# Technical Mathematics Grade 11 P1 November Mock 2026 Memo

> Grade 11 Technical Mathematics Paper 1 memorandum from the November Mock Exam 2026 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-11-technical-mathematics-2026-november-mock-exam-memorandum-paper-1/641dd9c5-7301-4c6a-90f0-49776520bda2
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 11

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## About this document

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

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## Matching paper and memo

- [Grade 11 Technical Mathematics P1 November 2026 Mock](https://www.hlayiso.com/library/grade-11-technical-mathematics-2026-november-mock-exam-question-paper-paper-1/b0b9c4e7-c2a0-4c11-9288-cafec0494a79) — 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 P1
                     NOVEMBER EXAMINATION 2026 (MOCK)
                             MARKING GUIDELINES

  MARKS: 150

                      These marking guidelines consist of 11 pages.

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### Page 2

Technical Mathematics/P1                         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
 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.
    • AO = answer only. Where AO is indicated, the stated marks are awarded for a correct answer
      without working.
    • Where a question says WITHOUT using a calculator, the working must be shown to earn full
      marks.

QUESTION 1

                           1
 1.1.1       (3a2 b)0 + 8 3
             = 1+2                                                    ✓ 1 and 2                 A
             =3                                                       ✓ answer                 CA
                                                                                               (2)

               3     1
             x2 · x2                                                    √
                     = x 2 + 2 −(−1)
                         3 1                                                    1
 1.1.2          −1
                                                                      ✓ x = x 2 / adding
              x
                                                                      exponents                 A
             = x3
                                                                      ✓ answer                 CA
                                                                                               (2)

             ( √      )2 ( √ )2 ( √ )
 1.1.3        2 5 − 1 = 2 5 − 2 2 5 (1) + 1
                     √                                                ✓ 20 √                    A
             = 20 − 4 5 + 1
                     √                                                ✓ −4 5                    A
             = 21 − 4 5                                               ✓ answer                 CA
                                                                                               (3)

             √        √       √
 1.1.4        25 × 2 + 9 × 2 − 4 × 2                                     √       √
                √     √     √
             = 5 2+3 2−2 2                                            ✓ 5√2 and 3 2             A
                √
             =6 2                                                     ✓2 2                      A
                                                                      ✓ answer                 CA
                                                                                               (3)

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### Page 3

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

             log3 34 − log2 2−3
 1.1.5
              log 103 + log5 1
                                                                ✓4                        A
                4 − (−3)                                        ✓ −3                      A
             =
                  3+0                                           ✓3                        A
                7
             =      OR 2 13                                     ✓0                        A
                3                                               ✓ answer                 CA
                                                                                         (5)

             2x · 22 − 2x
 1.2                                                            ✓ exponent laws           A
             3 · 2x · 2−1
                2x (4 − 1)                                      ✓ common factor           A
             = 3 x                                              ✓ simplification         CA
                   2 ·2                                         ✓ answer                 CA
                3 · 2x                                                                   (4)
             = 3 x
                2 ·2
             =2

             32x · 2x+1 · 3x+1
 1.3                                                            ✓ 32x and 33x              A
                  33x · 2x
                                                                ✓ 6x+1 = 2x+1 · 3x+1       A
             = 2x+1−x · 32x+x+1−3x
                                                                ✓ adding/subtracting
             = 21 · 31                                          exponents                CA
             =6                                                 ✓ answer                 CA
                                                                                         (4)

 1.4.1       X = 32 + 16 + 4 + 1 = 53                           ✓ 53                       A
             Y = 16 + 8 + 4 + 1 = 29                            ✓ 29                       A
                                                                                          (2)

 1.4.2       53 − 29 = 24 = 110002                              ✓ answer                 CA
                                                                                         (1)

                                                                                         [26]

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### Page 4

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

QUESTION 2

 2.1.1       log 6 = log(2 × 3)                                   ✓ product law               A
             = log 2 + log 3 = a + b                              ✓ answer                   CA
                                                                                             (2)

