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T970(E)(A4)T
APRIL EXAMINATION
NATIONAL CERTIFICATE
MATHEMATICS N5
(16030175)
4 April 2016 (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.
(16030175) -2- T970(E)(A4)T
Copyright reserved Please turn over
DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA
NATIONAL CERTIFICATE MATHEMATICS N5
TIME: 3 HOURS MARKS: 100
INSTRUCTIONS AND INFORMATION 1. 2. 3. 4. 5. 6. 7. 8.
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. Use ONLY blue or black ink. Write neatly and legibly.
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QUESTION 1 1.1 Determine the following limits:
1.1.1
(2)
1.1.2 (3)
1.2 Determine whether is continuous at .
(2)
[7] QUESTION 2 2.1 Determine the derivative of from first principles.
HINT:
(4)
2.2 Determine in each of the following cases (simplification is not required):
2.2.1
(4)
2.2.2
(5) 2.2.3 (2) 2.3 Determine with the aid of logarithmic differentiation if:
(4)
2.4 Given:
2.4.1 Determine the slope of the tangent at the point: (1;-5).
(3)
2.4.2 Hence, determine the equation of the tangent at this point. (2)
[24]
xxeex
x
x 2lim
0 -®
)tan(seclim2
xxx
-®p
327)(
3
--
=xxxf 3-=x
xxf cos)( =
1sinhlim; 01coshlim00
==-
®® hh hh
dxdy
4224 )4cos()4(cos -+-= xxy
xecxy4cosln2
1tan3 2 -=
)4.2sec( 3xarcy =
dxdy
)()arcsin(sinxexy =
xxyx 34 23 =-
dxdy
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QUESTION 3 3.1 Given: 3.1.1 Determine the co-ordinates of the turning points of (3) 3.1.2 Verify, using a table, that the equation has a root
between the points = 2 and = 3. Use values on the table:
(4) 3.1.3 Hence, make a neat sketch of the graph of the function (2) 3.1.4 If the positive root of is estimated as 2,7, use Taylor's/Newton's
method to determine a better approximation of this root.
(4) 3.2
Two sides of a rectangle are lengthened at a rate of 3 cm/s while the other two sides are being shortened in such a way that the figure remains a rectangle with a constant area of 50 cm2.
3.2.1 Calculate the rate of change of the perimeter of the rectangle 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 a straight line according to the distance formula
3.3.1 Calculate the velocity of the particle after 3,5 seconds. (4) 3.3.2 Calculate the acceleration after 2 seconds. (3)
[27]
4)5()( 2 --= xxxf
).(xf
4)5(0 2 --= xxx x
40 ££ x
).(xf
)(xf
).33()( 2tttts --=
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QUESTION 4 4.1 Determine:
(2) 4.2 Determine in each of the following cases:
4.2.1
(3) 4.2.2 (3) 4.2.3 (2)
4.2.4
(3)
4.3 Determine by resolving the integrand into partial fractions:
(5)
4.4 Determine:
(4) [22]
ò -- -+ dxeeee xxxx ).()( 4
ò ydx
xxycos1
sin+
=
xxy 2sec.=
xxy 2cos.6cos=
2432x
y+
=
ò ydx
12
4
3
--
=xxy
ò -dx
xx5
2
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QUESTION 5 5.1 Given: The curves and
Y y
dx X
5.1.1 Calculate the magnitude of the enclosed area. (3) 5.1.2 Calculate the volume generated when this area rotates about the x-axis. (4)
5.2 Prove that
(4) [11] QUESTION 6 6.1 Solve the differential equation:
(4) 6.2 Determine the particular solution of the differential equation
for which and when
(5) [9] TOTAL: 100
216)( xxf -= xxg -= 4)(
216)( xxf -=
xxg -= 4)(
ò¥
- -=-0
55s
dte st
xecy
dxdy
2
2
costan
=
p++-= xxdxyd
23
21 2
2
2
2=y 3' -=y .1=x
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FORMULA SHEET Any other applicable formulas may also be used. TRIGONOMETRY sin2 x + cos2 x = 1 1 + tan2 x = sec2 x 1 + cot2 x = cosec2 x sin 2A = 2 sin A.cos A cos 2A = cos2 A - sin2 A
tan 2A =
sin (A ± B) = sin A.cos B ± sin B.cos A cos (A ± B) = cos A.cos B sin A.sin B
tan (A ± B) =
sin A.cos B = ½ [sin(A + B) + sin (A - B)] cos A.sin B = ½ [sin(A + B) – sin(A - B)] cos A.cos B = ½ [cos(A + B) + cos(A - B)] sin A.sin B = ½ [cos(A - B) - cos(A + B)]
BINOMIAL THEOREM
AA2tan1
tan2-
AA 2cos21
21sin 2 -=
AA 2cos21
21cos2 +=
!
BABA
tan.tan1tantan
-±
xx
xx
xx
xsec1
=cos;cosec1
=sin;cossin
=tan
( ) ( ) ...!21. 221 +
-++=+ -- hxnnhxnxhx nnnn
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DIFFERENTIATION
PRODUCT RULE y = u(x).v(x)
QUOTIENT RULE
y =
CHAIN RULE y = f(u(x))
)()(
' afafe -=
ear +=
dxduv
dxdvu
dxdy .. +=
'.'. uvvu +=
)()(xvxu
2
..
vdxdvu
dxduv
dxdy -
=
2
'.'.v
vuuv -=
dxdu
dudy
dxdy .=
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______________________________________________________________________
f (x)
______________________________________________________________________
axn
a 0 ax + c ex ex ex + c
ax ax.lna
loge x __
loga x __
sin x cos x - cos x + c
cos x -sin x sin x + c
tan x sec2 x ln (sec x) + c
cot x -cosec2 x ln (sin x) + c
sec x sec x.tan. ln [sec x + tan x] + c
cosec x -cosec x.cot x ln [cosec x – cot x] + c
__
__
__
__
__
__
)(xfdxd ( )dxxfò
1na -nx cnaxn
++
+
1
1
caa x
+ln
x1
ax ln1
x-1sin21
1
x-
x-1cos21
1-
x-
x-1tan 2+11x
x-1cot 2x11
+
-
x-1sec1
12 -xx
x-1cosec1xx
12 -
-
(16030175) -4- T970(E)(A4)T
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__________________________________________________________________
f (x)
__________________________________________________________________
__
__
__
__
__
__
INTEGRATION
APPLICATIONS OF INTEGRATION AREAS
)(xfdxd ( )ò dxxf
22
1
xa -c
ax
+÷øö
çèæ1-sin
22 +1xa
cax
a+÷
øö
çèæ1-tan1
22
1
axx -c
ax
a+÷
øö
çèæ1-sec1
22 xa - cxaxaxa
+-+÷øö
çèæ 221-
2
2sin
2
221ax -
caxaxln
a+÷÷
ø
öççè
æ+-
21
221xa -
cxaxaln
a+÷÷
ø
öççè
æ-+
21
( ) ( ) ( ) ( ) ( ) ( )dxxgxfxgxfxgxfò ò-= .'.'.
( )( ) ( ) cxfdxxfxf
+=ò ln'
( )[ ] ( ) ( ) cnxfdxxfxfn
n+
+=
+
ò 1'.
1
dcxB
baxA
dcxbaxxf
++
+=)+)(+(
)(
( )( ) ( ) ( ) ( ) ( )nn ax
ZaxC
axB
axA
axxf
+++
++
++
+=
+...32
( )òò -==b
ax
b
ax dxyyAdxyA 12;
( )òò -==b
ay
b
ay dyxxAxdyA 12;
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VOLUMES
SECOND MOMENT OF AREA
MOMENTS OF INERTIA Mass = density × volume M = pV DEFINITION: I = m r2
GENERAL:
( )dxyyVdxyVb
ax
b
ax òò -== 21
22
2 ; pp
( )dyxxVdyxVb
ay
b
ay òò -== 21
22
2 ; pp
dArI;dArIbay
bax
22 òò ==
dVrdmrIba
ba
22 òò == r
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