (4)
5.1.3 If the shaded area revolves about the Y-axis, calculate the volume of the-
solid generated. G)
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(16030175) 1
MATHEMATICS N5/ WISKUNDE NS
FORMULA SHEET / FORMULEBLAD
Any other applicable formula may also be used.
Enige ander toepaslike formule kan ook gebruik word.
Trigonometry / Trigonometrie
sin? x+cos* x =1
1+ tan? x =sec? x
1+cot? x =cosec?x
sin 2.4 =2sin A.cos.4
cos2.4=cos? A—sin? A
2tan A
tan 24 =————-—
1—tan* 4
sin? 4-4 cos24
2 2
cos” A =F 450082
sin(A + B) = sin A.cos B t sin B.cos A
cos(A + B) = cos 4.cos BF sin A.sin B
tan. dt tan BE
tan(4+ B) = 94
1-tan A.tan B
sin A.cosB = WA [sin(4+.B) +sin(4—B)]
cosAsinB = Molsinc +B)-sin(4- 3B)
cos. d.cosB = ¥, [cos(4 +B) + cos(4~ B)]
sin A.sin B= Vo[eost — B)—cos(A+ B)]
cos x cosecx secx
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(16030175) -2-
BINOMIAL THEOREM / BINOMIAAL STELLING
(xh) =x" +nx"th ee +...
DIFFERENTIATION / DIFFERENSIASIE
--£@
f@
roate
PRODUCT RULE / PRODUKREEL
y= u(x)v(x)
=uv'tyvu'
QUOTIENT RULE / KWOSIENTREEL
aC)
v(x)
du
di & ak
a& v
CHAIN RULE / KETTINGREEL
v= fu)
yy _ dy du
de dud
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T970(E)(A1)T
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(16030175) 3- T970(E)(AL)T
d
f) rade) | Rice
a al La
ax RnaX
a 0 n+]
; x axt+e
[4 ée
; e+e
a @ Jna x
a
om +c
nda
log, x 1
x
log, x a _
xIna
si cOosx
nx ~sinx —cOosx+e
— secu sinx+¢
tan x 2 In(seex)+¢
cotx —cosec*x .
In(sin x) +c
secx secx, tan.
cosettcat Infsee x + tan x] +e
Cos eck In[cosecx —cotx} +e
sin? x 1
I-x? —
cos’ x =
1-x? ~
tan x a
1+x? =
-1
cot" x [4x2 —
sec x xv x? -1
cosec'x xx? —1
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(16030175) “4 T970(E)\AL)T
fe) < ft) (Reo
a _ sin'()+e
1 1 44g
= —tan™(—}+¢
a +x? a a
1 Leo (j4¢
xvx?—a’ _ a a
ax? © sin'(2) +2 Va? -¥ +e
_ 2 a 2
1 1 x-a
xa? — 2a Mao
5 ! 7 sin At *) be
a’ ~-x _ 2a -a-x
INTEGRATION / INTEGRASIE
[f@2'@) = f@.8@)-[f'@.g@ax
LR) 4 _
Vig Baa ete
#yyrt
[cor fear -29— +6
n+i
f@) A, B
(ai+bex+d) axt+b cxid
fe) 4, Bc, Z
(t+a)” (eta) (eta)? (x+0)° (x+a)"
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(16030175) “5 T970(E)(A1)T
Applications of integration / Toepassings van integrasie
AREAS
b bs
A, = [yds 4, = [02-year
b
4,=[,x05 4,=[@—ade
VOLUMES
V, = aff yd; V, =A) Oy - 9 bac
bs bs
V, =a] dy, V, =a] Ge a ay
SECOND MOMENT OF AREA / TWEEDE AREAMOMENT
I, = ['r7aa; 1, = [rad
MOMENTS OF INERTIA / TRAAGHEIDSMOMENTE
Mass = density « volume / Massa = digtheid x volume
M=pV
Definition / Definisie: J = mr?
General * Algemeen: J= [Pam re pf rar
la a
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0006000
A
higher « education
& training
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
_ MARKING & SUIDECINE |
|
i
NATIONAL ‘CERTIFICATE
APRIL/JU NE/AUGUST/NOV. EMBER EXAMINATION
MATHEMATICS NS
(EQURTH PAPER)
BACKUP 2014
—T
|
_t
ty This niarking guideline consists of 11 pages.
IM: T. Molcleln <GAe-
CM: RMatiener Libattlar
CM: S& Tow ar Apress
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MARKING GUIDELINE -2e
MATHEMATICS NS
QUESTION 1
Ll Determine the following limits:
LL . 3x
lim———->
sae x" — x42
vl
im =2% OW
xre2x—]
LL2 Prove (show ALL calculations) that:
12
Consider the function f(x) thie Fh ‘
oN 2k :
, Insecx
lim ———
x0 yer
LHS
secx.tanx J
lime
a0 1e°% — xe
=0
we
. tan x
= lim
290 @°* — xe
x
Determine whether 7 eS} is egntidtious for x=1
SQ=
v
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T..(E)BU)T
(2)
3)
@)
[8]
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MARKING GUIDELINE 3- T..AEXBU)T
MATHEMATICS NS
QUESTION 2
2.1 : oe ; 1-x ee
Determine the derivative of f(x) = aa from first principles
x
l-(xt+h) 1-x,1
= limf—-——— - —-].— Vv
fO)= eh Bee
+ _ -d-x~A3+x)-U-x)G+x+h), 1
=] _
FO)= ae (tx +hY3 +2) hy
—4h 1
PO lina hasn hv 4
-4
f@s= aay Vi " “
22 Determine 2 in each of the following cases,
2.2.1 ye giteasv=) +tan?(hy
x
ye cos x + tan’ ie Ye J
ay | iV | 21-1
ss = + 3 tans —. a
_ sin nr faa see > 3)
2.2.2
a oe (2)
2.2.3 _ Ssin2x
In(cosx)
v a Vv
o 10cos2x.In(cos x) -—5Ssin 2.
dx [Incosx?? y (3)
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MARKING GUIDELINE
-4- TeufEXBU)T
MATHEMATICS N5
rw
te
Given: y= arcsin x
2.3.1 Make a neat rough sketch of the graph of y for the range {0; 7]
1 mark for the shape
1 mark for any intercept
Q)
2.3.2 _. ay,
Det = if
etermine 74
y =aresin(3x)
(2)
24 Determine » if
xsinx+tysiny = 76,29;
4 Vo gv tv “ae
sinx + xcosx+—) sin y +.y COs ye 10
sin yr yoos y) =,
sinx~ xtosx
dy _—sing—xtoex
dx = siny+ ¥cos3
a rs 6)
25 Determine 2 with the aid of logarithmic differentiation if:
core
yex
Iny =cot3x.Inx
“i
~v
lay —3eosec’3x.Inx $cot3x- v
yax x
ay 2
= = y[-3cosec’3x.Inx+
ye x.
cot 3%) J
@)
[25}
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MARKING GUIDELINE ~5- T...E)(BU)T
MATHEMATICS NS
QUESTION 3
3.1 Given: f(x) =x’ -Sx~-8
3.1.1 Determine the coordinates of the turning points of f(x)
3x’-5=0 Vv
x= 4/1667
x=1291 or x=-1291 ¥
y=-12303 y=-3,697 Vv ‘ QB)
31.2 Draw up a table with values 0 < x <3 for the equation x —5x—8=0,
Use the table and turning points to make a néatsketch of the function
of f(x).
x {0
y_ |-8
1 Mark for shape
1 Mark for any
shown turning
point
, @)
3.1.3 Let ye 37, Eo an estimated value of the root. Use Taylor/Newton’s
method ONCE to determine a better approximation of this root to FOUR
decimal numbers.
; Vv
3D =53.)=8 __ 24153 _ 0.6696
3(3,7)° -5 36,07 ¥
root :3,7 — 0,696 = 3,0304 V (4)
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MARKING GUIDELINE 7 T..ACEMBUYT
MATHEMATICS NS
QUESTION 4
4.1 Determine f yde if
44d y=? —2xy
fx = 4x3 + 4x? dy
aia x' +40 B)
4.1.2 y= x.secx.tanx
f=2x g =secxtanxy
f=l g =Seex —
Vv
xseox~ [sec ke
Vv
)
4.1.3
a ee V7 ce 6)
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MARKING GUIDELINE 6- T...E)QBU)T
MATHEMATICS N5
3.2 Prove that a rectangular field with a perimeter of 100 m will have the greatest area,
when it is a square
P=2x+2y
100=2x+2y
50-x=y v
A= xy
A=x(50-x)
A=50x-x* b
50-2x =0V jh
x=25 WV os a
y=50-25=25 ¥
therefore: rectan gle is a square % Sinadt (5)
3.3 The distance of a particle is given by f(x) . f(x) gives the distance in
meters after x seconds a
3.3.1 ;
Determine the, velocity of the particle after = 4 seconds
vel0sectx ¥
v=l0see(T she
vs20mts ev (2)
3.3.2 Determitéthe acceleration of the particle at this moment
a=10.2secx,sec x tan x v
a= 20sec’ x.tanx
a= 20sec? (~).tan(=
Guten
a=40m/s? (3)
3.3.3 Determine the distance travelled by the particle in 4 seconds.
v
s=l0ian4=11,578 ¥ Q)
[22]
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MARKING GUIDELINE -8- T..(E)(BUIT
MATHEMATICS NS
4 N . @)
4.1.5. f tanx a
Incosx
derivative: * 2 ~tanx
J cose
-1 j —tanx x
Incosx
—In(Incosx)+e Vv %
42 Given
424
$x-11= aGrt2)+B0-4)
jorx = 2: 2 =B V
forx =4 a A a
oo Q)
4.2.2 Hence, determine
{ 5x-11
x? -2x-8
3 7
_f.2 2 a If the values of A and B are
“l@-4 + (+2) incorrect, mark with incorrect
3 7 J values
= ine 4) +7 ine +2) +e 6)
f24]
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MARKING GUIDELINE ~9- T...0E(BU)T
MATHEMATICS NS
QUESTION 5
3.1 The sketch shows the area bounded by the curves xy=6 and y+x=5
‘
aN
4
¥
s=5=5
x
x?-$x+6=0 ,
(x -3)(x-2)=0 * q
x=3 or xE'P ay on
ya2 or ya3o ws 3)
‘e rs
5.1.2 Caliah the aréa of the shaded region
=[Sy-7 lyk y
=(15— 4,5 -61n3)~(10—2-61n2)
=0,067 ¥ {4)
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MARKING GUIDELINE -10- T..AEXBU)T
MATHEMATICS NS
5.1.3 If the shaded area revolves about the Y-axis, calculate the volume of the
solid generated.
; 6
z[G-*-Cyay v
3 ¥
3
=2[25-l0y+y?~36y%dy V
2
3
=al(25y~5y” cae E V
= nl(95 ~45+9+12)—(50~20+5 +18] VB,
=0,3337 ¥ , @)
=1.047 ‘ :.
§.2 Determine the moment of inertia of a circular lampina’@f radius r and mass m about
an axis through its centre and perpendicular to the plane of the lamina
(4)
(16)
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MARKING GUIDELINE -H- T..{EMBU)T
MATHEMATICS N35
QUESTION 6
6 Solve the differential equation sin xe +cos2y=0
sinxdx =—cos2ydy
V_sin2y 4
—cosx =—-——+¢
(3)
6.2 : : : d’y 5 2 +
Determine the particular solution of sar =x’ +2x° — x, given that
dy
® oy ya22=1
“a
ay Ms
Change the subject to ds” before solving. ;
2
a
‘
1=02+1-l+e
O8sc ¥
1 $121 .
2=—+2-= +d i
303 2 rs Y
1333=d ¥v
x x? x? aq!
Yay tpg POG +1933 v 6)
[8]
TOTAL: 100
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Qe y higher education
le & training
afaols Department:
4 The Higher Education and Training
Se REPUBLIC OF SOUTH AFRICA
cooogooo00
4
S
‘
+
T960(E)(N17)T
NOVEMBER EXAMINATION
NATIONAL CERTIFICATE
MATHEMATICS N5
(16030175)
17‘Noyember 2014 (Y-Paper)
13:00-16:00
Scientific calculators may be used.
This question paper consists of 7 pages and a formula sheet of 5 pages.
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(16030175) -2- T960(E)ONI7)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N5
TIME: 3 HOURS
MARKS: 100
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.
4. Show ALL intermediate steps and simplify where possible.
5. ALL final answers must be rounded off to THREE decimal places (unless indicated
otherwise)
6. Questions may be answered in any)order, but subsections of questions must be kept
together.
7. Use only blue or black ink,
8. Write neatly and legibly.
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(16030175)
QUESTION |
LL
Determine the following limits:
LL lim sin.
290 In tan x
L12 lim sin(2x —12)
96 x G
Fay
1.2 Determine the value(s) of x for which f(x) is continuous if f(x) = ai
7
QUESTION 2
2.1 Determine the derivative of f(x) = 43x from first principles.
2.2 Determine 2 in each of the following cases;
(simplification of answers et required)
2.2.1 y=sin(cos/2@ 1)
2.2.2 y=e™ Infax? +br—c)
ox
a
» sec3x
2.3 Determine a with the aid of logarithmic differentiation ift
¥ = (cos5x)*
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@)
2)
@)
13]
4)
@G)
@
Q)
(4)
(16030175) -4- T960(E)(NI7)T
2.4 Given is the sketch of y = aresin x, x €[-1;7]
aA
A
B/ -z
2.4.1 Give the co-ordinates of A and B
42 If y = arcsin(tan 4x), give the derivative 2 of ¥
d
2.5 Determine the equation of the tangent to the curve xy? +3x? = x?y +12 at the point
(231)
QUESTION 3
3.1 Given: f(x) =2x°=4x*42x41
3.1.1 Determine the coordinates of the turning points of f(x)
Show ALL the calculations
3.1.2 Draw up a table of values of x and f(x) for values of x from -1 to 2 and
makea neat sketch of this part of the curve.
3.1.3 If the value of the negative root is estimated as -0,3 use Taylor/Newton's
method ONCE to determine a better approximation of THIS root correct
to FOUR decimal numbers.
3.2 A metal ball is covered by a layer of ice. The diameter of the ball is 100 mm. When
the layer of ice is 5 mm thick, it melts at a rate of 10 000 mm*/s.
At what rate is the thickness of the ice decreasing?
vesne
3
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(2)
Q)
(5)
[24]
(3)
(4)
4)
6)
(16030175) 5 T960(E\NLT
3.3 A rectangular wall is to be erected on a river bank. Building cost budgeted for is
R5 600-00. The building cost of the two parallel walls are R10-00 /m and the cost of
the wall parallel to the river bank is R15-00 / m (see the sketch).
river
x x
¥
Calculate the dimensions of the wall to enclose a maximum. area for the money
available. (5)
[21]
QUESTION 4
41 Determine f ydx in each of the followitig cases
41.1 y = tan 3x.sec 3x (3)
41.2 y= (8-5x¥ (2)
4.1.3 a
Fae 3)
4.14 v=sin(e yp (4)
415 y= x? In8x (4)
42 7 7 7 4x—2
: Use partial fractions to determine f ; ad
x’ -4x
Show ALL calculations. (5)
(21]
QUESTION 5
tx? —3x
5.1 Evaluate the definite integral [ dk
yal )
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(16030175) -6- T960CE)(NI7)T
5.2 The sketch shows the area bounded by the curves y=3x° and i =x,
intersecting at (0;0) , (-2:-8) and (2:8)
-
>
b 2
$.2.1 Calculate the area of the shaded region. 4
5.2.2 If the shaded area revolves about the X-axis, calculate the volume of the
solid generated, 6)
5.3 Determine the second momentiof area of a uniform rectangular lamina of area 'ab'
about an axis which bisects thédamina.
Make use of the information on the given sketch.
a
a,
4)
‘ (4)
[17]
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(16030175) -7- T960(E)(NI7)T
QUESTION 6
d’y
6.1 Determine the general solution of: mo xr+ert4
1)
6.2 Determine the particular solution of (1+ x)cos’ ydy = sin? yd, given that y(1) 7 (5)
9]
TOTAL: 100
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(16030175) Rk.
-1-
FORMULA SHEET
Any applicable formula may also be used.
Enige toepaslike formule kan ook gebruik word.
Trigonometry
sin* x+cos? x=1
1+ tan’ x = sec? x
1+cot? x =cosec?x
sin2.A = 2sin A.cos A
cos24 =cos? A-sin? 4
tan2A _——
1-tan‘ A
sin? A= Ere |
2 2
cos’ A= rey |
2 2
sin(A + B) = sin A.cos B+ sin B.cos.4
cos(A + B) = cos 4.cos BF sin Asin B
tan.d + tanB
tan(4 + B) =F nA tan B
sin A.cosE = Vy{sin(4 + B)+sin(A—B)]
cos A.sin B = 4 [sin(4 +.B) ~sin(A— B)]
cosid.cosB = ¥, [cos(4 + B) +cos(4— BY]
sin Asin B= V5 [cos(4- B) -cos(4+ BY]
sinx..
tan x = ssinx = 3cosx =
cosx cosecx secx
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(16030175) -
BINOMIAL THEOREM
(eth) =x" + nxn MAD 7 D rays vee
DIFFERENTIATION
iO)
F'(@)
F=at+e
PRODUCT RULE
y¥ = u(x).v(x)
wb ®t
a = Me +y. hk
=uy'tvu'
QUOTIENT RULE
= 4G)
v(x)
CHAIN RULE
v= fu)
Bh _h da
ad du dk
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(16030175) -10- T96O(E)(NI7)T
-3-
d
re) — fe) | fooee
dk
rd
ax” nL +c
nox n+l
a 0 ax+c
e e* 2 +0
a a’ .ina
+e
Ina
log, x 1
x _
log, x I
xing
sinx HOSS, —cosx+c
cosx = a sinxt+c
tanx = * In(secx) +¢
2 .
cotx —cosee’x In(sin x) +e
secx ox. tan. In[secx + tanx]+c¢
Cosecx —cosecx.cot x Infcosecx —cotx} +¢
sin’ x a
cos? x
tan“ x
cot x
sec! x
cosec 'x
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(16030175) “lle T960(E)\(N17)T
“4.
d
f@) 510 J Hosi
sin*(Sj +e
a’ -x? 7 a
1 1 pat (“the
a +x? a Cs
1 _ J se01(%) +¢
xvx? -a* a a
2
ax? © sin (4 Va? — x? +e
“wy 2 a 2
1
1, x-@
2 zn +e
* 7a — 2a & + 2
7 : 7 I (cs = =
a7x P— 2a -a-x
INTEGRATION
Jf@e'@=fa.8@)-| PO.s@ax
[Fea-inger+e
fU@orP ea -20™ 6
n+l
f@ __A B
(ax+bYex+d) axt+b cxt+d
f@)___A B Cc 4
(xtay’ (+a) (x+a)? “Gia? fete
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(16030175) 19-
5
Applications of integration
AREAS
4, = de; 4, =f, — yee
A, =f xdy; A, =[@ — dy
VOLUMES
_ fh yt. _ 2 2
V,=2[ yds, V, =a (7 ~ 9)
b ‘b
Vy, =a x dy; Vv, =ai) (x,? — x, dy
SECOND MOMENT OF AREA
o2 2
1,=[r dA: Le[r dA
MOMENTS OF INERTIA.
Mass = density x volume e
M=pV
Definition I = mr?
General 7 = [dm = pilav
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T960(EKN17)T
i
six yY higher education
== SF & eraining
no Sg
fee, |) DOspartment.
OO Higher Education and Training
SS REPUBLIC OF SOUTH AFRICA
gooce000
Sa
MARKING GUIDELINE
NATIONAL CERTIFICATE
NOVEMBER EXAMINATION
MATHEMATICS NS
17 NOVEMBER 2014
This marking guideline consists of 8 pages.
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MARKING GUIDELINE
2 T960(ENKNI7)T
MATHEMATICS NS
QUESTION 1
1.1 11.1 li Insin x
x0 In tan x
cosx
= lim-Sin
90 sec” x
tan x
= lim cos? x
=1
(4)
1.1.2 i -
lim sin(2x —12)
me x-6
Jim 2c0s(2x -12)
296 1
=2 2)
1.2 x-1
f=
(x-1)(x+1D =0
continuous : x € R;x # +1 2)
[8]
QUESTION 2
2.1 1
. V3(0+h)? 34x
SF @) = lim
0 h
1 a 3 1
3x2 43, 2h- 13. 2h? V3x2
=lim 2
h30 h
a 3
ned 43, 2 ht.)
~ lim —2. 8
0 h
_ 3
2vx (4)
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MARKING GUIDELINE 3. T96O(EXNI7)T
MATHEMATICS N35
2.2 2.2.1 yr sin(cos (2a? —)
D_ cos(cos 208 —1).-sin (20-F)L00-2)? - 6x? @)
a
2.2.2 y=e™ In(ax? + bx —e)
& _ be™ In(ax? +bx—c)+ em 2d
ad ax” +bx—c (2)
2.2.3 10>
y=
sec3x
dy _ 9.10°*.1n10.se03x — 10°*.3sec3x.tan 3x
dx (sec3x)? (2)
2.3 y =(cos5x)"
Iny = x.Incos5x
ly = Lincos5x+x,—> Si 5®
cos5x
& = y[Incos5x —5xtan 5x]
e co)
2.4 2.4.1 A(z) B(-1-2) (2)
2.4.2 ay _ 4sec? 4x
de (tan 4x)? Q)
2.5 3 2 dy 2d
y +3xy et OX PV TX de
® oxy" -x") = 2xp- y? -6x
dy _ 2xy -y? -6x
dk 3xy?-x?
Ue2Md)-1412
“3(-2)0)- (2)? = 10
ye a1, te
10
14
=i9t?
-O0,4=c
Equation : y = xT, -0,4
10 6)
[24]
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MARKING GUIDELINE -4- T960(E)(N17)T
MATHEMATICS N5
QUESTION 3
3.1 3.1.1 6x? -8x+2=0
3x? -4x41=0
(3x—-1)(x-1) =0
x=0,333 or x=]
y=1296 or y=l @)
@)
3.1.3 _0.3)3 — 7 =
e= _ 20,3)" — 4(-0,3)* +.2(-0,3) +1 _ 0,854 = 01729
6(-0,3)? — 8(—-0,3) +2 4,94
root : —0,3 + 0,1729 = -0,1271 (4)
3.2 Ve 4 nr
3
DY nag & NB: Radius =50+5= 55mm
dt dt
2dr
10000 = 42(55)° —
at
dr
0,263 = —
, dat G)
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MARKING GUIDELINE -5-
MATHEMATICS N5
2(10x) +15y = 5600
5600 20x
15 15
3.3 A = 373,333x -1,333x"
Derivative :373,333 — 2,666x = 0
x =140,035m
y = 186,62 m
QUESTION 4
41 41.1 y= J/tan3x.sec? 3x
u =tan3x
_ du
3sec” 3x
[vu-sec? 3x =
3sec* 3x
3 a
a2 == (tan3x)? +c
4.12 y=(8-5x)3
f@-5x)° ax
4 4
= B-5x)" 5... 8-59)"
4-5 20
4,13
=.
4+9x
(et
9 G2)
11, ix
oom re
3 3
alia? 46
6 2
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T960(E)(NI7)T
(5)
[21}
@
@)
8)
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MARKING GUIDELINE -6-
4.2
MATHEMATICS N5
4.14 y =sin(e™)e”
2x
u=e
_ du
_ 26"
. Me
fsinue™ <=
t a
=—-cos(e*) +e
2 (e"*)
4.1.5 y =x? In8x
f=in8x g =x?
= Pe
f=— gry
3 3
an Je?ingxdx = In 8x tra
3 3
~ nsx = +
3 nex 9 c
4x-2
oat
4x-2 A B
2 =—t
x?-dx x (x-4)
4x—2 = A(x—4) + B(x)
1
=0:-=4
Sorx 2
7
=4:-=B
Sforx :
i 7
2, 2
x (x-4
al inxs Zinge—4)+e
2 2
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A)
i)
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MARKING GUIDELINE -7- T960(ENI7)T
MATHEMATICS N5S
QUESTION §
5.1 aye
(2 3X oe
5 xt
x-4
x+1)x? -3x
x tx
-4x
Ax-4
4 - remainder
x? 3
= — 40+ Aine + OE
= (32-32 +41n9) —(12,5—20+41n6)
= 9,122 @)
[(2e—3e?e
5,2 $.2.1
i
12x? 3x",
es 4 To
=(24—12)—(0)
=12u?
4)
22
° V= al [a2x)? -@x°)? fee
V = af (44x? - 9x8 Je
144x° 9x’,
Vent -
am; 7 Io
V =a[219,429 -0]
V =689,355u°
(5)
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MARKING GUIDELINE -8-
MATHEMATICS NS
5.3 4
2
2] x? bd
o
34
xf
= 20,1
3
a
= 2-0
A ]
_a’b
12
QUESTION 6
6.1 2
Caw set ss
3
Bo® et paxse
a& 3
= et 42x tered
av)
6.2 cos*y,
sin’ y l+x
cot? ydy=— ae
l+x
—coty~y=Inl+x)+e
a Tn
—cot——— =In(l+f+e
a4 (1+)
0,215=In2+e
-0,478 =e
equation : —cot y — y = In(1 + x) -— 0,478
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T960(E)(N17)T
4)
[17]
(4)
G)
[9]
TOTAL: 100
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
Tl 040(E)(A9)T
APRIL EXAMINATION
NATIONAL CERTIFICATE
MATHEMATICS N5
(16030175)
9 April 2013 (X-Paper)
09:00~12:00
Calculators miay be used
This question Paper consists of 6 Pages anda 5-page formula Sheet.
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DHET “2 TLO40(E(A9)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N5
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1, Answer ALL the questions.
2. Read ALL the questions carefuily
3. Nuraber the answers according to the numbering system used in this question paper.
4. Questions may be answered in any order but subsections of questions must be kept
together.
5. Show ALL the intermediate steps. Siniplify where possible.
6. ALL graph work must be done in the ANSWER BOOK.
7. Questions must be answered in blue or black ink.
8. Write neatly and legibly.
t
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DHET 3.
QUESTION 1
Determine the values of the following limits:
T1040(E(A9)T
11 . Sa)
lim| SUBM42= 2e?
nl x-6 )
1.2 . (see ens)
lim] -—
xox x (A)
[7]
QUESTION 2
21 Calculate 2 with the aid of logarithmic differentiation if y = e' @)
2.2 2.2.1 Make a neat sketch of the graph y = arcsin x for the range - a 2)
2.2.2 Prove that if y=arcsin x, then
(3)
2.3 Determine 2 in each of the following cases: (Simplification is NOT required.)
23.1 y =sin(cosy¥x? +1 (3)
4
2.3.2 y= |
(5x-4) (3)
2.33 y =(arecot x)[In(l+ x7)] (3)
24 Determine 2 if ytJytx= xf by using implicit differentiation. (5)
[23]
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DHET 4 T1O40(EXAN)T
QUESTION 3
3.1 A rectangular tank to hold 342 m? is to be built with a square base and open top and is
to be lined with metal. If it costs R48/m? for the lining of the sides and R22/m* for the
bottom, what should the dimensions for the least cost for the lining be? (5)
3.2 A balloon is in the shape of a cylinder with a hemisphere at both ends. The height of
the cylinder is 4 times that of the radius. Air is pumped in the balloon at a rate of
60 cm/s. At what rate does the surface area change if the radius is 8 cm?
HINT: V =i + arth and A= 4ar?42mrh
(8)
33 Given:
F(Q=x -2x? — x43
3.3.1 Determine the co-ordinates of the turning points of f(x) (2)
3.3.2 Draw up a table of values for x and f(x), with x ranging from x=—-2
to x=3 (2)
3.3.3 Draw a neat graph of f(x) between these values and show the turning
points on it. (2)
3.3.4 Use the table and the graph to estimate a value for the root between
x=-2 and x=-1 of the equation x’ -2x? -x+3=0 and then use
Taylor’s/Newton’s method to determine a better approximation of this
root. (Root correct to THREE decimal figures.) cy)
[23]
QUESTION 4
4.1 Determine [y.dx if:
4b. y=e*.sin3x (4)
4.1.2 as
a 8)
4.1.3 y =sin? x.cos? x 4)
4.14 __ cosec*x
> Go cotx 3)
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DHET Se T1040(E)(A9)T
4.2 Determine f y.dx for the following by using partial fractions:
ys x71
4x? —20x4 25 (5)
[19]
QUESTION 5
i
2,
Hoe 1
5.1 Evaluate the definite integral Te
J 1x?
2 (5)
1
3.2 The sketch below shows the area bounded by the curves y=-x?+4x and y=e?"
The intersection points of these graphs are given as (0,173; 1,173) and (2,59 ; 3,652)
aes
-- yooxt4x
,
5.2.1 Draw the above sketch in the ANSWER BOOK and show on the sketch
the enclosed area, the representative strip and the intersection points. (4)
5.2.2 Calculate the magnitude of the area in QUESTION 5.2.1. (3)
5.2.3 Calculate the volume generated when the area in QUESTION 5.2.2
rotates about the y-axis. (4)
5.3 Determine, from first principles, the second moment of inertia of a rectangular lamina
with respect to a reference axis parallel to one side of the lamina that bisects the
lamina. (4)
[20]
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DHET 6-
QUESTION 6
61 Solve the differential equation:
awe
dx y(i+e")
6.2 Determine the particular solution of :
2
29 294-20
2
Given thar 2 =-2,y=2 and x = 1
be
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TLO4OCEMAI)T
16)
G)
[8]
TOTAL: 100
DHET T1040(E)(A9)T
MATHEMATICS NS
FORMULA SHEET
Any applicable formula may also be used.
TRIGONOMETRY
s*x = 1 -sin’x
sin? x + cos’ x= 1
sin*x = 1 —cos?x
1 +tan? x= sec? x ————-» tan’x = sec”x - 1
1 + cot” x =cosec” x ———+eot?x = cosec’x- 1
sin 2A =2 sin A cos A
cos 2A = cos"A - sin’A
tan 24 = 2 A
1-tan“ A
sin? A = %4- ¥2c08 2A =A(1 - cos2A)
cos” A =%+ ¥cos 24 =14(1 + cos2A)
sin(A+B)=sinAcosB+4sinBcosA
cos (A +B)=cosAcos B+sinA sin B
tan At tanB
1+ tan AtanB
tan(A+B)=
sin A cos B = % [sin (A+ B) + sin (A - B)]
cos A sin B =% [sin (A+ B)- sin (A- B)]
cos A cos B = % [cos (A + B) + cos (A - BY]
sin A sin B = % {cos (A - B) - cos (A + B)]
sinx , 1
tanx = ;sinx = $COsx =
cos x cosec x
cos.x 1 1
cotx =— 3 COS ecx = — 7 secx =
sin x sin x cos.x
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DHET TLO40(E){A9)T
BINOMIAL THEOREM
n(n-l(a—2) 3 3 +
nn) _,-
PAD TT yn? py? xn A tee
+h) ax" tnx"! ht
Cet hy" =a! + nx 2 1x2x3
DIFFERENTIATION
_£@
f(a)
roate
PRODUCT RULE
y = ulx) vx)
QUOTIENT RULE
v(x)
du dv
—--n.
yy a ae
CHAIN RULE
y = fue)
ay _ dy da
dx du dx
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DHET -10- T1040(EA9)T
a
Aw) 310 [fade
iat F')
sin” f(x) Farr ate _
. -f'()
4 f'@)
an t® 1+F@P =
cot” f(x) aor _
sec’ f(x) __ fo) _
FOWL @)P -1
cosec” f (x) _ =f) ~£@) _
FOYLF@)P -1
in (2
sin’(ax) _ ; - sna) +C
x _ sin(2ax)
cos*(ax) __ 7 +- a +C
1
tan*(ax) __ az tan(ax)—x+C
1
cot"(ax) — - = cot(ax) =x+C
— sin? (2) +C
a a
: — 1 tan” (2) +C
a+x a a
; —= 1 sec (2} +C
x x -a? a a
2
a ax — Ssinr(2)+3 ax? +C
* a uf x= 2) +C
a? 2a \xta
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a? — x? a a \a-x
DHET ~9- T1040(E)(A9)T
fx) ce f
reeke J fax
7 nl ad ( )
nx + #-
n+1 cu
ax” aft al *" dx
dx
ax+b
eth ere Slax +b) > ee £ +C
—(ax+b)
dx
dxte
ate a? Ina Stas +e) — +C
* Ina.—(dx+e)
ax
id
In(ax, —— In ax—x+C
(ax) nwa” adn ax —x
e ef) gy OLE
“dx
. d a £Q)
af? a™ Ina.— fx, — +C
dx fo) Ina
1 od
In fix) sa) —
f(x) dx
: cosax
sin ax acos ax = 4c
a
. sinax
COS ax -a sin ax ——— 4+
a
tan ax asec? ax _ntsec(ax)] +
a
cot ax - acosec” ax _ntsin(ax)] +C€
a
sec ax a sec ax-tan ax tin [sec ax + tan ax} +C
a
L ax
cosec ax -acosec axcot ax —In, taal 2 +C oralso
a
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fn [cosec ax - cot ax}+C
a
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DHET
INTEGRATION
[£02 g'(2) de = 10) ga) - | £'@) g0) de
"6 ‘ _@y _
[LAO © fade Sate (n#~1)
=In
epee inst
f(x) A, 2
(ax+bXcx+d) axtb ex+d
f@) _ oA ,_3 Cc Z
(xt+a)" (xta) (x+a)? "Gray "ct a)"
APPLICATIONS OF INTEGRATION
AREAS
A= [yar A= [Ox yddr
A= f adysAy= fm —ndey
VOLUMES
Vamp yae:ve=al (yt - y3)ax
Vy =a dy Vy =f (x? ~ 23 )ay
SECOND MOMENT OF AREA
I= ff Paa:t,= 0 aa
MOMENTS OF INERTIA
Mass = density x volume
Ages pv
DEFINITION: J=mr
GENERAL: [= [ r’dm= [ Pav
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T1040(E)(A9)T
a her education
& training
Department:
Hig ther Education and Training
REPUBLIC OF SOUTH AFRICA
MARKING GUIDELINE |
SETI RENE
NATIONAL CERTIFICATE
APRIL EXAMINATION
MATHEMATICS N5
9 APRIL 2013
This marking guideline consists of 10 pages.
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MARKING GUIDELINE 2:
MATHEMATICS N5
QUESTION 1
Ld in(2x~
in( 22==12)) 2 ow. indeterminate
196 x-6 0
=tin{ 25 we y
x6 x
=2c082(0) V
=27¥
12 __{ tan 4x all 0 0,4
lim| - =o N
x x 0 0
=-30)
._{ tan4x—sin4x
= lim| ————— |¥
10) x
3490
mtin( te e=tesste y
40)" 40) y
4
=0
QUESTION 2
2.4
y= eta
Iny=Ine*"®?
=(x* -Inx)eIne
Iny=x* —Inx
vov 4
1 gpl
eee) 4x3 -—
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T1040(E)(ANT
(3)
4)
[7]
(4)
Please turn over
MARKING GUIDELINE 3° TIOAEKANT
MATHEMATICS NS
2.2 2.2.1 Y
= y =arcsin x ¥ labels
(2)
2.2.2 y = aresin x
siny=x
cos ye ai Vv
ax
cam wherecos y=1—sin? y
dx cosy
¥
dy 1 :
>= where sin y= x
dx J1-sin? y
yd
a Vix? (3)
23 23.1 y =sin(cosv¥x? +1
yoy v
® —(oscosVx* +0) sinyx? +1 ! e2x
ax Ix? +1 (3)
2.3.2 x
y=ln| ——_;
(5x—4)
v v v
dy _{ Gx-4) | 42? (5x4)? — x4 (2Gx-4)95
dx x [6x-477
Also
yelnx* —In(5x-4)?V
y=4Inx—2In(5x—-4) V¥
yiotize Vv
x 5x-4 (3)
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MARKING GUIDELINE 4 T1040(E)(A9)T
MATHEMATICS N5
2.3.3 y =(arecot x)[In(i + x”)]
V
ay
dx
¥ Q)
24 ®
2.5 ytyytaca”
V Vv
1o24o(y4x) Meo] 12%
v Vv
atari area -t(y+ay
ci eee A
a4 x 7 (y +3)
dx 1 PA
l+-(y+x) 7
gt) (5)
[26]
QUESTION 3
3.1 C =48(4xh) + 22(x”)
A C =192xh + 22x"
= sas 2 422°
x
x
x c-(S), 22x ¥
x
dC (65664 !
Vaxrh a 2 } 440-09
=32 ¥ x= 1143 mV
x
hee 2.62 mV
11,43 @)
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-5-
MARKING GUIDELINE
MATHEMATICS NS
32 V -fa +mh
h=4r
aim + ar (4r)
veo ey
_ 6
167(8)"
oe 0,019 cm/s V
dt
3.3 3.3.1 f'@)=3x? ~4x-1=0
A
aA
a
T1O40E)(A9)T
A=4ar’ +2mmh
=4ar?+ 2m(4r)
=12m7V
=12”.2r ar
at
dr
= Mm
at
= 24n(8)(0,019) V
7 11,25 em?/s¥
(8)
_ 244 V4? -4@Cd _ 4+ V28
2(3) 6
x= 1,549 and x=-0,215
cif x= 1,549; y = 0,369 ¥
of x=—0,215; y =3,113 ¥
3.3.2
2 -1
“tl {dl
Z
3.3.3 y
120
« —90
L =
30, (1,167;1,093)
3 2 NY me x
NY3
me
Fal
(05:-0.57 i
—3
—-60
—-90
Q)
Ww
2)
Q)
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TLO4O(EM AQT
MARKIN -9-
(G GUIDELINE MATHEMATICS N5
5.2.3 OV =myi - yp ).&
vot,
OV = 2\ (-x7 +4x) —1e? a
Nv
2,590
uVon Ie) ~ 8x) +16x? —e7).dx
0,319
590
fe +x re]
10.319
yeti 300 unit (4)
54 y * , v
Aly =bx Axx x° = bx’ dx
a $5) 5
(-$2] 2°) Tyg =b fx? de
x a
— Pam ¥
259 a |= (£0| -as1.
, 4 a 2
3
b.
AA=bx Ax Y he = ts! V (4)
[20]
QUESTION 6
6.1 dy | e
dx y(I+e*)
e
dy = ax V
ray ae
Ly stndte) +
2
Voy 3)
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MARKING GUIDELINE
-8 T1040(E)(A9)T
MATHEMATICS N5
QUESTION 5
5.1 1
[eere
e
“iv1-x?
2
v V v
1
= |sin7 xP, =sin™ (3) ~sin7 (- |
+ 2 2
(2) _ (- 2) -Z
6 6) 3
Vv Vv ©)
5.2 5.2.1
¥
t Suip v
ay - (2,590; 3,652) ¥
Jah
ee
| x
y=-xtdx
4
5.2.2 dh = (y, —y,) 0 &
L
atc 40) ~(0 Joa
2,590 V 1,
A= f [-* +4x-e? Jee
0,319
4 L 2,590
AnD -20|
3 0,319
= 2,476 units* (3)
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MARKING GUIDELINE -7-
MATHEMATICS NS
4.1.3 sin’ x.cos? xdx
= [sin’ x.cos” x.cos x.dx
= sin’ x(1—sin? x).cos xdx V
= fosin? x—sin’ x).cosxdx
= flu? =u lu
4
a4 -24¢
4 6
. a
__ sin’ x_sin’ x, 6
4 6
v v
4.14 f cosec?x
(1-cot x)?
= fu? edu
e-lac y
ui
= icy
(1—cot x)
42 ys x-1
4x? -20x+25
x-l oA + B
(2x-5) (2x-5)’ (2x-5)
“.x—1= AC) + B(2x—5) ¥
x-l 3 1
Vee po a
2x-5)P ao lang 5)
3. 1
7 Gr Gaya din(ax- 5)+C
v V
T1O40(EA9)T
u=sinx
du=cosxdx V
(4)
u=1-cotx
du =cosec?xdx V
GB)
coefficient of x: B= A v
constant: -2=A-5 ..A= ; V
(5)
[19]
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6 T1L040(E)(A9)T
MARKING GUIDELINE
MATHEMATICS NS
3.3.4
Let x, = -i,] 1, EX - fv
fe)
fCL1) = 0,349 ¥ ==-11-{2349)y
7,03
f-L1) = 7,03 ¥ x2 =-1,150N (4)
[23]
QUESTION 4
J
at 411 Je* sin3xdx=e*.—4cos3x- fe*.— Leos 3xde
3 3
=-1e: cos3x+-4 fe*.cos3x.dx
3
=~ 1 6! cos3x4+{e7tsin3x- fe? 4 sin3xdx v
3 3 3 7 4
fet sindiude =—e" cos3x+ Ze" sins = e* sin 3x.dx
10 et.sinxdx=—e* cos3x+te? sinx
9 3 9
fe*.sindudx=-—2 e* cos3x+te* sinx+C
10 10
v Vv (4)
4.1.2 L
er l
edx =-—
Fs u x
od
= fe * o—edr —du=x? edxv
L
=— fe" ed,
1 fet edu
sae" +CV
1 a
=-e "+04
2° (3)
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MARKING GUIDELINE -10-
MATHEMATICS NS
6.2 d’y
~222=2- 4x-2x?
dx?
2
cies +2x-1 ¥
oa tx —x4C, V
=50° +O? -@M+e,
aa ip aly woe Ci,
12 3 2 3
1 1 1 7
—()* +=? -=()? -= M+
rhe 3 2) 3) 2
Bee
12
s.l.5 1. 7. ,53
YE +2 KX HT K+
12 3 2 30 «12
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TLO4O(E\(AN)T
6)
[8]
TOTAL: 103
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ey higher education
g training
== Bo; Department:
== UN )e es Higher Education and Training
== aoe” REPUBLIC OF SOUTH AFRICA
T1090(E)(N19)T
NOVEMBER 2012
NATIONAL CERTIFICATE
MATHEMATICS N5
(16030175)
19 November. (X-Paper)
09:00 — 12:00
yy
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(16030175) -2- T1O90(E)(N19)T
DEPARTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N5
TIME: 3.HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1. Answer ALL the questions.
2. Read ALL the questions carefully.
3. Number the answers correctly according to the numbering system used in this
question paper.
4. Questions may be answered in any order, but subsections of questions must be kept
together.
4. Show ALL the intermediate steps. Simplify where possible.
6. ALL graph work must be done in the ANSWER BOOK.
7. Questions must be answered only in BLUE or BLACK ink.
8. Write neatly and legibly.
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¥
(16030175) 3.
QUESTION 1
Determine the values of the following limits:
Ll
im | P+]
won| TF
T1090) N19)
(2)
1.2 lim f° = 4
3x6
(2)
13 in | 4x2 4x =]
oo i Se
| 3x? + x 4]
(3)
(7]
QUESTION 2
2.1 Prove that the derivative of » = arecot (2) + arecotx is zero, (4)
x
221 Make a neat sketch of the graph y = arccos ec x for the range [-2:2 . (2)
~ ody .
2.2.2 Derive a formula to determine a if y = arccosec x, (4)
23 Determine 2 in each of the following cases, (Simplification igs NOT required.)
23.1 Y = arctan (a ~x) (3)
2.3.2 y=tan fi—x : (3)
2.3.3 Y= (X+1(2r + Gx + 2) (3)
24 Calculate x ify= e with the aid of logarithmic differentiation, (4)
{23}
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4. T1090(E)(N19)T
QUESTION 3
3.1
Given: f(x) =- 4x3 + 10x? -7x+2
3.1.1 Determine the co-ordinates of the turning points of f(x).
3.1.2 Draw up a table of values for x and F(x) with x ranging from x =~-3 to
x= 3,
3.1.3 Draw a neat graph of f(x) between these values and show the tuming
points on it.
3.1.4 Use the table and the graph to estimate a value for the root between x =1,5
and x=2 of the equation -4x°+10x7-7x42=0 and then use
Taylor’s/Newton's method to determine a better approximation of this root
(root correct to THREE decimal figures).
3.2 Determine the cylinder of greatest volume. inscribed .in a given right circular cone
with a height of 34 cm and a base radius of 10 cm.
3.3 Two resistances R,and R, are connected in parallel, where the total resistance R is
given as aE +> If Ry and Rz are decreasing at rates of 0,02Q.5~' and
1 2
0,03 Q.s"' respectively, caleulate at what rate is R changing at the instant that
R, =40Q and R, =1009.
QUESTION 4
41 Determine the integrals of each of the following:
4.1.1 f 1
aV2x"
4.1.2 e*
J eas”
4.1.3 feria
4.14 i ax
7+16x?
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(2)
(2)
(2)
4)
(6)
(5)
[21]
@)
(3)
QB)
8)
(16030175) 5:
4.2 Determine foos2x.cos.xdx by using the following:
4.2.1 Cos A. Cos B= host + B)+cos(4~ B)]
4.2.2 “= sin x
4.3 Determine! yx for the following by tesolving the integrand into partial fractions:
Sx+4
“43x 10
QUESTION 5
5.1 Evaluate the definite integral 2 te
f yx ve +16) }
5.2 The sketch below shows the area bounded by the curvesx=2e-” 41 and
x=-y?+6y-5. The intersection points of x= 2e~’ +1 and x=-y?46~Sare
given as (1,4861; 1,415) and (1,018; 4,727).
x=~y?+6y-5
5.2.1 Draw the above sketch in the ANSWER BOOK and show on the sketch the
enclosed area, the representative strip and intersection points.
5.2.2. Calculate the magnitude of the area in QUESTION 5.2.1
5.2.3 Calculate the volume generated when the area in QUESTION 5.2.2 rotates
about the y-axis.
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TIOSOCE)N19)T
Q)
(3)
(3)
[23]
4)
(2)
GB)
(16030175) -6- TLO9O(E)X(N19)T
5.3 Determine, from first principles, the moment of inertia of a uniform circular disc of
radius 2 m and mass 10 kg about am axis through its centre and perpendicular to the
plane of the disc,
HINT: Consider a ring of thickness Ar a distance 'r' units from the zz'-axis.
The area of the ring is 27rAr and its mass is 27 rar. (6)
ki
[20]
QUESTION 6
Solve the following differential equations:
6.1 dy __ tx
da (3)
6.2 ld’y 1, 3
Qa 2” *Q**? 2)
IN
TOTAL: 100
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(16030175)
MATHEMATICS N5
FORMULA SHEET
Any applicable formula may also be used,
TRIGONOMETRY
; cos*x = | _ sintx
Sin’ + cos? = | <=
sin’x = 1 ~ cog’
i+tantxs sec’x —————>» tan’y = sec?x - |
2 2 2 2,
1 + cot*x = cosecy ———— 01" = cosec’x - |
sin 24 =2 sin 4 cos A
=cnetd . ot2
Cos 24 = cos" - sin?4
2 tan A
tan 24 = 7
1-tan* 4
sin’d =%- 4 cos 24 = “(1 ~ cos2A)
cost4 = 4 4 %cos2d=%(1+ cos2A)
sin (At B) = sin A cos Bt sin B cos A
cos (At B)=cos A cos B+ sin A sinB
tan (4p) = lands tan B
it tanAtanB
sin A cos B= % [sin (4+ B) + sin (4-B)]
cos A sin B= 4 [sin (A+B)-sin (4 - BJ
cos 4 cos B =¥, [cos (A + B)+ cos (4 - B)]
sin dA sin B = 4 [cos (A-B)-cos (4 + B)}
sinx _ 1
tanx = ;Sinx = 7 COSx = ——
cosx cosec x secx
cosx Lo
colx = —— ; cosecr = —— ; sec x = 1
sin x sin x cosx
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TO90(E\(Ntoyp
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(16030175)
BINOMIAL THEOREM
a(a-1) yr pra n(a—l)(n—2) as
(a+h)" =x" +x" he
2!
DIFFERENTIATION
=f
f (a)
rr=ate
PRODUCT RULE
¥ = u(x). v(x)
dy dv du
ered aerate a iene
dx a& dx
Surv tyew
QUOTIENT RULE
= He)
ve)
CHAIN RULE
¥ =fu))
dy yd
dx du dx
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1x2x3
h+
T1O90CE)(N19)T
(16030178)
TIO90(E)(N19)T
d
fo at”) [rear
sin” f(x) fe)
1-[/()P _
cos" f(x) F'@)
v!-[f@?P _
tan" f(x) f'@)
1+ @F ~
: -f'@)
efi) FO
10) ier
4 f'(x)
sec” f(x) AM
SOWIFC)P -1 —
4 -f'@)
cosec™ f(x} ———AM/
SO LFOoP =1 7
sin“(ax) #_sinltas) |
—— 2 4a.
cos*(ax)
tan’(ax)
cot*{ax) __
Va°-x
1
xg _.
1
2 —
a -x
Copyright reserved NOV 2012
% | sin(2ax)
z ——— +e
4.
gtan{ax)-—x+¢
— 2 cot(ax) -K+E
sin'(Z}nc
a
Phan (ax) -—x4€
a2] +C
a a
2
Soiw(2) 2a -24C
2 a
1 x-a
= Inf —— J+
2a x4]
ma S2*) +c
2a \a-x
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(16030175) TIOSOCEXNID)T
d
fe) AG J tara
n+]
x" nx"! +e (n=-1)
ntl
ax" at af <"dx
dx
arth
ete te gx +) sc
dx d
=~ (ax +b)
dx
dete
ate a“*** Ing Slax +e) —~— +#C
x
Ina.—- (de +e
ae ¢ }
1d
tn(ax) —.. ax xin ax-x+C
ax dx
ef ef ao Si)
dx _
a” a® Ina.— f(x) _
| od
ln fx, Se
Hy) Fo ax Si) —
sin ax a COS ax —— +C
a
COS ax -a sin ax me +C
a
tan ax asec’ ax dinfsec(ax)] +C
a
cot ax -a cosec? gx | infsin(ax)) +
a
sec ax asec ax-tan ax i In [see ax + tan ax] +C
a
1 ax
cosec ax -@ COSEC ax-cot ax — In} tan z +C oralso
a
Copyright reserved NOV 2012
1
~ In [cosee ax - cot ax} +C
a
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(16030175) TIO90(E)(N19)T
INTEGRATION
J x) a(x) de = fx) g(x) - 1 f°) gO) dx
[LPO]" « fi(xjdx = Yen". C (n#-15
(Fie -insle)ec
f(x) AB
(ax+bXex+d) axtb cx+d
fe) __A B c Z
= + Foe
(Gay Gre Gea? Gay Gray
APPLICATIONS OF INTEGRATION
AREAS
4,= [yer de= | Oi-ydee
fe a
A, = [954,= [Gime
VOLUMES
Vv, anf yrdx ; ye =a (2 — v3 ae
ra :
a =a] xy 2V, anf? (2 - day
SECOND MOMENT OF AREA
fea P Pda: t, = [Pada
MOMENTS OF INERTIA
Mass = density x volume
M= pv
DEFINITION: [= m7
GENERAL: J = [/ rdm= of ray
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ewe
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higher educati
& training venon
Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
| MARKING GUIDELINE |
| NATIONAL CERTIFICATE
MATHEMATICS N5
| 19 NOVEMBER 2012
This marking guideline consists of 9 Pages.
|_ Copyright reserved Please turn over
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MARKING GUIDELINE -2- T1O90(E)(NI9)T
MATHEMATICS N5
QUESTION 1
Lt xP 1
: xt ty?
. ‘F421 ye 1
1
(=:2
T
=\tv/
7 xt +7
SL Fay =1 Vv (2)
" aE i). had 1
mf 4
2m oc ae v Q)
1.3
7 4x3 x8
se 3x" FOr +t
k 4ytax—-1\_
oie? + et i): y (3)
{7}
QUESTION 2
21 xu =erecot (S}+ arccotx
; (4)
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MARKING GUIDELINE 3.
MATHEMATICS NS THOSOCEY(N19)7
2.2.1 ¥
arccosec x
we
Ma,
=
aN
]
d
ho
x
a
!
x
(2)
2.2.2 Y= arccosecx
cosecy =x
—~Ccasec 3, coty. =1 ¥
dy ~1
==
ax cosecysoty V where coty = ¥eosecty —]
ns
ax" cosec y.Jeosecty =F ’
». ec VY where cosec you
” ax” BRTST (4)
23.1 -mmy ta
ar 14 (aaa? ag F-* Vv
i.
=t¢@-y-? y
~i
=ir@-xyF y @B)
2.3.2 Y= tanvI—X
ayfdx= Usecl t2 fa — 29 ax G-wahin y
= Kseck "2 (v1 -2) “U2 (1 —aye—1y2) .-1 dv
sec*yT —&
=Wi-z y (3)
2.3.3 y=et+ 12x + 1)(3e + 2°
Z @
B= 6+ Mert Pert ase Nore aZex+14 2
Vv
= (2x7 + 3x4 198) 4 Gr? +axr4 2x + 22) + (x3 + 72+ 21)
= 6x? + Ov +3 +6x7 + 1Ox+44 6x2 47742
ay 2
dg 7 18x? + 26x +9 ¥ (3)
+1)(8x4 Qretr
ax(xt+T)
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MARKING GUIDELINE 4. TLO9O(EXN19)T
MATHEMATICS NS
24 Ps
Iny = lae*™*
Iny =x*” Ine
Inn y) = atinx
1
Iny.1 ay _ 1,
aden dale + ox vv
= = yiny[2zing + a] V
— oper [oxinx +x] ¥
(4)
[25]
QUESTION 3
31 fG) = ~tx? + 10x? - 7x42
3.11 f'@) = 12x? + 20x—7 =0
(6x -7\(2x—1) = 0
an
axa 1,167 x = 0,5 V
and
aye 1,093 y=O,5 resp. J @)
3.1.2 x 3-2 |-1 [0 1 2 3
f(x) | 221 | 88 | 23 (2 1 4 | -37
¥ (2)
3.1.3
T 120
V
(1,167;1,093)
<2 _i wea ,
t f +
(0,5:-0,3f[ 1 Xe
-30 Hr) = -4x" 4 1027 ~ 7242
+60
ro @)
3.1.4 stem #0)
Let/laat *1 — 16 Fy).
¢ 0,04
f0,6= 0,016 V¥ =1,6-\- 57207y
F°Q,0= - 5,720 ¥ %: =1,6037 (4)
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MARKING GUIDELINE a 6. TLO90(E)(NI OT
MATHEMATICS NS
4.1.2 oe _ lL pse* dx
Saxzs"* slams y q
jinte™* +5)+C y (3)
4.13 fo. elt-4) ay
= “<4 v dus-4x4dx Vy
age te y (3)
4.1.4 dx -{ az
J 7+ 16x? (WP + (4 y
1 _y fhe
Fee v
Ey faa?
=10583°™ *(S)+e y QB)
4.2 Jeosza. cosx. ax
421 _ 2
Cosa. CosB = 2Icos(4 + B}+ cask(A — BN
Bey
Lysin3x
=3\ 4 $sing]4+C
_ sin3x sine c
=e Fate vy (3)
4.2.2 f COS2X. COSK. GX
= fia — 2sin*x)cosx.dx J u=sine
= {ua ~ Ju?) du
2
=uw—-—yF
=u-gu +¢ Vv
du cosx. dx
; 2s
=siax-ssin?xt€ 3
Fy N ,
3)
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MARKING GUIDELINE +5. T1090(E)(N19)T
MATHEMATICS NS
3.
; (6)
/\ nh 34
40 -r To
34 .
Verrth
= rr?(34 ~ 3p)
A : ve 34rr4 - 37r8 v
—! yo _
1 Gy = O80r - onr7 = 0 v
n(68—9r}= 0
réfond y= 758m 7
“h=34-37 Vonrth
=11,32cm § =%7,567(11,32)
V=2032,544 cm* J
33 Bott
RU RTE,
apa rhe GO\(100) |
“RoR eas 204 ion = 28570
v
voy
~1,4R_ 2 GR, 1 ak,
1 L
_ 7 50F = 9,02) ~ 100*(-0,03)
1
aa “285702
—o, “1
Ten 0,0132. 5 v (5)
[21]
QUESTION 4
4.LL 1 if dx
2vaxt” 23 Bxt
1 =a
=—[x2.
aay 87 -8e
=3
1 oxztt c
“we 2yqt
z
a2
_t rz ae
ar a
wey ly
1 -2
Sree ete.
2 “2 et
Er tC
Ya to y (3)
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MARKING GUIDELINE
-7-
MATHEMATICS NS
43
a =f sx+4 ~ [_Adx Bde
+3e-10 J G-NG48 Jaa ley v
then x=-5 and x=2
Ifx=5 then
¥
Ffx=2 then
1
es |
Lo
v
QUESTION 5
5.1 4a “1
[rare f BE VE+ 4) dx usViFie
a 1
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(5)
(23)
MARKING GUIDELINE i -8- T1090(E)(N19)T
MATHEMATICS N3
4 du. i
= [ a du = ———— dx
no OY, 20+169= — orlof
1; :
= [ - V x+16—ui
oy 1
_ s[ (a? - LF du 2 fE=Vu? —16
_ 1
at - 10%
=4-—j— du
2.2 v
2 ay
= [-2eu - 187], v ifx = 1 then u =717 and
eS hall
~ [-2a - 10937] x= 4thenu = ¥20
=2 (4)
5.2.1 (2)
x=2e%4H1
(1,018;4,727)
Enclosed ¥
Strip ¥
(1,486; 1,415)
512.2 8A = ey — 1, )8y
a4 = EG y? + ey—5)- (2e-¥ + 1d
727
sae [(-y9] + 6 -6 = 2e ay J
L315,
a72F
ay
-| Z +3y? ~ 6+ 26]
, ats
A= 6.424 units’V (3)
5.2.3. ov= «(ad -aD ay
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MARKING GUIDELINE -9-
MATHEMATICS NS THOPOENIOYT
6 = r[C a 1 6y 5)? (2e-¥ 1 4)2]ay y
sven ats ‘T- 12y5 + 46y? — —60y 4 24~ ge-y ~ 4e-¥)ay V
ys 46 4727
Vey r- 3y4 to — 0p? + 24y 4 2e-e 4e~¥
V = 92,191 units? wee vy (5)
Mass per unit area
a
=m y
10
adm= te Vv
_5
QUESTION 6
6.1 dy
: dx 7?
~fRay= fom ax y
v¥
i ae
vhiy= Te + (3)
. id? 1 3
62 fast = 7gU ts gitEn
ye nat ay 47
dx? * 7 v
ay 3,
a ~ at pt +20x $C V
a Zz 2nx*
Rae ten Ft xe € Vv
1 x" atx?
S-Mt Set gree,
12° "2 v (4)
[7]
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‘DEPAHTMENT OF HIGHER EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
‘MATHEMATICS NS
TIME: 3 HOURS
MARKS: 100
3TRUCTIONS AND INFORMATION
Answer ALL the questions.
Read ALL the questions carefully,”
Number the answers correctly according to the numbering system used ir this
question paper.
Questions may be answered in any order, but subsections of questions must be kept
togeti:ex.
Show ALL the intermediate steps. Simplify where possible.
ALL ‘fraph work nmust be done in the ANSWER BOOK.
Ques:ions must be answered in blue or black ink.
Write seatly and legibly.
| Copytighteservea Please turn over
3 T1O90GEKAIYT
QUESTION 1
Detemmine the values of the following limits:
Lj - ( ef
al 7
2x? ~2x
@
12 1
‘a |_ x.
a Foremes )
> 17
QUESTION 2.
A sees 5x+4 aos.
21 Determine the derivative of y = from first principles.
2-3x )
22 Calculate & of the following, by meking use of the derivatives of sin x and cos.x as
well as the rules of differentiation: y= tan x * G6)
23° _ Determine 2 im each of the following cases. (Simplification is not required.)
L_,ft
231 yea+n(4)
:
2.3.2 =sec7| =
oes (=)
2.33 y= v¥veotx—x?
24 Calculate sic y= (ce? ~4)" with the aid of logarithinic differentiation.
25 Find % of the implicit function: yf +x? = fay
DHET Copyright reserved
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@)
8)
4)
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TIOIOCEMANT
\.THEMATICS N5
RMULA SHEET
y applicable formula may also be used.
IGONOMEFRY
. cos’x = 1 ~sin’x
x cos? x <<
. . sin’ = 1 - cos’
tan?x=sectx ——__». tans = sec*x- 1
cof x= cor::*x——» cot’x = cosee’r - |
2A = 2 sin A cos A
24 =cos74 - sin?
A=M4-Ycos 24 = VA(1 - cos2A)
A=%+% 20824 = %4 (1+ costa)
A+ B)=sin A cos Bt sin Bcos A
A+B)=c03A cos BF sind sini B
dapye fen A tanB
) 1¥ tanAtanB
cos B= ¥, [sin (4 +B) + sin (4 - BY]
sin B= % vin (A + B)-sin (4 - B)}
cos B* % (cos (A + B) + cos (4 - BY]
sin B= % [20s (4 » B)- cos (4 + B)]
sinx 1 i
= ;siax= 3 S0sx = ——
cosx -cosec x seex
cosx 1 1
=—— | cos cox = ——— ; seex =
sinz sin x cosx
Copyright reserved
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a
see ax @ sec ax-tan ax Ain {sec ax + tan ae] +C
a
cosec ax a cosec axcot ax 4 wo(=)] +C
a
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higher education
& training
Department:
iG) Higher Education and Training
Sd REPUBLIC OF SOUTH AFRICA
i
MARKING GUIDELINE
NATIONAL CERTIFICATE
AUGUST EXAMINATION
MATHEMATICS N5
1 AUGUST 2012
This marking guideline consists of 11 pages.
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MARKING GUIDELINE
QUESTION 1
La gv
#31] 5.2_9,
2x*° -2x
gt
| 5
2x" —2x
2+
MATHEMATICS N5
[5]
= («)
=)
1.2
1
x sin x
ie xe
zd
cotx— +
xCOS x— xsin x
a
o|o
1
=. tim x
°°! cotx—x
cos x
x(—sin x) + (cos x)(1) — xcos x + (sin x)(1)
QUESTION 2
5x+4
2-3x
S(xt+h)+4
+h)=———--
farm 2—3(x+h)
_Sx+5ht4
~ 2-3x—3h
21 f=
Sxt+Sh+4 Sxt+4
fOr W~ $0)= aan 23x
_ 22h
~ (2-3x—3h)(2 -3x)
_fathy-f@)_ 22
“ h ~ (2—3x— 3h M2 — 3x)
lim) f@+*D-f@)__ 22
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T1OSOE (ADT
(2)
(5)
@)
Please turn over
MARKING GUIDELINE 3.
MATHEMATICS N5
sin x
2.2 ystanx=
cos x
dy _ cos x(cos x) — sin x(—sin x)
ax cos? x
_ cos? xtsin? x
cos” x
do
dx cos? x
= sec” x
4 (tan x) = sec? x
dx
1 1
2.3 2.3.1 =—-+in =
one “3)
® 2 npg? Ah?
dx 1
x
-—t *
xén?x x”
et ane
xtn?x x
af x
2.3.2 y=sec (2)
2.3.3 y= veotx—x°
a
, 2,Jcot x .
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—cosec?x— »)
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T1O9O(EMANT
(3)
(3).
Q)
3)
Please turn over
MARKING GUIDELINE 4
MATHEMATICS N35
24 y=? aye
eny = (4— 2)én{x? — 4)
Viator
—.— =(-1é -4)+ » 2x (4—x)
y de (-l)én(x* — 4) yaa
dy 1
nm = n(x? - 4) + ya 2x (4—x)
di x 2x(4—x)
Bat -4) {- tnx? -9+(22)|
2.5 yenxt x? = fny
tnx, 42 43x? = lay
dx x ydx
2 tne) =-a0 -2
3x22
ys
dx
QUESTION 3
3.1 f(x) =3x° + 6x? 452-4
3.11 f(x) =9x? +1245
fil @x) =182 412
for point of inflection: O=18x+12
“= — 0,667
y=-5,556
(- 0,667;— 5,556)
3.1.2 x -2 |-1| 0 1 | 2 3
fi) | -14 |-6 | -4 | 10 | 54 | 146
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T1090CE)(ALT
@)
(4)
Q)
Q)
Please turn over
MARKING GUIDELINE -5-
MATHEMATICS NS
3.1.3
a‘
Bed
3
40 hy = Se + bx Sx -&
te
mac
ao we"
ae
ia
ar or war Sh.
3.14 letflaat x, =0,5 e= f@
F(0,5)=0,375 f'(@)
f'(0,5) =13,75
f(%)
=a
fi)
=05- 0,375
13,25
x, = 0,472
4 r=130+3
3.2 Vol fhemisphere V=—ar°
olume of hemisphere 3 r =133dm
WV 4 gt eager, &
dt 3 t dt
= 4m (133)? (-025)
“ = = —55571,632dm'Vhr
it
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T1090(EXALT
(2)
(4)
(5)
Please turn over
MARKING GUIDELINE -6-
MATHEMATICS N5
3.3
ala
xyal6 ays
S=ax’+y?
3
4096
x
S=x* +4096x7
ax +
SS Loy —12288.x- =0
dx
2x5 =12288
“.4= 5,724
16.
Y="y (given)
_ 16
5,724
“y= 2,795
QUESTION 4
41 4.4.1 free. dx f(x)=x
fiGx)=1
~~, 1,2 1 -2s
= he mae fl - ax
a_ly pt lp,
sare tale «dx
~ 1, ,2 il _1,»
Rage ta get te
—_1ly 4-2 _ 1-2
Bate qe.lfe
TIN9O(E (ALT
6)
gi(x)=e™
g'(z)=-fe*
(4)
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Please turn over
MARKING GUIDELINE -7-
4.1.4
Copyright reserved
MATHEMATICS NS
jase a
x
=ltnfz Lax
Let/Laat u = fnvx
wldu=t . dx
J2*- cot( *) de
= Joot(2*) . 2* , dx
Let/Laat u =2*
=, u=2*. dx
1
te foot. u. du
= = én(sin x) +¢
i er
=Fa" én(sin 2 ) +e
[¥25—927 «de
= 3f Baw as
25 cd
= 3 2-sint 4 +5 Bx +¢
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T1N90(E)(A 1)T
Please turn over
©))
@)
(4)
-8-
MARKING GUIDELINE
MATHEMATICS N5
4.1.5 Joos? 4x. dx
- f4ah
= [4+ fooser) a
1 sin8x
+5. te
8
hath
i dat bsin8x +e
42 x+1 xt
3 be = Serica) -x)
xtl - A + B
(24+ x\l-x) 2+x (l-x)
xtl =A(l—x)+ B+ x)
LetvLaat x=1 B= 4
x=? wA=-4
a1lpi 2p! 4
sean es lara
=z én(2+ x)= tn(t- x)te
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TIONNE\ADT
@)
(5)
Please turn over
MARKING GUIDELINE -9-
MATHEMATICS N5
QUESTION 5
31 f e .ftjat if f@)=4
= of ee". 4dt
k
e*
ai
_ im |=
= poe 5
a[-4.4
tule 5|
=o+4
5
-4
5
52 5.21 yeeFe yale
Intersection/Snyding: of =ve*
et ae"
4a x?
ox -4x=0
x(x-4)=0
“x=0 orof x=4
ficf: fO)=1 ff @=Ve*
f(a)=e? =7389 =f, (0)=V1=1
f(t)=e' =2,7183 (0) = 1,648
f,(2) = 2,718
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TLOSOEMADT
Please turn over
(4)
MARKING GUIDELINE
3.3
5.2.2
-10-
MATHEMATICS N5
3.2.3 dA=(y, 7 Ya dx
a= [e® ~ve*lax
-[re —264F — |
x
4
10
=[29,556—14,778-14,778|- [0-2-2]
=4 UNITS”
2
j Taser
2
eq tam
a
4fiu 2du
u
4
h
rope
=4
2+ |
=8./ve+4]
=8\V6 - V5|
=8(2,449 — 2,236)
= 1,704
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Let/Laat
TLO90E ADT
GB)
Please turn over
@
MARKING GUIDELINE -1l- T1O90(E)(ADT
MATHEMATICS NS
5.4 dV amy; dx—my} dx yy ae sy ax?
= mx ‘dx-mx'dx.
Vaz), (xt —x*)dx
=n7|—-—
5 7h
-a|i-4
= 0,057 UNITS® 6)
QUESTION 6
61 “ = —2y(4x?)
wdy=4x . dx
<n
fny=x'+e
=
SS
=(1)* +c
planta wie
(4)
62 a at
dx 9 10
10 6
yp=t_-2 48x? +exte
10 60 (3)
TOTAL: 100
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33
QUESTION 2
21 Determine * of the following, by making use of the derivatives of sinx and cosx, ‘we
as well as ihe rales of differentiation:
ys tans @)
22 2.21 Make a neat sketch of = arc cosx for the range [o, x). 2
223 . a
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2.3 Deiermine # in each of the following cases: (Simplification is NOT required.)
BB ne inet) @
Q)
23.2 vex tay
23.3 GB)
24 Deteanine Fa if » = (in z)" with the cid of logarithmic differentiation, y
25 Determine of the implicit function cos y- x = y? 4
a G)
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TIIIQ{EMA1T
QUESTION 3
A closed rectangular tank is constructed io Sccupy 70 dm? and its length is thrice its
breadih. Caiculate the height that will minimise the quantity of material io be used.
HINT: First draw the rectangular tank and then develop the formulas for volume and (6)
area.
A conical shaped ice artwork is exposed to ihe sun, The volume of this cone decrea-
ses at a rate of 100 cm?.s~! and the radins decreases at a rate of lems”. Caleulati
the rate ar which the height changes when the hight is 10 cm and the radius is3 cm .
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faj=-1y 1, +2x-] 6
24 —
331 Determine the coordinates of the turning points of f(x). rs (2)
33.2 Draw up a table of values of x and J(%), with x ra nging from x = -3 to (o)
x=3, = @
3.3.3 Draw a neat graph of f(x) between these values and show the suring
points on it. tg Q)
334 Use the table and the graph to eslimate a value for the root between x = rS
and x=2 of the equation -t 2 ae +2x-1=0 and then nse (OG
Taylor’s/Newton's method to determine 4 belter approximation of thi
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MATHEMATICS Ns
FORMULA SREET
Any applicable formula may also be used.
TRIGONOMETRY
sin? x + cos? x=
1 + tan? x= sec? x
1 + cot x= cosec? x
sin24™2sind cos 4
COs 24 = cost - sin2y
2 tan 4
1~1an
tan 24 =
sin? = 4% cox 2a
Cos! d=" 4% cos 24
sin (A+ B)=sin 4 cos Bt sin Bcos 4
cos (4 + B)= cos 4 cos BF sin A sin B
tan 4 + tanB
* =
BO adm
sin 4 cos B= % [sin (4 + BY + sin (4 - By}
cos4 sin B= % [sin (4 + 8)~ sin (4 - By
625 A cos B= % [eos (4 + B) +cos (A - 3)]
sind sin B= {cos (d - B)- eos (44+ B))
sinx 1 J
+819 X = —— cosy =
cos x cosec x secx
tanx =
PTO
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MOMENTS OF INERTIA
Mass = density x volume
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Mathematies(Wiskun:
Memorandum
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Lo SN apunijsi a /soaewaMEy
niemoranaum
August 2011
Sik
4
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: ; Memorandum August 2011
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a oon vu —E
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highei: education
Z & training
: Department:
Higher Education and Training
REPUBLIC OF SOUTH AFRICA
T1100(E)(A2) T
AUGUST 2010
NATIONAL CERTIFICATE
MATHEMATICS N5
(16030175)
2 August (X-Paper)
09:00 — 12:00
This question paper consists of 7 pages and a 5-page formula sheet.
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eM, ie or , 4) T1400(E)(A2)T
nad, : cn rn ae Ry : a 7
DEPARTMENT OF HIGH. R EDUCATION AND TRAINING
REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE
MATHEMATICS N5
TIME: 3 HOURS
MARKS: 100
INSTRUCTIONS AND INFORMATION
1.
2.
Answer ALL the questions.
Read ALL the questions carefully.
Number the answers correctly according to the numbering system used in this
question paper.
Questions may be answered in any order, but subsections of questions must be kept
together.
Show ALL the intermediate steps. Simplify where possible.
ALL graph work must be done in the ANSWER BOOK.
Questions must be answered in BLUE or BLACK INK.
“Write neatly and legibly.
PTO
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QUESTION 1 ‘
1 Determine the values of the following limits:
ne ;
im 1-tan? x
xoS
4 | zi x
8 2
1.1.2
jim | x-2
x32 i =)
2
12 Determine the value(s) of x where f (x) = —_—— is discontinuous.
x? 45x+6
QUESTION 2
2.1 Given: y= arccos x
2.1.1 Make a neat sketch of the graph y = arccos x for the range [o; n}.
. db 1
12 P that if y= arccosx, then —- = -—=——= =.
2.1 rove y x az oo
2.2 Calculate 2 of the following, by making use of the derivative of sin x and cos x,
as well as, the rules of differentiation:
ystanx
23 Determine 2 in each of the following cases:
(Simplification is NOT required.)
e
2.3.) = Tas)
In* x
2.3.2 y=x? arc cotx
2.33 y=afx* +sec 7x
PTO
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(2)
(2)
{8}
(2)
(3)
Q)
ie . . FAVOGCEMAQT
a ” ‘
24 Determine the slope of the tangent line at the point (1:2) of the graph:
devin ;
2.5 Calculate 2 if y=4" ‘with the logarithmic differentiation.
QUESTION 3
3.1 The sum of the slant height (4,) and the perpendicular height (A , ) of a right circular
cone is 20cm. Calculate the radius and the perpendicular height of the cone, if the
volume is to be a maximum.
HINT: Volume ofa cone smh, and h? = hk? +r?
3.2 The motion of a point on a spring is modeled according to the equation,
5" cos mt
s(t) =3e
where 5 is measured in centimetres and ¢ in seconds.
Determine the following:
3.2.1 The displacement after TWO seconds
3.3.2 The velocity after ¢ seconds
3.3 Given:
f(x) =~ -2x? 43x41
3.3.1 Determine the coordinates of the turning points of 7(x).
3.3.2 Draw up a table of values of x and f(x), with x ranging from
x=-4 to x=2. :
3.3.3 Draw a neat graph of f(x) between the values and show the turning
points on it.
3.3.4 Use the TABLE and the GRAPH to estimate a value for the root berween
x=-] and x=0 of the equation —x7-2x? +3x+1=0 and use
Taylor/Newton method to determine a better approximation of this root
(the root must be correct up to THREE decimal figures).
PTO
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(4)
(4)
25}
(6)
(2)
(2)
@)
2)
(4)
(20}
Bee ; Trro(Enaaysee! |
ny : i
‘QUESTION 4
4] Determine foae ift
4.1.1 yse™ sinx (4)
4 2 a
ae =——
a @)
4.1.3 -—1_
“ 7754 bax (3)
4.1.4 y# cos 7x.sin 5x (3)
cos 2x
4.1.5 y= ee
sin.x + cosx (4)
42 Determine fy.dr by resolving the integrals into partial fractions:
x+]
rr ers
x —10x+425 (5)
[22]
PTO
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QUESTION 5
ine [e~* fat if fy=—t
5. Determine [e Sat if N= 2. «)
5.2 The sketch below shows the area bounded by the curves:
Loge
yere, y= +1, x=3 and the y-axis.
2
\ fF 7
\ ‘ /
\
\
4
SSS i
Sr 2 a eT
-a}
5.2.1 Draw the sketch in the ANSWER BOOK and show on the sketch the
enclosed area, the representative strip, as well as, the lower and upper
limits. (3)
3.22 Calculate the magnitude of the area in QUESTION 5.2.] (3)
5.2.3 Calculate the volume generated when the area in QUESTION 5.2.2
rotates about the x-axis. (4)
5.3 Determine the second moment of mass of a rectangular lamina of mass m about an
axis parallel to ONE side of the lamina. (4)
(18)
PTO
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TH1OOKENAZIT:.
QUESTION 6
6.1 If the gradient of a curve is 2 = = , find the equation for the curve if it is known
y
that y=1 and x=. (4)
6.2 Determine the general solution of the following:
2
ay = fe" + iy
dx? (3)
7]
TOTAL: 100
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ad
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ae
n+l
a nax”! te
n+l
a axte
s x x
ete
x
: a
a dina te
Ina
i
logex = _
x
| ]
og ax
Bax xina =
sinx cos x -cosxte
cos x -sin x sinx+¢
tanx sec" x In (sec x) +c
cot x -cosec? x In (sin x) +e
sec x sec x-tan x In [secx + tanz] +s
coset x “cose x*cot x In (Cosecx = cotr) + ¢
: |
sin” x = _
Ti -x°
4 -l
cos” x =
1-x*
“ |
tan” x z
1+x* ~
. -f
cot" x 5 —_
J+x
sec’ x !
xx? -1 ~~
ty
. -)
cosec” x
PTO
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