                           3
 2.1.2       log 1,5 = log                                        ✓ 23                        A
                           2
             = log 3 − log 2                                      ✓ quotient law              A
                                                                  ✓ answer                   CA
             = b−a
                                                                                             (3)

                            (             )
 2.1.3       log 360 = log 22 × 32 × 10                           ✓ factors                   A
             = 2 log 2 + 2 log 3 + log 10                         ✓ product law              CA
             = 2a + 2b + 1                                        ✓ power law                CA
                                                                  ✓ answer (log 10 = 1)      CA
                                                                                             (4)

 2.2.1       2x+1 = 2−4                                           ✓ 2−4                       A
             x + 1 = −4                                           ✓ equating exponents       CA
             x = −5                                               ✓ answer                   CA
                                                                                             (3)

 2.2.2       5x · 5 + 5x = 150
             5x (5 + 1) = 150                                     ✓ common factor             A
             5x = 25 = 52                                         ✓ 5x = 52                  CA
             x=2                                                  ✓ answer                   CA
                                                                                             (3)

 2.2.3       2x − 1 = 32                                          ✓ exponential form          A
             2x = 10                                              ✓ 2x = 10                  CA
             x=5                                                  ✓ answer                   CA
                                                                                             (3)

                 [        ]
 2.2.4       log x(x − 3) = 1                                     ✓ product law               A
             x2 − 3x = 101                                        ✓ exponential form         CA
             x2 − 3x − 10 = 0                                     ✓ standard form            CA
                                                                  ✓ factors                  CA
             (x − 5)(x + 2) = 0
                                                                  ✓ answer with rejection    CA
             x = 5 or x ̸= −2 (log of a negative number is                                   (5)
             undefined)

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### Page 5

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

                               log 20
 2.3         x = log3 20 =                                             ✓ log form                  A
                                log 3
                                                                       ✓ answer                   CA
             x ≈ 2,73
                                                                                                  (2)

                  2E
 2.4.1       v2 =                                                      ✓ v2 = 2E
                                                                              m                    A
                 √m                                                    ✓ answer                   CA
                   2E
             v=                                                                                   (2)
                    m
                 √
                   2(1 250) √
 2.4.2       v=            = 25                                        ✓ substitution             CA
                      100
                                                                       ✓ answer                   CA
             v = 5 m/s
                                                                       NPU
                                                                                                   (2)

                                                                                                  [29]

QUESTION 3

 3.1.1       2x − 3 = 0 or x + 4 = 0
             x=
                 3
                     or x = −4                                         ✓ x = 23                     A
                 2                                                     ✓ x = −4                     A
                                                                                                   (2)
                               √
                     −(−1) ±       (−1)2 − 4(3)(−5)
 3.1.2       x=                                                        ✓ SF
                                                                         √                         A
                                    2(3)
                     √                                                 ✓ 61                       CA
                 1 ± 61
             x=
                     6                                                 ✓ x ≈ 1,47                 CA
             x ≈ 1,47 or        x ≈ −1,14                              ✓ x ≈ −1,14                CA
                                                                                                  (4)

 3.1.3       (x + 5)(x − 3) ≤ 0                                        ✓ factors                   A
             Critical values: x = −5 and x = 3                         ✓ critical values          CA

               −5          3
                                    x
                       −
                                                                       ✓ correct notation         CA
             −5 ≤ x ≤ 3        OR       x ∈ [−5 ; 3]
                                                                       AO: full marks
                                                                                                   (3)

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### Page 6

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

 3.1.4       × 2x: x2 − 8 = 2x                                  ✓ multiplying by LCD        A
             x2 − 2x − 8 = 0                                    ✓ standard form            CA
             (x − 4)(x + 2) = 0                                 ✓ factors                  CA
                                                                ✓ both answers             CA
             x = 4 or x = −2
                                                                                           (4)

 3.2         y = 2x + 1                                         ✓ y = 2x + 1                A
             x2 + (2x + 1)2 = 13                                ✓ substitution             CA
             x2 + 4x2 + 4x + 1 − 13 = 0
             5x2 + 4x − 12 = 0                                  ✓ standard form            CA
                                                                ✓ factors                  CA
             (5x − 6)(x + 2) = 0
                                                                ✓ both x-values            CA
                 6
             x=       or x = −2                                 ✓ both y-values            CA
                 5                                                                         (6)
                 17
             y=            y = −3
                  5

 3.3.1       A = w(w + 5)   OR    A = w2 + 5w                   ✓ answer                     A
                                                                                            (1)

 3.3.2       w2 + 5w = 84                                       ✓ equation                 CA
             w2 + 5w − 84 = 0
             (w + 12)(w − 7) = 0                                ✓ factors                  CA
                                                                ✓ w = 7 with rejection     CA
             w = 7 (w ̸= −12, a length cannot be negative)
                                                                ✓ length = 12 cm           CA
             Width = 7 cm and length = 12 cm
                                                                                           (4)

                  A
 3.3.3       r2 =                                               ✓ r2 = πA                   A
                 √π                                             ✓ answer                   CA
                    A
             r=                                                                            (2)
                    π

                                                                                           [26]

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### Page 7

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

QUESTION 4

 4.1         ∆ = b2 − 4ac = (−7)2 − 4(2)(3)                      ✓ SF                        A
             ∆ = 25                                              ✓ ∆ = 25                   CA
             Roots are real, rational and unequal.               ✓ nature                   CA
             (∆ > 0 and a perfect square)                                                   (3)

 4.2.1       k = −3                                              ✓ answer                     A
                                                                                             (1)

 4.2.2       k−2 < 0                                             ✓k<2                         A
             k < 2 (k ̸= −3)                                                                 (1)

 4.2.3       k−2 = 0                                             ✓ answer                     A
             k=2                                                                             (1)

 4.3         ∆ = 0: p2 − 4(1)(9) = 0                             ✓ ∆ = 0 with substitution A
             p2 = 36
             p = ±6                                              ✓ answer                   CA
                                                                                            (2)

                                                                                             [8]

QUESTION 5

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

                 4
 5.2         0=    −2                                            ✓ f (x) = 0                  A
                 x
             2x = 4
                                                                 ✓ answer                   CA
             x=2      (2 ; 0)
                                                                                            (2)

                 √
 5.3         0 = 16 − x2                                         ✓ g(x) = 0                   A
             x2 = 16
             x = ±4    (−4 ; 0) and (4 ; 0)                      ✓ answer                   CA
                                                                                            (2)

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### Page 8

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

 5.4         y = 4 OR      (0 ; 4)                                    ✓ answer                    A
                                                                                                 (1)

 5.5                            y                                     f:
                                                                      ✓ asymptotes             CA
                                                                      ✓ x-intercept (2 ; 0)    CA
                            4
                                                                      ✓ shape (both branches) A
                      g
                                                                      g:
                                                                      ✓ x-intercepts           CA
                                                  f
                                                      x               ✓ y-intercept            CA
                 −4                    2      4                       ✓ semicircle above x-axis A
                                                      y = −2                                   (6)

 5.6         −4 ≤ x ≤ 4     OR        x ∈ [−4 ; 4]                    ✓ answer                  CA
                                                                                                (1)

 5.7         0 ≤ y ≤ 4 OR           y ∈ [0 ; 4]                       ✓ answer                  CA
                                                                                                (1)

 5.8         y ∈ R , y ̸= −2                                          ✓ answer                  CA
                                                                                                (1)

 5.9         0<x<2         OR       x ∈ (0 ; 2)                       ✓ critical values         CA
                                                                      ✓ notation                CA
                                                                                                (2)

             4
 5.10          −2 = 2                                                 ✓ equation                  A
             x
             4
               =4                                                     ✓ answer                  CA
             x
             x=1                                                                                (2)

                                                                                                [20]

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### Page 9

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

QUESTION 6

 6.1         y = −9                                               ✓ answer                     A
                                                                                              (1)

 6.2         −6 = b1 − 9                                          ✓ substitution of P         A
             b=3                                                  ✓ answer                   CA
                                                                                             (2)

 6.3         y = 30 − 9 = 1 − 9 = −8                              ✓ substitution x = 0       CA
             C(0 ; −8)                                            ✓ answer                   CA
                                                                                             (2)

 6.4         0 = 3x − 9                                           ✓ f (x) = 0                CA
             3x = 9 = 32
             x=2        E(2 ; 0)                                  ✓ answer                   CA
                                                                                             (2)

 6.5         y > −9    OR    y ∈ (−9 ; ∞)                         ✓ answer                   CA
                                                                                             (1)

 6.6         g(x) = − (3x − 9)                                    ✓ − f (x)                  CA
             g(x) = −3x + 9                                       ✓ answer                   CA
                                                                                             (2)

 6.7         x<2      OR    x ∈ (−∞ ; 2)                          ✓ answer                   CA
                                                                                             (1)

                                                                                             [11]

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### Page 10

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

QUESTION 7

 7.1         −x2 + 2x + 8 = 0                                    ✓ f (x) = 0                  A
             x2 − 2x − 8 = 0
             (x − 4)(x + 2) = 0                                  ✓ factors                  CA
             A(−2 ; 0) and D(4 ; 0)                              ✓ both coordinates         CA
                                                                                            (3)

                    b       2
 7.2         x=−      =−         =1                              ✓x=1                        A
                   2a     2(−1)
                                                                 ✓ substitution             CA
             y = −(1)2 + 2(1) + 8 = 9                            ✓y=9                       CA
             C(1 ; 9)                                                                       (3)

 7.3         −x2 + 2x + 8 = x + 2                                ✓ equating f = g             A
             x2 − x − 6 = 0
             (x − 3)(x + 2) = 0                                  ✓x=3                       CA
             x = 3 (x = −2 at A)
                                                                 ✓y=5                       CA
             y = 3+2 = 5        B(3 ; 5)
                                                                                            (3)

 7.4         −2 ≤ x ≤ 3    OR    x ∈ [−2 ; 3]                    ✓ critical values          CA
                                                                 ✓ notation                 CA
                                                                                            (2)

 7.5         y ≤ 9 OR      y ∈ (−∞ ; 9]                          ✓ answer                   CA
                                                                                            (1)

 7.6         (1 ; 6)                                             ✓ answer                   CA
                                                                                            (1)

                                                                                            [13]

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### Page 11

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

QUESTION 8

 8.1         A = P(1 + ni)                                             ✓ formula                  A
                       (              )
             A = 12 000 1 + 4 × 0,075                                  ✓ SF                       A
             A = R15 600                                               ✓ answer                  CA
                                                                                                 (3)
                    (         )
                         0,09 4
 8.2         ieff = 1 +         −1                                     ✓ 0,09
                                                                          4                        A
                           4
                                                                       ✓ SF                        A
             = 0,093083 . . .
             ieff = 9,31%                                              ✓ answer                  CA
                                                                                                 (3)

 8.3         A = P(1 − i)n                                             ✓ formula                   A
             210 000 = P(1 − 0,12)5                                    ✓ SF                        A
                  210 000
             P=                                                        ✓ answer                  CA
                  (0,88)5
             P = R397 929,31                                                                     (3)

                       (        )
                            0,08 24
 8.4.1       A = 40 000 1 +                                            ✓ 0,08
                                                                          12                      A
                             12
                                                                       ✓ n = 24 & SF              A
             A = R46 915,52
                                                                       ✓ answer                  CA
                                                                                                 (3)

 8.4.2       Amount after withdrawal:                                  ✓ subtracting R10 000     CA
             46 915,52 − 10 000 = R36 915,52                           ✓ 0,09                     A
                            (          )                                  2
                                  0,09 6                               ✓n=6                       A
             A = 36 915,52 1 +                                         ✓ SF                      CA
                                    2
             A = R48 073,61                                            ✓ answer                  CA
                                                                                                 (5)
             OR
             A= (            )24 (          )6          (         )6
             40 000 1 + 0,08
                          12       1 + 0,09
                                        2      − 10 000  1 + 0,09
                                                              2
             A = R48 073,61

                                                                                                 [17]

                                                                                        TOTAL: 150

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