TECH MATHS P1 MEMO GR11 NOV2023_Afr+English_hlayiso.com_.pdf

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE / NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2023 TECHNICAL MATHEMATICS P1 MARKING GUIDELINE/ TEGNIESE WISKUNDE V1 NASIENRIGLYN MARKS / 150 PUNTE Marking codes in CAPS / Nasienkodes in KABV A Accuracy / Akkuraatheid CA Consistent Accuracy / Deurlopend 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 geen eenhede S Simplification / Vereenvoudiging SF Substitution in the correct Formula / Vervanging in die korrekte formule AO Answer Only / Slegs antwoord This marking guideline consists of 12 pages / Hierdie nasienriglyn bestaan uit 12 bladsy

Downloaded from hlayiso.com 2 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2023) QUESTION / VRAAG 1 1.1 ( ) =1 1 A 0 1.1.1 5 3 3 − x5 (1) 1.1.2 1 x (3 − x ) 2 1 1 1 1+ = 3.x − x2 2  3x 2 A 3 1 3  −x 2 = 3x − x 2 2 A (2) 1.1.3 ( 3 − 3)( 3 + 3) = ( 3 ) − ( 3) 2 2 3 A ‒9 A = 3−9 = −6 ‒6 CA (3) 1 1 1.1.4 5 3 log 2 32 + log 2 27 log 2 6 + log 7 x 0 1 1 = log 2 2 ( ) +log ( 3 ) 5 5 2 3 3  Prime factors / priemfaktore  log7 1 A A log 2 6+log 7 1 log 2 2 + log 2 3 =  Simplification / vereenvoudiging CA log 2 6+0  Log property / eienskap CA log 2 ( 2  3)  Log property / eienskap CA = log 2 6 log 2 6 = CA log 2 6 =1 1 CA (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 3 1.2.1 3x −1 − 7.3x +1 y=x 6 .9 x  Exponential property /eksponensiele A −1 3 .3 − 7.3 .3 x x eienskap y=x ( ) x 6 32  Prime factors / priemfaktore A y=x ( 3x 3−1 − 7.3 ) 2x 6 .3  Common factor / gemene faktor CA 1   − 21 y=   3 x x 6 .3 62 −  Simplification / vereenvoudiging CA y= x 3 6 .3 x 1  62  x y = − x   18.3  1 1  31  x y = . −   Simplification / vereenvoudiging CA 3  9 (5) 1.2.2 Even root of a negative number is  Imaginary or non-real / imaginêr of A imaginary and so x must be odd. nie-reëel Ewe wortel van ʼn negatiewe getal is imaginêr so x moet onewe wees (1) 1.3 1.3.1 X = 100 000 2 = 32  X = 32 A (2) Y = 1112 = 7 Y=7 A 1.3.2 X–Y = 100 0002  Method / metode A 1112 110 012  110012 A OR / OF X – Y = 32 – 7 = 25 Base Omitted / Basis weggelaat: 2/2 25 = 110012 (2) 1.4 1.4.1 1 cm 12 1  Conversion factor /herleidingsfaktor A =  10 mm 12 = 0,83 mm  0,83 NPU CA (2) Copyright reserved /Kopiereg voorbehou Please turn over /Blaai om asseblief

Downloaded from hlayiso.com 4 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2023) 1.4.2 = 0,83 mm = 8,3 10−1 mm  8,3 CA  10−1 NPU CA (2) [26] QUESTION / VRAAG 2 2.1 2.1.1 z=1 1 A (1) 2.1.2 x + y − 2z = log2 + log 7 − 2 log10  Substitution / vervanging A  Log property / eienskap A = log ( 2  7 ) − log10 2  Power rule / magreël A = log 14 − log100 S CA  14  = log    Log Property / eienskap CA  100  = log 0,14 (5) 2.2 2.2.1 3x +1.3x −3 = 1  Exponential prop. / eksponensiele 32 x −2 = 30 eienskap A 2x − 2 = 0  30 A x =1 Exponential prop. / eksponensiele CA eienskap S CA (4) 2.2.2 48 − x 2 3 = 27 24  3 − x 2 3 = 33  Prime factors / priemfaktore A 4 3 −3 3 − x 2 3=0  Standard form / standaardvorm A S CA 3−x 2 3=0  Factors / faktore CA ( 3 1 − x2 = 0 ) x=1  x = ‒1 CA x = 1 or/of x = −1 CA (6) 2.2.3 10 = 30 x  Logarithm / logaritme A x = log30 x = log (10  3) x = log10+log3  Log property / eienskap A x = 1 + 0,48  Substitution / vervanging CA x = 1,48  1,48 CA (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 5 2.3 LHS/LK = 22 x .7 x −1 − 5.28 x ( ) x = 22 x .7 x .7 −1 − 5. 22 .7  Prime factors / priemfaktore = 22 x .7 x .7 −1 − 5.22 x .7 x  Power rule / magreël A ( = 22 x .7 x 7 −1 − 5 )  Common factor / gemene faktor CA  −34  = 22 x .7 x   S CA  7  S = − 34.22 x .7 x −1 CA = RHS/RK CA (5) 2.4 2.4.1 5 ( F −  ) C= 9  9C 9C=5 ( F − 32 ) A 9C + 32 = F  F subject / onderwerp 5 CA (2) 2.4.2 9C + 32 = F 5 9 ( 2)  Substitution / vervanging + 32 = F CA 5  F = 35,6F NPU  35,6 CA (2) [28] Copyright reserved /Kopiereg voorbehou Please turn over /Blaai om asseblief

Downloaded from hlayiso.com 6 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2023) QUESTION / VRAAG 3 3.1 3.1.1 x ( x − 2) = 0 x=0 A x = 0 or/of x=2 A x=2 (2) 3.1.2  5 x2 + x  3 −  = 0  x  Standard form/ standaardvorm A x 2 + 3x − 5 = 0 − ( 3)  ( 3) − 4 (1)( −5 )  Substitution / vervanging 2 CA x= 2 (1)  x = 1,2 CA  x = 1,2 x = −4,2 R or/of  x = ‒ 4,2 CA (4) 3.1.3 −7 x 2 − 3x + 4  0  Substitution/ A 3 ( −3) − 4 ( −7 )( 4 ) 2 Vervanging / ( −7 x + 4 )( x + 1)  0 OR / OF x = 2 ( −7 ) Factors / Faktore 4 CVs/KWs: − 1 and/en  Critical values / CA 7 kritiese waardes 4 4 x  −1 and/en x  OR / OF − 1 < x < 7 7  Correct notation / A korrekte notasie  Number line / A getallelyn OR / OF OR / OF 7 x 2 + 3x − 4 < 0 3 ( −3) − 4 ( −7 )( 4 ) 2  Substitution/ ( 7 x − 4 )( x + 1) < 0 OR / OF x = 2 ( −7 ) Vervanging / A Factors / Faktore 4 CVs/KWs: − 1 and/en 7  Critical values / 4 4 kritiese waardes CA x  −1 and/en x  OR / OF − 1 < x < 7 7  Correct notation / korrekte notasie A  Number line / getallelyn A (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 7 3.2 x y− − 1 = 0 ................... (1) 2 x 2 + 3 y 2 = 2 xy + 4........... ( 2 ) x y= − 1..........................( 3)  y subject / onderwerp A 2 x  2 x   Substitution / vervanging x + 3  − 1 = 2 x  − 1 + 4 2 CA 2  2  2 3x  Standard form / x2 + − 3x + 3 = x 2 − 2 x + 4 4 standaardvorm 2 3x CA − x −1 = 0 4  Substitution/ Vervanging / (1) − 4   ( −1) 3 − ( −1)  2 Factors / Faktore CA 4 x= 3 2  4 2  Both x values / Beide x- CA  x = 2 or/of x=− waardes 3 4 y = 0 or/of y=−  Both y values / beide y- CA 3 waardes OR / OF OR / OF x y − − 1 = 0 ................... (1) 2 x + 3 y 2 = 2 xy + 4........... ( 2 ) 2  x subject / onderwerp A x = 2 y + 2.........................( 3)  Substitution / vervanging CA ( 2 y + 2) + 3 y2 = 2 ( 2 y + 2) y + 4 2 4 y2 + 8 y + 4 + 3y2 = 4 y2 + 4 y + 4  Standard form / CA 3y + 4 y = 0 2 standaardvorm y (3 y + 4) = 0  Substitution / vervanging / CA Factors / faktore 4  y = 0 or/of y=− 3  Both y-values / Beide y- CA 2 waardes x = 2 or/of x=− 3  Both x-values / beide x- CA waardes (6) Copyright reserved /Kopiereg voorbehou Please turn over /Blaai om asseblief

Downloaded from hlayiso.com 8 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2023) 3.3 3.3.1 Radius = 31 mm  31 A (1) 3.3.2 C = 2πr  62π CA C = 2π (31) = 62π (1) V = π ( 31) ( 73)  Substitution / vervanging 2 3.3.3 A V = 220392.25 mm3  Value of / waarde van V CA (2) 3.3.4 C = 2πr C r= r A 2π A = πr 2 2  Substitution / vervanging CA C  A= π   2π   C2  A= π 2   4π  C2  Simplification / A= vereenvoudiging CA 4π (3) 3.3.5 A = πr 2 5665,36 = πr 2  Substitution / vervanging A 5665,36 r=  Value of / waarde van r CA π r = 42,47 d = 2r = 84,94 mm d CA (3) [26] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 9 QUESTION / VRAAG 4 4.1 Real, Irrational and Unequal /  Real / Reëel A  Irrational/ irrasionaal A Reëel, irrasionaal en ongelyk  Unequal / ongelyk A (3) 4.2 4.2.1 x=1 1 A (1) 4.2.2 x +1 ∆=0 A +2=0 x −1 x +1 = −2  Simplification / CA x −1 vereenvoudiging x + 1 = −2 x + 2 3x = 1 1 CA x =  Value of / waarde van x 3 (3) 4.2.3 x +1  Less than / minder as – 2 A < − 2 x −1 (1) [8] QUESTION / VRAAG 5 2 f ( x ) = − − 2 and / en x g ( x ) = − 16 − x 2 5.1 x = 0 and /en x=0 A y=‒2 y=‒2 A (2) 5.2 x = 4 or/of x = − 4  x=4 A  x = −4 A (2) 5.3 2 y=0 A f ( x) = − − 2 x 2 0= − −2  Simplification / CA x vereenvoudiging 2 2=− x x = −1 x=‒1 CA (3) g ( 0 ) = − 16 − ( 0 ) = − 4 5.4 2 ‒4 A (1) Copyright reserved /Kopiereg voorbehou Please turn over /Blaai om asseblief

Downloaded from hlayiso.com 10 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2023) 5.5 f  Shape/ vorm A  x-intrcpt/afsnt CA  Asym / Asimp CA g  Shape/ vorm A  x-intrcpt/afsnt CA  y-intrcpt/afsnt CA (6) 5.6 See graph / sien grafiek  All points / alle punte A (1) 5.7 x  0 OR / OF  Solution set / A x  ( −  ; 0) ( 0 ;  ) OR / OF oplossingvers x  0 or / of x  0 OR / OF −   x  0 or / of 0  x   (1) 5.8 − 16 − x = −2 2  M Equating/ A gelykstelling 16 − x 2 = 4 12 = x 2 Accept/aanvaar: x = 2 3  3, 46  x =  12 CA  x =  12 − 12  x  12 OR / OF − 12  x and/en x  12 CA  Critical values / ( OR / OF x  − 12 ; 12 ) kritiese waardes  Correct notation / korrekte notasie A (4) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2023) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 11 QUESTION / VRAAG 6 6.1 y=‒4 y=‒4 A (1) 6.2 B (0 ; ‒3) x=0 A  y = ‒3 A (2) 6.3 ‒2=a ‒4 1  Substitution / vervanging A ‒2+4=a  Simplification / ⸫a=2 vereenvoudiging CA  Same exponents rule / dieselfde eskponente reël CA (3) 6.4 y  −4  y  −4 A (1) 6.5 0 = 2 −4 x  f(x) = 0 A 22 = 2 x  x-value / waarde CA x = 2 (2) 6.6 K (1; 2) x=1 A y=2 A (2) [11] QUESTION / VRAAG 7 7.1 A (‒ 1 ; 0) and/ en x-intercepts only /  A (‒ 1 ; 0) A D (3 ;0) slegs x-afsnitte: 1/2  D (3 ;0) A (2) 7.2 B (0 ; 6) x=0 A y=6 A (2) 7.3 −1 + 3  Method / Metode A Axis of symmetry = 2 Simmetriese − as = 1 x=1 CA Maximum turning = f (1) = (6 – 2)(1 +1) 8 CA Maksimum draaipt = 8 (3) 7.4 x = 0 and/en x = 3 x=0 CA x=3 CA (2) 7.5 h (x) = (6 – 2(x – 2)) (x– 2+1)  Substitution / vervanging A h (x) = (6 – 2x +4) (x – 1)  Simplification / CA h (x) =(10 – 2x) (x – 1) vereenvoudiging (2) 7.6 1  SF A Area of BOD = ( 6 )( 3) = 9 2 Area = 9 CA (2) [13] Copyright reserved /Kopiereg voorbehou Please turn over /Blaai om asseblief

Downloaded from hlayiso.com 12 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/NOVEMBER 2023) QUESTION / VRAAG 8 8.1 7,8 7,8% of/van R2 567 =  R2 567 100  R200,23 = R200,23 A (1) 8.2 ( ) −1 m ieff = 1 + i m  Formula / formule A ( 0,078 12 ) 12 ieff = 1 + −1  Substitution / vervanging CA ieff = 8, 01%  8,01% CA (3) 8.3 A = P (1 − i )  Formula / formule A n  i and n A 64  0,072  R800 000 = P 1 −   Substitution / vervanging A  4  R800 000 P= 64  0,072  1 −   P the subject / die onderwerp CA  4  P = R1 237 126,99 S CA (5) 8.4 A1 = R320 000 (1 + 0,05 )  Formula / formule 3 8.4.1 A  Substitution / vervanging A A1 = R370 440 A3 = R370 440 + R400 000  M + R400 000 CA A3 = R770 440 (3) 8.4.2 R950 000 = R770 440 (1 + 0,058 ) n  Substitution / vervanging CA R950 000 = 1,058n R770 440  Simplification / CA  R950 000  vereenvoudiging n = log1,058   A  R770 440   Logarithm / logaritme  n  3,72 years/jare CA Period = 3+3,72  6,72 years / jare  n = 3,72 years/ jare  It will take 7 years  Period = 7 years / jare CA (5) [16] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief