Technical Mathematics Grade 12 P1 Gauteng Mock 2020 Memo (Afr)

PREPARATORY EXAMINATION/ VOORBEREIDENDE EKSAMEN 2020 MARKING GUIDELINES/NASIENRIGLYNE TECHNICAL MATHEMATICS P1/ TEGNIESE WISKUNDE V1 11091 19 pages/bladsye Marking Codes/Nasienkodes A Accuracy/Akkuraatheid CA Consistent Accuracy/Volgehoue Akkuraatheid M Method/Metode R Rounding/Afronding NPR No Penalty for Rounding/Geen straf vir afronding NPU No Penalty for Units omitted/Geen straf vir weggelate eenhede SF Simplification/Vereenvoudiging F Substitution in correct formula/Vervanging in korrekte formule S Statement/Stelling AO Answer only/Slegs antwoord 1

11091/20 NOTE: • If a candidate answers a question TWICE, mark only the FIRST attempt. • Consistent accuracy applies in all aspects of the marking guidelines where appropriate. Stop marking at the second calculation error. LET WEL: • Indien 'n kandidaat 'n vraag TWEE keer beantwoord, sien slegs die EERSTE poging na. • Volgehoue akkuraatheid is deurgaans op alle aspekte van die nasienriglyne van toepassing. QUESTION/VRAAG 1 [25] 1.1.1 2( x + 3)( x −1) = 0 ✓ A x = -3 x =− 3 or x =1 ✓A x=1 (2) 1.1.2 𝑥 + 7 = −3𝑥 2 3𝑥 2 + 𝑥 + 7 = 0 ✓ M standard form/standard vorm −𝑏±√𝑏 2 −4𝑎𝑐 𝑥= 2𝑎 −1±√12 −4(3)(7) ✓ SF 𝑥= 2(3) −1 ±√−83 𝑥= 6 −1 ± √83𝑖 ✓ S √𝟖𝟑𝒊 𝑥= 6 1 √83 ✓CA x =1,35i 𝑥 = −6 ± 6 𝑖 ✓CA x = – 1, 69i 𝐎𝐑/𝑶𝑭 = −0,17 ± 1,52 𝑖 OR/OF OR/OF ✓ M standard form/standard vorm −3𝑥 2 − 𝑥 − 7 = 0 −𝑏±√𝑏 2 −4𝑎𝑐 ✓ SF 𝑥= 2𝑎 −1±√(−1)2 −4(−3)(−7) 𝑥= 2(−3) ✓ S √𝟖𝟑𝒊 −1 ±√−83 𝑥= 6 −1±√83𝑖 𝑥= ✓CA x =1,35i 6 1 √83 ✓CA x = – 1, 69i 𝑥= −6 ± 6 𝑖 (5) 𝐎𝐑/𝑶𝑭 = −0,17 ± 1,52 𝑖 1.1.3 𝑥 2 + 3𝑥 – 4 < 0 ✓ CA factors/faktore (x +4)(x– 1) < 0 ✓ CA critical values/kritiese waardes Critical values/Kritiese waardes: – 4 ; 1 ✓ CA solution/oplossing (3) –4 < x<1 OR/OF 𝑥 ∈ ( −4 ; 1) 2

11091/20 1.2 3 y + x = 2 & y 2 + x = xy + y ✓ M x = 2 – 3y x = 2 − 3 y ........(1) ✓ S y 2 + 2 − 3 y = y (2 − 3 y) + y ✓ M factors/faktore y2 + 2 − 3y = 2 y − 3y2 + y ✓ CA both values of y/beide waardes 4 y2 − 6 y + 2 = 0 2 y 2 − 3 y +1 = 0 (2 y −1)( y − 1) = 0 1 y = or y =1y = 1 or/of 2 1 x = or x or/of= −x1 = –1 ✓ CA both values of x/beide waardes 2 OR/OF 2− x y= 3 OR/OF 2− x 2− x 2− x 2   + x = x +  3   3  3 4x + 2x − 2 = 0 2 (2x −1)(x + 1) = 0 1 x = OR/OF x = −1 2 1 y= OR/OF y = 1 2− x 2 ✓ M y= 3 ✓ S ✓ M factors/faktore ✓ CA both values of x/beide waardes ✓ CA both values of y/beide waardes (5) 3

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11091/20 1.3 a − b log a + log b log   =  2  2 ✓ M multiplying by 2 a − b  = log a + log b  a −b  2 2log   2  ✓A    2   a −b  2  = log ( ab ) log  ✓ A ab  2   a −b  2  a −b  2  2  = ab ✓A   = ab    2  a − 2ab + b 2 2 = ab 4 ✓ CA a2 − 2ab + b2 = 4ab a 2 − 2ab + b2 = 4ab a 2 + b2 = 6ab (5) a 2 − 6ab + b2 = 0 1.4.1 1 1 1 = + f u v ✓M 1 1 1 = − 1 1 1 v f u = − v f u ✓ M : LCD 1 u− f = v fu ✓S fu v= u− f OR/OF OR/OF uv = fv − fu uv − fv = fu ✓M v(u − f ) = fu ✓ M factors/faktore fu v= u− f ✓ CA (3) 1.4.2 fu v= u− f 5 6 ✓ CA SF = 6−5 ✓ CA (2) = 30 AO Full marks/Vol punte 5

11091/20 QUESTION/VRAAG 2 (7) 2.1 p 3x 2 = 2 x − 4 p ✓ A standard form/standaard vorm 3x 2 − 2 x + = 0 4 ✓ S  = b − 4ac 2  p = (−2)2 − 4 ( 3)   ✓ CA ∆ = 4 – 3p 4 = 4−3p OR/OF OR/OF ✓ A standard form/standaard vorm 12x 2 − 8x + p = 0  = b 2 − 4ac ✓ S ✓ CA ∆ = 64 – 48p (3)  = 64− 48p 2.2.1 4 – 3𝑝  0 ✓ A for inequality/vir ongelykheid 4 – 3𝑝 > 0 4 𝑝  3 4 ✓ CA 𝑝 < 3 (2) OR/OF OR/OF ✓ A for inequality/vir ongelykheid 64 – 48p > 0 4 64 – 48p  0 ✓ CA 𝑝 < 3 (2) 4 𝑝 3 2.2.2 ∆<0 ✓M 0 4 4 − 3𝑝 < 0 ✓ CA p 3 4 (2) 𝑝 >3 6

11091/20 QUESTION/VRAAG 3 (10) 3.1 110101 − ✓M 101101 ✓ CA answer/antwoord 10002 AO: Full marks/Vol punte (2) 3.2.1 1 3x = 27 1 ✓ A 3−3 = 27 3x = 3−3 ✓ CA x = −3 x = −3 (2) 5 3.2.2  5 ✓ S 1 = (3 log 2 + log 4) log ( x + 2 ) + log ( 3x −1) =  3log 2 + log   4 ✓ A log property/log eienskap log ( x + 2 )( 3x −1) = 1 ( x + 2)(3x −1) =10 ( x + 2)(3x −1) = 10 3x 2 + 5 x − 2 = 10 ✓ CA 3x2 + 5x −12 = 0 3x 2 + 5x −12 = 0 ✓ CA factors/faktore (3x − 4)( x + 3) = 0 ✓ CA both values of x/beide waardes 4 x = or x = − 3 van x 3 ✓ CA for positive x -value/vir positiewe 4 ∴𝑥 = 3 waarde van x (6) 7

11091/20 QUESTION/VRAAG 4 (11) 4.1.1 r = x2 + y 2 = (−4)2 + (−3)2 = 25 ✓ A r=5 (1) r =5 −3 3 ✓ A tan  =   4.1.2 3 tan 𝜃 = ( ) = ( ) −4 4 4 ref. angle/Verwysings hoek = 36,87o ✓ CA ref . angle/verwysing hoek 𝜃 = 180° + 36,87° = 216,87°/1,20𝜋 = 36,87o ✓ CA 𝜃 = 216,87° OR/OF OR/OF −3 From/Van 4.1: r = 5 ✓ A 𝑠𝑖𝑛𝜃 = ( 5 ) −3 𝑠𝑖𝑛𝜃 = ( 5 ) ✓ CA ref . angle/verwysing hoek = 36,87o Ref.angle/verwysings hoek = 36,87o ✓ CA 𝜃 = 216,87° 𝜃 = 180° + 36,87° = 216,87°/1,20π OR/OF OR/OF −4 ✓ A 𝑐𝑜𝑠𝜃 = ( 5 ) From/Van 4.1: r = 5 −4 ✓ CA ref . angle/verwysing hoek 𝑐𝑜𝑠𝜃 = ( 5 ) = 36,87o Ref.angle/verwysings hoek = 36,87o ✓ CA 𝜃 = 216,87° 𝜃 = 180° + 36,87° = 216,87° = 1,20π (3) 4.2 5Cis216,87 or 5Cis1,20 ✓✓ CA (any form) or/ of 5(Cos216,87 + iSin 216,87) (2) or/ of 5(Cos1,20 + i sin 1,20 ) ) 8

11091/20 4.3 3 p + 2qi = ( 4 − 5i ) + −121 2 ✓ A 11i =16 − 40i + 25i 2 +11i ✓ S 16 − 40i + 25i 2 =16 − 29i + 25(−1) ✓ A (−1) 3 p + 2qi = − 9 − 29i − 29 ✓ CA q =  3 p = − 9 and/en and 2q = − 29 2 − 29 ✓ CA p=−3 p = − 3 and/en and q = 2 1 Accept/Aanvaar: 𝑞 = −14 2 ; 𝑞 = −14,5 (5) QUESTION/VRAAG 5 (17) 5.1 x =0 ✓ A x =0 y=2 ✓ A y =2 (2) 5.2 2 ✓S y=0 0= +2 𝑥 ✓ A x = −1 (2) 2𝑥 + 2 = 0 𝑥 = −1 5.3 y 1 ✓✓ A y 1 (2) OR/OF 𝑦 ∈ (1 ; ∞) 5.4 g(0) = 2.3° + 1 ✓✓A (0;3) (2) Accept/Aanvaar: x = 0 ; y = 3 =3 y intercepts : (0;3) 5.5 𝑥 ∈ℝ ✓✓ A 𝑥 ∈ ℝ (2) 5.6 Increasing function ✓ A increase (1) 9

11091/20 5.7 For/Vir f: For/Vir g: ✓ CA Asymptotes ✓ CA y- intercept (0 ; 3) x =0;y=1 ✓ CA Asymptote y = 1 ✓ CA x-intercept (– 1 ;0) ✓ A shape ✓ A shape (6) 10

11091/20 QUESTION/VRAAG 6 (15) 6.1 𝑓 ′ (𝑥) = −2𝑥 − 2 ✓M 𝑓 ′ (𝑥) = −2𝑥 − 2 𝑓 ′ (𝑥) = 0 ✓ CA x = – 1 ✓ CA 𝑓(−1) = 9 −2𝑥 − 2 = 0 ∴ 𝑥 = −1 𝑓 (−1) = −(−1)2 − 2(−1) + 8 ∴ 𝑓(−1) = 9 C( – 1 ; 9) OR/OF OR/OF −𝑏 𝑥 = 2𝑎 (−2) ✓ M 𝑥 = − 2(−1) (−2) 𝑥 = − 2(−1) ∴ 𝑥 = −1 ✓ CA x = – 1 Substitute 𝑥 = −1 in original equation/ ✓ CA 𝑓(−1) = 9 Vervang x = – 1 in oorspronklike vergelyking 𝑓 (−1) = −(−1)2 − 2(−1) + 8 ∴ 𝑓(−1) = 9 C( – 1 ; 9) OR/OF OR/OF −𝑥 2 − 2𝑥 + 8 = 0 ✓ M Factorisation/ 𝑥 2 + 2𝑥 − 8 = 0 (𝑥 + 4)(𝑥 − 2) = 0 ✓ CA x = – 1 𝑥 = −4 𝐎𝐑/𝐎𝐅 𝑥 = 2 ✓ CA 𝑓(−1) = 9 −4+2 𝑥 = = −1 2 𝑓 (−1) = −(−1)2 − 2(−1) + 8 ∴ 𝑓(−1) = 9 C( – 1 ; 9) (3) 6.2 𝑓 (𝑥) = −(𝑥 + 1)2 + 9 ✓ CA 𝑓 (𝑥) = −(𝑥 + 1)2 + 9 (1) 11

