# Mathematics Grade 11 P2 November Mock 2026 Memo (English)

> Grade 11 Mathematics Paper 2 memorandum from the November Mock Exam 2026 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-11-mathematics-2026-november-mock-exam-memorandum-paper-2/f37ed816-9fa5-4ba2-af0f-3cec6375b403
- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 11 · November Mock Exam · 2026 · English. Memorandum, 11 pages. Read online or download the PDF.

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### Page 1

hlayiso mock exams
                            Practice Series:
                            Grade 11 Exam Preparation
                            WWW.HLAYISO.COM

                                 NATIONAL
                            SENIOR CERTIFICATE

                                     GRADE 11

                                 MATHEMATICS P2
                     NOVEMBER EXAMINATION 2026 (MOCK)
                             MARKING GUIDELINES

  MARKS: 150

                      These marking guidelines consist of 11 pages.

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### Page 2

Mathematics/P2                                    2                                  Hlayiso/Mock 2026
                                        NSC – Marking Guidelines

NOTE:

    • If a candidate answers a question TWICE, only mark the FIRST attempt.
    • If a candidate has crossed out an attempt of a question and not redone the question, mark the
      crossed-out version.
    • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines. Stop marking at the
      second calculation error.
    • Assuming answers/values in order to solve a problem is NOT acceptable.

GEOMETRY

   S A mark for a correct statement (a statement mark is independent of a reason).
   R A mark for the correct reason (a reason mark may only be awarded if the statement is correct).
 S/R Award a mark if the statement AND reason are both correct.

QUESTION 1

                     370
 1.1         x̄ =        = 24,67 push-ups                            ✓ 370
                      15
                                                                     ✓ answer                      (2)

 1.2         Minimum = 12; Q1 = 18; Median = 23; Q3 = 28;            ✓ min and max
             Maximum = 49                                            ✓ Q1 and Q3
                                                                     ✓ median                      (3)

 1.3                                                                 ✓ box with Q1 , median, Q3
                                                                     ✓ whiskers to 12 and 49 (2)

              10          20       30        40       50

 1.4         Positively skewed (skewed to the right)                 ✓ answer                      (1)
             (mean 24,67 > median 23 / right whisker is longer)

 1.5         σ = 9,17                                                ✓ answer                      (1)

 1.6         (x̄ − σ ; x̄ + σ ) = (15,50 ; 33,84)                    ✓ interval
             Values from 17 to 33: 11 learners                       ✓ answer                      (2)

                                                                                                  [11]

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Mathematics/P2                                   3                                        Hlayiso/Mock 2026
                                       NSC – Marking Guidelines

QUESTION 2

 2.1         Cumulative frequencies: 6 ; 20 ; 42 ; 60 ; 67 ; 70           ✓ first three
                                                                          ✓ last three                  (2)

 2.2                                                                      ✓ grounding at (0 ; 0)
              70
                                                                          ✓ plotting at upper limits
              60
                                                                          ✓ smooth shape             (3)
              50

              40

              30

              20

              10

                                                                      t
                   0   30      60      90      120     150      180

             Points: (0 ; 0), (30 ; 6), (60 ; 20), (90 ; 42),
             (120 ; 60), (150 ; 67), (180 ; 70)

 2.3         Median at position 70
                                2 = 35                                    ✓ reading at 35
             Median ≈ 80 minutes (accept 78 – 82)                         ✓ answer                      (2)

 2.4         Midpoints: 15 ; 45 ; 75 ; 105 ; 135 ; 165                    ✓ midpoints
             ∑ f x = 90 + 630 + 1 650 + 1 890 + 945 + 495 = 5 700         ✓ ∑ f x = 5 700
                  5 700                                                   ✓ answer                      (3)
             x̄ =       = 81,43 minutes
                    70

                                                                                                       [10]

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### Page 4

Mathematics/P2                                     4                            Hlayiso/Mock 2026
                                         NSC – Marking Guidelines

QUESTION 3

                     √                           √         √
 3.1         AC =        (6 + 2)2 + (1 − 5)2 =       80 = 4 5       ✓ substitution
                                                                    ✓ answer                    (2)

                   7−5 1
 3.2         mAB =      =                                           ✓ mAB
                   4+2 3
                                                                    ✓ mBC
                   1−7
             mBC =      = −3                                        ✓ product = −1              (3)
                   6−4
             mAB × mBC = 13 × (−3) = −1
             ∴ AB ⊥ BC
                 (                 )
               −2 + 6 5 + 1
 3.3         M       ;                 = M(2 ; 3)                   ✓x=2
                 2      2
                                                                    ✓y=3                        (2)

                    1−5          1
 3.4         mAC =         =−                                       ✓ m = − 21
                    6+2          2
                                                                    ✓ substitution of B
             y − 7 = − 2 (x − 4)
                       1
                                                                    ✓ equation                  (3)
             y = − 12 x + 9

 3.5         Inclination of AB: tan α = 13 ⇒ α = 18,43◦             ✓ 18,43◦
             Inclination of AC:                                     ✓ 153,43◦
             tan β = − 12 ⇒ β = 180◦ − 26,57◦ = 153,43◦             ✓ method
                                                                    ✓ answer
             BÂC = 18,43◦ + (180◦ − 153,43◦ )
             BÂC = 45◦                                             ✓ AB and BC ✓ equal
             OR           √                                         ✓ method ✓ answer           (4)
             AB = BC = 40 and B̂ = 90◦ , so
                       ◦    ◦
             BÂC = 180 2−90 = 45◦

 3.6         M is the midpoint of BD (diagonals of a rectangle
             bisect each other)                                     ✓ xD = 0
             4 + xD                  7 + yD                         ✓ yD = −1                   (2)
                     = 2 ⇒ xD = 0;          = 3 ⇒ yD = −1
                2                       2
             D(0 ; −1)

                                                                                               [16]

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### Page 5

Mathematics/P2                                            5                            Hlayiso/Mock 2026
                                                NSC – Marking Guidelines

QUESTION 4

                    6+2
 4.1         mPQ =       =2                                                ✓ mPQ = 2
                    1+3
                                                                           ✓ mQR = mPQ
                     k−6
             mQR =       =2                                                ✓ answer                    (3)
                     5−1
             k − 6 = 8 ⇒ k = 14

 4.2         Inclination of PQ: tan θ = 2 ⇒ θ = 63,43◦                     ✓ 63,43◦
             Inclination of y = −x + 4: tan β = −1 ⇒ β = 135◦              ✓ 135◦
             Angle between the lines = 135◦ − 63,43◦ = 71,57◦              ✓ method ✓ answer           (4)
                                   (                      )
                                       −3 + 1 −2 + 6
 4.3         Midpoint of PQ:                 ;                = (−1 ; 2)   ✓ midpoint
                                         2      2
                                                                           ✓ m⊥ = − 21
                     1                                                     ✓ substitution
             m⊥ = −
                     2                                                     ✓ equation                  (4)
             y − 2 = − 12 (x + 1)
             y = − 12 x + 23

 4.4         T lies on the perpendicular bisector of PQ:                   ✓ T on perpendicular
             −1 = − 12 t + 32                                              bisector
                                                                           ✓ substitution
              1    5
              2t = 2⇒ t =5
                                                                           ✓ answer
             OR
             TP2 = TQ2 : (t + 3)2 + 1 = (t − 1)2 + 49
                                                                           ✓ TP2 =TQ2 ✓
             8t = 40 ⇒ t = 5
                                                                           simplification
                                                                           ✓ answer                    (3)

                                                                                                      [14]

QUESTION 5

                      3
 5.1.1       sin θ = − ; θ in the third quadrant                           ✓ x = −4 (diagram)
                      5
                                                                           ✓ answer                    (2)
                                        y
                                            θ
                                                 x
                               5

              (−4 ; −3)
             x2 = 52 − (−3)2 = 16 ⇒ x = −4
                     −3 3
             tan θ =     =
                     −4 4

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### Page 6

Mathematics/P2                                       6                            Hlayiso/Mock 2026
                                           NSC – Marking Guidelines

 5.1.2       sin(−θ ) + cos(180◦ − θ ) = − sin θ − cos θ              ✓ − sin θ
                  ( ) ( )
                      3        4                                      ✓ − cos θ
             =− −        − −                                          ✓ answer                   (3)
                      5        5
                7
             =
                5
               sin x . cos x . (− tan x)
 5.2         =                                                        ✓ sin x ✓ cos x
                  (− sin x) . cos x
                                                                      ✓ − tan x ✓ − sin x
             = tan x
                                                                      ✓ cos x
                                                                      ✓ answer                   (6)
                     ( )
                 (−1) 12
 5.3         =                + sin2 17◦ + cos2 17◦                   ✓ −1 and 12
                 − 12
                                                                      ✓ − 12
             = 1+1                                                    ✓ sin 73◦ = cos 17◦ and
             =2                                                       sin2 + cos2 = 1
                                                                      ✓ answer                   (4)

                            1             1 − cos2 x
 5.4.1       LHS =              − cos x =                             ✓ common denominator
                          cos x             cos x
                                                                      ✓ sin2 x
                 sin2 x                                               ✓ cos
                                                                        sin x
             =                                                              x = tan x      (3)
                 cos x
                         sin x
             = sin x .         = sin x . tan x = RHS
                         cos x

 5.4.2       Undefined when cos x = 0:                                ✓ 90◦
             x = 90◦ or x = 270◦                                      ✓ 270◦                     (2)

                                                                                                [20]

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### Page 7

Mathematics/P2                                             7                                                            Hlayiso/Mock 2026
                                                 NSC – Marking Guidelines

QUESTION 6

 6.1         (2 sin x + 1)(sin x − 1) = 0                                                                  ✓ factors
                        1                                                                                  ✓ both equations
             sin x = −      or sin x = 1
                        2
             Reference angle = 30◦                                                                         ✓ 210◦ ✓ 330◦
             x = 210◦ + k.360◦ or x = 330◦ + k.360◦                                                        ✓ 90◦
             or x = 90◦ + k.360◦ , k ∈ Z                                                                   ✓ +k.360◦ , k ∈ Z           (6)

 6.2         Reference angle: x − 20◦ = 56,31◦                                                             ✓ 56,31◦
             x − 20◦ = 56,31◦ + k.180◦                                                                     ✓ +k.180◦
             x = 76,31◦ + k.180◦
             x = 76,31◦ or x = 256,31◦                                                                     ✓ 76,31◦ ✓ 256,31◦          (4)

                                                                                                                                      [10]

QUESTION 7

 7.1                                              y                                                        f:
                                                                                          (180◦ ; 1)
                                                                                                           ✓ x-intercepts −180◦ ; 0◦ ;
              (−180◦ ; 1)                             (0◦ ; 1)     (90◦ ; 1)
                                                                                                           180◦
                                                                                      f                    ✓ turning points
                      g                                                                                    ✓ shape
                                                                                                           g:
                                                                                                       x
                 −180◦ −135◦              −45◦              45◦                135◦        180◦            ✓ x-intercepts ±45◦ ; ±135◦
                                                                                                           ✓ turning points
                                                                                                           ✓ end points                (6)
                            (−90◦ ; −1)                           (90◦ ; −1)

              f : solid line; g: dashed line

 7.2         180◦                                                                                          ✓ answer                    (1)

 7.3         30◦ < x < 150◦                                                                                ✓ critical values
                                                                                                           ✓ notation                  (2)

 7.4         −2 ≤ y ≤ 4        OR         y ∈ [−2 ; 4]                                                     ✓ −2
                                                                                                           ✓4                          (2)

                                                                                                                                      [11]

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### Page 8

Mathematics/P2                                    8                            Hlayiso/Mock 2026
                                        NSC – Marking Guidelines

QUESTION 8

 8.1         BC2 = 122 + 92 − 2(12)(9) cos 58◦                     ✓ cosine rule
             BC2 = 110,54 . . .                                    ✓ substitution
             BC = 10,51 m                                          ✓ answer                    (3)

             sin AB̂C sin 58◦
 8.2                  =                                            ✓ sine rule
                 9       10,51
                                                                   ✓ substitution
             sin AB̂C = 0,7260 . . .                               ✓ answer                    (3)
             AB̂C = 46,55◦

 8.3         Area ABCD = 12 (12)(9) sin 58◦ + 12 (9)(7) sin 40◦    ✓ area △ABC
             = 45,79 + 20,25                                       ✓ area △ACD
             = 66,04 m2                                            ✓ answer                    (3)

                                                                                               [9]

QUESTION 9

 9.1         s2 = 32 + 42 = 25                                     ✓ Pythagoras
             s = 5m                                                ✓ answer                    (2)

 9.2         V = π (3)2 (10) + 13 π (3)2 (4)                       ✓ volume of cylinder
             = 90π + 12π = 102π                                    ✓ volume of cone
             = 320,44 m3                                           ✓ answer                    (3)

