# Mathematics Grade 11 KwaZulu Natal June 2022 Paper & Memo

> Grade 11 Mathematics question paper and memo from the KwaZulu Natal June 2022, 16 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/maths-gr11-june-2022-qp1-and-memo/d76a57d7-c195-4d54-bd97-1188ce2e39ed
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2022
- Exam period: KwaZulu Natal June
- Curriculum: CAPS
- Pages: 16

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## About this document

Mathematics · Grade 11 · KwaZulu Natal June · 2022. Question paper and memorandum, 16 pages. Read online or download the PDF.

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## Document text

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### Page 1

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL

SENIOR CERTIFICATE

GRADE 11

: MATHEMATICS :
: COMMON TEST .
; JUNE 2022 :

MARKS: _ 100

TIME: 2 hours

This question paper consists of 6 pages and
2 DIAGRAM SHEETS.

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SS

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1. This question paper consists of 6 questions.

2. Answer ALL the questions.

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

OS

4. Clearly show ALL calculations, diagrams, graphs, etc. which you have used)
determining your answers. o:

5. Answers only will NOT necessarily be awarded full marks. x

6. You may use an approved scientific calculator (non- pg and non-graphical),
unless stated otherwise. 0

7. If necessary, round off answers correct to TWO oy places, unless stated otherwise.
8. Diagrams are NOT necessarily drawn to scale.

9. TWO DIAGRAM SHEETS for UBSTION 3.2, QUESTION 4, and QUESTION 5 are attached
at the end of this question paper.
Detach the DIAGRAM SHEETS and hand in together with your ANSWER BOOK.

@
10. Write neatly and ne
ww
ee

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Matherpatics

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wnloaded from Stanmoxephysic¢s.com —_‘1#"* 2022 Common Test

QUESTION 1

1.1

Consider the following linear number pattern: 11; 17; 23; 293 wow. ; 299,

peees| Determine 7, , the general term of this pattern, in the form 7, =an+b.

ell) Calculate the value of the 20" term in this pattern.

1.1.3 Calculate the number of terms in this pattern.

1.1.4 The terms of this linear number pattern is the sequence of 1“ differences of a

quadratic number pattern, i.e. the 1“ differences of the quadratic number pattern
are 11;17;23;...

If the fifth term of this quadratic number pattern is 100, what will value of
the second term? S

12 Consider the following quadratic number pattem: -16; -—12; e
1.2.1 Write down the next two terms in the pattern. Ss
1.2.2 Determine T,,, the general term of this rth form T, =an’ +bn+c.
1.2.3 Show that all the terms of this number are even numbers.
1.2.4 A new pattern with general rade formed such that P, = 7-128. How
many negative terms me e in this new pattern?
0
K\
QUESTION 2 S

oe
Given: f(x)=—x? +3x+ 10e

2.1 Calculate the cota of the turning point of f

2.2 Write range of f-

2.3 Calculate the x-intercepts of /:

24 Sketch the graph of f, clearly indicating all the intercepts with the axes and the coordinates of
the turning point.

2.5 For which values of k will ~x° +3x+10=k have two positive, unequal real roots?

2.6 The graph of f is translated 2 units to the right and 3 units down to obtain the graph
of g. Write down the equation of g in the form g(x) = a(x + py +q.

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(2)

(2)
(2)

(2)

(2)
(4)

(2)

(4)
[20]

(3)
(1)
(3)

(4)

(2)

(3)

[16]

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QUESTION 3

1 x
3.1 Given: ats)=(5) +4

3.1.1 Write down the equation of the asymptote of h. (1)
3.1.2 Draw a sketch graph of A, clearly indicating the asymptote and any intercepts with
the axes. (3)
3.1.3 Two transformations are applied to / to obtain the graph of mx)= -3* -4,
Write down the two transformations. (2)
3.2 The diagram below shows the graphs of (x) = 74 q and g(x) =mx-8.
X+p

e The asymptotes of / intersect at (1 od 2).
© Graphs fand g intersect the x-axis and y-axis at C and D respectively.

y

3.2.1 Write down the values of p and q. (2)
3.2.2 Write down the domain of f qy
3.2.3 Calculate the value of a. (3)
3.2.4 Calculate the value of m. (4)
3.2.5 Determine the equation of the axis of symmetry of g that has a positive gradient.

(2)
3.2.6 For which values of x will f (x)s g(x)? @)

[21]

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QUESTION 4

June 2022 Common Test

In the diagram below, A(2 ; 11), BC-10 ; -5), C and D(6 ; 9) are the vertices of a quadrilateral in the
Cartesian plane. BC is a horizontal line and CD a vertical line. E is the midpoint of AB.

F is a point on CD such that EF is parallel to AD.

*) ast
D6 ; 9)
E
oO x
F
BC-10; 5) c
41 Write down the coordinates of C. (2)
4.2 Calculate the coordinates Of E. ses (2)
.
43 Determin ‘ (4)
4.4 Calculate the C00 (2)
[10]

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NSC- Grade 11

QUESTION 5

In the diagram below, straight line GH cuts the x-axis at J(-6 ; 0). K is the point (1 5 4). KJ is
produced to M and GK is drawn. L(-3 i 2) is a point on GH and KL is drawn. Gik =a.
yd

5.1 Calculate the gradient of GH.

5.2 Calculate the angle of inclination of MK.

5.3 Determine the size of angle a.

5.4 Prove that KL 1 JL.

5.5 Calculate the area of AKJL.

QUESTION 6

6.1 Prove that (1—sin? @)(1+ tan? 6) =1.

6.2 Solve for x if sin(2x + 30°) = —0,4 and x €[—90° ;90°]

6.3 Determine the general solution of 3sinxcosx—4cos* x=0.

G

7

Kd; 4)

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(2)

(4)

(3)

(3)

(5)

{17]

(3)

©)
(6)

[16]

TOTAL: 100

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TEAR OFF

Mathematics

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NAME & SURNAME:

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June 2022 Common Test

DIAGRAM SHEET 1

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QUESTION 3.2

QUESTION 4

y A(2; 11)
D(6 ; 9)
E
oO x
F
B(-10;-5) [e

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NAME & SURNAME:

DIAGRAM SHEET 2

QUESTION 5

TEAR OFF

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|} KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

GRADE 11 |

MARKS: — 100

These marking guideline consist of 8 pages.

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Common Test June 2022

GRADE 11
Marking Guideline
QUESTION 1
11d T, =an+b
T, =6n+b
T, =6(1)+b=11 ¥
b=11-6=5 1
T, =6n+5 vv 6n4+5
(2)
1.1.2 T, =6n+5
Tyy = 6(20)+5 Y substitution
=125 vY answer
(2)
1.1.3 299 =6n+5 Y substitution
6n = 294
n=49 Y answer
There are 49 terms in the pattern. (2)
1.1.4 Quadratic number pattern:
tee PR PR vo a
i 17 23 29
Term 2 =100—(17+ 23+29) ¥ 100-(17+23429)
=31 Y answer
(2)
1.2.1 24; 44 v 24
v 44
Q)
1.2.2 — 16 -12 -8 —4
\, 4 \ : 4OoO™N » a
N\ SN 4
4 4
Second difference = 4
a= 2nd difference
2
=2 ¥ value of a
first Ist difference = 3a+b=4
3(2)+b=4
b=-2 v value of b
Term 1=a+b+c=-16
2-2+c=-16
c=-16 Y value of c
Therefore: 7, =2n>—2n-16 v answer
(4)

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Common Test June 2022

GRADE 11
Marking Guideline
1.23 T, =2n* —2n-16
= An? -n-8) Y¥ An? -n-8)
Because n is a natural number, n? —2—8 will also be a
natural number.
And: 2(a natural number) will always be even. Y reasoning: multiples of 2
(2)
OR ; OR
The given terms in the number pattern are even. The 1“ | / given terms are even
differences between terms are also even. An even
number plus an even number will always give an even
number. Therefore all terms are even numbers. ¥ reasoning
Q)
124 P,=T, -128=2n* —2n—-144 Y expression for P,
Let P, <0
.2n? -2n-144 <0 Y 2n? -2n-144<0
nv —n-72<0
(n—-9Xn+8)<0
\-8 yl 9
n
o —8<n<9 Y¥ solution for inequality
But neN, son>d: «. O<n<9
.. there are 8 negative terms. v answer
(4)
Answer only: | mark only
[20]

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Common Test June 2022

GRADE 11
Marking Guideline
QUESTION 2
2A yee
2a
-3
= ~2E0 Y substitution
3
=> v x-value of TP
2 ; -
2
3 3 3
=J=4 =] +3)—]+10
a)G) “G)
=P!
4
Turning point: 3, vi
g point: 3° °°4 @)
2.2 ys vi OR y €(—00 ; 12,25] Y answer
4 )
23 -x° +3x+410=0 ¥ f(x)=0
x —3x-10=0 y
(x+2\(x-5)=0 factors
y=? or x=5 v x-values
(3)
24
y (1,5 ; 12,25)
10 ¥ shape
¥ x-intercepts
oy
¥ y-intercept
= 5\ x
¥ turning point
(4)
1 1
2.5 10<k<127 OR ke 10127 Y VY answer (2)
2 ¥ —(value of a=-1)
2.6 es)=-{x-2) vot 7 7
2 4 v 5 (value of p=-—)

v +90 (value of ¢=95) (3)

[16]

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Common Test June 2022

GRADE 11
Marking Guideline
QUESTION 3
3.1.1 y=4 Y answer
dQ)
3.1.2 y
¥ shape
h
¥ asymptote
5
i) ¥ y-intercept
< (3)
3.1.3 reflection in x-axis ¥ reflection in x-axis
reflection in y-axis ¥ reflection in y-axis (2)
3.2.1 p=-l v¥ p=-l
q=? vq=2 2)
3.2.2 xeR,but x #1 ¥xeER,but x41 qd)
OR OR
xe(-0 ; 1) or (1; «) ¥ xe(-o ; 1) or (1; «) qd)
3.2.3 Coordinates of D: (0 - 8) v D0 ; -8)
Equation of fi y= 1-2
x-1
Substitute (0 7 8): -8= ra -2 Y substitution
a=8-2
a=6 Y answer (3)
3.2.4 Cis the x-intercept of f
Substitute y=0: 0= a Y substitution
x-1
y= 8%
x-l
Ix-2=6 ;
x=4 Y x-coordinate of C
Va -8-0
ma 222i 8 ¥ substitution
x,-x, 0-4
=2 ¥ answer
(4)
3.2.5 yHxte
Substitute (1;-2): —2=1+e
c=-3
pex-3 vv x-3
(2)

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Common Test June 2022

GRADE 11
Marking Guideline
3.2.6 O<x<l or x24 vv 0sx<l
VY x>4 (3)
OR OR
x €[0; 1) or [4;00) ¥¥xe[0;1)
v [450) (3)
[21]
QUESTION 4
41 C(6 ;-5) v6
v -5
(2)
4.2 E Xi +X) V+.
2° 2
_ “10+2 =5+1 Y substitution
2 2
= (-4 ; 3) ¥ answer
(2)
Answer only: full marks
43 Myy = v2 71
x, ~%
aoc Y substitution in gradient
6-2 formula
1
2
Mop = May =~ Y Mey
Substitute m,, an
y=mx+e
3=(— 1 (- 4) m3 v substitution in formula for
2 equation of line
3=2+¢
c=1
1
ya-pxtl ¥ answer
(4)
44 AtF: x=6
y= 61 Y substitution of x=6 in
2 equation for EF.
ya-2 Y yp=2
F(6 ; —2) (2)
[10]
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Common Test June 2022

GRADE 11
Marking Guideline
QUESTION 5
0-(-2)
5.1 Mon = ~6-(3) Y substitution
2
“3 v answer (2)
5.2 Tyg = 22!
x2 —%
= 4-0 Y substitution
1-(-6)
4 ,
=a answer
* 4 2 4
tan KJO = my =5 v tankJO=—
KJO = 29,74° v answer (4)
53 tan GJO = mg, --3
GJO = 180° — 33,69”
- 146,31 . Y size of GIO
a= GJO-KIO Y subtracting
= 146,31 —29,74
=116,57° v answer (3)
5.4 m., = 4- (- 2) ¥ substitution in gradient formula for
KL
1-(-3) KL
_3
_) Y gradient of KL
Mg, X My = 2 x 3 -l
ee 302 Y multiplying gradients and getting a
Therefore KL 1 JL. product of — 1 (3)
5.5 JL= (x, -m) +0. -H)

= \(2-0F +E 3-COF

= +13 units

KL=,\(-2-4) +(-3-1)

=V52 OR 2v13 units
1 .
Area of AKJL= 5 x base x height

¥ substitution in distance formula for
JL
Y length of JL

Y length of KL

=] ox KL
2

= 1 x VIB x 2713 ¥ substitution in formula for area of
2 triangle

=13 square units v answer (5)

{17]

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GRADE 11

Marking Guideline

QUESTION 6

Common Test June 2022

6.1 | LHS=(1-sin? 4) + tan? 4)

= cos’? 6+sin’ 6

=1
= RHS

Y 1-sin* @=cos’@

a)
Y tan? 9 = 5 o

cos’ 0
Y cos? @+sin’ 9

(3)

62 sin(2x +30°)=-0,4
reference 7 = 23,58°
2x+30° =180° + 23,58° +k.360°,k EZ
2x =173,58° +k.360°
x = 86,79" +k.180°
or
2x +30° = 360° — 23,58° + k.360°, ke Z
2x = 306,42° + k.360°
x =153,21° +k.180°

v reference 7 = 23,58°
Y 2x+30° =180° + 23,58° +4.360°

VY x =86,79° +k.180°
¥ 2x+30° =360° —23,58° +k.360°

Y x=153,21 +k.180°

In the interval [-90°;90°|: x = 86,79° or — 26,79° Y 86,79"
¥ —26,79°
(7)
6.3 3sin x cos x—4cos” x =0
cos x(3sin.x—4cos x)= 0 Y factorisation
cosx=0 or 3sinx—4cosx=0 Y both equations
x=90°+4.180°>,keEZ or 3sinx=4cosx Y x¥=90° +k.180°
sinx 4
cosx 3
tanx= 4 Y tanx= 4
3a 3
oe 4 Y x=53,13°
x =5313°+k.180, kez Y +k.180°sk EZ
\ (6)
! [16]

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TOTAL: 100

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