11091/20 6.3 −5 −5 2 −5 ✓S 𝑓( 2 ) = −( 2 ) − 2( 2 )+ 8 27 27 = 4 ✓ CA (2) 4 6.4 27  5 ✓ M m = tan1350 =− 1 y− = tan1350  x +  4  2 ✓ S CA 𝑥1 = −5 and/en 2  5  27 27 y =− 1 x +  + 𝑦1 = 4  2 4 17 ✓CA 𝑐 = 4 17 y =− x + 17 4 ✓CA g ( x) =− x + 4 17 𝑔(𝑥) = −𝑥 + 4 OR/OF OR/OF ✓ M m = tan1350 =− 1 y = mx + c −5 ✓ S CA 𝑥1 = and/en 2 27 5 27 = (−1)(− ) + c 𝑦1 = 4 4 2 17 ✓CA 𝑐 = 4 17 17 c= ✓CA g ( x) =− x + 4 4 17 (4) 𝑔(𝑥) = −𝑥 + 4 6.5 17 ✓ S CA x = −1 g (−1) =− (−1) + 4 21 ✓ CA 21 4 = 4  21  H  −1;  (2)  4 6.6 (𝑥 + 4)(𝑥 − 2) = 0 ✓ M Factorisation ✓ A B(2 ;0) E(– 1 ; 0) ; B(2 ;0) ✓CA EB = 3 units/eenhede ∴ EB = 33 𝑢𝑛𝑖𝑡𝑠 units/eenhede (3) 12

11091/20 QUESTION/VRAAG 7 (14) A = P (1− i ) ✓F A = P(1 − i)n n 7.1  A log    A n = P log   log (1− i ) ✓F n = P log (1− i )  4  log   =  12,6  ✓SF log (1− 0,05) n  22,37 ✓S It will take 23 days/Dit sal 23 dae neem (4) Accept method of inspection, i.e.: Substitute values for n>0 7.2.1 Loan amount required: R6 000 – R600 = R5 400 Total owing : A = P (1 + in) ✓A P = 5 400 = 5 400 (1+ (1,5)(0,09) ✓ SF A = 6 129 ✓ CA A = R6 129 6 129 Monthly payment: = R340,50 18 ✓ CA: R340,50 R340,50 will need to be budgeted. (4) 7.2.2 R6 129 – R5 400 = R729,00. ✓M ✓CA R729,00 OR/OF (2) R6 129+600=R6 729 R6 729 – R6 000=R729 13

11091/20 7.3  0,096  12  0,096  10  0,096  4 ✓M A = 15 000 1 +  − 5 0001 +  + 3 5001 +  0,096 12  4   4   4  1 500 (1 + ) 4 A = R17 448,46 ✓M 0,096 10 −5 000 (1 + ) 4 ✓M 0,096 4 +3 500 (1 + ) 4 ✓ CA R 17 448,46 OR/OF OR/OF 0,096 2 ✓M A= 15 000(1 + ) = R15 728,64 4 ✓M R15 728,65 – R5 000=R10 728,64 0,096 6 ✓M A = 10 728 (1 + ) = R12 369,27977 + R3 500 4 ✓ CA A = R17 448,46 0,096 4 A R15 869,27977(1 + ) = R17 448,46 (4) 4 14

11091/20 QUESTION/VRAAG 8 (17) 8.1 f ( x ) = − 2 + 3x f ( x + h)− f ( x) ✓ M definition/definisie f / ( x ) = lim h →0 h ✓ SF = lim −2 + 3( x + h) − (−2 + 3x) h →0 h ✓ S simplification/ −2 + 3x + 3h + 2 − 3x vereenvoudiging = lim h →0 h ✓S simplification/ 3h = lim Vereenvoudiging h →0 h = lim(3) ✓CA 𝑓 ′ (𝑥) = 3 h →0 (5)  f / ( x ) =3 8.2.1 x3 − x 2 +  y= ✓ S  y = x2 − x +  x−1 x x x2  3 ✓ CA 2x = − + ✓ CA −1 x x x ✓ CA − x −2 y = x2 − x +  x −1 dy (4)  = 2x −1 −  x−2 dx 8.2.2 x 1 f ( x) = − 3 1 12 2 6x ✓S x 1 12 1 −3 2 = x − x 1 2 6 ✓ S − x−3 1 1 − 1 6  f / ( x) = x 2 + x −4 1 −1 4 2 ✓ CA x 2 4 1 −4 ✓ CA x 2 15

11091/20 8.3 s = 3t 2 − 4t − 6 f ( x) = 3t 2 − 4t − 6 ✓ A s1 = − 2 s2 = f (5) = 3(5)2 − 4(5) − 6 ✓ A s2 = 49 s1 = f (2) = 3(2) − 4(2) − 6 2 and/en = 75 − 20 − 6 =12 − 8 − 6 = 49 =−2 s −s  Average gradient = 2 1 ✓SF substitution t2 − t1 average gradient 49 − (−2) = − 2 − 49 formula 5−2 or/of 2−5 51 ✓CA 17m / s = 3 = 17 m / s (4) QUESTION/VRAAG 9 (16) 9.1 C(0; −12) ✓ Ax= 0 ✓A y = –12 (2) 9.2 f ( x ) = x + 5 x − 8 x − 12 3 2 f ' ( x ) = 3 x 2 +10 x − 8 ✓ M f / ( x) = 3x2 +10 x − 8 f '( x ) = 0 ✓ A f / ( x) = 0 ( 3x − 2)( x + 4) = 0 ✓ A x =− 4 or x = 2 2 3 x= or x = −x 4= –4 OR/OF ✓ SF 3 2 400 f   =− 3 27 f ( − 4 ) = 36 ✓ CA A( − 4;36)  2 400  2 400  ✓ CA B  ; − A ( − 4;36) and/en and B ;−  3 27   3 27  (6) 2 9. 3 x  − 4 OR/OF or x ✓ A x− 4 3 OR OR/OF 2 2  ✓ A x x  ( − ; − 4 ) OR/OF or x  ;   3 3  (2) 9.4 2 ✓ CA critical values −4  x  ✓ CA notation 3 (2) 16

11091/20 9.5 f / ( x) = 3x 2 +10 x − 8 and/en and x ==00 at/by CC ✓ M f / ( x) = 3x2 +10 x − 8 mt = f / (0) = 3(0)2 + 10(0) − 8 =− 8 ✓ CA mt = − 8 y − y1 =− 8( x − x1 ) y + 12 =− 8( x − 0) ✓CA subst (0; −12)  y =− 8x − 12 ✓ CA y =− 8x −12 (4) 17

11091/20 QUESTION/VRAAG 10 (6) 10.1  (t ) = 256t − 24t 2 ✓ M derivative/ d v(t ) = = 256 − 48t dt v(t ) = 256− 48t ✓ t=3 ✓ CA v(3) =112 v(3) = 256− 48(3) ✓ CA 112rad/s v(3) = 112 (4)  Instantaneous Angular velocity after 3 seconds is 112 rad/s / 10.2 𝜃(𝑡) = 256𝑡 − 24𝑡 2 ✓ M second derivative/ 𝑣(𝑡) = 256 − 48𝑡 𝑎(𝑡) = –48 ✓ A -48rad/s2 (2) Instantaneous Angular acceleration after 3 seconds is –48rad/s2/ 18

11091/20 QUESTION/VRAAG 11 (12) 11.1.1  3 7 1 4   2x − 3x − x  dx ✓A 2 x 2 1 ✓ A − x3 = x4 − x3 − 7ln x + c 2 ✓ A − 7ln x ✓A c (4) 11.1.2 3 ✓ A 3𝑥 ∫ (3𝑥 + ) 𝑑𝑥 𝜋 ✓ A ln 3 𝑥 3 3 = + 𝑥+𝑐 3x ln 3 𝜋 ✓A +c  (3) 11.2 3 ✓M Area notation Ax = −  ydx using 2 integrals/oppervlakte 3 notasie deur gebruik 1 11  = −  x 4 − 2 x3 + x 2 − 6 x  van integrasie 4 2 2 ✓A  1 11  1 11  =−  (3)4 − 2(3)3 + (3)2 − 6(3)  −  (2)4 − 2(2)3 + (2)2 − 6(2)  integral/integraal  4 2  4 2   9   ✓CA = −  −  − ( −2) substituting/  4   vervanging 1 = ==0,25square 0,25 squareunits/vierkantseenheid units/vierkantseenheid 9 4 ✓S − 4 ✓CA 1 square units / 4 (5) 19

11091/20 OR/OF OR/OF ✓M Area notation A =  ydx 3 2 using 3 integrals/oppervlakte 1 11  =  x 4 − 2 x 3 + x 2 − 6 x notasie deur gebruik 4 2 2 van integrasie = 1 4 11 2  1 4 11 2  ✓A  4 (3) − 2(3) + 2 (3) − 6(3) −  4 (2) − 2(2) + 2 (2) − 6(2) 3 3 integral/integaal 1 ✓CA =− 4 substituting/ 1 vervanging  A = sq units 4 9 ✓S − 4 ✓ CA 1 square units / 4 (5) TOTAL/TOTAAL: 150 20