# Technical Mathematics Grade 12 P1 2026 Memo (Eng)

> Grade 12 Technical Mathematics Paper 1 memorandum from the NSC November Prediction Exam 2026 (English), 13 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-technical-mathematics-2026-nsc-november-prediction-exam-memorandum-paper-1/c8769b48-6903-4a0a-9753-5e5c569c3cb0
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Prediction Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 13

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## About this document

Technical Mathematics · Grade 12 · NSC November Prediction Exam · 2026 · English. Memorandum, 13 pages. Read online or download the PDF.

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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 P1
                      PREDICTION EXAMINATION 2026
                            MARKING GUIDELINES

MARKS: 150

                     These marking guidelines consist of 13 pages.

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### Page 2

Technical Mathematics/P1                        2                              Hlayiso/Prediction 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
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 where indicated.
    • No penalty for rounding (NPR) for ALL questions, unless stated otherwise.
    • AO = answer only. Where AO is indicated, the stated marks are awarded for a correct answer
      without working.

QUESTION 1

 1.1.1       (5 − 2x)(x + 3) = 0                                      ✓ x = 25                    A
             x=
                 5
                     or x = −3                                        ✓ x = −3                    A
                 2                                                                               (2)

 1.1.2       3x2 − 2x − 7 =p
                           0                                          ✓ standard form             A
                 −(−2) ± (−2)2 − 4(3)(−7)                             ✓ SF                       CA
             x=
                     √       2(3)
                 2 ± 88                                               ✓ x ≈ 1,90                 CA
             x=                                                       ✓ x ≈ −1,23                CA
                     6
             x ≈ 1,90 or x ≈ −1,23                                    NPR
                                                                                                 (4)

 1.1.3       x2 + x − 12 > 0                                          ✓ standard form & factors A
             (x + 4)(x − 3) > 0                                       ✓ both critical values  CA
             Critical values: x = −4 and x = 3

                                x
               −4          3
                                                                      ✓ correct notation         CA
             x < −4 or         x>3   OR   x ∈ (−∞ ; −4) ∪ (3 ; ∞)                                (3)

 1.2.1       y = 2x + 3                                               ✓ y = 2x + 3                A
                                                                                                 (1)

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### Page 3

Technical Mathematics/P1                       3                        Hlayiso/Prediction 2026
                                     NSC – Marking Guidelines

 1.2.2       x2 + x(2x + 3) − (2x + 3)2 = −1                    ✓ substitution              CA
             x2 + 2x2 + 3x − 4x2 − 12x − 9 + 1 = 0
             −x2 − 9x − 8 = 0                                   ✓ standard form             CA
             x2 + 9x + 8 = 0
             (x + 1)(x + 8) = 0                                 ✓ factors                   CA
             x = −1 or x = −8                                   ✓ both x-values             CA
             y=1          y = −13                               ✓ both y-values             CA
                                                                                            (5)

                  2Ek
 1.3.1       v2 =                                               ✓ v2 asqsubject               A
                 rm                                                      2Ek
                   2Ek                                          ✓v=       m                 CA
             v=
                    m                                           AO: full marks
                                                                                             (2)
                     s
                   2(2,4 × 105 ) √
 1.3.2       v=                 = 320                           ✓ substitution              CA
                     1,5 × 103
                                                                ✓ answer                    CA
             v ≈ 17,89 m/s
                                                                NPU NPR
                                                                                             (2)

 1.4         1101012 = 32 + 16 + 4 + 1 = 53                     ✓ answer                      A
                                                                                             (1)

 1.5         53 + 27 = 80                                       ✓ 80 OR binary addition M
             80 = 64 + 16                                       ✓ answer               CA
             = 10100002                                                                 (2)

                                                                                            [22]

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### Page 4

Technical Mathematics/P1                         4                        Hlayiso/Prediction 2026
                                       NSC – Marking Guidelines

QUESTION 2

 2.1.1       2m + 6 = 0
             m = −3                                               ✓ m = −3                   A
                                                                                            (1)

 2.1.2       9 − 3m < 0                                           ✓ 9 − 3m < 0               A
             m>3                                                  ✓m>3                      CA
                                                                                            (2)

 2.2         2x2 − 8x + p + 4 = 0                                 ✓ standard form            A
             Non-real roots: ∆ < 0                                ✓∆<0                       A
             (−8)2 − 4(2)(p + 4) < 0                              ✓ SF                      CA
             64 − 8p − 32 < 0
             −8p < −32
             p>4                                                  ✓ p>4                     CA
                                                                                            (4)

                                                                                            [7]

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### Page 5

Technical Mathematics/P1                           5                        Hlayiso/Prediction 2026
                                         NSC – Marking Guidelines

QUESTION 3
                        1−x
             52x+1 · 52
 3.1.1                                                              ✓ prime bases              A
                    53
             =5 2x+1+2−2x−3                                         ✓ adding/subtracting
             = 50 = 1                                               exponents                 CA
                                                                    ✓ answer                  CA
                                                                                              (3)

               √         √      √
 3.1.2       = √ 25 × 2√
                       − 9 ×√2 + 4 × 2                                 √     √ √
             = 5√2 − 3 2 + 2 2                                      ✓ 5 2−3 2+2 2              A
             =4 2                                                   ✓ answer                  CA
                                                                                              (2)

               log2 25 + log 10−2
 3.1.3       =
                     log 103
                                                                    ✓ log2 32 = 5              A
               5 + (−2)
             =                                                      ✓ log 0,01 = −2            A
                   3
             =1                                                     ✓ answer                  CA
                                                                                              (3)

                                
 3.2         log3 (x + 1)(x − 5) = 3                                ✓ product law             A
             (x + 1)(x − 5) = 33                                    ✓ exponential form      CA
             x2 − 4x − 32 = 0                                       ✓ standard form & factors
             (x − 8)(x + 4) = 0                                     CA
             x = 8 (x ̸= −4, since x − 5 > 0)
                                                                    ✓ answer with rejection CA
                                                                                            (4)

 3.3.1       z = (2 − 4) + (−3 + 1)i = −2 − 2i                      ✓ answer                   A
                                                                                              (1)

 3.3.2                      Im                                      ✓ point (−2 ; −2) in
                                                                    quadrant 3               CA
                       −2
                                    Re                              ✓ line from origin & labels
                                                                    CA
                               −2
                                                                                              (2)
              z(−2 ; −2)

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### Page 6

Technical Mathematics/P1                         6                         Hlayiso/Prediction 2026
                                       NSC – Marking Guidelines

                  p                  √      √
 3.3.3       r=     (−2)2 + (−2)2 = 8 = 2 2 ≈ 2,83                ✓r                         CA
                     −2                                           ✓ tan θ / ref. angle       CA
             tan θ =     = 1; reference angle = 45◦
                     −2
                     ◦     ◦       ◦
                                                                  ✓ θ = 225◦                 CA
             θ = 180
                   √ + 45 ◦= 225                                  ✓ polar form               CA
             z = 2 2 cis                        ◦
                       √225 OR ◦ 2,83 cis 225                                                (4)
             (Accept 2 2 cis(−135 ))

 3.4         (3 − 2i)2 − 4i = 9 − 12i + 4i2 − 4i                  ✓ expansion                  A
             = 9 − 12i − 4 − 4i                                   ✓ i2 = −1                    A
             = 5 − 16i
             m = 5 and n = −16                                    ✓ both values              CA
                                                                                             (3)

                                                                                             [22]

QUESTION 4

 4.1.1       −4 ≤ x ≤ 4    OR    x ∈ [−4 ; 4]                     ✓ critical values            A
                                                                  ✓ notation                   A
                                                                                              (2)

 4.1.2       x=0     and y = −2                                   ✓x=0                         A
                                                                  ✓ y = −2                     A
                                                                                              (2)

                   6
 4.1.3       0 = − −2                                             ✓ g(x) = 0                   A
                   x
             2x = −6
             x = −3   (−3 ; 0)                                    ✓ answer                   CA
                                                                                             (2)

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### Page 7

Technical Mathematics/P1                              7                        Hlayiso/Prediction 2026
                                            NSC – Marking Guidelines

 4.1.4                              y                                  f:
                                                                       ✓ intercepts ±4 and 4     A
                                g
                                                                       ✓ semicircle shape (above
                                        4
                                              f
                                                                       x-axis)                   A
                                                                       g:
                                                                       ✓ both asymptotes       CA
                                                                       ✓ x-intercept −3        CA
                     −4    −3                      4
                                                               x       ✓ shape (2nd & 4th quad.) A
                                                                                                (5)
                                                           y = −2
                                                       g

 4.1.5       −3 < x < 0     OR      x ∈ (−3 ; 0)                       ✓ critical values         CA
                                                                       ✓ notation                CA
                                                                                                 (2)

 4.2.1       h(x) = a(x + 1)(x − 5)                                    ✓ a(x + 1)(x − 5)         A
             Substitute C(0 ; 5): 5 = a(1)(−5)                         ✓ substitution of C       A
             a = −1                                                    ✓ a = −1 & simplification A
             h(x) = −(x2 − 4x − 5)
             h(x) = −x2 + 4x + 5
             OR                                                        OR
             c = 5: h(x) = ax2 + bx + 5                                ✓c=5                        A
             A: a − b + 5 = 0; B: 25a + 5b + 5 = 0                     ✓ both equations            A
             Solving: a = −1 and b = 4                                 ✓ a = −1, b = 4             A
             h(x) = −x2 + 4x + 5                                                                  (3)

                    b        4
 4.2.2       x=−      =−         =2                                    ✓x=2                      CA
                   2a     2(−1)
                      −1 + 5
             OR x =          = 2 OR h′ (x) = −2x + 4 = 0, x = 2        ✓y=9                      CA
                        2
             y = −(2)2 + 4(2) + 5 = 9                                                            (2)
             D(2 ; 9)

 4.2.3       k(2) = −2 + 5 = 3    E(2 ; 3)                             ✓ yE = 3                  CA
             DE = 9 − 3 = 6 units                                      ✓ answer                  CA
                                                                                                 (2)

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### Page 8

Technical Mathematics/P1                               8                          Hlayiso/Prediction 2026
                                             NSC – Marking Guidelines

 4.2.4       0≤x≤5         OR       x ∈ [0 ; 5]                         ✓ critical values            A
                                                                        ✓ notation                   A
                                                                                                    (2)

 4.3                       y                                            ✓ asymptote y = −4           A
                                                                        ✓ intercepts −3 and 2        A
                                         p                              ✓ increasing shape           A
                                                                                                    (3)

                                             x
                                     2

                               −3

                                         y = −4
             (p(x) = 2x − 4)

                                                                                                   [25]

QUESTION 5
                             
                         0,108 12
 5.1         ieff = 1 +           −1                                    ✓ 0,108
                                                                           12                        A
                           12
                                                                        ✓ SF                         A
             = 0,113509 . . .
             ieff ≈ 11,35%
                                                                        ✓ answer                    CA
                                                                                                    (3)

 5.2.1       A = P(1 − i)n                                              ✓ SF                         A
             A = 185 000(1 − 0,18)3
             A = R102 003,08                                            ✓ answer                    CA
                                                                                                    (2)

 5.2.2       40 000 = 185 000(0,82)n                                    ✓ SF                         A
             (0,82)n =
                         40 000
                                = 0,216216 . . .                        ✓ simplification            CA
                        185 000                                         ✓ use of logs               CA
                                                   log 0,2162
             n = log0,82 (0,216216 . . .) OR n =
                                                    log 0,82            ✓ 8 years                   CA
             n ≈ 7,72
                                                                        (After 7 years the value is
             The compressor will be replaced after 8 years.
                                                                        R46 117,81; after 8 years
                                                                        R37 816,61.)
                                                                                                    (4)

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### Page 9

Technical Mathematics/P1                        9                           Hlayiso/Prediction 2026
                                      NSC – Marking Guidelines

                                    
                              0,072 24
 5.3         A2 = 50 000 1 +             = R57 719,36               ✓ i = 0,072
                                                                            12 ; n = 24      A
                                12
                                                                  ✓i= 4 0,084
                                                                                             A
                                  0,084 4
             A3 = 57 719,36 1 +            = R62 722,66             ✓ −8 000 with n = 4 (or
                                    4
                                                                  n = 12)                 CA
                                           0,084 4                  ✓ +10 000 with n = 8 CA
             A4 = (62 722,66 − 8 000) 1 +            = R59 466,20
                                              4
                                                  
                                             0,084 8                ✓ R82 031,28              CA
             A6 = (59 466,20 + 10 000) 1 +
                                                4
             A6 = R82 031,28
             OR
             A6 = 50 000(1,006)24 (1,021)16 − 8 000(1,021)12 +      ✓ conclusion              CA
             10 000(1,021)8                                         Accept answers in the range
             = R82 031,28                                           R82 031 – R82 032 (NPR for
             R82 031,28 > R80 000, so Lerato WILL have enough       intermediate values, e.g.
             money to pay the deposit.                              R62 722,66 or R62 722,67)
                                                                                              (6)

