# Technical Mathematics Grade 12 P2 2026 Memo (Eng)

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

- Page: https://www.hlayiso.com/library/grade-12-technical-mathematics-2026-nsc-november-mock-exam-memorandum-paper-2/39f6f984-9be0-4fa9-b7d4-7fbff9fbd659
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 12

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## About this document

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

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## Document text

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### Page 1

hlayiso mock exams
                           Practice Series:
                           Grade 12 NSC Preparation
                           WWW.HLAYISO.COM

                                NATIONAL
                           SENIOR CERTIFICATE

                                    GRADE 12

                        TECHNICAL MATHEMATICS P2
                          MOCK EXAMINATION 2026
                            MARKING GUIDELINES

  MARKS: 150

                     These marking guidelines consist of 12 pages.

Copyright reserved                                                   Please turn over

### 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
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 7, 8 and 9 (Euclidean geometry) a statement without the correct reason earns
      no mark for the reason. Accept any valid alternative proof.

QUESTION 1

                 √
 1.1         AB√= (3 −
                     √(−3)) + (4 − 2)
                            2        2                               ✓ SF                       A
             = 40 = 2 10 units                                       ✓ answer                  CA
                                                                                               (2)
                (               )
                −3 + 3 2 + 4
 1.2         M        ;                                              ✓x=0                        A
                    2    2
                                                                     ✓y=3                        A
             M(0 ; 3)
                                                                                                (2)

                     4−0
 1.3         mBC =       =2                                          ✓ SF                       A
                     3−1
                                                                     ✓ answer                  CA
                                                                                               (2)

 1.4         tan θ = 2                                               ✓ tan θ = mBC             CA
             θ = 63,43◦                                              ✓ answer                  CA
                                                                                               (2)

 1.5         m = 2 (parallel lines have equal gradients)             ✓m=2                      CA
             y − 2 = 2(x − (−3))                                     ✓ substitution of A       CA
             y = 2x + 8                                              ✓ equation                CA
                                                                                               (3)

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### Page 3

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

                    2−0         1
 1.6         mAC =         =−                                        ✓ mAC                      A
                   −3 − 1       2
                            1                                        ✓ product = −1            CA
             mAC × mBC = − × 2 = −1                                  ✓ conclusion              CA
                            2
             ∴ AC ⊥ BC, i.e. AĈB = 90◦                                                        (3)

                                                                                               [14]

QUESTION 2

 2.1.1       r2 = (−2)2 + (4)2 = 20                                  ✓ substitution of T        A
             x2 + y2 = 20                                            ✓ equation                CA
                                                                                               (2)

                     √         √
 2.1.2       OS =        20 = 2 5 units                              ✓ answer                  CA
                                                                                               (1)

                      4−0
 2.1.3       mOT =           = −2                                    ✓ mOT                      A
                      −2 − 0
                        1                                            ✓ mtangent                CA
             mtangent =     (tangent ⊥ radius)                       ✓ substitution of T       CA
                        2
                      1                                              ✓ equation                CA
             y − 4 = (x + 2)
                      2                                                                        (4)
                  1
             y = x+5
                  2

 2.1.4       K(0 ; 5)                                                ✓ answer                  CA
                                                                                               (1)

              4x2 y2
 2.2.1            +  =1                                              ✓ dividing by 16           A
              16 16
              x2 y2                 x2 y2                            ✓ standard form           CA
                 + = 1 OR             +   =1                                                   (2)
              22 42                 4 16

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### Page 4

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

 2.2.2                 y                                                   ✓ x-intercepts ±2        CA
                           4                                               ✓ y-intercepts ±4 and shape
                                                                           CA
                                                                                                    (2)

                                        x
              −2                    2

                           −4

                                                                                                    [12]

QUESTION 3

                     180◦
 3.1.1       2,3 ×        = 131,78◦                                        ✓ answer                   A
                      π
                                                                                                     (1)

                          1
 3.1.2       sec 47◦ =                                                     ✓ cos147◦                 A
                       cos 47◦
             = 1,47                                                        ✓ answer                 CA
                                                                                                    (2)

 3.1.3       sin(131,78◦ − 47◦ ) = sin 84,78◦                              ✓ 84,78◦                 CA
             = 1,00 (0,9959 . . .)                                         ✓ answer                 CA
                                                                                                    (2)

 3.2.1                          y                                          ✓ x = −5, y = 12, r = 13 in
                                                                           quadrant II                A

                       13
              12
                                                                           ✓ answer                 CA
                                                                                                    (2)
                                    θ
                                            x
                −5

             r2 = (−5)2 + 122               r = 13
                        5
             cos θ = −
                       13

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### Page 5

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

 3.2.2       1 − sec(θ · sin)
                            θ( )
             = 1− −
                         13   12                                  ✓ sec θ = − 13
                                                                               5              CA
                          5   13                                  ✓ sin θ = 13
                                                                            12
                                                                                              CA
                    12
             = 1+
                     5                                            ✓ answer                    CA
                17
             =        OR 3,4                                                                  (3)
                5

 3.3         sin x = −0,6                                         ✓ sin x = −0,6                A
             reference angle = 36,87◦
             x = 180◦ + 36,87◦ = 216,87◦                          ✓ 216,87◦                   CA
             or x = 360◦ − 36,87◦ = 323,13◦                       ✓ 323,13◦                   CA
                                                                                              (3)

                                                                                              [13]

QUESTION 4

 4.1.1       cos x                                                ✓ answer                      A
                                                                                               (1)

 4.1.2       − sin x                                              ✓ answer                      A
                                                                                               (1)

             (− sin x)(cos x)
 4.1.3                        + (− sin x)2                        ✓ − sin x                    A
                 − tan x
               sin x cos x                                        ✓ cos x                      A
             =             + sin2 x                               ✓ − tan x                    A
                  sin x
                                                                  ✓ sin2 x                     A
                  cos x
             = cos2 x + sin2 x                                    ✓ tan x = cos
                                                                            sin x
                                                                                x             CA
             =1
                                                                  ✓ answer                    CA
                                                                                              (6)

 4.2.1       1 − cos2 θ = sin2 θ                                  ✓ answer                      A
                                                                                               (1)

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### Page 6

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

                        1
                            − cos θ
 4.2.2       LHS = cos θ                                                    ✓ sec θ = cos1 θ              A
                          sin θ                                                        sin θ
                                                                            ✓ tan θ = cos  θ              A
                          cos θ
                                                                              1−cos2 θ
               1 − cos2 θ cos θ                                             ✓ cos θ                      CA
             =             ×
                  cos θ       sin θ                                         ✓ sin2 θ                     CA
               sin2 θ                                                                                    (4)
             =
                sin θ
             = sin θ = RHS

                                                                                                         [13]

QUESTION 5

 5.1                 y                                                      f:
                         f                                                  ✓ intercepts at 90◦ , 270◦ and
                 2
                                                                            (0 ; 2)                       A
                 1
                             g
                                                                            ✓ turning points (0◦ ; 2);
                                                                            (180◦ ; −2); (360◦ ; 2)       A
                                  90◦      180◦        270◦      360◦
                                                                        x   ✓ shape                       A
                                                                            g:
              −1
                                                                            ✓ x-intercepts at 0◦ ; 90◦ ;
              −2                                                            180◦ ; 270◦ ; 360◦            A
                                                                            ✓ turning points at 45 ,◦

                                                                            135◦ , 225◦ , 315◦ (y = ±1) A
                                                                            ✓ shape                       A
                                                                                                         (6)

 5.2.1       180◦                                                           ✓ answer                       A
                                                                                                          (1)

 5.2.2       2                                                              ✓ answer                       A
                                                                                                          (1)

 5.2.3       x = 90◦         or     x = 270◦                                ✓ 90◦                        CA
                                                                            ✓ 270◦                       CA
                                                                                                         (2)

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### Page 7

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

 5.2.4       −1 ≤ y ≤ 3     OR     y ∈ [−1 ; 3]                   ✓ answer                    CA
                                                                                              (1)

                                                                                              [11]

QUESTION 6

 6.1         D̂ = 180◦ − 35◦ − 115◦ = 30◦                         ✓ answer                      A
                                                                                               (1)

               BD           40
 6.2                ◦
                      =                                           ✓ sine rule                  A
             sin 35      sin 30◦
                     40 sin 35◦                                   ✓ SF                        CA
             BD =
                      sin 30◦
             BD = 45,89 m                                         ✓ answer                    CA
                                                                                              (3)

                       AB
 6.3         tan 28◦ =                                            ✓ trig ratio                 A
                       40
             AB = 40 tan 28◦ = 21,27 m                            ✓ answer                    CA
                                                                                              (2)

               CD            40
 6.4                 ◦
                       =                                          ✓ rule                       A
             sin 115       sin 30◦
                    40 sin 115◦                                   ✓ SF                        CA
             CD =                 = 72,50 m                       ✓ answer                    CA
                       sin 30◦
             OR                                                                               (3)
             CD2 = 402 + 45,892 − 2(40)(45,89) cos 115◦
             CD = 72,50 m

                          1
 6.5         Area △BCD = (BC)(BD) sinCB̂D                         ✓ area rule                  A
                          2
               1                                                  ✓ SF                        CA
             = (40)(45,89) sin 115◦                               ✓ answer                    CA
               2
             = 831,74 m2                                          Accept 831,7 – 831,8
                                                                                               (3)

                                                                                              [12]

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### Page 8

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

QUESTION 7

 7.1         …twice the angle subtended by the same arc at the   ✓ answer                  A
             circumference (of the circle).                                               (1)

                  1
 7.2.1       Â = BÔD = 65◦                                     ✓S                        A
                  2
             (∠ at centre = 2 × ∠ at circumference)              ✓R                        A
                                                                                          (2)

 7.2.2       BĈD = 180◦ − 65◦ = 115◦                            ✓S                      CA
             (opp ∠s of cyclic quad)                             ✓R                       A
                                                                                         (2)

 7.3.1       PM = MQ = 8 cm                                      ✓S                        A
             (line from centre ⊥ to chord)                       ✓R                        A
             OM 2 = OP2 − PM 2 (Pythagoras)
             OM 2 = 102 − 82 = 36
             OM = 6 cm                                           ✓ answer                CA
                                                                                         (3)

 7.3.2       ON = OP = 10 cm (radii)                             ✓ ON = 10 (radii)        A
             MN = 10 − 6 = 4 cm                                  ✓ answer                CA
                                                                                         (2)

                                                                                         [10]

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### Page 9

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

QUESTION 8

 8.1.1       PR̂Q = T P̂Q = 62◦                                   ✓S                        A
             (tan-chord theorem)                                  ✓R                        A
                                                                                           (2)

 8.1.2       PÔQ = 2 × 62◦ = 124◦                                ✓S                      CA
             (∠ at centre = 2 × ∠ at circumference)               ✓R                       A
                                                                                          (2)

 8.1.3       OP̂T = 90◦ (radius ⊥ tangent)                        ✓ S/R                    A
             OP̂Q = 90◦ − 62◦ = 28◦                               ✓ answer                CA
             OR                                                                           (2)
             OP̂Q = OQ̂P (∠s opp equal radii)
                    180◦ − 124◦
             OP̂Q =             = 28◦ (sum of ∠s in △)
                         2

 8.1.4       Ŝ = 180◦ − 75◦ = 105◦                               ✓S                        A
             (opp ∠s of cyclic quad)                              ✓R                        A
                                                                                           (2)

 8.2.1       DĈE = BÂD = 80◦                                    ✓S                        A
             (ext ∠ of cyclic quad)                               ✓R                        A
                                                                                           (2)

 8.2.2       AB̂D = AD̂B (∠s opp equal sides)                     ✓ S/R                    A
                    180◦ − 80◦                                    ✓ answer                CA
             AB̂D =            = 50◦ (sum of ∠s in △)
                        2                                                                 (2)

 8.2.3       DÂC = DB̂C = 35◦                                    ✓S                        A
             (∠s in the same segment)                             ✓R                        A
                                                                                           (2)

 8.2.4       BĈD = 180◦ − 80◦ = 100◦ (opp ∠s of cyclic quad)     ✓ BĈD = 100◦ S/R       CA
             BD̂C = 180◦ − 100◦ − 35◦ = 45◦ (sum of ∠s in △)      ✓ answer                CA
                                                                                          (2)

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### Page 10

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

 8.2.5       AĈB = AD̂B (∠s in the same segment)                   ✓ S/R                        A
             AĈD = AB̂D (∠s in the same segment)                   ✓ S/R                        A
             But AD̂B = AB̂D (∠s opp equal sides)                   ✓ S/R                        A
             ∴ AĈB = AĈD (= 50◦ )
             ∴ AC bisects BĈD                                      ✓ conclusion                 A
                                                                                                (4)

                                                                                               [20]

QUESTION 9

             AD AE
 9.1.1            =                                                 ✓S                           A
             DB EC
             (line ∥ one side of △ / prop theorem; DE ∥ BC)         ✓R                           A
                                                                                                (2)

             6   9
 9.1.2         =                                                    ✓ substitution             CA
             4 EC
             EC = 6 units                                           ✓ answer                   CA
                                                                                               (2)

 9.1.3       AC = 9 + 6 = 15 units                                  ✓ answer                   CA
                                                                                               (1)

 9.2         In △ADE and △ABC:
             Â = Â (common)                                       ✓ S/R                        A
             AD̂E = AB̂C (corresp ∠s; DE ∥ BC)                      ✓ S/R                        A
             AÊD = AĈB (corresp ∠s; DE ∥ BC)                      ✓ S/R                        A
             ∴ △ADE ||| △ABC (∠, ∠, ∠)                                                          (3)

             DE AD
 9.3            =         (||| △s)                                  ✓ ratio                    CA
             BC    AB
             DE     6
                =                                                   ✓ answer                   CA
             15    10
             DE = 9 units                                                                      (2)

                                                                                               [10]

QUESTION 10

                     6 000
 10.1.1      n=            = 100 rev/s                              ✓ answer                     A
                       60
                                                                                                (1)

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### Page 11

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

 10.1.2      ω = 2π n = 2π (100)                                 ✓ SF                    CA
             ω = 628,32 rad/s                                    ✓ answer                CA
                                                                                         (2)

 10.1.3      D = 180 mm = 0,18 m                                 ✓ D = 0,18 m             A
             v = π Dn = π (0,18)(100)                            ✓ SF                    CA
             v = 56,55 m/s                                       ✓ answer                CA
                                                                                         (3)

 10.1.4      40 = π D(100)                                       ✓ SF                    CA
             D=
                   40
                       = 0,1273 m = 127,32 mm                    ✓ answer in mm          CA
                 100π                                                                    (2)

                       π      3π
 10.2.1      135◦ ×       ◦
                            =    rad = 2,36 rad                  ✓ answer                  A
                      180      4
                                                                                          (1)

                          3π
 10.2.2      s = rθ = 6 ×                                        ✓ SF                    CA
                           4
             s = 14,14 m                                         ✓ answer                CA
                                                                                         (2)

                    r2 θ   62 × 34π
 10.2.3      Area =      =                                       ✓ SF                    CA
                     2        2
             = 42,41 m2                                          ✓ answer                CA
                          rs 6 × 14,14                                                   (2)
             OR Area = =               = 42,42 m2
                           2        2

 10.2.4      Perimeter = s + 2r = 14,14 + 2(6)                   ✓ s + 2r                CA
             = 26,14 m                                           ✓ answer                CA
                                                                                         (2)

 10.3        4h2 − 4dh + x2 = 0
             4(0,4)2 − 4d(0,4) + (1,6)2 = 0                      ✓ SF                     A
             0,64 − 1,6d + 2,56 = 0                              ✓ simplification        CA
             1,6d = 3,2
             d=2m                                                ✓ answer                CA
                                                                                         (3)

                                                                                        [18]

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### Page 12

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

QUESTION 11

                     12
 11.1.1      a=         =2m                                        ✓ answer                   A
                      6
                                                                                             (1)

                  3,0 + 3,4
 11.1.2      m1 =           = 3,2; m2 = 3,6; m3 = 3,7              ✓ any THREE mid-ordinates
                      2
             m4 = 3,4; m5 = 3,0; m6 = 2,6                          A
             AT = 2(3,2 + 3,6 + 3,7 + 3,4 + 3,0 + 2,6)             ✓ all mid-ordinates    A
             AT = 2(19,5) = 39 m2                                  ✓ SF                 CA
                                                                   ✓ answer             CA
                                                                                         (4)

 11.1.3      Cost = 39 × R185                                      ✓ multiplying by 185     CA
             = R7 215                                              ✓ answer                 CA
                                                                                            (2)

 11.2.1      V = π (2)2 (6) = 24π                                  ✓ SF                      A
             V = 75,40 m3                                          ✓ answer                 CA
                                                                                            (2)

 11.2.2      s2 = 22 + 1,52 = 6,25                                 ✓ Pythagoras              A
             s = 2,5 m                                             ✓ answer                 CA
                                                                                            (2)

                      1
 11.2.3      Vcone = π (2)2 (1,5) = 2π = 6,28 m3                   ✓ SF cone                 A
                      3
             Vtotal = 75,40 + 6,28                                 ✓ adding volumes         CA
             = 81,68 m3                                            ✓ answer                 CA
                                                                                            (3)

 11.2.4      Area = 2π rh + π rs                                   ✓ cylinder curved area   CA
             = 2π (2)(6) + π (2)(2,5)                              ✓ cone curved area       CA
             = 75,40 + 15,71
             = 91,11 m2                                            ✓ answer                 CA
                                                                                            (3)

                                                                                            [17]

                                                                                  TOTAL:     150

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