TECH MATHS P1 MEMO GR12 JUNE 2023 (E+A).pdf

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2023 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 MARKING CODES/NASIENKODES A Accuracy/Akkuraatheid CA Consistent accuracy/Volgehoue akkuraatheid M Method/Metode R Rounding/Afronding NPR No penalty for rounding/Geen penalisering vir afronding nie NPU No penalty for units omitted Geen penalisering vir eenhede weggelaat nie S Simplification/Vereenvoudiging SF Substitution in correct formula/Vervanging in korrekte formule This marking guideline consists of 15 pages./ Hierdie nasienriglyn bestaan uit 15 bladsye.

Downloaded from hlayiso.com 2 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2023) NOTE: ▪ If a candidate answers a question TWICE, only mark the FIRST attempt. ▪ If a candidate has crossed out an attempt to a question and not attempt the question again, then mark the crossed-out version should be marked. ▪ Consistent accuracy (CA) applies to ALL aspects of 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 en nie poog om die vraag weer te beantwoord dan moet die gekanselleerde antwoord gemerk word. • Volgehoue akkuraatheid (CA) is van toepassing op ALLE aspekte van die nasienriglyn. • Die aanvaarding van antwoorde / waardes om ʼn probleem op te los is NIE aanvaarbaar NIE. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 3 QUESTION/VRAAG 1 1.1 1.1.1 x ( 3x − 1) = 0 x=0 A 1  x = 0 or/of x= 1 3 x= A 3 (2) 1.1.2 2 x 2 + 13 = 5 x 2 x 2 − 15 x + 3 = 0  Standard Form / standaardvorm A −b  b 2 − 4ac x= 2a  Formula / Formule A −(−15)  (−15) 2 − 4(2)(3) x=  Substitution /Vervanging 2(2) CA x = 7, 29 or / of x = 0, 21  both values of x / beide waardes van x CA (4) 1.1.3 ( x − 3)( x+ 4 )  0 C.V / KW : − 4 and / en 3  Critical Values /Kritiese Solution / Oplossing x  −4 or / of x  3 waardes A  x  −4 CA  or / of x  3 CA OR/OF OR/OF  Correct number line / Korrekte getal lyn CA (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2023) 1.2 y = x2 − 11x + 36 and/en y = 2 x − 6  Equating equations / gelykstelling van x 2 − 11x + 36 = 2 x − 6 vergl A x 2 − 13 x + 42 = 0  Simplification/Std form ( x − 7 )( x − 6 ) = 0 /vereenvoudiging /std vorm CA  Factors/ Substitution /faktore/ x = 7 or/of x = 6 y = 2 ( 7 ) − 6 = 8 or / of y = 2 ( 6 ) − 6 = 6 vervanging CA  x-values/waardes CA  y-values/waardes CA OR/OF OR/OF y = x − 11x + 36 2 y+6  Making x the subject =x 2 and substitution /maak x die onderwerp 2  y+6  y+6 y =  − 11  + 36 en vervanging A  2   2   y 2 + 12 y + 36   11y + 66  y = −  + 36  4   2  4 y = y 2 + 12 y + 36 − 22 y − 132 + 144 0 = y 2 − 14 y + 48  Simplification/Std form / vereenvoudiging/std vorm CA  Factors/ Substitution ( y − 8)( y − 6 ) = 0 /faktore/vervanging CA y = 8 or/of y = 6  y-values /waardes CA 8+6 6+6  x-values/waardes CA x= = 7 or/of x = =6 2 2 (5) 1.3 1.3.1 1 A = h( a + b) 2 2A  h the subject /die onderwerp A =h ( a + b) (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 5 1.3.2 2A h= ( a + b) 2 (1,8064 100 )  Conversion /herleiding A = (15, 24 + 20,32)  Substitution /vervanging A = 10,16cm  Simplification/ vereenvoudiging A OR/OF OR/OF 1 A = h( a + b) 2 1  Conversion /herleiding A 1,8064 100 = h (15, 24 + 20,32 ) 2  Substitution /vervanging A 180, 64  2 =h 35,56  Simplification /vereenvoudiging CA 10,16 = h (3) 1.4 K = 89 − 16  Difference /verskil A = 73 73 = 26 + 0 + 0 + 23 + 0 + 0 + 20  Method / metode A K = 10010012  10010012 CA (3) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2023) QUESTION/VRAAG 2 2.1 2.1.1 2 roots / wortels  Answer /antwoord A (1) 2.1.2  = b 2 − 4ac = 0 − 4 (1)( −121)  SF A = 484  Simplification / vereenvoudiging CA (2) 2.1.3 Roots are real, unequal, and irrational/  Real, unequal and irrational / Wortels is reeël,ongelyk en irrasionaal reël,ongelyk en irrasionaal CA (1) 2.2 x 2 + px + 4 = 0  = b 2 − 4ac  SF A = p 2 − 4 (1)( 4 ) = p 2 − 16  Simplification/vereenvoudiging 0 CA p − 16  0 2 ( p + 4 )( p − 4 )  0  0 A p  −4 or / of p  4  Value(s) of p /waarde(s) van p CA (4) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 7 QUESTION/VRAAG 3 3.1 3.1.1 m6 n 7 m6 n7 = (m n) 2 3 ( m 6 n3 )  Exponential property/ eksponensiële eienskap A = n4  n4 CA (2) 3.1.2 98 x 2 + 32 x 2 = ( 49  2 ) x 2 + (16  2 ) x 2  Factors /faktore A = 7x 2 + 4x 2  Simplification / vereenvoudiging CA = 11x 2  Simplification / vereenvoudiging CA OR/OF OR/OF  Prime factors/ priemfaktore A 98 x 2 + 32 x 2 = (2 7 ) x + (2 2 ) x 2 2 4 2  Simplification/ vereenvoudiging CA = 7 x 2 + 22 x 2  Simplification / vereenvoudiging CA = 11x 2 (3) 3.1.3 1 log 2 16 + log 3 27 2  Log property/eienskap A 1 = log 2 24 + log 3 33  Log property /eienskap A 2 4  Simplification / vereenvoudiging = log 2 2 + 3log 3 3 CA 2 = 2 1 + 3 1 =2 (3) ( x + 1) = 64 3 3.2 3.2.1  Exponential property / eksponensiële ( x + 1) = 43 3 eienskap A x +1 = 4  Equal exponent /gelyke eksponent A x =3  x=3 CA OR/OF OR/OF  Expanded form / uitgebreide vorm A ( x + 1) = 64 3  Factors / faktore A ( x + 1) ( x 2 + 2 x + 1) = 64  x = 3 CA x 3 + 2 x 2 + x 2 + x + 2 x + 1 − 64 = 0 x 3 + 3 x 2 + 3 x − 63 = 0 ( x − 3) ( x 2 + 3x + 21) = 0 x = 3 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2023) 3.2.2 log x + 1 = log ( x + 9 )  log law/wet A log x + log10 = log ( x + 9 ) log10 x = log ( x + 9 )  log law /wet A 10 x = x + 9  Simplification / vereenvoudiging 9x = 9 CA x =1  x =1 CA OR/OF OR/OF log x + 1 = log ( x + 9 ) log x − log ( x + 9 ) = −1log10 x log = log10−1 ( x + 9)  log law /wet A x 1 =  log law/wet A ( x + 9 ) 10 10 x = x + 9  Simplification / vereenvoudiging 9x = 9 CA x =1  x =1 CA (4) 3.3 m − ( 3 − i ) = ni + 5  Simplification / vereenvoudiging A m − 3 + i = ni + 5 m − 3 = 5 and/en i = ni  Value of m /waarde van m CA m=8 and/en n = 1  Value of n /waarde van n CA (3) 3.4 z = 1 − 5i r= (1) + ( −5) 2 2  SF A = 26  r = 5,1 CA = 5,1 tan  = − 5  tan  = −5 CA 1 ref / verw. :  = 78, 7  Ref /verw  = 78,7 CA ο  = 360 − 78, 7 = 281,3  z = 5,1cis 281,3 CA  z = 3, 74 cis 281,3 (5) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 9 QUESTION /VRAAG 4 4.1.1 y=4 y=4 A 0 = − x2 + 4  Equating to 0 /gelyk stel x2 = 4 aan 0 A x = 2 x = 2 CA (3) 4.1.2 x = 0 and/en y = 4  x=0 A  y=4 A (2) 4.1.3 y − int/ afsnit y =1 y=1 A x − int/ afsnit 0 = − 2x +1 y=0 A −1 = − 2 x 1 x= 1 x= CA 2 2 (3) 4.1.4 f:  x – intercepts/afsnitte CA  y – intercept /afsnit CA  Turning point/draaipunt CA  Shape /vorm A g:  x and/en y- intercepts/afsnitte CA  Shape/vorm A (6) 4.1.5 x OR/OF −   x   OR/OF x   − ;    Critical values /kritiese waardes CA  Correct notation / korrekte notasie A (2) 4.2 4.2.1 r = 7 OR / OF 2,65  Value of r /waarde van r A (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2023) 4.2.2 y =1  Equation of asymptote/vergelyking van asimptoot A (1) 4.2.3 h ( 0 ) = 2 +1 0  y= 2  y-intercept /afsnit A (1) 4.2.4 − 7  x  7 OR / OF x   − 7; 7   Critical values /kritiese waardes CA OR / OF − 2, 65  x  2, 65 OR / OF x  −2, 65; 2, 65  Correct notation / korrekte notasie A (2) 4.2.5 h:  Asymptote /asimptoot CA  Shape /vorm A  y-intercept /afsnit CA k:  x-and/en y- int/afsnit CA  Shape /vorm A (5) 4.2.6 Shaded area on the graph / Geskakeerde gedeelte  Shaded area / op die grafiek Geskakeerde gedeelte (1) 4.3 4.3.1 8 0= +2 y=0 A x 8 −2 = x  Simplification / − 2x = 8 vereenvoudiging CA x = −4  x = −4 CA (3) 4.3.2 y=2 y=2 A (1) 4.3.3 x , x  4 x , x  4 CA (1) [32] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 11 QUESTION /VRAAG 5 5.1 A = P(1 − i ) n F CA = 300 000 (1 − 0,5 ) 10 = 292 000  SF CA There will be 292 000 people in the town./ Daar sal 292 000 mense in die dorp wees.  292 000 CA (3) 5.2 A = P(1 + i) n F CA 75 000 = 5 000 (1 + 9,5 ) n  SF CA 75 000 = (1 + 9,5 ) n 5 000  Simplification / vereenvoudiging CA 15 = (1,95 ) n log1,9515 = n  log form /vorm CA 4, 0550 = n  n= 4 CA (5) 5.3 A = P(1 + i) n 36  SF CA  7,5%  = 200 000 1 +   12  = R250 289, 23  R250 289,23 CA Amount after withdrawal / Bedrag na onttrekking R250 289, 23 − R50 000 = R 200 289, 23  M - R50 000 A Value of the investment at the end of 5 years / Waarde van die belegging aan die einde van 5 jaar  SF CA n A = P(1 + i) 4×2  R225 757,96 CA  6% = R200 289, 23 1 +   4  = R225 624, 33 (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2023) QUESTION /VRAAG 6 6.1 f ( x + h) − f ( x) f  ( x ) = lim  Definition /definisie A h →0 h 1 − 3 ( x + h ) − (1 − 3x ) = lim h →0  Substitution /vervanging CA h 1 − 3 x − 3h − 1 + 3 x = lim  Simplification / vereenvoudiging CA h →0 h −3h = lim h →0 h  Simplification / vereenvoudiging CA = lim − 3 h →0 = −3  f  ( x ) = −3 CA (5) 6.2 6.2.1 2 y= − 15 x+ 7m x3 = 2 x −3 − 15 x+ 7m  2x −3 A dy = −6 x −2 − 15  −6x −2 CA dx (3)  −15 CA 6.2.2  3  Dx 9 + 2 x −1 + x 27    27  x 9 OR / OF x 3 A = Dx 9 + 2 x −1 + x 9  −2  −2x CA −2 = −2 x + 9 x 8  9x 8 CA (3) 6.3 6.3.1 x2 7 x 1 g ( x) = − + 15  x A 20 20 10 1 7 g ( x) = x − 7 10 20 − A 20 (2) 6.3.2 1 7 g  ( 5) = ( 5) −  Substitution /vervanging CA 10 20 3 =  Simplification / vereenvoudiging CA 20 (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 13 QUESTION / VRAAG 7 7.1 h( x) = x 3 − 16 x  Substitution by 0 /vervanging deur 0 0 = x( x 2 − 16) A 0 = x( x − 4)( x + 4) x = 0 or/of x = − 4 or/of x = 4  Factors/ Substitution / faktore/vervanging CA A(−4;0) B(4;0)  Coordinates of A /koördinate van A CA  Coordinates of B /koördinate van B CA (4) 7.2 h( x) = x 3 − 16 x  Derivative / afgeleide A h( x) = 3 x 2 − 16 0 = 3 x 2 − 16  h( x ) = 0 A 16 = x2 3  Both x values /beide x-waardes CA x =  2,31 h ( 2,31) = ( 2,31) − 16 ( 2,31) = −24, 63 3 h ( −2,31) = ( −2,31) − 16 ( −2,31) = 24, 63 3 D ( −2,31; −24, 63)  y-coordinate of D / y-koördinaat van D CA E ( 2,31; 24, 63)  y -coordinate of E / y-koördinaat van E CA (5) 7.3 x  − 2,31 or / of x  2,31 OR / OF  − 2,31 CA −  x  −2,31 or/of 2,31  x    2,31 CA OR / OF x  ( −; −2,31 or/of x   2,31;  )  correct notation / korrekte notasie A (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 14 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2023) QUESTION / VRAAG 8 8.1 2b + 2h = 80  Formula /formule A 2b = 80 − 2h  Simplification / b = 40 − h vereenvoudiging CA (2) 8.2 V = l  hb  Formula / formule A = ( 20 − 2h )( 40 − h )  h  ( 20 − 2h ) CA = 2h3 − 100h 2 + 800h  SF CA (3) 8.3 V = 2h3 − 100h 2 + 800h dV = 6h 2 − 200h + 800  Derivative / afgeleide A dh 0 = 6h 2 − 200h + 800  Derivative / afgeleide = 0 A − ( −200 )  ( −200 ) − 4 ( 6 )(800 ) 2 h=  SF CA 2 ( 6) h = 28, 69 or / of h = 4, 64  Both values of h / beide V ( 28, 69 ) = 2 ( 28, 69 ) − 100 ( 28, 69 ) + 800 ( 28, 69 ) 3 2 waardes van h CA = −12129, 21cm3 V ( 4, 64 ) = 2 ( 4, 64 ) − 100 ( 4, 64 ) + 800 ( 4, 64 ) 3 2  Calculating both V / = 1758,83cm3 berekening beide V CA The value of h is / Die waarde van h is 4,64  Choosing/Kies h = 4,64 CA (6) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 15 QUESTION 9 9.1 9.1.1  ( 2 x + x ) dx 2 2 3 2 x3 x 2  x A = + +C 3 3 2 x2  A 2 C A (3) 9.1.2 16 x 6 − 4 x 2  2 x dx  (8 x − 2 x ) dx  8x 5 A 5  −2x A 8x6 = − x2 8x6 6  CA 6 4 x6 = − x2 + C  − x 2 +C CA 3 (4) 2  x dx 9.1.3 3 0 2 x4 = 4 0 x4  A 24 4 =  Substitution / vervanging CA 4  Simplification / vereenvoudiging =4 CA (3) ( x + 3x ) dx −2  Definite integral formula / Bepaalde 9.2 A = 2 − 0,5 integral formule CA −2  x3 3 2  = + x   Integral /Integraal CA  3 2  − 0,5  (− 0,5)3 3   (−2)3  = + (− 0,5) 2  −  + 2(−2) 2   Substitution / vervanging CA  3 2   3  1 16  Area CA A = − = −5 3 3 A = 5 square units / vierkante eenhede (5) [15] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief