# Mathematics Grade 11 KZN March Test 2026 Paper & Memo

> Grade 11 Mathematics question paper and memo from the KZN March Test 2026, 22 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-11-mathematics-2026-kzn-march-test-question-paper-and-memorandum/b2630e82-3795-48ed-bbdb-76d63218c02a
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2026
- Exam period: KZN March Test
- Curriculum: CAPS
- Pages: 22

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## About this document

Mathematics · Grade 11 · KZN March Test · 2026. Question paper and memorandum, 22 pages. Read online or download the PDF.

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## Document text

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### Page 1

| KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL

SENIOR CERTIFICATE

GRADE 11

MARKS: 75

TIME: 1% hours

This question paper consists of 4 pages and an answer book of 8 pages.

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INSTRUCTIONS AND INFORMATION

Read the following instructions and information carefully before answering the questions.

iF

Va

This question paper consists of 3 questions.
Answer ALL the questions in the SPECIAL ANSWER BOOK provided.

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

Answers only will NOT necessarily be awarded full marks.

An approved scientific calculator (non-programmable and non-graphical) may be
used, unless stated otherwise.

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

Write neatly and legibly.

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QUESTION 1

L2 Solve for x and y simultaneously:

V36-24m

L3 The roots of a quadratic eg A

Calculate the values of m-

18 | fig—-2N2

2.1.3 a

2.2 Solve for x, WITHOUT using a calculator:

221 a= 92
2.22 12.3" +1=0

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ts of the equation will be real.

March 2026

(2)
(4)
(4)

(4)

(6)

(6)

(2)
[28]

(3)

(4)

(3)

(2)

(5)
[17]

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QUESTION 3

3.1 oll If 3tan0+2= 0 and sin@>0, evaluate the following with the use of a diagram and
) > without using a calculator:

312 — 1Bcos?@-Vi3sind

3.2 Simplify the following expression to a single term:
cos (360° — x). tan (—x).cos(90°+x)
sin (x +720°) ,
33) Evaluate without the use

tan 120°.sin 115°
cos 240°.cos 205°"

3.4 Given: — (cosx+ tan)
tan %

of x in the interval x €[0°;360°) is

in x) undefined?

for Be [1 80° ; 270°]. Calculate, without the use
sible e(s) of k, leaving your answer in surd form.

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(4)

(3)

(6)

(7)

(4)

(2)

(4)
[30]

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

PROVINCIAL STANDARDISED
ASSESSMENT

GRADE 11

Marks: 75

Time: 1% hours

Question Marks
1

2
3
TOTAL

N.B. This answer book consists of 8 pages.

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### Page 6

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Marks
Q)
(4)
1.1.3
; fmorephfales com
(4)
1.1.4
(4)

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March 2026

Solution Marks
Ts —
(6)
1.2
1.3:

[28]

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QUESTION 2
Solution Marks

211

GB)
212

(4)
23

G3)

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2.2.1
2)
2.2.2
(5)
(17]

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March 2026

QUESTION 3
Solution Marks

3.1.1

(4)
3.12

GB)
3.2

(6)

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3.3

(7)
3.4.1

(4)
3.4.2

(2)
38.

(4)

[30]

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March 2026

Additional space

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¥
‘i
> 4 J KWAZULU-NATAL PROVINCE
ss fL. EDUCATION
ea REPUBLIC OF SOUTH AFRICA

NATIONAL
TOR CERTIFICATE

MARKS: 75

These marking guidelines consist of 10 pages.

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QUESTION 1

KZN/March 2026

Tle

(x+1)(2x-6)=0

x=-l or x=3 vA answer YA answer
(2)
1.1.2 | 3x°+8x=4
3x7 +8x-4=0 YA standard form
—= =b+Vb° —4ac
YCA correct subst. into quadratic
formula
YCA answer ¥CA answer
(4)
YA correct factors
YCA critical values
(accept if shown on a sketch or
number line)
imorephysics.com
s-ll<x<-8 OR xe(-11;-8) VCAYCA answer
(4)
1.1.4 | ¥36—5x? -2x=0
VAG ,
36-5" aie A isolation of surd
2 * 2
(V36-5x7) =(2x)
36-5x? =4x° ¥YCA squaring both sides
36 =9x°
9x* -36=0
gin hal YCA standard form
(x+2)(x-2)=0 OR x =4
x#—2 or x=2 OR x#-2 or x=2 ¥CA answer with selection
(4)

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KZN/March 2026

LL.

x(¥>4)+ l(x-2) =2

i re re
x=2=x+4 x°-6x+8
x 1 2

+ =
=D X-4 (x-2)(x-4)

¥-2)(x-4)

X(x—4)+1(x—2) _

(x-2)(x-4)

x? —4x+%-2 _

(=2)(=4)

x -3x-2 2

(x-2)(x-4) (x-2)(x-4)

wx? -3x-2=2

Award mark for standard form

x -3x-4=0 aan ; '
only if it is a quadratic equation

(x+1)(x-4)=0

x=-l or x#4

vA factoring denominator on RHS
YA writing as a single fraction

YCA simplification
YCA standard form
YCA both x-values
¥CA valid answer

(6)

OR

YA factoring denominator on RHS

vA LCD on LHS

¥YCA equating numerators
YCA standard form

¥CA both x-values

¥YCA valid answer

(6)

1.2

2x+y=3
VHB 2WNvrrrrccrssereeeed equ.(3)

2 Hy y

vay
x? -(3-2x)° =-x—-(3-2x)
x? =(9-12x-4+4x7) = -v-3 42x

x? -94+12x-4x? =x-3
3x? +11x-6=0

3x? —11x+6=0]| B/D if standard form is NOT quadratic

(3x-2)(x-3)=0

x== or x=3

yo or y=-3

wilnr wlry

vA expressing y in terms of x

YCA substitution

YCA simplification (after removing
all brackets)

YCA standard form

YCA values of x

YCA values of y

(6)

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KZN/March 2026

—3y° -6y+9=-2y +6

3y? +4y-15=0

YA expressing x in terms of y

YCA substitution

YCA simplification (after removing
all brackets)

YCA standard form

YCA values of y

or x=
2
x= 7 or x=3 YCA values of x
(6)
1.3 36-—24m20 YA 36-24m>0
—24m = -36
msa Y¥CA answers
Q)
[28]

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QUESTION 2

KZN/March 2026

2 RSP

=3'2%54
3

“3

ss
125

Answer only, award 0 marks

VA writing 0,125 as =
1000

YA applying exponent / surd law

YCA answer
(3)
OR

VA rewriting 0,125 as (0,5)

“A applying exponent / surd law

¥CA answer

(3)

= BPH2PHl gp pt2tp-2-2p 5 p-2-p-l

vA rewriting as prime bases

VA applying the law of exponents

YCA simplification

¥CA answer
(4)

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KZN/March 2026

2.1.3 | —LOghk -2./2
v2
18-2) -=
ti Wa) N2 =x? 22 YA rationalising denomin. i.e. x A
= a -3V2 -2,2 VA 3V2
=9/2-52
= 4/2 Answer only, award 0 marks || “CA answer
(3)
OR
VA 3V2
YA LCD
= 38 Answer only, award 0 marks || /, GA answer
a GB)
2.2.1 | 2°* =32
2 =2° YA 32=2°
-3x=5
— OR exile vA answer
3 @)
OR OR
2332
1 PR, Pres
9% Be 25* oF
3x=-5
a OR x=-l= vA answer
3 2)

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KZN/March 2026

2.2.2

2h" +1=0
9*

VA 9 =37*

YA correct standard form
¥CA factors

¥CA both equations with base 3
vA both answers

(5)
27 _193*41=0
9
27, =
gr 7123" +1=0 VA 9° =3"
27.37 -12.3* +1=0 “A correct standard form
(9.3 -1)(3.3* -1)=0 CA factors
9.3*-1=0 or 3.3*-
3° =i or 3*= u
9 3
3°%=3° or 3*=3" YCA both equations with base 3
x=2 or x=1 YA both answers (5)
[17]
QUESTION 3
3.1.1 | 3tand+2=0
tan@=—2 vA —
3 3
Paxgy? v A diagram in the correct
3 quadrant
=(-3) +(2)
=13
r=VB VA r=Ni3
cos @ = _s ve
Vis CA answer
(4)
3) D
3.1.2 | 13cos?0-Vi3 sino=13(-5] -Vi3) (F) YCA substitution
13 vi3
= (2 -2 ¥CA simplification
=7 ¥CA answer
(@)

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3.2. | £98(360°-x).tan(—x).cos(90°+ x)

sin(x+720°)

_ (cos.x)(—tan x)(—sin x)
> (sin x)

cosx sinx sinx

¥A cos(360° — x) = cosx
vA tan(—x) =—tanx

VA cos(90°+ x) =—sinx
YA sin(x + 270°) =sinx

OR

eee ee eee sinx
soll “cosx’ 1 "A
sinx
=sinx YCA answer
(6)
3.3. | tan120°.sin115°
cos 240°.cos 205°
vA tan120° =—tan 60°
SK es Ot °
(—tan 60°)(sin 65°) StS Sens
~ (—cos 60°)(—cos 25°) vA cos 240° =—cos 60°
¥ Acos 205° =—cos 25°
__ xi 65°) OR - Y CA special angles
z (sin 65°) Y CA co-ratio
=-2y3 ¥CA answer
2)
3.4.1
LHS= | (cosx+ tan x.sin x)
tanx
= Sts S088 sins sins) VA tany = Si
sinx\ 1 cosx 1 cos x
cos’x  sinx .
= = v A removing brackets
sinx 1
= cos* x+sin* x YA writing as a single fraction
sin x
—s v A cos’ x+sin? x=1
sinx (4)
=RHS

OR

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### Page 21

KZN/March 2026

—— (cos x#tan x;

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1

v A removing brackets

sin
YA tany = 0*
cos x

YA writing as a single fraction

Y A cos’ x+sin? x=1

=RHS (4)
3.4.2 | (i) If sinx=0
“x =0° or x =180° YA 0° and 180°
and
(ii) If. tan. x is undefined
“x =90° or x =270° ¥A 90° and 270°
(2)
3.5 | cosB=k and sinf=
sin’ B+cos’ B=1 YA sin? B+cos* B=1
2 2 2
(2k) +(k) =(1) ¥ A substitution
4k +k? =1
5k’ =1
2_l 1
== VA k?=—
5
ik 1 fi
5 5
(4)
OR OR

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### Page 22

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KZN/March 2026

sin B =2cos 8
4
tan Z ==
an 2 {
sx=s1 and y=-2

Pax y

=i) 4-2)

r=NV5

k=cos Pp

cosB=k and sinf=2k

v Asin B =2cos 8

VA tan p ==

v A diagram in the correct

quadrant

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TOTAL MARKS: 75

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