Technical Maths P1 Memo June 2022 a E_hlayiso.com_.pdf

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2022 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 21 pages./ Hierdie nasienriglyn bestaan uit 21 bladsye.

Downloaded from hlayiso.com 2 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2022) NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • The method of consistent accuracy marking must be applied to all aspects of the marking guideline where applicable as indicated with the marking code CA. • If a candidate strikes off a response to a question and does not attempt the question again, then the struck off question should be marked LET WEL: • Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, sien slegs die EERSTE poging na. • Die metode van volgehoue akkuraatheid-nasien moet waar moontlik op alle aspekte van die nasienriglyne toegepas word soos aangedui deur die nasienkode CA. • Indien ʼn kandidaat ʼn antwoord deurhaal en nie poog om die vraag weer te beantwoord dan moet die deurgehaalde antwoord gemerk word. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 3 QUESTION/VRAAG 1 1.1.1 (2 x + 3)(3 x + 1) = 0  both values of /beide waardes 3 1 x − = or/of x − = van x CA 2 3 (1) 1.1.2 5 x + 1 = −2 x 2 2 2x + 5 x + 1 = 0  standard form/standaardvorm A 2 − b ± b − 4ac x= 2a 2  SF CA − ( 5) ± ( 5) − 4(2)(1) = 2(2) − 5 ± 17 =   each x-value/elke waarde 4 CA ∴ x ≈− 0,2 or/of x ≈ − 2,3 R (4) 1.1.3 2 x − 5x − 6 > 0 ( x − 6)( x + 1) > 0  factors/formula / faktore/formule A Critical values/kritiese waardes: −1 and / en 6  both critical values/ CA x < − 1 or / of x > 6 OR/OF x ∈ ( 0; 6 ) albei kritiese waardes  correct notation/notasie A (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 2022) 1.2 y = 8x −1 and / en y = 3x 2 + 4 2 8 x − 1 = 3x + 4  substitution/vervanging A 2 3 x − 8 x + 5=0 correct standard form/ (3 x − 5)( x − 1) = 0 korrekte standaardvorm CA 5  factors/formula / ∴x= ≈ 1,67 or/of x = 1 3 faktore/formule. CA 5 y 8 =   − 1 or/of y 8 = (1) − 1  both x-values/-beide waardes CA   3 37 ∴ y= ≈ 12,33 or/of y= 7  both y-values/-beide waardes CA 3 OR/OF OR/OF y +1 x= 8 2  y +1 y = 3  +4  8   substitution/vervanging A  y2 + 2 y + 1  y = 3 +4  64  2 64 y = 3 y + 6 y + 3 + 256 2 0 = 3 y − 58 y + 259 (3 y − 37)( y − 7) = 0  correct standard form/ 37 korrekte standaardvorm CA ∴ y= ≈ 12,33 or/of y= 7 3  factors/form./faktore/vorm. CA 37 +1 3 7 +1 =x = or/of x  both y-values/-waarde CA 8 8 5 ∴ x= ≈ 1, 67 or/of x = 1 3  both x-values/-waardes CA NPR (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 5 1.3.1 R1 R 2 3 R= where / waar R 1= x and / en R 2 = x 2  SF A R 1+ R 2 3 x × x 2  Exp. property/eienskap = 3 CA x + x 2 3 1 x × x2 2 = 3 1 x2 + x2 3 1 x ×x2 2 = 1 S CA x ( x + 1) 2 3 x 2 = x +1 OR/OF OR/OF x3 × x  SF A = x3 + x  surd property/ x3 × x wortelvorm eienskap = x ( x2 + 1 ) CA x3 = ( x2 + 1 ) S CA (3) 1.3.2 3 x 2 R= x +1  SF CA 3 25 2 = 25 +1  value of/waarde van R CA 125 = 26 = 4,81 Ω NPR NPU (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2022) 1.4 75 ÷ 5 = 15  15 A 23 22 21 20 1 1 1 1 =75 15 = 1111 2  binary form/binêre vorm CA AO: full marks/volpunte (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 7 QUESTION/VRAAG 2 2.1 2 ∆= b − 4 a c  SF ∆ ( 5) − 4 ( 2 )(8) 2 = S = − 39  non-real/nie-reeël Roots are Non-real/ Wortels is nie-reeël Accept Imaginary/Aanvaar Imaginêr (3) 2.2 5 x 2 − kx − 2 = 0 2 b − 4ac> 0  ∆ >0 A  SF CA ( −k ) − 4 (5)( − 2) > 0 2 2 k + 40 > 0 S CA 49 is least perfect square after 40 and so k = ±3 49 is kleinste volkome kwadraat na 40 en so k = ±3  k=3 CA ∴k=3 (4) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2022) QUESTION/VRAAG 3 3.1.1 32 x − 3x − 6 3x + 2  Factors/faktore A = ( 3 + 2 )( 3 − 3) x x (3 + 2) x x  3x − 3 CA =3 −3 (2) 3.1.2 3 3 81 − 3 27 2 + 4 625 4 3 2 1  (3 ) 4 4 A = (3 ) − (3 ) + (5 ) 4 4 3 3 4 4  (3 ) 2 3 3 A 3 2 = 3 −3 +5 1 = 23  (5 ) 4 4 A  23 CA OR/OF OR/OF 3 12 81 − 3 27 2 + 4 625 4 12 6 4  ( 3) 4 A = ( 3) − ( 3) + ( 5 ) 4 3 4 6 = 27 − 9 + 5  ( 3) 3 A 4 = 23  ( 5) 4 A  23 CA (4) 3.1.3 log 216 − log 2 8 log 2 4  log property/eienskap A 4 log 2 2 - 3log 2 2  S CA = 2 log 2 2 1  CA log 2 2 2 = 2log 2 2 1 = 2 OR/OF OR/OF log 216 − log 2 8  S CA log 2 4 1 log 2 24 - log 2 23  CA = 2 log 2 22 4 log 2 2 − 3log 2 2 = 2 log 2 2 log 2 2 (4 − 3)  log property/eienskap A = 2log 2 2 1 = 2 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 9 OR/OF OR/OF 16 log 2  log property/eienskap A = 8 log 2 22  S CA log 2 2 1 =  CA 2log 2 2 2 1 = 2 (3) 3.2 ( log x − 4 ) × log ( x − 4 ) = 0  Factors/faktore A log x − 4= 0 or/of log ( x − 4 ) 0 =  exponential/log form/eksponensiële 0 CA log = x 4 or/of (x − = 4 ) 10 /log vorm 𝑥𝑥 = 104 = 10 000 or/𝑜𝑜𝑜𝑜 𝑥𝑥 = 1 + 4 = 5  x =10 000 CA  x =5 CA (4) 3.3.1 z = − 3 + 4i z = r= x 2 + y 2  SF A (4 ) + ( − 3 ) 2 2 r=  value of/waarde van r CA = 25 = 5 (2) 3.3.2 4  Ratio/verhouding A tan θ = −3  ref. angle/verw. hoek CA o ref .angle / verw. hoek     = 53,13  value of/waarde van θ CA 𝜃𝜃 = 180𝑜𝑜 − 53,13𝑜𝑜 = 126,87𝑜𝑜 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2022) 3.3.3 ο  z in polar form/polêre ∴z =5 cis 126,87 vorm CA correct quadrant/korrekte kwadrant CA OR/OF  correct angle/korrekte = o ( ∴ z 5 cos126,87 + isin 126,87 o ) hoek CA OR/OF  z in polar form/polêre vorm CA correct quadrant/korrekte kwadrant CA  correct angle/korrekte hoek CA (3) 3.4 2 p − qi − 8i = − 2i ( 3i + 7 )  Product/produk A 2 2 p − qi − 8i = − 6 i − 14i 2  substituting i with – 1/ 2 p − ( q + 8 ) i = − 6( −1) − 14i 2 vervang i met – 1 A 2 p − ( q + 8) i = 6 − 14i  value of/waarde van p ∴2p = 6 CA p=3 and/en ∴ − ( q + 8) i = − 14i  value of/waarde van q CA q =6 OR/OF OR/OF 2 p − qi − 8i = − 2i ( 3i + 7 )  Product/produk A 2 2  substituting i with – 1/ 2 p − qi − 8i = − 6 i − 14i 2 vervang i met – 1 A 2 p − ( q + 8 ) i = − 6( −1) − 14i 2 p − ( q + 8) i = 6 − 14i  value of/waarde van p 2 p − 6 = ( q + 8) i − 14i CA ∴ 2 p − 6 = 0 and / en ( q + 8) i − 14i = 0 p=3 − ( q + 8) i = − 14i  value of/waarde van q q =6 CA Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 11 OR/OF OR/OF 2 p − qi − 8i = − 2i ( 3i + 7 ) 2  Product/produk A 2 p − qi − 8i = − 6 i − 14i 2 p − ( q + 8 ) i = − 6( −1) − 14i 2  substituting i with – 1/ 2 vervang i met – 1 A 2 p − 6 = qi − 6i ∴ 2 p − 6 = 0 and / en q − 6 =0 p = 3 and / en q =6  value of/waarde van p CA  value of/waarde van q CA (4) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2022) QUESTION/VRAAG 4 4.1.1 f ( x ) = x 2 − 8 x − 20 x − ints/afsn..:  factors/form./faktore/vorm. A 2 x − 8 x − 20 = 0 ( x − 10)( x + 2) 0 = OR/OF − ( −8) ± ( −8) − 4 (1)( −20 ) 2 x=  coordinates of/koördinate van A CA 2 (1)  coordinates of/koördinate van B x = 10 or / of x = − 2 CA A ( − 2;0) and/en B (10;0) (3) 4.1.2 −b − ( − 8 ) x = =  SF A 2a 2 (1) =4  x-value/waarde CA OR/OF OR/OF 10 − 2 M CA =x = 4  x-value/waarde CA 2 OR/OF OR/OF M CA f ′ ( x) = 2x − 8 = 0  x-value/waarde CA x=4 (2) 4.1.3 y= 2 ( 4) − 8( 4) − 20 = − 36 M CA C (0; − 36)  y-value/ waarde CA OR/OF OR/OF 4 (1)( −20 ) − ( −8) 2 M CA y= = −36 4 (1)  y-value/waarde CA (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 13 4.1.4 A ( − 2;0) and/en D (7; − 27) M CA y2 − y1  m CA m= x2 − x1  equation of/vergelyking van AD −27 − 0 CA = 7+2 = −3 y mx + c = 0= −3 ( −2 ) + c c = −6 ∴ y =−3 x − 6 OR/OF OR/OF y2 − y1 y− = y1 ( x − x1 ) M CA x2 − x1  m CA −27 − 0 y = −0 ( x + 2)  equation of/vergelyking van AD CA 7+2 y=−3 x − 6 (3) 4.1.5 x < 0 or x > 8  critical values/kritiese waardes CA  correct notation/ OR/OF korrekte notasie A x∈( − ∞ ;0) or / of x∈( 8; ∞ )  critical values/kritiese waardes CA  correct notation/ OR/OF korrekte notasie A − ∞< x < 0 or / of 8 < x < ∞ (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 14 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2022) 4.2.1 ∴x =3  eq. of asymptote/vergl van asimptoot A (1) 4.2.2 1 g( = x) + 2  SF A x−3 1 =0 + 2 x−3 S A −2( x − 3) = 1  value of/waarde van x CA −2 x + 6 = 1 5 ∴x = 2 (3) 4.2.3 For y-int/Vir y-afsnit. Let/laat x = 0 Subst/Vervang. x = 0 A 1 g = (0) + 2 0−3 5 ∴y =  y-int.of/y-afsnit van g A 3 And/en h ( 0 ) = 30 1 ∴y =  y-int. Of/y-afsnit van h A (3) 4.2.4 g:  shape/vorm A  x and y- int./afsn CA  asymptotes/asimptote CA h:  shape/vorm A  y int../afsn CA  asymptote/asimptote A (6) 4.2.5(a) x ∈  ,x ≠ 3  x≠3 A (1) 4.2.5(b) x = 0 x=0 A (1) 4.2.5(c) k ( x )= 3x − 2  eq. of /vergl. van k A (1) [28] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 15 QUESTION/VRAAG 5 5.1 A = P(1 + i ) n  SF A i CA  5,5%  24×10 n CA 1 200 = P 1 +   24  1 200 =P  P the subject/ 240  0 ,055  die onderwerp CA 1 +   value of/waarde van ʼn  24  CA 693 = P (5) 5.2.1 Value of investment after 24 months/  SF A Waarde van belegging na 24 maande 24  7 , 5%   1 +  A A = P(1 + i ) n  12  24  7,5%  value of/waarde van A = R500 000 1 +   12  CA ≈ R580646, 0091  M subtraction/aftrekking A Value of investment after the withdrawal / value of/waarde van A CA Waarde van belegging na onttrekking =A R580 646, 0091 − R365 000,00 = R215 646, 01 OR/OF  SF A 24  7,5%  24 A = R500 000 1 +  − R365 000  7 , 5%   12   1 +  A  12  ∴ A = R215 646,01 value of/waarde van A CA  M subtraction/aftrekking A M A (5) 5.2.2 Value of investment at the end of 5 years/ 24 Waarde van belegging aan die einde van 5 jaar  7 , 5%   1 +  A  12  A = P(1 + i ) n 4×5 SF CA  6, 75%  = R215 646,01 1 +   4  value of/waarde van A ≈ R301 365, 01 CA (3) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 16 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2022) QUESTION/VRAAG 6 6.1 f ( x) = − 6x + 3 f ( x + h)− f ( x)  definition/definisie A f ′ ( x ) = lim h→0 h  SF CA − 6( x + h ) + 3 − ( − 6x +3) S CA = lim h→0 h S CA − 6 x − 6h + 3 + 6 x − 3 = lim –6 CA h→0 h AO: 1 mark/punt − 6h = lim h→0 h Penalty of one mark for = lim ( − 6) incorrect notation h→0 Penaliseer een punt indien ∴ f ′ ( x) − = 6 notasie foutief is. (5) 6.2.1 2  x − x − 12  Dx    x−4   Factors/faktore A  ( x − 4 )( x − 3)  Dx   S CA  x−4  D x ( x − 3)  1 CA =1 (3) 6.2.2 dy 2 if / as y = 3 − 7 x 2 + x dx 3x S A 2 x −3 y= − 7 x2 + x −2 3  − 2x CA  −14x CA dy −2 1 CA = − 2 x − 14 x + 1 dx (4) 6.3 2 g ( x) = x − 2 x + 1 y − y1  SF A Ave. grad./Gem grad = 2 x2 − x1  mave/gemid value/waarde CA 4− 0 = 3 −1 =2 (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 17 QUESTION/VRAAG 7 3 2 7.1 f ( x) = x − 2x − 7 x − 4  y-intercept/afsnit A y = 6 OR/OF ( 0 ; 6 ) (1) 7.2 3 2 f (1) = (1) − 2(1) − 5(1) + 6  substitution/vervanging A =0 0 A ∴ ( x −1) is a factor of/ is ʼn faktor van f (x) (2) 7.3 x-intercepts/afsnitte; y = 0 2 ( x − 1)( x − x − 6) = 0  quadratic factor/kwdr.faktor A ( x − 1)( x + 2)( x − 3) = 0  factors/faktore CA ∴ x= 1 or/of x = − 2 or/of x =3  x-intercepts/afsnitte CA OR/OF OR/OF 2 (x + 2)(x − 4x + 3) 0 =  quadratic factor/kwdr.faktor A ( x + 2)( x − 1) ( x − 3) = 0  factors/faktore CA ∴x= − 2 or/of x = 1 or/of x =3  x-intercepts/afsnitte CA OR/OF OR/OF  quadratic factor/kwdr.faktor A 2 (x − 3)(x + x − 2) = 0  factors/faktore CA ( x − 3)( x − 1)( x + 2) 0 =  x-intercepts/afsnitte CA ∴ x= 3 or/of x = 1 or/of x =−2 AO: Full marks/Volpunte (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 18 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2022) 7.4 2 f ′ ( x )= 3 x − 4 x − 5= 0  derivative/afgeleide A − ( − 4) ± ( −4 ) − 4 ( 3)( −5) 2  equating derivative to 0/ x= stel afgeleide gelyk aan 0 A 2 ( 3)  SF CA ∴ x =2 ,12 or/of x = − 0 , 79  both values of/beide 3 2 waardes van x CA f ( 2 ,12 ) =( 2 ,12 ) − 2 ( 2 ,12 ) − 5 ( 2 ,12 ) + 6 ≈ − 4 ,06 3 2 f ( − 0 ,79 ) ( =− 0 , 79 ) − 2 ( − 0 , 79 ) − 5 ( − 0 , 79 ) + 6  both values of/beide waardes van y CA ≈ 8, 21 ∴ ( 2 ,12 ; − 4 , 06 ) and /en ( − 0,79 ;8, 21) (5) 7.5  shape/vorm y- A  y-intercept/afsnit CA  both x-intercepts/beide x- afsnitte CA  both turning points/beide draaipunte CA (4) 7.6 2 f ′ ( x ) = 3x − 4 x − 5  derivative/afgeleide A 2 f ′ (3 = ) 3 (3) − 4 (3) −=5 10  substitution/vervanging CA tanθ = 10  m of tangent/van raaklyn =θ 84 ,3° CA θ 84 ,3° = CA (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 19 QUESTION/VRAAG 8 8.1 Wages = time×hourly  SF A Loon = tyd × uurliks 2 ( 75 ) 21750 = × 145  CA x x 21750 = x NPU (2) 8.2 Petrol costs = time x fuel consumption x fuel price  Setting up a formula/ Brandstofkoste = tyd x brandstofverbruik x brandstofprys Opstelling van formule A 2 ( 75)  2 x  = ×2 +  × 18  SF CA x  100  S CA 5400 = + 27 x x NPU (3) 8.3 C ( x ) = Wages + Petrol costs  Setting up a formula/ = Loon + brandstof kostes Opstelling van formule A 21750 5400 C ( x) = + + 27 x x x  SF CA 27150 C = ( x) + 27 x x NPU (2) 8.4 27150 C = ( x) + 27 x x  derivative/afgeleide A −1 = 27150 x + 27 x −2  equating derivative to/stel Dx = − 27150 x + 27 = 0 afgeleide aan 0 A −2 −27150 x − 27 = 0 −2 S CA −27150 x − 27 = S CA −2 27 x = 21750  x ≈ 28,38 km/h CA 1 27 2 = x 21750 2 27 x = 21750 2 ∴x = 805,56 ∴ x ≈ 28,38 km/h Ethans speed to minimise costs is Ethans se spoed om koste te verminder is x ≈ 28,38 km/h NPU (5) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 20 TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 (EC/JUNE/JUNIE 2022) QUESTION/VRAAG 9 9.1.1 1   ln x A ∫  x + 1 dx  x A C A = lnx + x + C (3) 9.1.2 ∫ ( x + 6 x ) dx 6 5 x  CA 6 6 x = + 3x 2 + C  3x 2 +C CA 6 (2) 9.2 2  M Area notation using integrals/Area-notasie A = ∫ f (x) dx met gebruik van integrale A m 2 = ∫ x3 dx x4 m  A x  4 2 4 = 4  m  SF CA (2) 4 (m) 4 = − 4 4 m 4  equating area to/stel oppervl.gelyk aan 3,75 ∴4− = 3,75 CA 4 S CA ∴m 4 1 = ∴m =± 41 value of/waarde van m CA ∴m = −1 OR/OF OR/OF 2 A = ∫ f (x) dx 0  M Area notation using integrals/Area-notasie 2 met gebruik van integrale A = ∫ x dx 3 0 4 x4 x  4  A = 4 4  0 ( 2) 4 (0) 4  SF CA = − = 4 4 4 OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/JUNE/JUNIE 2022) TECHNICAL MATHEMATICS P1/TEGNIESE WISKUNDE V1 21 0 A = ∫ f (x) dx m x4 ]m 0 = S CA 4 ( 0) 4 (m) 4 m4 = − = 4 4 4 m4 M CA ∴ 4− =3,75 4 ∴m 4 1 = ∴m =± 41 ∴m = −1  value of/waarde van m CA (7) [12] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief