# Mathematics Grade 11 KZN March Test 2024 Paper & Memo

> Grade 11 Mathematics question paper and memo from the KZN March Test 2024, 14 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/kzn-maths-grade-11-march-2024-qp-and-memo/b10b0793-952b-491d-91c8-a4681b2eca8a
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2024
- Exam period: KZN March Test
- Curriculum: CAPS
- Pages: 14

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## About this document

Mathematics · Grade 11 · KZN March Test · 2024. Question paper and memorandum, 14 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

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KWAZULU-NATAL PROVINCE

EDUCATION ee Se
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 75

TIME: 1% hours

This question paper consists of 6 pages.

### Page 2

Mathematics 2
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Common Test March 2024

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

tw

This question paper consists of 4 questions.
Answer ALL the questions.

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

Clearly show ALL calculations, diagrams, graphs, etc. which you have used in
determining your answers.

Answers only will NOT necessarily be awarded full marks.

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

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

Write neatly and legibly.

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### Page 3

Mathematics j
Downloaded from Stanmorepfysics.com
QUESTION 1

| Solve for x:
at 2? +5r=0
12 4x? +3x+6=0 (answer correct to TWO decimal places)
13 Wx-2=x

12 py x? -x- 20-205

*x-20<0.

determine the sum of all the integers satisfying the
0<0.

13

1.4 The roots of a quadratic equation are x = feb 5
Calculate the value(s) of k for which the roots are equal.

1.5 Given: (x+5)° =1-p?
Calculate the values of p for which the roots of the equation are non-real.

(3)
(3)
(4)

@)

(2)

(6)

(2)

(5)
[28]

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### Page 4

Mathematics 4 Common Test March 2024
NSC

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QUESTION 2

2.1 Simplify fully without using a calculator. Your answer must be free of negative
exponents and irrational denominators.

26h G3)
322 _ ort
2
ae ey
V3N6 (4)
2.2 Solve for x, without using a calculator:
2.2.1 9/32 =128 (4)
a eo
2.2.2 x? —3x4-10=0 4)
[20]

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### Page 5

Mathematics 5

} ey
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QUESTION 3

DO NOT-USE A CALCULATOR WHEN ANSWERING QUESTION 3.

3.1 In the diagram below, P(-6 ; 8) is a point in the Cartesian plane, A is a point on the

positive x-axis. AOP = 0.

Determine, with the aid of the diagram, the following:

344 The length of OP. (2)

3.1.2 sin0+cos@ @)
32 If sin25° = 7 determine the value of tan 25° in terms of k. (3)
33 Simplify the following expression fully, without the use of a calculator:

sin 698°.cos 300°
tan 135°.cos(—248°) 2)
[15]
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### Page 6

6
NSC

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QUESTION 4

Mathematics

4l Simplify the following expression fully:
sin(180° + x).sin(90° +, 7
NY ld wn lal + sin. x.cos(90°— x) (7)
tan(-x)
4.2 Given; —Si08
1+cos0
4.2.1 Prove that 899 __ 1-cosé

(3)

1+cos0 sin?

42.2 Forwhich value of @ in the interval 8 <[ 0"; 360°] is : am?
+cos

7 undefined? — (2)

[12]

TOTAL: 75

### Page 7

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|} KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL

SENIOR CERTIFICATE

GRADE 11 |

These marking guidelines consist of 8 pages.

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### Page 8

Mathemewnloaded from Stanmorephysics.com Common Test March 2024
GRADE 11
Marking Guideline
QUESTION 1
1.1.1 2x7 +5x=0
x(2x+5)=0 Y factors
5
x=0 orx= “3 Y answer Y answer
GB)
1.1.2 | -4x?+3x+6=0 OR 4x’ —3x-6=0
yn Tbe vb? ~4ac ye DD END? ~4ac
2a 2a
2
= 343° ~4(-4)(6) = 34 (-3) -4(4)(-6) | v substitution
2(-4) 2(4)
=—0,91 or x =1,66 Y answer Y answer
QB)
1.1.3 3/ x-2=x
(3 |p 2) =x? Y squaring both sides
9(x-2) =x?
x? —-9x+18=0 Y standard form
(x-6)(x-3)=0 Y factors
x=3 or x=6 Y both answers
(4)
1.2.1 x’ —x-20<0
(x+4)(x-5) <0 Y factors
v Y answer
(3)

12:2)

Sum of integers

= (-3)+(-2)+(-1)+04+14+24+3+4

=4

Y adding integers between —4
and 5
Y answer

(2)

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### Page 9

Mathemawnloaded from Stanmorephysics.com

GRADE 11
Marking Guideline

Common Test March 2024

1.3 4x+y=7
V=7H4X  eeeeceeeeee line 1 v making y the subject of the
Bx? +.2KV = V2 eeeeceeeeeeee line 2 formula
Substitute line 1 into line 2:
3x* + 2x(7-4x) =(7- 4x)’ Y substitution
3x° +14x —8x° = 49 —56x + 16x"
21x’ + 70x-49 =0
3x? -10x+7=0 Y standard form
(3x-7)(x-1) =0 Y factors
x -2 or x=1 Y x-values
7
yerg or y= 3 Vv y-values
(6)
1.4 For equal roots: 13—2k =0 v 13-2k=0
2k =13
13 1
k= > or 85 v answer
(2)
1.5 (x+5)" =1-p°
x+5=4/1- p’ Y square root taken on both sides
x=-5+,/1-p* Y x subject of formula
For non-real roots:
1-p’ <0 v¥1-p’<0
“.p’-1>0
OR +o oO
—_ >
-1 1
=} 1
p<-lor p>1 OR pe(-%;~-1) or (1; ~) v ¥ answer
(5)
OR OR

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### Page 10

Mathemewnloaded from Stanmorephysics.com Common Test March 2024
GRADE 11
Marking Guideline

(x+5)' =1-p?
x +10x+25=1-p*
x? +10x+24+ p? =0
b* —4ac = 10° — 4(1)(24+ p*)
=100—96—4p”
=4-4p° Y b? -4ac=4-4p*
For non-real roots: 4—4p’ <0 v¥ 4-4p’ <0
p-1>0
(p-1)(p+1)>0 Y factorisation
OR +o oo
mm
-1 1
=] 1
p<-lor p>1 OR pe(-#;~-1) or (1; ») v ¥ answer
()
[28]
QUESTION 2
2
2.1.1 64 3
2 2
=(2°) 3 Vv (2°)
=2* vo
-t
16 Y answer
G3)
OR OR
2
64 3
2 2
=(4) 2 v (aya
=4? v4?
at
16 v answer
(3)
OR OR

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### Page 11

Mathemewnloaded from Stanmorephysics.com Common Test March 2024
GRADE 11
Marking Guideline
ae
1 1
wIEIE! v 2
64° 643
1 1
la 3; 2 v
a er
_t
= Co
‘ Answer only:
7 0 marks Y answer
(3)
gen _ gmt
21.2 | Sa
gene -(¥ y"
~ BgBe
genre gir Vv ge = (eas
v 23") =6.3
Y factorising numerator
Y simplification
Y answer
(S)
2
3-3
2.1.3 (3-v8)
V3.6
= 9-63 +3 Y multiplying out numerator
V3.6
_ 12-6y3
V3.6
_ 6(2 ~ v3) Y factorising numerator
Vile
S28), V6
Be “V6
_ ¥6(2-v3) V6(2-V3)
~ V3 aoe only: ve iB
= ¥2(2-3) OR =2/2-J6 marks Y answer
(4)

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### Page 12

Mathemewnloaded from Stanmorephysics.com

GRADE 11
Marking Guideline

Common Test March 2024

2.2.1 | ¥32=128
1
324 +128 ¥ 32
Pc
(2°): £2’ ¥ 32=2° and 128=2’
E
=)’
—— Y equating exponents
x
x= 5 v answer
7 (4)
i 1
2.2.2 x? —3x4 -10=0
1 1 ¥Y one mark for each correct
[« +2 -s}=0
factor
1 1 1
x? 4-2 or xt =5 v xt #-2
x=5*
=625 Y answer
x=-1 (4)
OR OR
i i
x? —3x4 -10=0
1
Let k = x?:
k* —3k-10=0
(k+2)(k-5)=0 ¥V one mark for each correct
factor
k#-2 or k=5
1 1 A
ox4#-2 or xt =5 V xt #-2
x=5* y
=625 answer
(4)
[20]

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### Page 13

Mathemewnloaded from Stanmorephysics.com Common Test March 2024
GRADE 11
Marking Guideline
QUESTION 3
3.1.1 OP? =(-6)' +8 Y substitution in r? = x? + y”
=100
“OP =10 Y answer
(2)
3.1.2 | sin@+cosé
_ 8 (8) vi _=8
10 10 10 10
—2 21
“0 5 Y answer
(3)
3.2 x’ =r?—y? [Pythagoras] Y substitution in Pythagoras or
=k2-1 k sketch
x=Vk? -1 v Veo
25°
° 1 2
tan 25° = —— kl Y answer
33 sin 698°. cos 300°
, tan135°.cos(—248°)
v —si Oo
_ _—sin 22°.cos 60° v ceo
—tan 45°. — cos 68° Y —tan 45°
1 1 v —cos68°
—sin 22°.— —cos 68°. —
= ——_2. OR = ——_—. Y special angle values
~1.—sin 22 —1.—cos 68 Y cos68° = sin 22° or vice versa
1
“9 Y answer
(7)
[15]

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### Page 14

Mathemewnloaded from Stanmorephysics.com

QUESTION 4

Common Test March 2024
GRADE 11
Marking Guideline

4.1 sin(180° + x).sin(90° + x)

+ sin x.cos(90" - x)

v sin (180° + x) =—sin x

tan(-x)
¥ sin(90° + x) =Cosx
—- +sinx.sinx Y tan(—x)=-tanx
—tanx
Y cos (90° - x) =sinx
pg SUERTE sin x.sin x
sinx F ta
cos x cos x
sinx.cosx_cosx a
= x —* + sinx.sinx
1 sinx
=cos’ x + sin’ x Y simplification
=1 ¥ answer
(2)
4.2.1 sind
1+cosO
. i= _
__ sind A cos) y m cos 0)
1+cos@ (1—cos@) (1-cos)
sin @(1—cos@)
Y 1-cos’ @ in denominator
¥ 1-cos’ @=sin’ 6
_ 1-cosé
sin (3)
4.2.2 | Undefined when 1+ cos@ =0
cos9=-1 Y¥ cos9=-1
“0 =180° Y answer
(2)
[12]
TOTAL: 75

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