national senior certificate examination ......mathematics paper 2 grade 11 june 2015 5 3.4 prove...

MATHEMATICS PAPER 2 GRADE 11 JUNE 2015

1

NATIONAL SENIOR CERTIFICATE EXAMINATION

MATHEMATICS

JUNE EXAMINATION

2015

GRADE 11

PAPER 2

MARKS: 100

TIME : 2 HOURS

This paper consists of 7 pages including diagram sheet


2

INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions.

1 This paper consists of 5 questions.

2 Answer ALL the questions

3 Clearly show ALL calculations, diagrams, graphs et cetera which you have used in

determining the answers.

4 Answers only will not necessarily be awarded full marks.

5 You may use an approved scientific calculator (non-programmable and non-graphical),

unless stated otherwise.

6 If necessary, round off answers to TWO decimal places, unless stated otherwise.

7 Diagrams are NOT necessarily drawn to scale.

8 Number the answers correctly according to the numbering system used in this question paper.

9 Write neatly and legibly.

10 One diagram sheet for QUESTION 5.1 is attached at the end of this question paper. Please

write your NAME in the diagram sheet.


3

QUESTION 1 In the diagram below, ABCD is a quadrilateral with vertices A (2; 6), B (4; 0); C (7; 1) and

D(8; 8) in a Cartesian plane. E and F are on AD and CD respectively. AD // BC.

1.1 Determine

1.1.1 the coordinates of F, the mid-point of DC. (2)

1.1.2 gradient of BC (2)

1.1.3 the value of E, the angle of inclination of BC (3)

1.1.4 the equation of AD in the form y = …… (4)

1.1.5 the equation of CE if CE A AD (4)

1.2 Calculate the length of AD and AB (4)

1.3 Show that AD is perpendicular to AB (4)

1.4 Calculate the area of 'ABD (3) [26]

x

y

A (2; 6)

B (4; 0)

D (8; 8)

C(7; 1)

0

F

E

E


4

QUESTION 2

In the diagram, ABCD is a quadrilateral with A(4; –1) and B(6; 9), The diagonals of ABCD intersect at M(0; 3), The gradient of AB = 5 and E = 45q 2.1 What type of a quadrilateral is ABCD? Give a reason for your answer. (2) 2.2 Determine the coordinates of C (2) 2.3 Determine the value of T (2) 2.4 Hence or otherwise, determine the value of D (3) [9] QUESTION 3

3.1 Reduce the following to one trigonometric ratio:

TTT 222 sintantan � (3)

3.2 If 5

62sin A and > @360;90q�A

With the aid of a diagram, calculate the value of AA cos.tan15 (5)

3.3 If cos 20q = t, express the following in terms of t:

3.3.1 tan 20q (3)

3.3.2 sin 340q (2)

3.3.3 2cos 110q (2)

x T

D

B(6; 9)

M(0; 3)

C

A(4; –1) 0

y

E

D


5

3.4 Prove that

TT

TTcos

1sin1

costan �

� (5)

3.5 Simplify:

»¼º

«¬ª

��

� 1sin1

1sin1cos2

xxx (6)

[26]

QUESTION 4 4.1 Simplify the expressions fully

)90sin()180cos()360tan()180sin()cos()180sin(xxx

xxx�q�qq��

�q��q (7)

4.2 Without the use of a calculator, calculate

qq

qqq135tan.170sin

80cos.225sin.180cos (7)

4.3 Determine the general solution of 0sin3cos3 � TT (5)

[19] QUESTION 5 Consider the functions defined by:

]180;90[tan21)(2cos)( qq�� xforxxgandxxf

5.1 Sketch the graphs of f and g on the same system of axes. (6)

5.2 Determine

5.2.1 the period of f (1)

5.2.2 the amplitude of f (1)

5.2.3 the range of f (2)

5.3 For which values of x will:

5.3.1 f(x) t 0 (4)

5.3.2 g(x) > f(x) (2)


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5.4 If the graph of f is shifted 2 units downwards and 45q to the left, write down

the equation of f in the form h(x) = ….. (2)

5.5 If g(x) is reflected about the y-axis, write down the image of g(x) in the form y = …. (2)

[20] TOTAL: 100


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DIAGRAM SHEET FOR QUESTION 5 NAME:………………………………….

NATIONAL SENIOR CERTIFICATE EXAMINATION NASIONALE SENIOR SERTIFIKAAT EKSAMEN

MATHEMATICS / WISKUNDE

JUNE EXAMINATION / JUNIE EKSAMEN

2015

GRADE/ GRAAD11

PAPER/VRAESTEL 2

MARKING MEMORANDUM

MARKS/PUNTE: 100 TIME/ TYD: 2 hours/uur

This memorandum consists of 7 pages / Hierdie memorandum bestaan uit 7 bladsye

Mathematics P1 / Wiskunde V1NSC Grade /Graad 12 memorandumMDoE/June/Junie 2015

2

QUESTION/ VRAAG 1�1.1.1

� �5,4;5,7/29;

215

218;

278);(

FOFOR

yxF

¸¹·

¨©§

¸¹·

¨©§ ��

9 2

15 x

9 29 y

(2)�1.1.2

31

4701

��

BCm

9 substitution/ vervang �9 answer/ antw. (2)�

1.1.3 Etan BCm

31tan ? E

¸¹·

¨©§ �

31tan 1E

q 44,18E

9 Etan BCm

9

31tan E

9 q 44,18E (3)�

1.1.4 BCADmAD //,31

cmxy � � � )6;2(/2

316 Aasforc�

316 c

316/

316

31

�

�

xyofor

xy

931 ADm

9substitution/vervang �9value of c/ waarde van c 9correct equation/ Korrekte vergelyking (4)�

1.1.5 31 ADm

,/ CEADmaarBut A 3� ? CEm cmxy �

� � )1;7(,731 Cforc�� 22 ?c

223 �� xy

9 3� CEm �9substitution/ vervang �9value of c/ waarde van c �9correct equation/ Korrekte vergelyking (4)�


3

1.2

9substitution for AD/ Vervang vir AD �9answer /antw. ��9substitution for AB Vervang vir AB �9answer Antw. (4)

1.3

�9substitution into gradient formula Vervang in gradient formule �9gradient of/van AB �9method/ metode �9 1� u ADAB mm (4) �

1.4 �9substitution into area formula/ vervang in oppervlak formule ��9simplification/ vereenvoudig �9answer/antw. (3)

QUESTION/ VRAAG 2�2.1 Parallelogram

Both pairs of opposite sides are parallel/ albei pare teenoorst. sye //

99 answer and reason Antw. en rede (2)�

2.2

4420

2

� ?� u

�

C

C

ACm

xx

xxx

9 substitution into/ vervang in

2AC

m

xxx � ��

And/ en 2

ACm

yyy � �

� � � �

units

AD

102

40

364

2868 22

�

��

ABAD

mm

m

m

ADAB

AD

AB

A?�

�u u?

� �

��

1

331

31

32

64206

enhedevierkanteeunitssquare

hbABDoppervlofArea

eenhedeunitsABenandAD

/202104

10210221

21/

/102/102

u

uu

uu '?

� � � �

units

AB

102

40

3640624 22

�

��


4

2AC

M

yyy �

)1(23 �� u Cy

)7;4(7

�? ?

Cyc

�9x = –4 and/ en y = 7 (2)�

2.3

RTT

69,785tan

5

ABm

9 tan T = 5 �9answer/ antw. (2)�

2.4 E + D = T (ext. angle of Δ) / (buite hoek van Δ) ?D = 78,69q – 45q = 33,69q OR/ OF 45q + 101,31 + D = 180q (sum of angles in a ')/ ( hoeke van Δ)

?D = 33,69q

9E + D = T 9reason/ rede 9answer/ antw. OR/ of 945q+101,31+D=180q 9reason/ rede 9answer/ antw. (3)

QUESTION/VRAAG 3�3.1 TTT 222 sintantan �

)sin1(tan 22 TT � TT 22 cos.tan

TTT 2

2

2

cos.cossin

T2sin

�9 factorization/ faktore 9 T2cos 9

TT

2

2

cossin

(3)�3.2

� �

11

2425562

2

222

222

� ?r

�

�

�

xxxx

ryx

¸¹·

¨©§�¸̧¹

·¨̈©

§�

51

16215costan15 AA

66

�9 sketch in correct quadrant/ skets in korrekte kwadrant �9 value of x/ waarde van x �9

162tan

� A

951cos � A

9 answer/ antw. (5)

3.3.1 22 1 ty � 21 ty �

tt2120tan � q

�9 sketch/ skets 9 value of y/ waarde van y 9 answer / antw. OR/ of

5630

y

0 x

5

–1

A 62

y

x

1

20q t

y


5

OR/ of

qq q

20cos20sin20tan

sin220q = 1 – cos220q ? 2120sin t� q

tt 2120tan � q?

��9 identity for tan 20q Identiteit vir tan20˚ 9 identity for sin2 20q Identiteit vir sin220˚ 9 answer/ antw. (3)�

3.3.2 )20360sin(380sin q�q q q� 20sin

21 t�� 9 q� 20sin 9 21 t�� (2)�

3.3.3 2cos 110q = 2cos (90q+20q) = –2sin 20q = –2t

9–sin 20q 9–2t (2) �

3.4 TTT

sin1costan/�

� LKLHS

TT

TT

sin1cos

cossin

��

)sin1(cos

cos)sin1(sin 2

TTTTT

��

)sin1(cos

cossinsin 22

TTTTT

��

)sin1(cos

1sinTT

T��

Tcos

1

RKRHS /

9TT

cossin

�9 TTT 2cos)sin1(sin �� 9Denominator/noemer, (cosT (1+sinT )) 9 1cossin 22 � TT 9simplification/ vereenvoudig (5)�

3.5 »¼º

«¬ª

��

� 1sin1

1sin1cos2

xxx

»¼

º«¬

ª��

)1)(sin1(sin1sin1sincos2

xxxxx

»¼º

«¬ª

�

1sinsin2cos 2

2

xxx

»¼º

«¬ª�

x

xx 22

cossin2cos

xsin2�

9Numerator/ teller �9Denominator/ noemer �9 2sin x �9 1sin2 �x �9 x2cos� 9answer/ antw. (6)

QUESTION/ VRAAG 4�


6

4.1

9 sin x 9 cos x 9 –tan x 9 –cos x 9 cos x �9

xx

cossin

9 answer / antw. (7)�

4.2

9 –sin 45q �9 sin 10q 9 –tan 45q 9 –1 9

22

9 1 9 answer/ antw. (7)�

4.3 0sin3cos3 � TT

TT

TT

cossin3

coscos3

Ttan33

33tan T

q � 30ref Zkk �q�q ,.18030T OR Zkk �q�q ,180210T

�9division by cos T / deling deur cosT�9

33tan T

9 kq�q 18030T 9 k.180210 q�q T 9 Zk� (5)�

OUESTION/VRAAG 5�

.22

122

45tan45sin.1

45tan10sin10sin.45sin.180cos

)45180tan()10180sin()1090cos()45180sin(.180cos

�

�

�

��

��

q

q

qq

qqq

qqqq

qqqqq

.1sinsin

coscos.sin

sincos.cos.tan

cos.sin)90sin()180cos()360tan(

)cos()180sin(

��

��q

qqq

xx

xxx

xxxx

xxxxx

xx


7

5.1

9 shape of/ vorm van f 9 x-intercepts/ afsnitte of f 9 turning points of/ draaipunte van f �9 shape of/ vorm van g 9 x-intercepts/ afsnitte of g 9 asymptote of/ asimptote van f (6)

5.2.1 Period = 180q 9answer/ antw. (1)�5.2.2 Amplitude = 1 9answer/ antw. (1)�5.2.3 y � [–1; 1] or

–1 d y d 1 9critical values/ kritiese punte 9notation/ notasie (2)�

5.3.1 –45q d x d 45q or/ of x t 135q 99–45q d x d 45q�9or/ of�9 x t 135q (4)�

5.3.2 45q d x < 90q 9critical values/ kritiese waardes 9notation/ notasie (2)�

5.4 h(x) = cos 2(x + 45q) – 2 h(x) = cos(2x + 90q) – 2

9cos 2(x + 45q) 9– 2 (2)�

5.5 )tan(21 xy �

xy tan21�

9 )tan(21 xy � �

9 xy tan21�

(2)�

f

g

national senior certificate examination ......mathematics paper 2 grade 11 june 2015 5 3.4 prove...

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