                                                                                              [15]

QUESTION 6

 6.1         f (x + h) = 8 − 3(x + h) = 8 − 3x − 3h                 ✓ f (x + h)                A
                            f (x + h) − f (x)                       ✓ formula                  A
             f ′ (x) = lim
                       h→0          h                               ✓ SF                      CA
                      8 − 3x − 3h − (8 − 3x)                        ✓ −3h                     CA
             = lim                                                      h
                 h→0             h
                      −3h                                           ✓ answer                  CA
             = lim
                 h→0 h                                              Penalty of 1 mark for
             = lim (−3)
                h→0                                                 incorrect notation (no lim).
              f ′ (x) = −3                                          AO: 0 marks
                                                                                               (5)

 6.2.1       y = 2x5 − 3x−1 + 7                                     ✓ −3x−1                    A
             dy
                = 10x4 + 3x−2 OR
                                              3
                                       10x4 + 2                     ✓ 10x4                     A
             dx                              x                      ✓ +3x−2 (and 0 for 7)     CA
                                                                                              (3)

             x3 − 4x x2                                                2
 6.2.2                = −2                                          ✓ x2                       A
                
                2x
                    2   2                                          ✓ −2                       A
                  x
             Dx       −2 = x                                        ✓ answer                  CA
                   2
                                                                                              (3)

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### Page 10

Technical Mathematics/P1                       10                      Hlayiso/Prediction 2026
                                     NSC – Marking Guidelines

                        2
 6.2.3        f (x) = 4x 3                                      ✓ exponential form        A
                       8 1
              f ′ (x) = x− 3   OR
                                     8
                                     √                          ✓ answer                 CA
                       3            33x                                                  (2)

 6.3.1       g(−1) = −1 − 2 + 3 = 0      g(2) = 8 − 8 + 3 = 3   ✓ g(−1) and g(2)          A
                                g(2) − g(−1) 3 − 0              ✓ SF                     CA
             Average gradient =               =
                                  2 − (−1)        3
             =1                                                 ✓ answer                 CA
                                                                                         (3)

 6.3.2       g′ (x) = 3x2 − 4x                                  ✓ g′ (x)                  A
             m = g′ (2) = 3(2)2 − 4(2) = 4                      ✓m=4                     CA
             Point: (2 ; 3)
             y − 3 = 4(x − 2)                                   ✓ substitution of m and
             y = 4x − 5                                         (2 ; 3)                 CA
             OR 3 = 4(2) + c                                    ✓ equation              CA
             c = −5                                                                     (4)
             y = 4x − 5

                                                                                        [20]

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### Page 11

Technical Mathematics/P1                           11                       Hlayiso/Prediction 2026
                                         NSC – Marking Guidelines

QUESTION 7

 7.1         f (4) = 0                                              ✓ f (4) = 0                   A
             (4)3 − 12(4) + k = 0                                   ✓ substitution &
             64 − 48 + k = 0                                        simplification                A
             k = −16                                                                             (2)

 7.2         y = −16 OR         (0 ; −16)                           ✓ answer                      A
                                                                                                 (1)

 7.3         f (x) = (x − 4)(x2 + 4x + 4)                           ✓ quadratic factor          CA
             (by inspection, OR synthetic/long division by x − 4:   ✓ (x + 2)2                  CA
             coefficients 1 0 −12 −16 → 1 4 4, remainder 0)
             = (x − 4)(x + 2)2                                      ✓ both x-intercepts         CA
             x = 4 or x = −2                                                                    (3)
             (4 ; 0) and (−2 ; 0)

 7.4         f ′ (x) = 3x2 − 12 = 0                                 ✓ f ′ (x)                    A
             x2 = 4                                                 ✓ f ′ (x) = 0               CA
             x = 2 or x = −2                                        ✓ both x-values             CA
             f (2) = 8 − 24 − 16 = −32
             f (−2) = −8 + 24 − 16 = 0                              ✓ y = −32                   CA
             Turning points: (2 ; −32) and (−2 ; 0)                 ✓y=0                        CA
                                                                                                (5)

 7.5                       y                                        ✓ x-intercepts              CA
                                                  f                 ✓ y-intercept               CA
                                                                    ✓ turning points            CA
               (−2 ; 0)
                                                                    ✓ shape                      A
                                                      x                                         (4)
                                              4

                               −16

                                 (2 ; −32)

 7.6         −2 < x < 2    OR        x ∈ (−2 ; 2)                   ✓ critical values           CA
                                                                    ✓ notation                  CA
                                                                                                (2)

                                                                                                [17]

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### Page 12

Technical Mathematics/P1                           12                           Hlayiso/Prediction 2026
                                         NSC – Marking Guidelines

QUESTION 8

 8.1         2(x · x + x · h + x · h) = 600                         ✓ SF in surface area = 600 A
             2x2 + 4xh = 600
             4xh = 600 − 2x2                                        ✓ h as subject                 A
                   600 − 2x2 300 − x2                                                             (2)
             h=                =
                      4x              2x
                                            
                                    300 − x2
 8.2         V = x·x·h = x     2                                    ✓ substitution of h            A
                                       2x
                                                                    ✓ simplification               A
                x(300 − x ) 2
             =                                                                                    (2)
                      2
                                1
             V (x) = 150x − x3
                                2
                             3
 8.3         V ′ (x) = 150 − x2 = 0                                 ✓ V ′ (x)                      A
                             2
             x2 = 100                                               ✓ V ′ (x) = 0                 CA
             x = 10 (x > 0)                                         ✓ x = 10                      CA
                   300 − (10)2
             h=                = 10                                 ✓ h = 10                      CA
                      2(10)
             Dimensions: 10 cm × 10 cm × 10 cm                                                    (4)
             (Maximum volume = 1 000 cm3 ; not required)

                                                                                                  [8]

QUESTION 9

                           3x
 9.1.1       = 6 ln x −        +C                                   ✓ 6 ln x                       A
                          ln 3                                             x
                                                                    ✓ − ln3 3                      A
                                                                    ✓ +C                           A
                                                                                                  (3)

             Z            
                     3                                                  3
 9.1.2             x − 4x dx
                     2                                              ✓ x 2 − 4x                     A
                                                                           5
              2 5                                                   ✓ 52 x 2                      CA
             = x 2 − 2x2 +C                                         ✓ −2x2 +C                     CA
              5
                                                                                                  (3)

 9.2.1       x2 − 4x = 0                                            ✓ factors                      A
             x(x − 4) = 0
             x = 0 and x = 4                                        ✓ both values                 CA
                                                                                                  (2)

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### Page 13

Technical Mathematics/P1                         13                           Hlayiso/Prediction 2026
                                       NSC – Marking Guidelines

             Z 6                3        6
                                x
 9.2.2           x − 4x dx =
                     2
                                   − 2x 2
                                                                      ✓ integration              A
              0                  3         0
             = (72 − 72) − 0 = 0
                                                                      ✓ = 0 shown               CA
                                                                                                (2)

 9.2.3       Region A lies BELOW the x-axis, so its integral is       ✓ reason (A below x-axis) A
             negative and cancels the integral of B. Area cannot be                            (1)
             negative, so Thabo’s claim is NOT valid.

             Z 4                           
                                   64                32
 9.2.4           x − 4x dx =
                     2
                                        − 32 − 0 = −
                                     3                 3
              0
                                                                      ✓ Area A = 32
                                                                                  3             CA
                           32      32
             Area A = −         =
                            3       3                                 ✓ Area B = 32
                        Z 6                                                     3              CA
             Area B =        x2 − 4x dx                               ✓ total                   CA
                         4                                                                      (3)
                                        32
             = (72 − 72) − 64  3 − 32 = 3
                           32 32 64
             Total area =     +       =     ≈ 21,33 square units
                            3     3      3

                                                                                               [14]

                                                                                      TOTAL:     150

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- [Technical Mathematics P1 Nov 2025 Afr](https://www.hlayiso.com/library/grade-12-technical-mathematics-2025-nsc-november-exam-question-paper-paper-1/00c4af24-67df-4872-a5e8-a2928fb0fe99) — QP · 2025 · Afr
- [TECH NATHS P2 QP ENG JUNE 2025](https://www.hlayiso.com/library/grade-12-technical-mathematics-2025-nw-june-exam-question-paper-paper-2-english/1c89aab5-ff62-4aaa-8067-21c06324be7b) — QP · 2025 · Eng
- [TECH MATHS P2 ANSWER BOOK JUNE 2024 Afr+Engl](https://www.hlayiso.com/library/tech-maths-p2-answer-book-june-2024-afr-engl/b5a5c28f-caaa-41b0-b1a1-136115668877) — Addendum · 2024 · Afr

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- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

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Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
