# Technical Mathematics Grade 11 P2 November Mock 2026 Memo

> Grade 11 Technical Mathematics Paper 2 memorandum from the November Mock Exam 2026 (English), 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-11-technical-mathematics-2026-november-mock-exam-memorandum-paper-2/a37148d9-e08a-443a-8872-c30e985311d8
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 12

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## About this document

Technical Mathematics · Grade 11 · November Mock Exam · 2026 · English. Memorandum, 12 pages. Read online or download the PDF.

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

## Matching paper and memo

- [Grade 11 Technical Mathematics P2 November 2026 Mock](https://www.hlayiso.com/library/grade-11-technical-mathematics-2026-november-mock-exam-question-paper-paper-2/2bc4e4a3-d29d-4b4c-bc10-f35c140e3085) — QP · 2026 · Eng

## Document text

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### Page 1

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

                                 NATIONAL
                            SENIOR CERTIFICATE

                                     GRADE 11

                         TECHNICAL MATHEMATICS P2
                     NOVEMBER EXAMINATION 2026 (MOCK)
                             MARKING GUIDELINES

  MARKS: 150

                      These marking guidelines consist of 12 pages.

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### Page 2

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

MARKING CODES
 A   Accuracy                       CA     Consistent accuracy
 M   Method                         R      Rounding
 NPR No penalty for rounding        NPU    No penalty for units omitted
 S   Simplification                 SF     Substitution in correct formula
 S/R Statement/Reason               AO     Answer only
NOTE:

    • If a candidate answers a question TWICE, only mark the FIRST attempt.
    • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines where indicated.
    • No penalty for rounding (NPR) for ALL questions, unless stated otherwise.
    • In QUESTIONS 6, 7 and 8 (Euclidean geometry) a statement without the correct reason earns
      no mark for the reason. Accept any valid alternative method.

QUESTION 1

 1.1         …equal.                                                ✓ answer                     A
                                                                                                (1)

                 √
 1.2         AB = (4 − (−2))2 + (2 − (−1))2                         ✓ SF
                                                                      √                         A
               √
             = 36 + 9                                               ✓ √ 45                     CA
               √      √
             = 45 = 3 5 units                                       ✓3 5                       CA
                                                                                               (3)

                     2 − (−1) 3 1
 1.3         mAB =           = =                                    ✓ SF                        A
                     4 − (−2) 6 2
                                                                    ✓ answer                   CA
                                                                                               (2)

                   8−2    6
 1.4         mBC =      =   = −2                                    ✓ SF                        A
                   1 − 4 −3
                        1                                           ✓ mBC = −2                 CA
             mAB × mBC = × (−2) = −1                                ✓ product = −1             CA
                        2
             ∴ AB ⊥ BC                                                                         (3)

                (              )
                 −2 + 1 −1 + 8
 1.5         M         ;                                            ✓ x-value                    A
                   2      2
               (       )                                            ✓ y-value                    A
                  1 7
             M − ;       OR M(−0,5 ; 3,5)                                                       (2)
                  2 2

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### Page 3

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

 1.6         y − y1 = m(x − x1 )                                        ✓ substitution of m         CA
             y − 2 = −2(x − 4)                                          ✓ substitution of point     CA
             y = −2x + 10                                               ✓ answer                    CA
                                                                                                    (3)

                               1
 1.7         tan θ = mAB =                                              ✓ tan θ = 12                CA
                               2
             θ = 26,57◦                                                 ✓ answer                    CA
                                                                                                    (2)

 1.8         Diagonals of a rectangle bisect each other, so M is also   ✓ method                      M
             the midpoint of BD:
             4 + xD     1          2 + yD 7                             ✓ x = −5                    CA
                    =−       and           =
                2       2             2       2                         ✓y=5                        CA
             D(−5 ; 5)                                                                              (3)
             OR B → A: (x − 6 ; y − 3), so C(1 ; 8) → D(−5 ; 5)

                  √                       √
 1.9         BC =    (1 − 4)2 + (8 − 2)2 = 45                           ✓ SF √                      A
                            1                                           ✓ BC = 45                 CA
             Area △ABC = × AB × BC (B̂ = 90◦ )
                            2                                           ✓ area formula / substitution
               1 √        √                                             CA
             = × 45 × 45
               2                                                        ✓ answer                  CA
             = 22,5 square units                                                                   (4)

                           1
 1.10        m = mAB =                                                  ✓ m = 21                    CA
                           2
                    1                                                   ✓ substitution of C         CA
             y − 8 = (x − 1)                                            ✓ answer                    CA
                    2
                  1   15                                                                            (3)
             y = x+        OR      y = 0,5x + 7,5
                  2   2

                                                                                                    [26]

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### Page 4

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

QUESTION 2

 2.1.1       cos(56,3◦ + 23,8◦ ) = cos 80,1◦                       ✓ 80,1◦                     A
             = 0,17                                                ✓ answer                   CA
                                                                                              (2)

              tan 56,3◦
 2.1.2                     = tan 56,3◦ × sin 23,8◦                 ✓ cosec B = sin1 B           A
             cosec 23,8◦
                                                                   ✓ substitution               A
             = 1,4994 . . . × 0,4035 . . .
             = 0,61                                                ✓ answer                   CA
                                                                                              (3)

                        5
 2.2.1       sin θ =      , θ in the 2nd quadrant                  ✓ sketch in correct quadrant
                       13
                            y                                      A
                       13
              5                 θ
                                    x
                     −12
                                                                   ✓ x = −12                  CA
             x2 = 132 − 52 = 144, x = −12                          ✓ answer                   CA
                       12                                                                     (3)
             cos θ = −
                       13
                                 13   5
 2.2.2       sec θ + tan θ =        +                              ✓ sec θ = − 13             CA
                                −12 −12                                        12
                18                                                 ✓ tan θ = − 12
                                                                                5
                                                                                              CA
             =−                                                    ✓ answer                   CA
                12
                3                                                                             (3)
             =−
                2

 2.3         cos θ = −0,65                                         ✓ cos θ = −0,65             A
             Reference angle = 49,46◦                              ✓ reference angle          CA
             θ = 180◦ − 49,46◦ = 130,5◦                            ✓ 130,5◦                   CA
             or θ = 180◦ + 49,46◦ = 229,5◦                         ✓ 229,5◦                   CA
                                                                                              (4)

                                                                                              [15]

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### Page 5

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

QUESTION 3

               (− sin θ )(− tan θ )
 3.1         =                                                                        ✓ − sin θ                   A
                (− cos θ )(tan θ )
                                                                                      ✓ − tan θ                   A
                sin θ · tan θ
             =                                                                        ✓ − cos θ                   A
               − cos θ · tan θ                                                        ✓ tan θ                     A
                 sin θ
             =−                                                                       ✓ simplification           CA
                 cos θ                                                                ✓ answer                   CA
             = − tan θ
                                                                                                                 (6)

                         1
 3.2         LHS =            − cos x
                       cos x
                                                                                      ✓ common denominator A
               1 − cos2 x
             =                                                                        ✓ sin2 x              A
                  cos x                                                               ✓ splitting         CA
               sin2 x
             =                                                                        ✓ tan x = cos
                                                                                                 sin x
                                                                                                     x    CA
                cos x                                                                                      (4)
                        sin x
             = sin x ·
                       cos x
             = sin x · tan x = RHS

                                                                                                                 [10]

QUESTION 4

 4.1                 y                                                                f:
                             (90◦ ; 2)                                                ✓ intercepts 0◦ , 180◦ , 360◦ A
                 2
                         f                                                            ✓ turning points              A
                                                                                      ✓ shape                       A
                                                                                      g:
                                                                                  x   ✓ end points (0◦ ; 0),
                               90◦                 180◦         270◦       360◦
                                                                                      (360◦ ; 0)                    A
                                          g                                                                  ◦
                                                                                      ✓ turning point (180 ; −2) A
              −2
                                                                                      ✓ shape                       A
                                         (180◦ ; −2)         (270◦ ; −2)                                          (6)

 4.2         2                                                                        ✓ answer                     A
                                                                                                                  (1)

 4.3         360◦                                                                     ✓ answer                     A
                                                                                                                  (1)

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### Page 6

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

 4.4         −2 ≤ y ≤ 0     OR    y ∈ [−2 ; 0]                     ✓ critical values            A
                                                                   ✓ notation                   A
                                                                                               (2)

 4.5         180◦ < x < 360◦     OR    x ∈ (180◦ ; 360◦ )          ✓ critical values          CA
                                                                   ✓ notation                 CA
                                                                                              (2)

                                                                                              [12]

QUESTION 5

 5.1         BD2 = AB2 + AD2 − 2(AB)(AD) cos                     ✓ cosine rule                A
             BD2 = 502 + 702 − 2(50)(70) cos 65◦                   ✓ SF                         A
             BD2 = 4 441,67 . . .
             BD = 66,65 m                                          ✓ answer                   CA
                                                                                              (3)

 5.2         BD̂C = 180◦ − 72◦ − 48◦ = 60◦                         ✓ answer                     A
                                                                                               (1)

               BC          BD
 5.3                ◦
                      =                                            ✓ sine rule                CA
             sin 60      sin 48◦
                     66,65 × sin 60◦                               ✓ SF                       CA
             BC =                                                  ✓ answer                   CA
                         sin 48◦
             BC = 77,67 m                                                                     (3)

                             1
 5.4         Area △ABD = (50)(70) sin 65◦ = 1 586,04 m2            ✓ area △ABD                 A
                             2
                             1                                     ✓ SF area △BCD             CA
             Area △BCD = (66,65)(77,67) sin 72◦ = 2 461,67 m2      ✓ area △BCD                CA
                             2
             Total area = 4 047,71 m2                              ✓ total                    CA
                                                                                              (4)
             (Accept 4 047,43 m2 if unrounded values were used.)

                                                                                              [11]

QUESTION 6

 6.1         …a right angle    OR 90◦                              ✓ answer                     A
                                                                                               (1)

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### Page 7

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

 6.2.1       PQ̂R = 90◦                                          ✓S                        A
             (∠ in semicircle)                                   ✓R                        A
                                                                                          (2)

 6.2.2       PR̂Q = 180◦ − 90◦ − 34◦ = 56◦                       ✓S                      CA
             (sum of ∠s in △)                                    ✓R                       A
                                                                                         (2)

 6.2.3       QÔR = 2 × QP̂R = 68◦                               ✓S                        A
             (∠ at centre = 2 × ∠ at circumference)              ✓R                        A
                                                                                          (2)

 6.2.4       PŜQ = PR̂Q = 56◦                                   ✓S                      CA
             (∠s in the same segment)                            ✓R                       A
                                                                                         (2)

 6.3         …supplementary      OR     add up to 180◦           ✓ answer                  A
                                                                                          (1)

 6.4         SR̂Q = 180◦ − 100◦ = 80◦                            ✓S                        A
             (opp ∠s of cyclic quad)                             ✓R                        A
                                                                                          (2)

                                                                                         [12]

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### Page 8

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

QUESTION 7

 7.1         …the angle in the alternate segment.               ✓ answer                  A
                                                                                         (1)

 7.2.1       BÂC = F ĈB = 48◦                                 ✓S                        A
             (tan-chord theorem)                                ✓R                        A
                                                                                         (2)

 7.2.2       AB̂C = AĈE = 70◦                                  ✓S                        A
             (tan-chord theorem)                                ✓R                        A
                                                                                         (2)

 7.2.3       AĈB = 180◦ − 70◦ − 48◦ = 62◦                      ✓S                      CA
             (∠s on a straight line) OR (sum of ∠s in △)        ✓R                       A
                                                                                        (2)

 7.2.4       BÔC = 2 × BÂC = 96◦                              ✓S                      CA
             (∠ at centre = 2 × ∠ at circumference)             ✓R                       A
                                                                                        (2)

 7.2.5       OĈF = 90◦ (radius ⊥ tangent)                      ✓ S/R                    A
             OĈB = 90◦ − 48◦ = 42◦                             ✓ answer                CA
             OR                                                                         (2)
             OĈB = OB̂C (∠s opp equal radii)
                    180◦ − 96◦
             OĈB =            = 42◦
                         2

 7.3         …equal in length.                                  ✓ answer                  A
                                                                                         (1)

 7.4.1       PA = PB (tangents from same point)                 ✓ S/R                    A
                           180◦ − 50◦                           ✓ answer                CA
             PÂB = PB̂A =            = 65◦
                               2                                                        (2)

 7.4.2       AĈB = PÂB = 65◦                                  ✓S                      CA
             (tan-chord theorem)                                ✓R                       A
                                                                                        (2)

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### Page 9

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

 7.4.3       AÔB = 2 × AĈB = 130◦                               ✓S                      CA
             (∠ at centre = 2 × ∠ at circumference)               ✓R                       A
             OR OÂP = OB̂P = 90◦ (radius ⊥ tangent);                                     (2)
             AÔB = 360◦ − 90◦ − 90◦ − 50◦ = 130◦ (sum of ∠s in
             quad)

 7.4.4       OÂP = 90◦ (radius ⊥ tangent)                        ✓ S/R                    A
             OÂB = 90◦ − 65◦ = 25◦                               ✓ answer                CA
                                                                                          (2)

                                                                                          [20]

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### Page 10

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

QUESTION 8

 8.1         …bisects the chord.                                ✓ answer                   A
                                                                                          (1)

                     1
 8.2.1       AM = AB = 12 cm                                    ✓S                         A
                     2
             (line from centre ⊥ to chord)                      ✓R                         A
                                                                                          (2)

 8.2.2       OM = r − 8                                         ✓ answer                   A
                                                                                          (1)

 8.2.3       OA2 = AM 2 + OM 2 (Pythagoras)                     ✓ Pythagoras S/R          A
             r2 = 122 + (r − 8)2                                ✓ substitution           CA
             r2 = 144 + r2 − 16r + 64                           ✓ expansion              CA
             16r = 208
                                                                ✓ answer                 CA
             r = 13 cm
                                                                                         (4)
             OR 4h2 − 4dh + x2 = 0: 4(8)2 − 4d(8) + 242 = 0
             32d = 832, d = 26, r = 13 cm

 8.2.4       OM = 13 − 8 = 5 cm                                 ✓ answer                 CA
                                                                                         (1)

 8.2.5       d = 2 × 13 = 26 cm                                 ✓ answer                 CA
                                                                                         (1)

                                                                                         [10]

QUESTION 9

                450
 9.1.1       n=      = 7,5 r/s                                  ✓ n = 7,5                 A
                 60
             ω = 2π n = 2π (7,5)                                ✓ SF                     CA
                                                                ✓ answer                 CA
             ω = 47,12 rad/s
                                                                                         (3)

 9.1.2       D = 300 mm = 0,3 m                                 ✓ conversion to m         A
             v = π Dn = π (0,3)(7,5)                            ✓ SF                     CA
             v = 7,07 m/s                                       ✓ answer                 CA
                                                                                         (3)
             OR v = ω r = 47,12 × 0,15 = 7,07 m/s

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### Page 11

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

                       π                                                π
 9.2.1       110◦ ×        = 1,92 rad                              ✓ × 180 ◦                 A
                      180◦
                                                                   ✓ answer                 CA
                                                                                            (2)

 9.2.2       s = rθ = 1,8 × 1,92                                   ✓ SF                     CA
             s = 3,46 m                                            ✓ answer                 CA
                                                                                            (2)

                    r2 θ   (1,8)2 (1,92)
 9.2.3       Area =      =                                         ✓ SF                     CA
                     2           2
                                                                   ✓ answer                 CA
             = 3,11 m2
                                                                                            (2)

 9.2.4       P = 2r + s = 2(1,8) + 3,46                            ✓ 2r + s                 CA
             P = 7,06 m                                            ✓ answer                 CA
                                                                                            (2)

 9.3         4h2 − 4dh + x2 = 0                                    ✓ formula                 A
             4(0,4)2 − 4d(0,4) + (1,2)2 = 0                        ✓ SF                      A
             0,64 − 1,6d + 1,44 = 0                                ✓ simplification         CA
             1,6d = 2,08
                                                                   ✓ answer                 CA
             d = 1,3 m
                                                                                            (4)

 9.4.1       V = π r2 h = π (0,15)2 (0,8)                          ✓ SF                      A
             V = 0,06 m3 (0,05655 m3 )                             ✓ answer                 CA
                                                                                            (2)

 9.4.2       Two hemispheres = one sphere                          ✓ SF                       A
                 4
             V = π (0,15)3
                 3                                                 ✓ answer                 CA
             V = 0,01 m3 (0,01414 m3 )                                                      (2)

 9.4.3       Total volume = 0,05655 + 0,01414 = 0,07069 m3         ✓ adding volumes         CA
             Capacity = 0,07069 × 1 000                            ✓ ×1 000                  A
             = 70,69 litres                                        ✓ answer                 CA
                                                                                            (3)

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### Page 12

Technical Mathematics/P2                       12                           Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

 9.4.4       A = 2π rh + 4π r2                                  ✓ 2π rh + sphere          A
             A = 2π (0,15)(0,8) + 4π (0,15)2                    ✓ SF cylinder            CA
             A = 0,754 + 0,283                                  ✓ SF sphere              CA
                                                                ✓ answer                 CA
             A = 1,04 m2
                                                                                         (4)

                                                                                         [29]

QUESTION 10

                     30
 10.1        a=         =5m                                     ✓ answer                   A
                      6
                                                                                          (1)

                  8 + 10
 10.2        m1 =        = 9 ; m2 = 10,5 ; m3 = 11,75           ✓ mid-ordinates            A
                     2
             m4 = 12,25 ; m5 = 11,25 ; m6 = 9,75                ✓ mid-ordinates            A
             AT = a(m1 + m2 + . . . + m6 )
                                                                ✓ SF                     CA
             AT = 5(9 + 10,5 + 11,75 + 12,25 + 11,25 + 9,75)
             AT = 5(64,5) = 322,5 m2                            ✓ answer                 CA
                                                                                         (4)

                                                                                          [5]

                                                                                   TOTAL: 150

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