# Mathematics Grade 12 P2 NSC November Mock 2026 Memo (Eng)

> Grade 12 Mathematics Paper 2 memorandum from the NSC November Mock Exam 2026 (English), 9 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2026-nsc-november-mock-exam-memorandum-paper-2/494a607b-247c-41db-9a5e-a57ea496e3c4
- Updated: 2026-10-29
- Subject: Mathematics
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 9

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## About this document

Mathematics · Grade 12 · NSC November Mock Exam · 2026 · English. Memorandum, 9 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

                               MATHEMATICS P2
                          MOCK EXAMINATION 2026
                            MARKING GUIDELINES

  MARKS: 150

                     These marking guidelines consist of 9 pages.

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### Page 2

Mathematics/P2                                  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.

GEOMETRY

   S A mark for a correct statement (a statement mark is independent of a reason).
   R A mark for the correct reason (a reason mark may only be awarded if the statement is correct).
 S/R Award a mark if the statement AND reason are both correct.

QUESTION 1

 1.1         a = 32,07                                               ✓a
             b = 4,32                                                ✓b
             ŷ = 32,07 + 4,32x                                      ✓ equation                    (3)

 1.2         ŷ = 32,07 + 4,32(11)                                   ✓ substitution
             ŷ = 79,62% (accept 79,59% – 79,62%)                    ✓ answer                      (2)

 1.3         r = 0,97                                                ✓ answer                      (1)

 1.4         Very strong positive relationship                       ✓ answer                      (1)

 1.5         25 hours lies far outside the range of the data         ✓ outside the data range
             (extrapolation).
             The predicted score would be                            ✓ score > 100%                (2)
             32,07 + 4,32(25) ≈ 140%, which is not possible.

                                                                                                   [9]

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### Page 3

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

QUESTION 2

 2.1.1       30 ≤ t < 40                                          ✓ answer                     (1)

 2.1.2       Midpoints: 15 ; 25 ; 35 ; 45 ; 55                    ✓ midpoints
             ∑ f x = 4(15)+9(25)+14(35)+8(45)+5(55) = 1 410       ✓ ∑ f x = 1 410
             x̄ =
                  1 410
                        = 35,25 minutes                           ✓ answer                     (3)
                    40

 2.1.3       Cumulative frequencies: 4 ; 13 ; 27 ; 35 ; 40        ✓ cumulative frequencies
                   CF                                             ✓ plotting at upper limits
              40
                                                                  ✓ grounded at (10 ; 0) and
                                                                  smooth shape               (3)
              30

              20

              10

                                                         t
                        10   20   30    40    50    60

 2.1.4       At t = 45: cumulative frequency ≈ 31                 ✓ reading from ogive
             40 − 31 = 9 learners (accept 8 – 10)                 ✓ answer                     (2)

 2.2         New mean = 19                                        ✓ mean
             New standard deviation = 2,5                         ✓ standard deviation         (2)

                                                                                              [11]

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### Page 4

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

QUESTION 3

                     √                           √         √
 3.1         AB =        (5 + 3)2 + (8 − 4)2 =       80 = 4 5       ✓ substitution
                                                                    ✓ answer                    (2)

                   8−4 1
 3.2         mAB =      =                                           ✓ mAB
                   5+3 2
                   −4 − 4
             mAC =        = −2                                      ✓ mAC
                    1+3
             mAB × mAC = −1 ∴ AB ⊥ AC                               ✓ product = −1              (3)
                 (               )
               5+1 8−4
 3.3         M    ;                  = M(3 ; 2)                     ✓ x-value
                2   2
                                                                    ✓ y-value                   (2)

 3.4         BÂC = 90◦ ∴ BC is a diameter (converse: ∠ in          ✓ reason                    (1)
             semi-circle), so its midpoint M is the centre.

                    8+4
 3.5         mBC =        =3                                        ✓m=3
                    5−1
             y − 4 = 3(x + 3)                                       ✓ substitution of A
             y = 3x + 13                                            ✓ equation                  (3)

 3.6         tan β = 3 ⇒ β = 71,57◦ (inclination of BC)             ✓ 71,57◦
             tan θ = 12 ⇒ θ = 26,57◦ (inclination of AB)            ✓ 26,57◦
             AB̂C = 71,57◦ − 26,57◦ = 45◦                           ✓ subtraction ✓ answer
             OR          √                                                 √
             AB = AC = 80 and BÂC = 90◦                            ✓ AC = 80 ✓ isosceles
                            180◦ − 90◦                              ✓ method ✓ answer    (4)
             AB̂C = AĈB =             = 45◦
                                 2
                   √                       √
 3.7         AC = (1 +   3) 2 + (−4 − 4)2 = 80                      ✓ AC
                        √      √
             Area = 12 × 80 × 80 = 40 square units                  ✓ answer                    (2)

 3.8         D=B+C−A
             D(9 ; 0)                                               ✓x=9 ✓y=0                   (2)

                                                                                               [19]

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### Page 5

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

QUESTION 4

 4.1         x2 − 6x + 9 + y2 + 4y + 4 = 12 + 9 + 4               ✓ completing the square (x)
             (x − 3)2 + (y + 2)2 = 25                             ✓ completing the square (y)
             M(3 ; −2)                                            ✓ centre
             r=5                                                  ✓ radius                 (4)

                      2+2 4
 4.2         mMP =           =                                    ✓ mMP
                      6−3 3
                        3                                         ✓ mtan
             mtan = −                                             ✓ substitution of P
                        4
             y − 2 = − 34 (x − 6)                                 ✓ equation                  (4)
             y = − 34 x + 13
                           2

              (       )
 4.3         T 0 ; 13                                             ✓ coordinates of T
                   √
                   2
                            (       )2 √                          ✓ substitution
             PT = (6 − 0)2 + 2 − 13   = 56,25
                                  2                               ✓ answer                    (3)
             PT = 7,5 units

 4.4         tan θ = − 34                                         ✓ tan θ = − 34
             θ = 180◦ − 36,87◦ = 143,13◦                          ✓ answer                    (2)

 4.5         p>3     or    p < −7                                 ✓ p>3
                                                                  ✓ p < −7                    (2)

 4.6         Circles touch externally: distance between centres   ✓ sum of radii = 10
             =
             √5 + 5 = 10                                          ✓ distance formula
                (3 + 5)2 + (−2 − k)2 = 10
             64 + (k + 2)2 = 100                                  ✓ (k + 2)2 = 36
             (k + 2)2 = 36                                        ✓ both values               (4)
             k = 4 or k = −8

                                                                                             [19]

QUESTION 5

 5.1.1       sin 203◦ = sin(180◦ + 23◦ ) = − sin 23◦ = −p         ✓ reduction
                                                                  ✓ answer                    (2)

                       √             √
 5.1.2       cos 23◦ = 1 − sin2√23◦ = 1 − p2                      ✓ identity/diagram
             (or by diagram: x = 1 − p2 , r = 1)                  ✓ answer                    (2)

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### Page 6

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

 5.1.3       cos 46◦ = cos 2(23◦ ) = 1 − 2 sin2 23◦ = 1 − 2p2         ✓ double angle
                                                                      ✓ answer                     (2)

             (− sin x)(sin x)(− tan x)
 5.2                                                                  ✓ − sin x ✓ sin x
                 (cos x)(− sin x)
               − sin x tan x                                          ✓ − tan x ✓ cos x
             =                                                        ✓ − sin x
                   cos x
                         sin x                                        ✓ answer                     (6)
             = − tan x ·       = − tan2 x
                         cos x

                                                                                                  [12]

QUESTION 6

                      1 − (1 − 2 sin2 x)
 6.1.1       LHS =                                                    ✓ 1 − 2 sin2 x
                         2 sin x cos x
                 2 sin2 x                                             ✓ 2 sin x cos x
             =                                                        ✓ 2 sin2 x
               2 sin x cos x
               sin x                                                  ✓ cos
                                                                        sin x
                                                                            x                      (4)
             =        = tan x = RHS
               cos x

 6.1.2       sin 2x = 0 or cos x = 0                                  ✓ 0◦ and 180◦
             x = 0◦ ; 90◦ ; 180◦                                      ✓ 90◦                        (2)

 6.2         1 − 2 sin2 x + 3 sin x − 2 = 0                           ✓ 1 − 2 sin2 x
             2 sin2 x − 3 sin x + 1 = 0                               ✓ standard form
             (2 sin x − 1)(sin x − 1) = 0                             ✓ factors
             sin x = 12 or sin x = 1
             x = 30◦ + k.360◦ or x = 150◦ + k.360◦                    ✓ 30◦ + k.360◦
             x = 90◦ + k.360◦         k∈Z                             ✓ 150◦ + k.360◦
                                                                      ✓ 90◦ + k.360◦ , k ∈ Z       (6)

 6.3         cos 15◦ − cos 75◦ = cos(45◦ − 30◦ ) − cos(45◦ + 30◦ )    ✓ splitting into special
             = (cos 45 cos 30 + sin 45 sin 30) − (cos 45 cos 30 −     angles
             sin 45 sin 30)                                           ✓ both expansions
                                    ( √ ) ( ) √2                      ✓ special values             (3)
             = 2 sin 45◦ sin 30◦ = 2 22 12 =
                                                    2

                                                                                                  [15]

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### Page 7

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

QUESTION 7

 7.1         2 sin x cos x − cos x = 0                                                      ✓ double angle
             cos x(2 sin x − 1) = 0                                                         ✓ factors
             cos x = 0: x = −90◦ or x = 90◦                                                 ✓ ±90◦
             sin x = 12 : x = 30◦ or x = 150◦                                               ✓ 30◦ and 150◦               (4)

 7.2                                                y                                       f:
                       (−135◦ ; 1)                      (45◦ ; 1)                           ✓ shape
                                                1
                                                                                            ✓ turning points
                                                                                            ✓ intercepts
                                                                                    f
                                                                                        x   (0◦ ; ±90◦ ; ±180◦ )
              −180◦           −90◦                              90◦          180◦
                                                                                            g:
                                                                                            ✓ shape and y-int (0; 1)
                                      (−45◦ ; −1)                     (135◦ ; −1)   g
                                                                                            ✓ x-intercepts and end points
                                                                                            (±180◦ ; −1)              (5)

 7.3         30◦ ≤ x ≤ 90◦           or   150◦ ≤ x ≤ 180◦                                   ✓ 30◦ ≤ x ≤ 90◦
                                                                                            ✓ 150◦ ≤ x
                                                                                            ✓ ≤ 180◦                     (3)

 7.4         −1 ≤ 2 cos x + 1 ≤ 3                                                           ✓ −1
             y ∈ [−1 ; 3]                                                                   ✓ 3 (with notation)          (2)

                                                                                                                        [14]

QUESTION 8

                     h
 8.1         tan α =                                                                        ✓ trig ratio
                    BC
                    h                                                                       ✓ BC                         (2)
             BC =
                  tan α
                        BD     BC
 8.2         In △BCD:       =                                                               ✓ sine rule
                       sin x sin y
                  BC sin x     h sin x                                                      ✓ substitution of BC         (2)
             BD =          =
                   sin y     tan α . sin y
                     30 sin 65◦
 8.3         BD =                                                                           ✓ substitution
                  tan 40◦ . sin 50◦
             BD = 42,30 m                                                                   ✓ answer                     (2)

                                                                                                                         [6]

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### Page 8

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

QUESTION 9

 9.1         Construction: Join OB and OD.                        ✓ construction
             Ô1 = 2Ĉ [∠ at centre = 2 × ∠ at circumference]     ✓ S/R
             Ô2 = 2Â [∠ at centre = 2 × ∠ at circumference]     ✓S
             Ô1 + Ô2 = 360◦ [∠s around a point]                 ✓ S/R
             2Â + 2Ĉ = 360◦
             ∴ Â + Ĉ = 180◦                                     ✓ conclusion                (5)

 9.2.1       AĈB = 90◦   [∠ in semi-circle]                      ✓S ✓R                       (2)

 9.2.2       TĈA = AB̂C = 28◦    [tan-chord theorem]             ✓S ✓R                       (2)

 9.2.3       TĈB = 90◦ + 28◦ = 118◦                              ✓ 118◦
             T̂ = 180◦ − 118◦ − 28◦ [sum of ∠s in △]              ✓R
             T̂ = 34◦                                             ✓ answer                    (3)

                                                                                             [12]

QUESTION 10

 10.1        Construction: Join DC and BE. Draw heights h (from   ✓ construction
             E to AB) and k (from D to AC).
             area △ADE       1
                               AD.h AD                            ✓ AD
                                                                    DB ratio
                         = 21       =
             area △DBE       2 DB.h   DB
             area △ADE       1
                               AE.k AE                            ✓ AE
                                                                    EC ratio
                         = 21       =
             area △EDC       2 EC.k
                                      EC
                                                                  ✓S
             Area △DBE = area △EDC
                                                                  ✓R
             [same base DE, same height; DE ∥ BC]
                AD AE                                             ✓ conclusion                (6)
             ∴      =
                DB EC
              AE AD 3
 10.2.1         =  =          [line ∥ one side of △]              ✓S ✓R                       (2)
              EC DB 2

 10.2.2      In △ADG and △ABF: DÂG common; AD̂G = B̂
             [corr ∠s; DE ∥ BC]                                   ✓ ||| with reason
             △ADG ||| △ABF [∠∠∠]                                  ✓ DG
                                                                     BF = 5
                                                                           3
             DG AD 3
                  =     =                                         ✓ answer                    (3)
              BF    AB 5
             DG = 35 (10) = 6 units

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### Page 9

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

 10.2.3      △ADE ||| △ABC [∠∠∠]                                     ✓ ratio
             DE 3
                = ⇒ DE = 18                                          ✓ DE = 18
             BC 5                                                    ✓ answer                   (3)
             GE = 18 − 6 = 12 units

                          1
             area △ADE       AD.AE. sin Â
 10.2.4                 = 21                                         ✓ area rule
             area △ABC     2 AB.AC. sin Â
                3 3    9
             = × =                                                   ✓ 53 × 35
                5 5 25                                               ✓ answer                   (3)

                                                                                              [17]

QUESTION 11

 11.1        In △PTA and △PBT:
             P̂ = P̂ [common]                                        ✓S
             PT̂A = TB̂A [tan-chord theorem]                         ✓S ✓R
             ∴ △PTA ||| △PBT [∠∠∠]                                   ✓R                         (4)

             PT PA
 11.2           =     [||| △s]                                       ✓ S with R                 (1)
             PB PT
             ∴ PT = PA.PB
                 2

 11.3        PT2 = 4 × 9 = 36                                        ✓ substitution
             PT = 6 units                                            ✓ answer                   (2)

 11.4        PT̂A = TB̂A [tan-chord theorem]                         ✓ S/R
             CT̂B = TB̂A [alt ∠s; TC ∥ PB]                           ✓S ✓R                      (3)
             ∴ PT̂A = CT̂B

 11.5        TB̂A = CT̂B [alt ∠s; TC ∥ PB]                           ✓S ✓R
             TB̂A subtends chord TA and CT̂B subtends chord CB       ✓S
             ∴ TA = CB [equal ∠s subtend equal chords]               ✓R                         (4)
                            (        )2     ( )2
              area △PTA         PT           6     4                   ( )2
 11.6                   =                 =      =                   ✓ 96
              area △PBT         PB           9     9
                                                                     ✓ answer                   (2)

                                                                                              [16]

                                                                                    TOTAL:      150

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