Mathematics Grade 11 P1 November Mock Exam 2 2026 Memo (Eng)

Mock hlayiso mock exams Practice Series: Grade 11 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS P1 NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 150 These marking guidelines consist of 11 pages. Copyright reserved Please turn over

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. • The mark for substitution is awarded for substitution into the correct formula. QUESTION 1 1.1.1 x2 − 3x − 28 = 0 (x − 7)(x + 4) = 0 ✓ factors x = 7 or x = −4 ✓ both answers (2) 1.1.2 4x2 − 4x − 5 =√ 0 ✓ standard form −(−4) ± (−4)2 − 4(4)(−5) ✓ correct substitution x= 2(4) √ ✓ x = 1,72 4 ± 96 x= ✓ x = −0,72 (4) 8 x = 1,72 or x = −0,72 Penalise 1 mark for incorrect rounding. 1.1.3 CV: x = −2 or x = 5 ✓ critical values (x + 2)(5 − x) < 0 ⇔ (x + 2)(x − 5) > 0 ✓ method (sign/graph) −2 5 ✓ answer (3) x < −2 or x>5 √ 1.1.4 x−1 = x+5 ✓ isolating the surd x2 − 2x + 1 = x + 5 ✓ squaring both sides x2 − 3x − 4 = 0 ✓ standard form/factors (x − 4)(x + 1) = 0 x = 4 or x = −1 (rejected) ✓ answer with selection (4) √ Check x = −1: −1 − 4 = −3 ̸= 1 Copyright reserved Please turn over

Mathematics/P1 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 1.1.5 LCD: (x − 1)(x + 1) ✓ multiplying by LCD 3(x + 1) − 2(x − 1) = (x − 1)(x + 1) ✓ standard form x + 5 = x2 − 1 x2 − x − 6 = 0 ✓ factors (x − 3)(x + 2) = 0 ✓ both answers (4) x = 3 or x = −2 (both satisfy x ̸= ±1) 1.2 y = 2x − 3 …(1) ✓ y the subject x2 + x(2x − 3) = 6 ✓ substitution 3x2 − 3x − 6 = 0 ✓ standard form x2 − x − 2 = 0 (x − 2)(x + 1) = 0 ✓ factors x = 2 or x = −1 ✓ both x-values y=1 y = −5 ✓ both y-values (6) y+3 OR: x = and substitute for x. 2 1.3 ∆ = b2 − 4ac = [−2(k + 1)]2 − 4(1)(k2 + 5) ✓ substitution into ∆ = 4k2 + 8k + 4 − 4k2 − 20 ✓ 8k − 16 = 8k − 16 ✓∆≥0 Real roots: ∆ ≥ 0 8k − 16 ≥ 0 ✓ answer (4) k≥2 [27] Copyright reserved Please turn over

Mathematics/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 3x+1 . 3x−1 . 5x−1 2.1.1 = ✓ 15x−1 = 3x−1 .5x−1 5x . 32x−2 ✓ 9x−1 = 32x−2 = 3x+1+x−1−2x+2 . 5x−1−x ✓ adding/subtracting = 32 . 5−1 exponents 9 = ✓ answer (4) 5 ( ) 5x 52 − 1 2.1.2 = ✓ common factor 5x 4 . 5x . 5 24 ✓ 5x+1 = 5x .5 = ✓ answer (3) 20 6 = 5 5 5 2.2.1 x 3 = 32 ✓ x 3 = 32 3 ( )3 ✓ raising to 53 x = (32) 5 = 25 5 ✓ answer (3) x = 23 = 8 23 8 2.2.2 2x + =6 ✓ 23−x = x 2x 2 ✓ standard form Multiply by 2x : (2x )2 − 6 . 2x + 8 = 0 ✓ factors (2x − 2) (2x − 4) = 0 ✓x=1 2x = 2 or 2x = 4 ✓x=2 (5) x = 1 or x = 2 OR: let k = 2x : k2 − 6k + 8 = 0 … √ √ ( 5 + 2) − ( 5 − 2) 2.3 LHS = √ √ ✓ common denominator ( 5 − 2)( 5 + 2) ✓ numerator = 4 4 ✓ denominator 5 − 4 = 5−4 ✓=4 (4) = 4 = RHS √ √ OR: rationalise each term: ( 5 + 2) − ( 5 − 2) = 4 [19] Copyright reserved Please turn over

Mathematics/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 3.1.1 d = −6 ✓ −6n Tn = −6n + 53 ✓ +53 (2) 3.1.2 −6n + 53 < 0 ✓ inequality 53 ✓n=9 n> = 8,83 . . . 6 n = 9: T9 = −6(9) + 53 = −1 ✓ T9 = −1 (3) The first negative term is −1; it is the 9th term. 3.2.1 18 ; 22 ; 26 ; . . . ✓ 4n Tn = 4n + 14 ✓ +14 (2) 3.2.2 4n + 14 = 94 ✓ equating n = 20 ∴ 20 rows ✓ answer (2) 3.2.3 T3 = 4(3) + 14 = 26 ✓ T3 = 26 and equation 4k + 14 = 3(26) = 78 ✓ answer (2) k = 16 [11] Copyright reserved Please turn over

Mathematics/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 First differences: −13 ; −9 ; −5 Second difference: 4 ✓a=2 2a = 4 ⇒ a = 2 ✓ b = −19 3a + b = −13 ⇒ b = −19 ✓ c = 57 a + b + c = 40 ⇒ c = 57 ✓ Tn (4) Tn = 2n2 − 19n + 57 4.1.2 First differences: Tn+1 − Tn = 4n − 17 ✓ 4n − 17 4n − 17 = 107 ✓ n = 31 n = 31 ✓ answer (3) Between T31 and T32 OR: Tn+1 − Tn = 2(2n + 1) − 19 = 4n − 17 … b 19 4.1.3 Turning point at n = − = = 4,75 ✓ n = 4,75 2a 4 ✓ testing T4 and T5 n must be a natural number: test n = 4 and n = 5 T4 = 13; T5 = 2(25) − 19(5) + 57 = 12 ✓ answer (3) Smallest value = 12 OR: first differences 4n − 17 < 0 for n ≤ 4 and > 0 for n ≥ 5 ⇒ T5 is the smallest. 4.2 First differences: p − 2 ; 20 − p ; 18 ✓ first differences (20 − p) − (p − 2) = 18 − (20 − p) ✓ equal second differences 22 − 2p = p − 2 ✓ simplification 3p = 24 ✓ answer (4) p=8 [14] QUESTION 5 5.1 f (x) = a(x + 4)(x − 2) ✓ substitution of x-intercepts Subst. C(0 ; 8): 8 = a(4)(−2) ✓ substitution of C a = −1 ✓ a = −1 (3) f (x) = −(x2 + 2x − 8) = −x2 − 2x + 8 b −2 5.2 x=− =− = −1 ✓ x = −1 2a 2(−1) ✓y=9 (2) f (−1) = −1 + 2 + 8 = 9 D(−1 ; 9) OR: x = −4+2 2 = −1 Copyright reserved Please turn over

Mathematics/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines 5.3 y ≤ 9 OR y ∈ (−∞ ; 9] ✓ answer (1) 5.4 −x2 − 2x + 8 = x + 4 ✓ equating x2 + 3x − 4 = 0 ✓ standard form/factors (x + 4)(x − 1) = 0 x = 1 (x = −4 at A) y = 1+4 = 5 E(1 ; 5) ✓ coordinates of E (3) f (2) − f (−1) 0 − 9 5.5 Average gradient = = = −3 ✓ substitution 2 − (−1) 3 ✓ answer (2) 5.6 g(x) > 0 for x > −4 and g(x) < 0 for x < −4; x ̸= −4 f (x) ≤ 0 and g(x) > 0: x ≥ 2 ✓ critical value 2 x ≥ 2 OR x ∈ [2 ; ∞) ✓ x ≥ 2 (only) (2) 5.7 f (x) − g(x) = −x2 − 2x + 8 − (x + 4) ✓ −x2 − 3x + 4 = −x2 − 3x + 4 b 3 ✓ x = − 23 x=− =− ✓ answer (3) 2a 2 25 Max distance = − 49 + 92 + 4 = = 6,25 units 4 5.8 Roots opposite in sign: the line y = k cuts f on both ✓ critical value 8 sides of the y-axis ✓k<8 (2) k<8 OR: product of roots of x2 + 2x + (k − 8) = 0 is negative: k − 8 < 0 [18] Copyright reserved Please turn over

Mathematics/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 p = −3 and q = −1 ✓ p = −3 ✓ q = −1 (2) a 6.2 f (x) = −1 ✓ substitution of K x−3 a ✓ a = −2 (2) Subst. K(1 ; 0): 0 = −1 1−3 a =1 −2 a = −2 −2 1 ( ) 6.3 f (0) = −1 = − 0 ; − 13 ✓ answer (1) 0−3 3 6.4 y = −x + c through P(3 ; −1): ✓ substitution of (3 ; −1) −1 = −3 + c ⇒ c = 2 ✓ answer (2) y = −x + 2 6.5 Reflection in y = −x + 2: (x ; y) → (2 − y ; 2 − x) ✓x=2 K′ (2 ; 1) ✓y=1 (2) OR: K′ lies on f and on y = x − 1 (through K, ⊥ to y = −x + 2): −2 x−3 − 1 = x − 1 ⇒ x − 3x + 2 = 0 2 x = 1 (K) or x = 2: K′ (2 ; 1) Answer only: full marks 6.6 x≤1 or x>3 ✓x≤1 ✓x>3 (2) −2 −2 6.7 f (x) = − 1 → g(x) = +3 ✓ 4 units left x−3 x+1 ✓ 4 units up (2) f is translated 4 units to the left and 4 units upwards. [13] Copyright reserved Please turn over

Mathematics/P1 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1.1 y = −6 ✓ answer (1) 7.1.2 y-intercept: g(0) = 2(1) − 6 = −4 (0 ; −4) ✓ (0 ( 1 ); x−4) ( )x ( )x ( )−1 ✓ 3 =3 x-intercept: 2 13 = 6 ⇒ 13 = 3 = 13 ✓ (−1 ; 0) (3) x = −1 (−1 ; 0) 7.1.3 y ✓ asymptote ✓ intercepts ✓ shape (decreasing) (3) (−1 ; 0) x (0 ; −4) y = −6 7.1.4 x ≤ −1 OR x ∈ (−∞ ; −1] ✓ answer (1) ( )x ( )x 7.1.5 2 13 − 6 = 12 ✓ 31 = 9 ( 1 )x ( )−2 ✓ answer (2) 3 = 9 = 13 x = −2 ( )−x 7.1.6 h(x) = 2 13 − 6 = 2 . 3x − 6 ✓ answer (1) 7.2 Asymptote y = 2: q = 2 ✓q=2 (0 ; 5): a . b0 + 2 = 5 ⇒ a = 3 ✓a=3 (1 ; 8): 3b + 2 = 8 ⇒ b = 2 ✓b=2 (3) p(x) = 3 . 2x + 2 [14] Copyright reserved Please turn over

Mathematics/P1 10 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1 A = P(1 − i)n ✓ substitution 190 000 = 380 000(1 − i)4 (1 − i)4 = 0,5 ✓ fourth root √ 1 − i = 4 0,5 = 0,8408 . . . ✓ answer (3) i = 15,91% p.a. ( ) inom 12 8.2 1 + 0,085 = 1 + ✓ substitution 12 ✓ twelfth root inom √ = 12 1,085 − 1 12 (√ ) ✓ answer (3) inom = 12 12 1,085 − 1 = 0,08185 . . . inom = 8,19% p.a. compounded monthly 8.3 Option A: A = 12 000(1 + 0,092 × 4) = R16 416,00 ✓ R16 416 ( )16 ✓ substitution Option B Option B: A = 12 000 1 + 0,084 4 = R16 733,74 ✓ R16 733,74 Option B gives R317,74 more, so Sizwe should choose ✓ conclusion (4) Option B. 8.4 46 418,61 = ✓ 25 000(1 + 0,08 12 ) 36 [ ( )36 ( )24 ] ( )6 0,08 0,08 ✓ 8 000(1 + 0,08 12 ) 24 25 000 1 + 12 + 8 000 1 + 12 1 + 0,09 2 ✓ ×(1 + 0,09 2 ) 6 ( )4 − x 1 + 0,09 ✓ −x(1 + 0,09 2 ) 4 2 ✓ answer (5) 46 418,61 = 53 573,72 − 1,19251 . . . x 1,19251 . . . x = 7 155,11 x = R6 000 Accept R5 999,99 – R6 000,01 (rounding in steps). OR: Value after 3 years: 25 000(1,006̇)36 + 8 000(1,006̇)24 = R41 139,03; [ ] 41 139,03(1,045)2 − x (1,045)4 = 46 418,61 … [15] Copyright reserved Please turn over

Mathematics/P1 11 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 9.1.1 a = 32; b = 70; c = 120 ✓ a and b ✓c (2) 9.1.2 No. n(Grade 10 and Walk) = 32 ̸= 0 (the events can ✓ No happen at the same time). ✓ reason (2) 120 118 9.1.3 P(Grade 11) = = 0,6; P(Taxi) = = 0,59 ✓ P(Gr 11) and P(Taxi) 200 200 ✓ product 0,354 P(Grade 11) × P(Taxi) = 0,6 × 0,59 = 0,354 ✓ P(Gr 11 and Taxi) = 0,35 70 P(Grade 11 and Taxi) = = 0,35 ✓ conclusion (4) 200 0,354 ̸= 0,35 ∴ the events are NOT independent. 9.2.1 P(A or B) = P(A) + P(B) − P(A) × P(B) ✓ P(A and B) = P(A).P(B) 0,8 = 0,6 + P(B) − 0,6P(B) ✓ substitution 0,4P(B) = 0,2 P(B) = 0,5 ✓ answer (3) 9.2.2 S ✓ P(A and B) = 0,3 A B ✓ A only 0,3; B only 0,2; 0,3 0,3 0,2 outside 0,2 (2) 0,2 9.3.1 0,15 ✓ first branches with Late (Bus, L) probabilities 0,6 Bus ✓ second branches with 0,85 On time (Bus, OT) probabilities 0,25 Late (Taxi, L) 0,4 ✓ outcomes (3) Taxi 0,75 On time (Taxi, OT) 9.3.2 P(late) = P(Bus and L) + P(Taxi and L) = 0,6 × 0,15 + 0,4 × 0,25 ✓ 0,6 × 0,15 = 0,09 + 0,1 ✓ 0,4 × 0,25 = 0,19 ✓ answer (3) [19] TOTAL: 150 Copyright reserved