# Mathematics Grade 11 Paper 1 Limpopo November 2019 Memo > Grade 11 Mathematics Paper 1 memorandum from the Limpopo November Exam 2019, 12 pages. Read online or get the free PDF download, question paper linked when available. - Page: https://www.hlayiso.com/library/gr11-p1-nov-2019-limpopo-memo/5f81ee6d-7b37-4add-98bb-a259b31c4b6d - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Document type: Memorandum - Year: 2019 - Exam period: Limpopo November Exam - Paper: 1 - Curriculum: CAPS - Pages: 12 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 Paper 1 Limpopo November 2019 Memo](https://www.hlayiso.com/library/gr11-p1-nov-2019-limpopo-memo/5f81ee6d-7b37-4add-98bb-a259b31c4b6d) ## About this document Mathematics · Grade 11 · Limpopo November Exam · 2019. Memorandum, 12 pages. 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Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 1 Limpopo DoE/November 2019 NSC NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS P1 MEMORANDUM NOVEMBER 2019 MARKS: 150 This memorandum consists of 12 pages Please turn over ### Page 2 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 2 Limpopo DoE/November 2019 NSC QUESTION/VRAAG 1 1.1.1 5 x 2 − 33 x − 14 = 0 ( 5 x + 2 )( x − 7 ) = 0 ✓Factors/faktore x=− 2 or x = 7 ✓both x − values 5 beide x -waardes (2) 1.1.2 x2 + 7 x + 4 = 0 −b  b 2 − 4ac x= 2a ✓Correct substitution in −7  49 − 4 (1)( 4 ) = correct formula/ 2 (1) korrekte vervanging in = −6,37 / −0, 63 korrekte formule ✓✓correct answers/ korrekte antwoorde (3) 1.2.1 8− x  0 and x−20 − x  −8 x2 ✓8 − x  0 x8 ✓ x−2 0 2  x  8 ✓2 x 8 (3) 1.2.2 8− x = x−2 8 − x = x2 − 4x + 4 ✓ 8 − x = x 2 − 4x + 4 0 = x 2 − 3x − 4 ✓standard form/ 0 = ( x − 4 )( x + 1) Standaardvorm ✓factors/faktore  x = 4 or x  −1 ✓both answers/beide antwoorde ✓selection/seleksie (5) 1.2.3 x = −1 ✓✓ answer/antwoord (2) 1.3 x + 2 x  15 2 x2 + 2 x − 15  0 ✓standard form ( x + 5)( x − 3)  0 ✓factors Critical values are – 5 and 3 ✓ critical values  x  −5 or x  3 ✓  x  −5 or x  3 (4) Please turn over ### Page 3 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 3 Limpopo DoE/November 2019 NSC 1.4 2 y = 32 x 2 y = 25 x ✓ 2 y = 25 x ✓ y = 5x  y = 5 x.....(1) en y = x 2 − 2 x − 18.....(2) ✓sub (1) in (2) Sub (1) in (2) : 5 x = x 2 − 2 x − 18 0 = x 2 − 7 x − 18 ✓standard 0 = ( x − 9 )( x + 2 ) form/standaard vorm  x = 9 or x = −2 ✓ x − values sub into (1): y = 45 y = −10 ✓ y − values (6) 1.5.1 for equal roots: ✓  = 7 + 3p = 0 =0 7 + 3p = 0 7 −7 p=− ✓p= (2) 3 3 1.5.2 for non-real roots: ✓ 0 0 −7 7 ✓p< p− 3 3 (2) [29] QUESTION /VRAAG 2 2.1.1 98 + 8 50 ✓ 7 2 +2 2 7 2 +2 2 Simplified = 5 2 surd/vereevoudigde 9 2 wortels = ✓5 2 5 2 9 = 9 2 5 ✓ 5 2 9 ✓ (4) 5 3 ### Page 4 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 4 Limpopo DoE/November 2019 NSC 5 x + 3 − 3.5 x + 2 2.1.2 5 x +1 ✓ = ( 5 x +1 52 − 3.5 ) factorisation/faktorisering 5 x +1 ✓( 5 2 − 3.5 ) = 25 − 15 ✓25 − 15 = 10 ✓10 (4) ( x − 3) −3 2.2.1 = 27 3 ( x − 3) 2 = 33 − −3 2 ✓ (x − 3) 2 = 33 x − 3 = (3 ) − 3 3 ( ) 2 x−3 = 1 3 −3 ✓ 3 9 1 28 ✓ x= 9 9 28 ✓x = (4) 9 2.2.2 4 x − 3  2 x = 40 ✓standard 22 x − 3  2 x − 40 = 0 form/standaardvorm ( 2 − 8)( 2 + 5) = 0 x x ✓factors/faktore 2 x = 8 or 2 x  −5 ✓ 2 x  −5 2 x = 23 x = 3 ✓x=3 (4) [16] 4 ### Page 5 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 5 Limpopo DoE/November 2019 NSC QUESTION/VRAAG 3 h ( x ) = − x2 + 4x ( = − x2 − 4 x )  ✓ completing the squares 1  1  = −  x 2 − 4 x +  (−4) 2  −  (−4) 2    2  2  ✓simlification = − ( x − 2 ) − 4  2   ✓ Writing in the form = − ( x − 2) + 4 2 a ( x + p) + q 2  maximum value of h( x)is 4 units ✓ maximum value OR h ( x ) = − x2 + 4x Maximum height at turing point. ✓correct substitution in −b x= the simmetry formula of 2a parabola/korrekte −4 vervanging in die x = 2 ( −1) simmetries formule van x = 2 parabool ✓x =2  h ( 2) = − ( 2) + 4 ( 2) 2 ✓substitution/vervangin =4 ✓answer/antwoord (4)  Maximum heigth = 4units [4] QUESTION/VRAAG4 4.1 4.1.1 19 ; 24 ✓19 ✓ 24 (2) 4.1.2 Tn = 5n − 1 ✓✓ 5n − 1 (2) 4.1.3 Tn = 1099 ✓equating/stel gelyk 5n − 1 = 1099 ✓simplification/vereenvoudiging 5n = 1100 5n = 1100 n = 220 ✓answer/antwoord (3) T220 = 1099 it is in the sequence 4.2 4.2.1 30 ; 46 ✓30 (2) ✓46 5 ### Page 6 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 6 Limpopo DoE/November 2019 NSC 4.2.2 Tn = an 2 + bn + c ✓second difference = 4/tweede 2a = 4 konstante verskil = 4 a = 2 3a + b = 3 ✓ a=2 3( 2) + b = 4  b = −2 ✓ b = −2 a+b+c = 6 2−2+c = 6 ✓c = 6 (4) c=6  Tn = 2n 2 − 2n + 6 4.2.3 T12 = 2 (12 ) − 2 (12 ) + 6 ✓ correct substitution in Tn 2 korrekte vervanging in Tn = 270 ✓answer/antwoord (2) 4.2.4 Tn = 766 ✓ equating /gelyk stelling 2n 2 − 2n + 6 = 766 2n 2 − 2n − 760 = 0 ✓standard form =0/standaard n 2 − n − 380 = 0 formule =0 ( n − 20 )( n + 19 ) = 0 ✓factors/faktore n = 20  it is term number 20 ✓ n = 20 (4) 4.3 T1 ; T2 ; T3 ; T4 ✓ first differences/eerste k ; 14 ; 24 ; 7k verskille 14 − k ; 10 ; 7 k − 24 ✓ second differences/tweede verskille − 4 + k = 7 k − 34 ✓ equating/gelykstelling 6 k = 30 ✓ 6k = 30 k =5 ✓answer/anwoord (5) [24] 6 ### Page 7 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 7 Limpopo DoE/November 2019 NSC QUESTION/VRAAG 5 5.1.1 x=2 ✓x=2 y =1 ✓ y =1 (2) 5.1.2 k f ( x) = + 1 through ( 0; −2 ) x−2 ✓ ✓substitute /vervang (0;−2) k −2 = +1 0−2 k −3 = . −2 k =6 ✓k = 6 (3) 6  f ( x) = +1 x−2 5.1.3 xR ; x  2 ✓ x2 (1) 5.1.4 x − intercept ( y = 0 ) 6 0= +1 x−2 ✓ substitute /vervang = 0 −x + 2 = 6 −x = 4 ✓simplification/vereenvoudiging x = −4 ✓ x = −4  D ( − 4;0 ) ✓ D ( −4;0 ) (4) 5.1.4 h ( x ) = a ( x − R1 )( x − R2 ) ✓substitution in correct h ( x ) = a ( x − 2 )( x + 4 ) through ( 0; −2 ) formula/vervanging in korrekte −2 = a ( 0 − 2 )( 0 + 4 ) formule −2 = −8a ✓substitute /vervang (0;−2) 1 1 a = ✓a = 4 4 1 h ( x) = ( x − 2 )( x + 4 ) 4 ✓ h( x ) = 1 (x − 2)(x = 4) 4 = 1 2 4 ( x + 2x − 8 ) ✓answer in quadratic 1 1 form/antwoord in kwadratiese = x2 + x − 2 vorm (5) 4 2 7 ### Page 8 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 8 Limpopo DoE/November 2019 NSC 5.2.1 y ✓ shape/vorm 4 g ✓ y − intercept/afsnit 0 x ✓ asymptote /asimptoot y=0 (3) 5.2.2 g ( x − 3 ) = 2 x − 3+ 2 . x −1 =2 ✓✓answer/antwoord (2) [20] QUESTION /VRAAG 6 6.1 D ( −3;0 ) ✓ (− 3;0) (1) 6.2 y = − x2 + b x + c y = a ( x − R1 )( x − R2 ) but D ( −3;0 ) ; B (1;0 ) and a = −1 ✓substitute /vervang . D ( −3;0 ) , B (1;0 ) y = − ( x + 3)( x − 1) ✓substitute/vervang a = −1 ( y = − x2 + 2x − 3 ) ✓ multipliation 2  y = − x2 − 2x + 3 ✓ y = −x − 2x + 3 (4) 6.3.1 f ( x ) = − x2 − 2x + 3 f ( −1) = − ( −1) − 2 ( −1) + 3 2 =4 . ✓ f (−1) = 4 A ( −1; 4 )  Range : y  4 ✓y4 OR y  (−  ; 4] (2) 6.3.2 A ( −1; 4 ) and D ( −3;0 ) 4−0 m= ✓ m=2 −1 + 3 =2 . y − yc = m ( x − xc ) y − 0 = 2 ( x + 3) ✓substitute A (− 1;4) OR  y = 2x + 6 C (− 3;0) 8 ### Page 9 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 9 Limpopo DoE/November 2019 NSC ✓ y = 2x + 6 (3) 6.4 x  −3 or 0  x  1 ✓ x  −3 ✓ or ✓ critical values ✓ 0  x  1 correct notation (4) 6.5 − x2 − 2x + 3 = k − x 2 − 2 x + (3 − k ) = 0 ✓ standard form   0 for non-real roots b 2 − 4ac  0 ✓ b2 − 4ac  0 4 − 4 ( −1)( 3 − k )  0 4 + 12 − 4k  0 ✓ k 4 − 4k  −16 (answer only, full marks) (3) k  4 6.6 PM = f ( x ) − g ( x ) = − x2 − 2x + 3 − 2x − 6 ✓ − x 2 − 2x + 3 − 2x − 6 = −x − 4x − 3 2 ✓ simplifiction ( = − x2 + 4 x + 3 ) = − ( x + 4 x + 4 − 4 + 3) 2 = − ( x + 2 ) − 1 ✓ − [( x + 2 ) − 1] 2 2   = − ( x + 2) + 1 2 ✓ maximum PM = 1 unit (4)  max PM = 1 unit [21] QUESTION/VRAAG 7 y ✓ shape/vorm f 1 3 x ✓asymptote/asimptote ✓ intercepts of the axis/ afsnitte van die asse (3) [3] 9 ### Page 10 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 10 Limpopo DoE/November 2019 NSC QUESTION/ VRAAG 8 8.1 A = P (1 − i ) n ✓correct substitution in 150000 = 250000 (1 − i ) 6 correct formula/korrekte 3 vervanging in korrekte i = 1− 6 formule 5 = 0, 0816 3 r = 8,16% p.a. ✓6 5 ✓ r = 8,16% (3) 8.2 f  i nom  i eff = 1 +  −1 ✓correct substitution in  f  correct 12 formula/korrekte  0, 089  vervanging inkorrekte = 1 +  −1  12  formule ✓ f = 12 = 0, 092721727 ✓ r = 9,27% (3) r = 9, 27% p.a. compounded yearly 8.3 1,52 3,512  0.08   0, 065  A = 35000 1 +  1 +   0,08   2   12  ✓ 1 +   2  = R 49 397,32 ✓ n = 1,5  2  0,065  ✓ 1 +   12  ✓ n = 3,5  12 ✓R49 397,32 (5) 8.4 712 412 212  0.16   0,16   0,16  A = 15000 1 +  + 10000 1 +  − 8000 1 +  84  12   12   12   0,16  ✓ 150001 +  = R53 524,85  12  48  0,16  ✓ 100001 +   12  24  0,16  ✓✓ − 80001 +   12  ✓ R53524,85 (5) [16] 10 ### Page 11 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 11 Limpopo DoE/November 2019 NSC QUESTION /VRAAG 9 9.1 9.1.1 No, A and B are not mutually exclusive: P( A and B)  0 ✓No Nee, A en B is nie onderling uitsluitend is want ✓ P( A and B)  0 P ( A en B )  0 P( A en B)  0 (2) P( A and B) = 0,1 9.1.2 And/en P ( A)  P ( B ) = 0, 2  0,5 = 0,1  P ( A )  P ( B ) = P ( AenB ) ✓ 0,2  0,5 = 0,1 Ja, die twee gebeurtenisse is onafhanklik P( A)  P(B) = P( A and B) ✓comparing Yes, they are independent ✓conclusion/gevolgtrekking (3) 9.2 P( A and B) = P( A)  P(B) / P ( A en B ) = P ( A)  P ( B ) ✓independant rule/ Onafhanklikheidsreël 0, 2 = ( x + 0, 2 )  0,5 ✓substitution of correct values/vervanging van 0, 4 = x + 0, 2 korrekte waardes x = 0, 2 ✓ x = 0,2 ✓ y = 0,3 (4) y = 0,3 9.3 9.3.1 2 Green :  60 = 24 5 Blue :15% of 60 = 9 Red : 60 − 24 − 9 = 27 2 ✓✓ 60 − 24 − 9 = 27 (2) Groen:  60 = 24 5 Blou : 15% van 60 = 9 Rooi: 60 − 24 − 9 = 27 11 ### Page 12 Downloaded from hlayiso.com Mathematics Gr 11 P1 Memorandum 12 Limpopo DoE/November 2019 NSC 9.3.2 26 ✓ 59 27 ✓ 59 23 ✓ (3) 59 9.3.3 9 24 P(B and G ) = 9 24 24 9  +  ✓  60 59 60 59 60 59 = 0,122 24 9 ✓  9 24 24 9 60 59 P ( B en G ) =  +  60 59 60 59 = 0,122 ✓0,12 (3) [17] TOTAL: [150] 12 ## Related documents - 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