# Mathematics Grade 11 April Test 2021 Paper & Memo

> Grade 11 Mathematics question paper and memo from the April Test 2021, 13 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/maths-april-2021-qp-memo-grade-11/12ff8d6c-9950-4d4f-a85e-702d3476966a
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Question paper and memo
- Year: 2021
- Exam period: April Test
- Curriculum: CAPS
- Pages: 13

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## About this document

Mathematics · Grade 11 · April Test · 2021. Question paper and memorandum, 13 pages. Read online or download the PDF.

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

## Document text

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                  testpapers.co.za

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NSC-Grade 11

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions

.

es

10.

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This question paper consists of 5 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.
Diagrams are NOT necessarily drawn to scale.

TWO DIAGRAM SHEETS for QUESTION 4.1, QUESTION 4.2, QUESTION 4.3
AND QUESTION 5 are attached at the end of this question paper. Detach the
DIAGRAM SHEETS and hand in together with your ANSWER BOOK.

Write neatly and legibly.

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Mathematics ommon Test April 2021
NSC-Grade 11
QUESTION 1
il Solve for x:
T11 (2x+3\6—x)=0 (2)
1.1.2 5x? +x-—7=0 (correct to two decimals) QB)
113 x74+8x+15>0 (3)
1 Solve for x and y:
4x—y=3 and y’—2xy+1=0 (6)
13° Given: kx*-5x-1=0, with k #0.
1.3.1 For which value(s) of k will kx? —5x—1=0 have real roots? (3)

1.3.2. Determine two integral values of k for which the roots of kx” —5x-1=0
will be rational. (2)

[19]

QUESTION 2

2.1 Without using a calculator, rationalise the denominator and simplify:

v3 —V2
V3 +2
Q)
22 Solve for x:
x-2Vx-8=0 (5)
23 Simplify:
1
3x 412.377 fe
7.9%
(4)
[12]

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Common Test April 2021

Mathematics

QUESTION 3
371 Simplify, without the use of a calculator:
sin{- 20°
cos 430° (4)

3.2 Without “nl a calculator, simplify to a single trigonometric ratio:
i

1+sin x. tan{i80° — x).cos(~ x) (5)

3.3 If 3tan@+1=0 and cos@ > 0, calculate without the use of a calculator, and
with the aid of a diagram, the value of:

30sin? 6+/40 cos 0. (6)
[15]

GIVE REASONS FOR YOUR STATEMENTS AND CALCULATIONS IN
QUESTIONS 4 and 5.

QUESTION 4

41 In the diagram, O is the centre of the circle. A, B and D are points on the circle.
OA, OB, OD and AB are drawn. AB and OD intersect at E.
AE=EB=6cm. DE=2 cm.
If the radius of the circle is x cm, calculate the numerical value of x.

A

Nye

1)

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Mathematics Common Test April 2021

NSC-Grade 11

4.2 In the diagram, O is the centre of the circle. F, G and H are points on the circle.
FH, FO and GO are drawn.

Prove the theorem which states that FOG =2FHG.

(5)
4.3 Qis the centre of circle ABCDE. BE || CD and O, = 130°.

4.3.1 Determine, with reasons, the size of the following angles:
(a) B (2)
(b) A (2)
(c) D (2)
43.2 Prove that B; =B3. (4)
[22}

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Common Test Anril 7071

Mathematics

wtoururaue ot

QUESTION 5

In the diagram, M is the centre of the circle and diameter KL is produced to N.
MP is drawn perpendicular to KN such that PRN is a tangent to the circle at R.

MP and chord KR intersect at S. MR and LR are drawn. Let R 4=Xx.

M
K N
[3N af
2
Uo
Sp \3
y R
P
5.1 Write down with reasons two other angles each equal to x. @)
5.2 Prove that KM is a tangent at M to the circle passing through M, P and R. (4)
7]
TOTAL: 75

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Mathematics Common Test April 2021
NSC-Grade 11

NAME & SURNAME:

supeuneeeeusnseneunenane:

: DIAGRAM SHEET 1

QUESTION 4.1

SY

TEAR OFF

i QUESTION 4.2

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Mathematics Common Test Anril 2091

NAME & SURNAME:

DIAGRAM SHEET 2

QUESTION 4.3

A

B

(\

QUESTION 5

TEAR OFF

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MATAS Ge wh 4 )0°

——=————— ———
Mathematics 2 Common Test April 202!
Grade 1 1-Marking Guideline
KWAZULU-NATAL PROVINCE |
EDUCATION . a GEOMETRY * MEETKUNDE
REPUBLIC OF SOUTH AFRICA
A mark for a correct statement
(A statement mark is independent of a reason)
s
; " ‘n Punt vir'n korrekte bewering
NATIONAL ('n Punt vir 'n bewering is onafhanklik van die rede)
SENIOR CERTIFI A mark for the correct reason
: (A reason mark may only be awarded if the statement is correct)
'n Punt vir 'n korrekte rede
('n Punt word slegs vir die rede toegeken as die bewering korrek is)
Award a mark if statement AND reason are both correct
sm
peace ccsscssssseeeces : : ;
2 Ken 'n punt toe as die bewering EN rede beide korrek is
" MATHEMATICS
"
a COMMON TEST
a QUESTION 1
APRIL 2021 a ;
7 : = ye-> or x=6 V answer “ answer
MARKING GUIDELINE é @
eee eErreerr Preceerrrer eerTEee 4 ee —b+ Vb? ~ 4ac
2a
: a
eal eyF -46)C7) ¥ substitution
2(5)
MARKS: 75 _ nie vial
es
=1,09 or -1.29 v answer ¥ answer
i : : (3)

° T13 | x? 48r415>0
(x+5Xx+3)>0

This marking guideline consists of 9 pages.

Y critical values

v ¥ answer

Q)

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_————

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Mathematics                                                                              Common Test April 2021        Mathematics                                                                             Common Test April 2021
                                               Grade I I-Marking Guideline                                                                                           Grade I I-Marking Guideline

1.2         y=4x-3                                                           ✓ makingy the subject of                  QUESTION 2

                                                                                                                                                  r-
                                                                                                                                         ✓3 -✓2
            Substitute in y1 -2.:\J' +I = 0:                                 the formula
                                                                                                                     2.1
            {4x-3)2 -2x(4x-3)+1 =0                                           ✓ substitution
                                                                                                                                         ✓3 +✓2
                                                                                                                                         ✓2) x
                                                                                                                                  (../3 +✓2 ) 3✓ -✓2J
            16x 2 -2 4x+9-8x2 +6x+I =0
                                                                                                                                 JJi-                       2

                                                                                                                                                                                                            ✓3 -✓
                                                                                                                                                                                                           r-Jzl
            8x2 -18x+l0=0                                                                                                                         ✓                                                    x

                                                                                                                                 = 3-2.Ji✓J +2
                                                                             ✓ standard fonn                                                                                                       ✓
            4x2 -9x+5 =0                                                                                                                                                                                         2
            (4x-5Xx-1)=0                                                     ✓ factorisation                                                                                                       ✓ multiplying out

                                                                                                                                 =5-2✓6
                                                                                                                                            3-2                                                    denominator correctly
            x=1or x=I                                                        ✓ values of x                                                                                                         ✓ answer
                                                                                                                                                                                                                                     (3)
            y=2 or r= I                                                      ✓values of y                            2.2                  x-2-h-8 = 0
                                                                                                     (6)
1.3.1            h 2 -4ac�0                                                  ✓ condition for real roots                                    x-2x2 -8=0                                              ✓ exponential form
            (-5} -4k(-1)�0
                  2                                                          ✓ substitution
                                                                                                                                 () -4)() +2)=0                                                    ✓ factorisation
                  25+4k � O
                                25                                                                                                   I            I

                          k�-                                                                                                    x2 =4 or x=i =-'.!                                                                      .!.
                                                                                                                                                                                                           I

                                                                                                                                                                                                   ✓ x2 =4        or           =-2
                                4                                                                                                 x=l6     no solution
                                                                                                                                                                                                                       _y 2
                                                                             ✓ answer:                                                                                                             ✓ x=l6
                          k�-6.!_                                                   25
                                                                                                                                                                                   I

                                                                             k� -                                                                                                                  ✓ no solution
                               4                                                          or k�-6 _!_                                                           OR
                                                                                                                                                                                                                                     (5)
                                                                                    4                4                                                                                                            OR
                                                                                                            (3)
1.3.2       E.g.: k=-4:-6:6:14 ...                                                                                                    x-2-h-8=0                                                    ✓ isolate surd
                                                                             ✓ any correct value of k
                                                                             ✓ another correct value of k                                   1-h =x-8                                               ✓ squaring both sides
            Any two values of k that will result in b 2 -4ac
                                                                                                                                             4x=x! -16x+64                                         ✓ standard form
            being a perfect square.                                                                         (2)
                                                                                                                                    x 1 -20x+64=0
                                                                                                     1191                           (x-16Xx-4)=0
                                                                                                                                 x=16 or x=4                                                       ✓ x=l6      or x=4
                                                                                                                                                      NIA                                          ✓ rejecting x=4    (5)

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Mathematics s Common Test April 2021 Mathematics 6 Common Test April 2021
Grade 11-Marking Guideline Grade 11-Marking Guideline
23 i QUESTION 3
371 412.381 )e
195 4
: tu 3.1
_/3*B+12.37)]* Y factorising numerator
ee en Y ~sin20”
A Y cos 70"
_[ 73°}
73s Y co-function
i Y answer
=b-F ¥ simplification (4)
=3! OR
1
*3 v answer
(4)
¥ -sin20°
OR OR
¥ cos 70°
Y¥ co-function
¥ answer
(4)
¥ factorising numerator 3.2 T+sin x. tan{{80° —x).cos(- x)
aoe = I+sinx.—tanx.cosx v—tanx ¥ cosx
i sinx
=1-sinx 2 cosx Ce
i COs x cos
32) Y I-sin? x
= - Y answer
73 . (5) |
= Y simplification
eu Y answer
° (4)
[2]

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Mathematics fe
Grade 11-Marking Guideline

Common Test April 2021

3.3 tang=—4

cal.
vio
reJray
= Ae +P
=4i0
30sin? 6+ /40 cos 8

=3+6

Y standard form

¥ 4° quadrant

v value of r

Y substitution
Y simplification

Y answer
(6)

{15}

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Mathematics g Common Test April 2021
Grade 1 |-Marking Guideline
QUESTION 4
41 | OELAB [line from centre to midpoint of YSYR
chord] ¥ OE=x-2
OE=x-2 YR

OA? =OE? +AE* [Pythagoras]

xax?-4x44436

¥ substitution
¥ simplification

Y answer
: (ea)
42 | Contruction: Draw line HOJ Y construction
H
iG
F
[radii Zs opp. = sides] v SIR
ext. Z of AHFO] v SIR
[radii: Zs opp. = sides]
[ext. Z of AHGO] “Ss
vs
(5)
4.3.1 [Z atcentre= 2x Z at circumference} |v S YR
(a) (2)
43.1 | A=65° [ Zs in the same segment] YSYR
(b) (2)
OR OR
A=65° [ext. Z of AAOE: Zs opp.=radii] |VS VR
(2)
43.1 | D=1is® [opp. Zs of acyclic quadrilateral] YSVYR
(c) (2)

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### Page 13

Mathematics 9

Grade 11-Marking Guideline

Common Test April 2021

~~ Downloaded from hlayiso.com

Copyright Reserved

4.32
YSVYR
[4s opp. = radii] ae
(4) |
[22)
{tan-chord-theorem] YSYR
[radii Zs opp. = sides] vsiR
@)
52 [radius L tangent) YSYR
My [both = 90°} “S
-. KM isa tangent at M to the circle through M. P and R
[converse: tan-chord-theorem] VR
(4)
OR OR
MRP =90° [radius L tangent) YSYR
M, =2x [ext Z of AMKR]
My =180°-(0°+2x) [Zs ona straight line]
=90" -2x
6 =180" - (90° +90" - 2x) [sum of Zs of AMPR]
vs
=2y
M, =
©. KM isa tangent at M to the circle through M. P and R YR
(i: [converse: tan-chord-theorem] (4)
= 17)
TOTAL MARKS: 75

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