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T890(E)(N22)T

NOVEMBER EXAMINATION

NATIONAL CERTIFICATE

MATHEMATICS N5

(16030175)

22 November 2016 (X-Paper) 09:00–12:00

Scientific calculators may be used.

This question paper consists of 6 pages and a formula sheet of 5 pages.

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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. Questions must be answered in blue or black ink. Write neatly and legibly.

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QUESTION 1 1.1 Determine the following limits: 1.1.1

(3) 1.1.2

(2) 1.2 Determine whether is continuous at .

(2)

[7] QUESTION 2 2.1 Determine the derivative of from first principles. (5) 2.2 Determine if:

2.2.1

(4) 2.2.2 (3) 2.2.3 (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.

(5)

2.4 Given:

Make use of implicit differentiation to calculate .

(4)

[24]

xxx

x ln1

lnlim --

¥®

xxxx

x 252lim 2

3

0 ---

®

214)( -= xxf 1-=x

3 2)( xxf =

dxdy

22 3ln3ln xexy x -=

)2sin(arccos5 xxecy =

)1(cos4 23 --= xxy

ttts cos)( =

4p

=t

2sinsin)cos( =+++ yxyx

dxdy

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QUESTION 3 3.1 Given: 3.1.1 Determine the coordinates of the turning points of (3) 3.1.2 Verify by using a table that the equation has a root

between the points =1 and =2. Use values on the table:

(2)

3.1.3 Hence, make a neat sketch of the graph of the function (3) 3.1.4 If the positive root of is estimated as 1,7, use Taylor's/Newton's

method to determine a better approximation of this root.

(2) 3.2

A farmer wants to enclose a field with length and breadth x meter and y meter respectively. The cost of fencing is R36,00/m for the length and R45,00/m for the breadth. He has an amount of R56 000,00 available.

3.2.1 Give the formula of the area and the cost in terms of x and y. (2) 3.2.2 Calculate the dimensions of the field with maximum area. (5) 3.3 A fluid flows into a cylindrical tank of radius 1,5 m at a rate of 3 m3/s.

Calculate how fast the surface is rising. HINT:

(4) [21]

2243)( 23 +--= xxxxf

).(xf

22430 23 +--= xxxx x

21 ££- x

).(xf

)(xf

mr 5,1=

hrV .. 2p=

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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

(5)

4.2.3

(3) 4.2.4 (3) 4.3 Determine by resolving the integrand into partial fractions:

(5) 4.4 Determine: (2) [23] QUESTION 5 5.1 Given: The curves and

5.1.1 Sketch the TWO graphs. Shade the area enclosed by the two graphs and

show the representative strip with dimensions

(3) 5.1.2 Calculate the magnitude of the enclosed area towards the Y-axis. (4) 5.1.3 Calculate the volume generated when this area rotates about the Y-axis. (4)

ò +--

xxx2sin22cos1

ò ydx

dxex xò 2.2

ò +- dx

xx142 3

dxxx

ò 3sin3sin3

dxxxò + .125 32

ò ydx

ò -+-14134

2

2

xxx

ò - dxx )sin(arcsin 1

4yx 2 +-= yx 24 -=

. and dyx

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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. Y X

(4) [15] QUESTION 6 6.1 Solve the differential equation

(4)

6.2 Determine the general solution of the differential equation:

(6)

[10]

TOTAL: 100

x

yxyx edydxe 8532 . +-+ =

42sin110.2 2

42

2

-+=xdx

yd x

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MATHEMATICS N5 FORMULA SHEET Any applicable formula 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 = cos2A - sin2A

tan 2A =

sin2 A = ½ - ½ cos 2A cos2 A = ½ + ½ cos 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)]

AA2tan1

tan2-

!

BABA

tantan1tan±tan

!

xx

xx

xx

xsec1

=cos;cosec1

=sin;cossin

=tan

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BINOMIAL THEOREM

DIFFERENTIATION

PRODUCT RULE y = u(x) . v(x)

QUOTIENT RULE

y =

=

CHAIN RULE y = f(u(x))

!+-

++=+ -- 221!2)1()( hxnnhnxxhx nnnn

)()(

' afafe -=

ear +=

dxdu.v

dxdv.u

dxdy

+=

'' uvvu ×+×=

)()(xvxu

2vdxdv.u

dxdu.v

dxdy -

=

2v'vu'uv ×-×

dxdu

dudy

dxdy

×=

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______________________________________________________________________

f(x)

______________________________________________________________________

axn

a 0 ax + c ex ex ex + c

ax ax.lna

logex __

logax __

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 x ln [secx + tanx] + c

cosec x -cosec x·cot x ln (cosecx - cotx) + c

__

__

__

__

__

__

)(xfdxd

dxxf )(∫

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 -

-

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__________________________________________________________________

f(x)

__________________________________________________________________

__

__

__

__

__

__

INTEGRATION

f(x) g(x) - f '(x) g(x) dx

APPLICATIONS OF INTEGRATION AREAS

)(xfdxd

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

=ò xxx d)(g)f( ' ò

cxfdxxfxf

+)(ln=)()(

∫'

cnxf

dxxfxfn

n +1+)]([

=)()]([∫1+

'

dcxB

baxA

dcxbaxxf

++

+=)+)(+(

)(

nn axZ

axC

axB

axA

axxf

)+(+

)+(+

)+(+)+(

=)+()(

32 !

dxyyAydxAb

ax

b

ax )(; 21∫∫ -==

dyxxAxdyAb

ay

b

ay )(; 21 -== òò

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VOLUMES

SECOND MOMENT OF AREA

MOMENTS OF INERTIA Mass = density × volume M = DEFINITION: I = m r2

GENERAL:

( )dxyyV;dxyV bax

bax

22

21

2 -== òò pp

( )dyxxV;dyxV bay

bay

22

21

2 -== òò pp

dArI;dArIbay

bax

22 òò ==

Vr

dVrdmrIba

ba

22 òò == r