Technical Mathematics Grade 12 P1 Eastern Cape Mock 2020 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2020 TECHNICAL MATHEMATICS P1 MARKING GUIDELINE/ TEGNIESE WISKUNDE V1 NASIENRIGLYN MARKS/PUNTE: 150 MARKING CODES / NASIENKODES A Accuracy/ Akkuraatheid AO Answer only / Slegs antwoord CA Consistent accuracy / Volgehoue akkuraatheid M Method / Metode R Rounding / Afronding NPR No penalty for rounding / Geen penalisering vir afronding NPU No penalty for units omitted / Geen penalisering vir weglating van eenhede nie S Simplification / Vereenvoudiging F Correct formula / Korrekte formule SF Substitution in the correct formula / Vervanging in die korrekte formule This marking guideline consists of 20 pages./ Hierdie nasienriglyn bestaan uit 20 bladsye.

Downloaded from hlayiso.com 2 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) 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) to be applied as indicated on the marking guideline.  Assuming answers/values to solve a problem is NOT ACCEPTABLE. LET WEL:  Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, sien slegs die EERSTE poging na.  Indien ʼn kandidaat ʼn poging kanselleer, maar nie die vraag herdoen nie, merk die gekanselleerde poging.  Volgehoue akkuraatheid (CA) sal toegepas word soos op die nasienriglyne aangedui.  Aanvaarding van antwoorde / waardes om ʼn probleem op te los is ONAANVAARBAAR. QUESTION/VRAAG 1 1.1 1.1.1 0   3x  2  x  1 2 A  x 2 3 x or / of x  1  x  1 A 3 (2) 1.1.2 7 x2  9 x  4 = 0  Formula /Formule A –b  b 2  4ac x= 2a – (9)  (9) 2  4(7)(4)  Substitution/Vervanging CA x= 2(7) x = 1, 64 or / of x = 0,35  x = – 1,64  x = 0,35 R CA -1 MARK for incorrect rounding / -1 PUNT vir verkeerde afronding (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 3 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 EC/SEPTEMBER 2020 1.1.3  6 x2 < 5x  6 x2  5x < 0 x  6 x + 5 > 0 Factors/Substitution A Faktore/vervanging M Critical Values / Kritiese waardes : 5 Both critical Values CA 0 and / en  =  0,83 6 Beide kritiese waardes NPR 5 x or / of 0 < x Correct Notation for CA 6 each interval Korrekte notasie vir CA elke interval (4) 1.2 x + 3 y = 1..................(1) 2 x 2  2 y 2   5 xy.....(2) x = 1  3 y...................(3)  Equation/vergelyking 3 M A 2 1  3 y  + 2 y   5 1  3 y  y 2` 2  Substitution/vervanging CA 2 1  6 y + 9 y 2  + 2 y 2   5 y  15 y 2  Simplification/ S 2  12 y  18 y 2  2 y 2  5 y  15 y 2  0 Vereenvoudiging 5 y2  7 y  2  0  Standard form/ S Standaardvorm   7    7   4  5 2  2 y 2  5  Substitution/vervanging SF y  0, 4 or / of y 1 x = 1 3y  Both y values / x  1  3  0, 4  or / of x  1  3 1 CA Beide y-waardes x  0, 2 or / of x  2  Both x values CA Beide x-waardes NPR OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) x + 3 y = 1..................(1) 2 x 2  2 y 2   5 xy.....(2) 1 x y= ...................(3) 3  1 x   1 x  2 2 x 2` + 2     5x    3   3   1  2 x  x 2  5 x  5 x 2  2 2x + 2   9  3 18 x 2  2  4 x  2 x 2  15 x  15 x 2 18 x 2  15 x 2  2 x 2  4 x  15 x  2  0 5 x 2  11x  2  0  11  11  4  5  2  2 y 2  5 x  2 or / of x  0, 2 1 x y= 3 1   2  1   0, 2  y= or / of y= 3 3 y  1 or / of y  0, 4 (7) 1.3 1.3.1 5, 425 rev 5, 425 rev / min = 60sec 5, 425 rev / min  0, 09rev / sec  0,09 rev/sec A (1) 1.3.2 5,425 rev / min  9 102 rev / sec  9 102 rev / sec NPU CA (1) 1.4 1011112  1 25  0  24  1 23  1 22  1 21  1 20  Method/Metode A  32  8  4  2  1  47 47 A AO: Full Marks/ Volpunte (2) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 5 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 EC/SEPTEMBER 2020 QUESTION/VRAAG 2 2.1  x2  9 x  2  0  Substitution/Vervanging SF b 2  4ac  92  4  1 2  b 2  4ac  73 ∆ =73 CA Roots are real, irrational and unequal  Nature of roots/Aard CA Wortels is reël, irrasionaal en ongelyk van die wortels (3) 2.2  x2  9x  c  0 ∆<0 A b 2  4ac  0 92  4  1 c   0  Substitution/Vervanging CA 4c   81 81  81  c 4 CA 4  Correct notation / Korrekte notasie CA (4) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 3 3.1 3.1.1 2.8n  2  3.8n 1 23 n  5 2.23 n  2  3.23 n 1  Prime bases/Priem-  A 23 n  5 grondtalle 2.23n 6  3.23n 3  23 n  5   23n 2.26  3.23  23n.25  Factorisation / CA 1 Faktorisering  24  32 1 32 Simplification/ CA  1     24  32 Vereenvoudiging  32   1  768  769  Simplification/ Vereenvoudiging CA (4) 3.1.2 3 (27) x 6  25 x 4  3 (3)3 x 6  52 x 4  Prime bases/Priem- A grondtalle  3x 2  5 x 2  2 x2 Simplification/ CA Vereenvoudiging Simplification/ CA Vereenvoudiging (3) 3.1.3 log 2 32  2 log 5 25 log 2 25   Prime bases/Priem A 2 log 5 52 grondtalle 5log 2 2  2  2 log 5 5  Log property/wet CA 5  4  Simplification/ CA Vereenvoudiging (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 7 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 EC/SEPTEMBER 2020 3.2 log 2 (8 x 3  1)  log100  log 2 (4 x 2  2 x  1) log 2 (8 x 3  1)  log 2 (4 x 2  2 x  1)  log100  8 x3  1   Log property/Wet A log 2  2   log100  4x  2x 1      2 x  1 4 x 2  2 x  1  log 2    log102 Factors/Faktore A  4x  2x 1 2    log 2  2 x  1  2 log10 Log property / wet A log 2  2 x  1  2 Exponential form / CA 22  2 x  1 Eksponensiale vorm 5 x 2 Simplification/ CA OR/OF Vereenvoudiging log 2 (8 x 3  1)  log100  log 2 (4 x 2  2 x  1) log 2 (8 x 3  1)  log102  log 2 (4 x 2  2 x  1) log 2 (8 x 3  1)  2  log 2 (4 x 2  2 x  1) log 2 (8 x 3  1)  log 2 4  log 2 (4 x 2  2 x  1)  8 x3  1    log 2 (4 x  2 x  1) 2 log 2   4  8 x3  1  4 x2  2 x  1 4   2 x  1 4 x 2  2 x  1 4 4x2  2x  1 2x 1  4 5 x 2 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) 3.3 z  3  i 3  3   3  2 z  2 Substitution/Vervanging A z  12  Modulus CA z 2 3  tan ratio/verhouding A 3 tan   3  3   tan 1    3  Reference angle/ CA   30  Reference Angle / Verw  Verwysingshoek   180  30  150  z  2 3 cos150  i sin150  Polar form/Polêre vorm CA (5) 3.4 y i  3 xi 2i y  i  (3 xi )(2  i ) y  i  6 xi  3xi 2  Simplification / A Vereenvoudiging y  i  6 xi  3x(1) y  i  3x  6 xi  i 2  1 CA y  3 x and / en  1  6 x 1 y  3 x and / en x   6  1 1 x 1 y  3    and / en x   CA  6 6 6 1 1 y   and / en x    y 1 CA 2 6 2 OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 9 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 EC/SEPTEMBER 2020 y i 2i   3 xi  Conjugate product / A 2i 2i Toegevoegde produk 2 y  yi  2i  i 2  3 xi 22  i 2 2 y  yi  2i  (1)  i 2  1 CA  3 xi 4  (1) 2 y  1  ( y  2)i  3 xi 5 2 y  1  ( y  2)i  15 xi 2 y  1  0 and / en y  2  15 x 1 1 y and / en   2  15 x 2 2 1 5 y and / en   15  x x 1 2 2 CA 6 1 1 y and / en   x 2 6 1  y CA 2 (4) [24] QUESTION/VRAAG 4 4.1 4.1.1 6  Equating to 0 / A 0 2 Gelykstel aan 0 x 6 2 x A x3 x=3 S (2) 4.1.2 0 = x2 – x – 6  Factors/faktore SF A 0 = (x –3) (x +2) x = 3 or/of x = –2  x =3 or/of CA  x = –2 CA (3) 4.1.3 y=–6 y=–6 A (1) 4.1.4 x = 0 and/en y = 2 x=0 A y = 2 A (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) 4.1.5 Axis of symmetry / simmetriese - as : M A 2  3 1 x  1 2 2  x 2 2 CA 1 1 1 g     6 2 2 2  1  25 g   2 4 CA y-coordinate/koördinaat  1 25  Turning point / draaipunt  ;  2 4  OR/OF g ( x)  2 x  1 0  2x 1  g ( x) =0 A 1 x 1 CA 2  x 2 2 1 1 1 g     6 2 2 2  1  25 g   2 4  1 25  CA y-coordinate/koördinaat Turning point / draaipunt  ;  2 4  (3) 4.1.6 CA  Shape of / vorm van g CA  Shape of / vorm van f Asymptotes/ asimptote CA x-intercepts / afsnitte y-intercept / afsnit CA turning point / draaipunt CA CA (6) 4.1.7 x  0 OR/OF x < 0  x > 0  x 0 CA (1) 4.1.8 x ≥ 3 OR/OF x  [3 ; ∞) x≥3 CA (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 11 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 EC/SEPTEMBER 2020 4.2 4.2.1 r 2  (0)2  (2)2  4  r2 = 4 A y   4  x2  Equation / vergelyking CA (2) 4.2.2 y = 2x + q  SF A −2 = 20 + q −2 = 1 + q ⸫q = −3 q = −3 CA (2) 4.2.3 y  2x  3 0  2x  3  Substitution / A 32 x vervanging SF x  log 2 3 CA x  1,58  Logarithm Property/wet A 1,58;0  CA  1,58 (3) 4.2.4 y > −3 y>−3 A (1) [27] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 5 5.1 Depreciation value 8,8   800 000  R70 400 Verminderde waarde 100  R729 600 A Current Asset value = R800 000 R70 400 = R729 600 Waarde van bate OR / OF Current Asset value  R729 600 A = 91, 2%  R800 000 = R729 600 Waarde van bate (1) 5.2 5.2.1 Infected people / besmette mense = 9 9 A (1) 5.2.2 A  P(1  i ) n  Substitution/Vervanging SF 13  9(1  i )7 1  13  7    1 i 9 1  13  7 i    1 i CA 9 i  0, 0539 Daily infection rate  5,39% CA  = 5,39% Daaglikse besmettingskoers (3) 5.2.3 2 P  P(1  0, 0339)n  Substitution/Vervanging A 2  (1, 0339)n n  log1,0339 2 A  n  21 days / dae  Log law/wet CA  n =21 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 13 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 EC/SEPTEMBER 2020 5.3 A4  P (1  i ) n 12 n  i  A4  P 1    12   Substitute/Vervang x SF A 12 4  0, 065  A4  x 1    12  2n  Substitute i and n A  i A  A4 1   Vervang i en n SF  2 124 26  0, 065   0, 07  800 000  x 1   1    Substitution/ CA  12   2  Vervanging SF 800 000 x 12 4 26  0, 065   0, 07   Simplification / CA 1   1   Vereenvoudiging  12   2  S  x  R408 501, 77  R 408 501,77 CA (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 14 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 6 6.1 1 f ( x) =  x+ π 2 f  x  h  f  x f   x   lim h0  Definition/Definisie A h 1  1    x  h  + π    x+π   Substitution / A f   x   lim h0 2  2  Vervanging SF h 1 1 1  x  h+π  x  π f   x   lim h0 2 2 2  Simplification/ CA h Vereenvoudiging S 1  h f   x   lim h0 2 h 1 f   x   lim h0   Simplification / CA 2 Vereenvoudiging S 1  f  x   2 1  f  x   2 CA (5) 6.2 6.2.1  14  D x  9    x  14   Simplification/ A  D x 9  1 Vereenvoudiging S    x2     1 D x  9  14 x 2   Simplification / A   Vereenvoudiging S 14  12 1  0+ x 2 3   7x 2  Derivative/Afgeleide CA (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 15 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 EC/SEPTEMBER 2020 6.2.2 2 xy + x = x 9 x 9  x y  y Subject/Onderwerp A 2x 10 x 1 y   Simplification/ 2 2 CA Vereenvoudiging S dy 10 x 11  CA dx 2 dy CA   5 x 11 11 dx  5x (4) 6.3 y y2  y1  Definition / definisie A Average gradient   x x2  x1 3  Substitution/ Gemiddelde gradient  1 A 2  x1 Vervanging 3 = 2  x1  x1  1 1 CA or / of 3 Average gradient  1 x1  2 3  x1  2  x1  5 −5 CA (4) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 16 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 7 7.1 y=2 2 A (1) 7.2 1 1 1 3 2 1 1 A h    2       5   2  0  h   0 2  2  2  2 2 1 1 1 A Since/omdat h    0 , then/dan x  is the  x 2 2 2 x-intercept/ is die x-afsnit (2) 7.3  2 x  2 x  4   x  12   0 2  2x2 + 2x – 4 A  2 x  2  x  2   x  1 0  (x + 2)(2x – 2) CA  2 x  2 or / of x  1  x = – 2 or/of x = 1 CA (3) 7.4 h  x   6 x 2  2 x  5  0  h  x   6 x 2  2 x  5 A 2  4  4  6  5   h  x   0 A x 12 x  1, 09 or / of x  0, 76  CA Substitution/Vervanging y  2  1, 09    1, 09   5  1, 09   2  6, 05 3 2 or / of  Both x values / CA Beide x-waardes y  2  0, 76    0, 76   5  0, 76   2  0,34 3 2  1 09  6 05 and/en  0 76 0 34   Both y-values / CA Beide y-waardes (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 17 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 EC/SEPTEMBER 2020 7.5  Shape/Vorm A  x-intercepts/afsnitte CA  y- intercept/afsnit CA  Turning Points/ CA Draaipunte (5) 7.6 h  x   6 x 2  2 x  5 =mtangent/raaklyn  A mtangent/raaklyn = − 5 mtangent/raaklyn  6 x2  2 x  5 A y – y1 = m(x –x1)  mtangent/raaklyn = − 5 at/by x=0 y – 2 = − 5(x – 0) CA y = − 5x + 2 Substitution/Vervanging CA SF  y = − 5x + 2 (4) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 18 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 8 8.1 3 AO: FULL MARKS S  t   3  6t  t 2 AO: VOLPUNTE 4 3 2 Truck Height = S  0   3  6  0   0 4 A Trok hoogte = 3m  3m (1) 8.2 3 S  t   3  6t  t 2 AO: FULL MARKS 4 AO: VOLPUNTE 3 2  Substitution/Vervanging A S  2  3  6  2   2 4 S  2   12m  12 m CA (2) 8.3 3 S  t   3  6t  t 2 4 3 S  t   6  t  S  t  A 2 3 6 t  0  S  t   0 A 2 t  4s  4s CA (3) 8.4 3 Maximum Height = S  4   3  6  4   4  Substitution/Vervanging 4 A Maksimum hoogte = 15m  15 m A No, maximum height that can be reached Conclusion/Gevolgtrekking is 15m. CA  Reason / Rede CA Nee, die maksimum hoogte behaalbaar is 15m OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 19 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 EC/SEPTEMBER 2020 3 3  6t  t 2  16  Equating to 16/ A 4 3 Gelykgestel aan 16 13  6t  t 2  0 4  6   4     13 3 6  2  4 t  3 2    4 t  Non - real / nie - reeel  Non-real/nie-reëel A This indicates that it is not possible for the height of water ejected by the fire truck to reach 16 m, Conclusion/gevolgtrekking CA and so this truck is not suitable. Dit dui aan dat dit nie moontlik is vir water om  Reason/rede CA 16 m te bereik nie en dus is die brandblussertrok nie geskik daarvoor nie. (4) [11] QUESTION/VRAAG 9 9.1 9.1.1   y x  dx 5  15     y x  dx  Exponential form / A   Eksponensiële vorm S 1 1 5 x y A  c 1 1 c 5 6 x5 y  c 6 5 6 5x y 5 6 CA  c  5x 5 y 6 6 (3) 9.1.2 1 5    x  x  dx 2  Fraction/Breuk S A 1 2    x  5 x  dx  ln x A ln x  5 x 1  c  5x-1 + c CA (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 20 TECHNICAL MATHEMATICS P1 (EC/SEPTEMBER 2020) 9.2 0   x  4  dx 2  Area notation/notasie A 2 0  x3   Integral/Integraal A    4x  c 3  2   2 3   Substitution/Vervanging CA  0  4  2    3     Substitution/Vervanging CA  8       8 Simplification/  3  CA Vereenvoudiging 16  Area =  5,33. 3  Correct value of the CA bounded area/ Korrekte waarde van die begrensde area (6) [12] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief