# Mathematics Grade 12 P1 NSC November Mock 2026 Memo (Eng)

> Grade 12 Mathematics Paper 1 memorandum from the NSC November Mock Exam 2026 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2026-nsc-november-mock-exam-memorandum-paper-1/9a3dbbcc-f7fa-4864-acaf-d4cc70481fd4
- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Mock Exam
- Paper: 1
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 11

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## About this document

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

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### Page 1

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

                                NATIONAL
                           SENIOR CERTIFICATE

                                    GRADE 12

                                MATHEMATICS P1
                          MOCK EXAMINATION 2026
                            MARKING GUIDELINES

  MARKS: 150

                     These marking guidelines consist of 11 pages.

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### Page 2

Mathematics/P1                                     2                                 Hlayiso/Mock 2026
                                         NSC – Marking Guidelines

NOTE:

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

QUESTION 1

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

 1.1.2       3x2 − 5x − 4 =√
                           0
                 −(−5) ± (−5)2 − 4(3)(−4)                               ✓ correct
                                                                          √       substitution
             x=
                     √       2(3)                                       ✓ 73
                 5 ± 73                                                 ✓ x = 2,26
             x=                                                         ✓ x = −0,59                 (4)
                     6
             x = 2,26 or x = −0,59
                                                     Answer only: 2/4

 1.1.3       x2 + 2x − 15 ≤ 0                                           ✓ factors
             (x + 5)(x − 3) ≤ 0                                         ✓ critical values
             CV: x = −5 or x = 3

             −5         3                                               ✓ answer                    (3)
             −5 ≤ x ≤ 3 OR           x ∈ [−5 ; 3]

             √
 1.1.4         x+7 = x+1                                                ✓ isolating the root
             x + 7 = x2 + 2x + 1                                        ✓ squaring both sides
             x2 + x − 6 = 0                                             ✓ standard form
             (x + 3)(x − 2) = 0      √
             x = 2 or x ̸= −3 (check: 4 ̸= −2)                          ✓ answer with selection (4)

 1.1.5       3x .3
                (  + 3x ).3−1 = 90                                      ✓ exponential laws/factor
             3x 3 + 13 = 90
             3x = 27 = 33                                               ✓ 3x = 27
             x=3                                                        ✓ answer                    (3)
                                                     Answer only: 1/3

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### Page 3

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

 1.2         y = 2x − 3 …(1)
             x2 + xy − y2 = 5 …(2)
             x2 + x(2x − 3) − (2x − 3)2 = 5                         ✓ substitution
             x2 + 2x2 − 3x − 4x2 + 12x − 9 = 5                      ✓ expansion
             x2 − 9x + 14 = 0                                       ✓ standard form
             (x − 2)(x − 7) = 0                                     ✓ factors
             x = 2 or x = 7                                         ✓ both x-values
             y=1        y = 11                                      ✓ both y-values              (6)

 1.3         For non-real roots: ∆ < 0
             (−k)2 − 4(1)(k + 3) < 0                                ✓ ∆ < 0 with substitution
             k2 − 4k − 12 < 0
             (k − 6)(k + 2) < 0                                     ✓ factors/critical values
             −2 < k < 6                                             ✓ answer                     (3)

                                                                                                [25]

QUESTION 2

 2.1.1       48                                                     ✓ answer                     (1)

 2.1.2       First differences: 5 ; 9 ; 13   Second difference: 4
             2a = 4 ⇒ a = 2                                         ✓a=2
             3a + b = 5 ⇒ b = −1                                    ✓ b = −1
             a+b+c = 4 ⇒ c = 3                                      ✓c=3
             Tn = 2n2 − n + 3                                       ✓ Tn                         (4)

 2.1.3       2n2 − n + 3 = 1 038                                    ✓ equating
             2n2 − n − 1 035 = 0                                    ✓ standard form/factors
             (2n + 45)(n − 23) = 0
             n = 23 (n ̸= − 452)                                    ✓ answer                     (3)
             T23

 2.1.4       Tn+1 − Tn = 4n + 1                                     ✓ general term of first
             4n + 1 = 300                                           differences
             n = 74,75 ∈
                       /N
             ∴ no two consecutive terms differ by 300               ✓n∈
                                                                      / N with conclusion (2)

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### Page 4

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

 2.2         a=3; d=4                                             ✓ a and d
                  n[               ]
             Sn = 2(3) + (n − 1)(4) = 1 830                       ✓ substitution into Sn
                  2
             n(2n + 1) = 1 830
             2n2 + n − 1 830 = 0                                  ✓ standard form
             (2n + 61)(n − 30) = 0
             n = 30 (n ̸= − 61                                    ✓ answer                     (4)
                              2)

                                                                                              [14]

QUESTION 3

                 24 1
 3.1.1       r=     =                                             ✓ r = 21
                 48 2
                    ( )7 3                                        ✓ answer                     (2)
             T8 = 48 12 = = 0,375
                         8
                      48
 3.1.2       S∞ =           = 96                                  ✓ substitution
                     1 − 12                                       ✓ answer                     (2)
                       (   ( )n )
                     48 1 − 21               ( )n
 3.1.3       Sn =                   = 96 − 96 12                  ✓ Sn
                          1 − 21
                              ( )n
             S∞ − Sn = 96 12 < 0,01                               ✓ inequality
             ( 1 )n       1
               2    <
                       (9 600
                           1
                              )                                   ✓ use of logs
                   log 9 600
             n>                  = 13,23
                      log 12                                      ✓ answer                     (4)
             n = 14

 3.2         a=2; r=3                                             ✓ a and r
             2 (3 p − 1)                                          ✓ substitution into Sn
                         = 6 560
                 3−1
             3 − 1 = 6 560
               p

             3 p = 6 561 = 38                                     ✓ answer                     (3)
             p=8

 3.3         r = 2x − 1
             −1 < 2x − 1 < 1                                      ✓ −1 < r < 1
             0 < x < 1 (also accept x ̸= 12 )                     ✓ answer                     (2)

                                                                                              [13]

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### Page 5

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

QUESTION 4

 4.1         9 = b−2                                                    ✓ substitution
                  1
             b2 =
                  9                                                     ✓ b2 = 91                   (2)
                 1
             b=      (b > 0)
                 3
                           ( 1 )y
 4.2         f −1 : x =     3                                           ✓ swop x and y
             y = log 1 x                                                ✓ answer                    (2)
                     3

 4.3                        y                                           f:
                                                                        ✓ shape (decreasing,
                                                                        asymptote y = 0)
                                                                        ✓ (0 ; 1) and one other point
              (−1; 3)
                                                                        f −1 :
                                (0; 1)
                                                                        ✓ shape (decreasing,
                                    (1; 0)         f                    asymptote x = 0)
                                                       x
                                                                        ✓ (1 ; 0) and one other point
                                      (3; −1)                           (4)
                                                f −1

 4.4         log 1 x ≤ −2                                               ✓ critical value 9
                 3
             x≥9                                                        ✓ inequality                (2)

 4.5         g(x) = 3x − 3
             y > −3 OR              y ∈ (−3 ; ∞)                        ✓ answer                    (1)

                                                                                                   [11]

QUESTION 5

 5.1         p = −1                                                     ✓ p = −1
             q=2                                                        ✓q=2                        (2)

                  a
 5.2         3=     +2                                                  ✓ substitution of (4; 3)
                4−1
                a
             1=                                                         ✓ simplification            (2)
                3
             a=3

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### Page 6

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

                  3
 5.3         0=       +2                                      ✓y=0
                x−1
             −2(x − 1) = 3
                  1     ( 1 )                                 ✓ answer                    (2)
             x=−          −2 ; 0
                  2

 5.4         y = −(x − 1) + 2                                 ✓ substitution of (1; 2) &
             y = −x + 3                                       m = −1
                                                              ✓ answer                   (2)

                         3
 5.5         −x + k =        +2                               ✓ equating
                       x−1
             (x − 1)(−x + k − 2) = 3
             x2 − (k − 1)x + (k + 1) = 0                      ✓ standard form
             No intersection: ∆ < 0
             (k − 1)2 − 4(k + 1) < 0                          ✓ ∆ < 0 with substitution
             k2 − 6k − 3 <√0
                      6 ± 48           √
             CV: k =            = 3±2 3                       ✓ answer                    (4)
                  √       2       √
             3 − 2 3 < k < 3 + 2 3 (−0,46 < k < 6,46)

                                                                                        [12]

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### Page 7

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

QUESTION 6

 6.1         −x2 + 2x + 8 = 0                                         ✓y=0
             x2 − 2x − 8 = 0
             (x − 4)(x + 2) = 0                                       ✓ factors
             A(−2 ; 0) B(4 ; 0)                                       ✓ both coordinates          (3)

                    b        2
 6.2         x=−       =−        =1                                   ✓x=1
                   2a      2(−1)
                                                                      ✓y=9                        (2)
             y = −(1)2 + 2(1) + 8 = 9
             C(1 ; 9)
             OR f ′ (x) = −2x + 2 = 0 ⇒ x = 1

 6.3         −x2 + 2x + 8 = 2x − 1                                    ✓ equating
             x2 = 9                                                   ✓ x2 = 9
             x = 3 or x = −3                                          ✓ both values               (3)
             D: x = 3 E: x = −3

 6.4         −3 ≤ x ≤ 3                                               ✓ critical values
                                                                      ✓ notation                  (2)

 6.5         f (x) − g(x) = −x2 + 9                                   ✓ f (x) − g(x)
             Maximum distance = 9 units (at x = 0)                    ✓ answer                    (2)

 6.6         f ′ (x) = −2x + 2 = 2                                    ✓ f ′ (x) = 2
             x=0; y=8                                                 ✓ point of contact (0; 8)
             y = 2x + 8                                               ✓ equation                (3)

                                                                                                 [15]

QUESTION 7

 7.1         A = P(1 − i)n
             A = 320 000(1 − 0,14)5                                   ✓ substitution
             A = R150 536,65                                          ✓ answer                    (2)
                        (            )
                                 0,09 12
 7.2         1 + ieff = 1 +                                           ✓ 0,09
                                                                         12
                                  12
             ieff = 0,0938 . . .                                      ✓ substitution into formula
             ieff = 9,38% p.a.                                        ✓ answer                  (3)

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### Page 8

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

                   x [1 − (1 + i)−n ]
 7.3.1       Pv =
                           i[     (             )−240 ]              ✓ i = 0,1125
                                                                             12
                                         0,1125
                           x 1 − 1 + 12                              ✓ n = 240
             1 200 000 =              0,1125                         ✓ substitution into correct
                                         12                          formula                     (3)
             x = R12 591,07
                                    [   (          )−180 ]
                                            0,1125
                           12 591,07 1 − 1 + 12
 7.3.2       Balance =                    0,1125
                                                                     ✓ n = 180
                                            12                       ✓ substitution into Pv
             = R1 092 647,98                                         ✓ answer
             OR                   (           )60
             Balance = 1 200 000 1 + 0,1125                          ✓ 1 200 000(1 + i)60
                           [(              12
                                          )60    ]
                                   0,1125                            ✓ Fv with n = 60
                 12 591,07 1 + 12             −1                     ✓ answer                    (3)
               −              0,1125
                                  12
             = R1 092 648,34

                  x [(1 + i)n − 1]
 7.4         Fv =
                         i[(            )72    ]                     ✓ n = 72 and i = 0,072
                                                                                         12
                                  0,072
                        x 1 + 12            −1                       ✓ substitution into Fv
             250 000 =            0,072                              ✓ answer                    (3)
                                    12
             x = R2 786,30

                                                                                                [14]

QUESTION 8

 8.1         f (x + h) = 2(x + h)2 − 3(x + h) =                      ✓ f (x + h)
             2x2 + 4xh + 2h2 − 3x − 3h                               ✓ 4xh + 2h2 − 3h
             f (x + h) − f (x) = 4xh + 2h2 − 3h                      ✓ formula with substitution
                            4xh + 2h2 − 3h                           ✓ lim(4x + 2h − 3)
             f ′ (x) = lim
                       h→0        h                                  ✓ answer                  (5)
                     = lim (4x + 2h − 3)
                         h→0
                  = 4x − 3
             Penalise 1 mark for incorrect notation.

             y = 4x 2 − 3x−2
                     1
 8.2.1                              (       )                        ✓ exponential form
                                                                     ✓ 2x− 2
                                                                           1
             dy                       2   6
                = 2x− 2 + 6x−3
                       1
                                     =√ + 3
             dx                        x x                           ✓ +6x−3                     (3)

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### Page 9

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

                       (x − 2)(x2 + 2x + 4)
 8.2.2        f (x) =                                               ✓ factorisation
                               x−2
                                                                    ✓ simplification
              f (x) = x2 + 2x + 4
                                                                    ✓ answer                       (3)
              f ′ (x) = 2x + 2

                                                                                                  [11]

QUESTION 9

 9.1         f (3) = 27 − 27 − 27 + 27 = 0 ∴ (x − 3) is a factor    ✓ f (3) = 0
             f (x) = (x − 3)(x2 − 9) = (x − 3)2 (x + 3)             ✓ factorisation
             x-intercepts: (3 ; 0) and (−3 ; 0)                     ✓ both intercepts              (3)

 9.2         f ′ (x) = 3x2 − 6x − 9 = 0                             ✓ f ′ (x) = 0
             x2 − 2x − 3 = 0
             (x − 3)(x + 1) = 0                                     ✓ factors
             x = 3 or x = −1                                        ✓ x-values
             Turning points: (3 ; 0) and (−1 ; 32)                  ✓ y-values                     (4)

 9.3                           y                                    ✓ x-intercepts and
                                                                    y-intercept
                        (−1 ; 32)                                   ✓ turning points
                                   27                               ✓ shape                        (3)
                                                  f

                                                      x
               −3                       (3 ; 0)

 9.4         f ′′ (x) = 6x − 6 > 0                                  ✓ f ′′ (x) > 0
             x>1                                                    ✓ answer                       (2)

 9.5         0 < k < 32                                             ✓ critical values
                                                                    ✓ notation                     (2)

                                                                                                  [14]

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### Page 10

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

QUESTION 10

 10.1        Length = 24 − 2x ; breadth = 15 − 2x ; height = x    ✓ dimensions
             V = x(24 − 2x)(15 − 2x)                              ✓ V = x(24 − 2x)(15 − 2x)
             V = x(360 − 78x + 4x2 )                              (2)
             V = 4x3 − 78x2 + 360x

 10.2        V ′ (x) = 12x2 − 156x + 360 = 0                      ✓ V ′ (x) = 0
             x2 − 13x + 30 = 0
             (x − 3)(x − 10) = 0                                  ✓ factors
             x = 3 (x ̸= 10, since 15 − 2x > 0)                   ✓ x = 3 with selection
             Vmax = 3(18)(9) = 486 cm3                            ✓ maximum volume              (4)

                                                                                                [6]

QUESTION 11

 11.1        P(A and B) = P(A) + P(B) − P(A or B)
             = 0,4 + 0,35 − 0,61 = 0,14                           ✓ P(A and B) = 0,14
             P(A) × P(B) = 0,4 × 0,35 = 0,14                      ✓ P(A) × P(B) = 0,14
             P(A and B) = P(A) × P(B) ∴ A and B are independent   ✓ conclusion                  (3)

 11.2.1                              0,1                          ✓ first branches with
                                              Late
                                                                  probabilities
                            Taxi
                     0,7                                          ✓ second branches (taxi)
                                    0,9      Not late
                                                                  ✓ second branches (walk)
                                                                  with outcomes            (3)
                                     0,25     Late
                     0,3
                           Walk
                                    0,75     Not late

 11.2.2      P(late) = (0,7)(0,1) + (0,3)(0,25)                   ✓ sum of products
             = 0,07 + 0,075 = 0,145                               ✓ answer                      (2)

 11.3.1      8! = 40 320                                          ✓ answer                      (1)

 11.3.2      Treat O and E as one unit: 7!                        ✓ 7!
             Vowels can swop: 2!                                  ✓ ×2!
             7! × 2! = 10 080                                     ✓ answer                      (3)

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### Page 11

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

 11.3.3      Consonants: P, R, B, L, M, S (6) ; vowels: O, E (2)
             n(E) = 6 × 6! × 2 = 8 640                             ✓ 6×...×2
             P=
                  8 640
                        =
                            3
                               ≈ 0,21                              ✓ 6!
                 40 320 14                                         ✓ answer              (3)

                                                                                        [15]

                                                                               TOTAL:    150

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