 9.3         Curved surface area = π (3)(5) = 15π                  ✓ substitution
             = 47,12 m2                                            ✓ answer                    (2)

                                                                                               [7]

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### Page 9

Mathematics/P2                                 9                              Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

QUESTION 10

 10.1                A
                     1 2

                       O
                     1 2
              B                C

                           D
                                                                ✓ construction
             Construction: Draw AO and produce it to D.
             In △AOB: OA = OB (radii)
                                                                ✓ S/R
             ∴ Â1 = B̂ (∠s opp equal sides)                    ✓ S/R
             Ô1 = Â1 + B̂ (ext ∠ of △)
             ∴ Ô1 = 2Â1                                       ✓ similarly
             Similarly, in △AOC: Ô2 = 2Â2                     ✓ adding                    (5)
             BÔC = Ô1 + Ô2 = 2(Â1 + Â2 ) = 2 BÂC

 10.2.1      Ô1 = 2D̂ = 128◦ (∠ at centre = 2 × ∠ at           ✓S ✓R                       (2)
             circumference)

 10.2.2      AB̂C = 180◦ − 64◦ = 116◦ (opp ∠s of cyclic quad)   ✓S ✓R                       (2)

 10.2.3      OA = OC (radii)                                    ✓ S/R
             OÂC = OĈA (∠s opp equal sides)
                    180◦ − 128◦                                 ✓R
             OÂC =             = 26◦ (sum of ∠s in △)          ✓ answer                    (3)
                         2

 10.2.4      BÂC = BĈA (∠s opp equal sides; AB = BC)          ✓ S/R
                    180◦ − 116◦                                 ✓ answer                    (2)
             BÂC =             = 32◦ (sum of ∠s in △)
                         2

                                                                                           [14]

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### Page 10

Mathematics/P2                                 10                          Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

QUESTION 11

 11.1.1      AM = MB = 12 (line from centre ⊥ chord)            ✓S ✓R
             OM2 = OA2 − AM2 (Pythagoras)                       ✓ Pythagoras with
             = 132 − 122 = 25                                   substitution
             OM = 5 units
                                                                ✓ answer                   (4)

 11.1.2      ON = 13 (radii)                                    ✓ ON = 13 (radii)
             MN = 13 − 5 = 8                                    ✓ MN = 8
             AN2 = AM2 + MN2 = 122 + 82 = 208 (Pythagoras)      ✓ substitution
                  √          √
             AN = 208 = 4 13 = 14,42 units                      ✓ answer                   (4)

 11.2.1      AB̂C = TÂC = 70◦ (tan-chord theorem)              ✓S ✓R                      (2)

 11.2.2      BÂC = AB̂C = 70◦ (∠s opp equal sides; AC = BC)    ✓S ✓R                      (2)

 11.2.3      PÂB = 180◦ − 70◦ − 70◦ = 40◦ (∠s on a str line)   ✓S ✓R                      (2)

 11.2.4      AĈB = PÂB = 40◦ (tan-chord theorem)              ✓ AĈB = 40◦ with reason
             [or: sum of ∠s in △ABC]
             AD̂B = 180◦ − 40◦ = 140◦ (opp ∠s of cyclic quad)   ✓ answer with reason       (2)

                                                                                        [16]

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### Page 11

Mathematics/P2                                 11                               Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

QUESTION 12

 12.1        AĈB = x (tan-chord theorem)                         ✓S ✓R
             AD̂E = x (alt ∠s; DE ∥ TS)                           ✓S ✓R                       (4)

 12.2        AD̂E = BĈE = x (both equal to x)                    ✓S
             AD̂E is an exterior angle of quad DBCE               ✓ S (ext ∠ = int opp ∠)
             ∴ DBCE is a cyclic quad (converse: ext ∠ of cyclic   ✓R                      (3)
             quad)

 12.3        AÊD = AB̂C (ext ∠ of cyclic quad DBCE)              ✓S ✓R                       (2)

 12.4        AD̂E = AĈB (proved in 12.1)                         ✓ S (both substitutions)
             AÊD = AB̂C (proved in 12.3)
             AD̂E = AÊD (given)
                                                                  ✓ S B̂ = Ĉ
             ∴ AB̂C = AĈB
                                                                  ✓R                          (3)
             ∴ AB = AC (sides opp equal ∠s)

                                                                                             [12]

                                                                                 TOTAL:       150

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