Technical Mathematics Grade 12 IEB Paper 1 2019 Memo (Afrikaans)

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT-EKSAMEN NOVEMBER 2019 TEGNIESE WISKUNDE: VRAESTEL I NASIENRIGLYNE Tyd: 3 uur 150 punte Hierdie nasienriglyne is opgestel vir gebruik deur eksaminators en hulp- eksaminators van wie verwag word om almal ’n standaardiseringsvergadering by te woon om te verseker dat die riglyne konsekwent vertolk en toegepas word by die nasien van kandidate se skrifte. Die IEB sal geen bespreking of korrespondensie oor enige nasienriglyne voer nie. Ons erken dat daar verskillende standpunte oor sommige aangeleenthede van beklemtoning of detail in die riglyne kan wees. Ons erken ook dat daar sonder die voordeel van die bywoning van ’n standaardiseringsvergadering verskillende vertolkings van die toepassing van die nasienriglyne kan wees. IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 2 van 12 VRAAG 1 1.1 1.1.1 6 – x2 + x = 0 x2 – x – 6 = 0 (x – 3)(x + 2) = 0 x = 3 of x = –2 1.1.2 f(x) > 0 vir x 𝜖𝜖 (–2; 3) x 𝜖𝜖  OF deur getallelyn te gebruik OF –2 < x < 3 1.2 3y = 34 x of y = x2 – 6x + 9 ……… (2) ∴y = 4x ……… (1) Vervang (1) → (2): 4x = x2 – 6x + 9 0 = x2 – 10x + 9 0 = (x – 9)(x – 1) x = 9 of x = 1 Vervang in (1) y = 36 of y = 4 1.3.1 − ( x − 2 ) + 3 = 2 1.3 5 −x 2 + 4x − 4 − 2 =0 x − 4x + 6 = 2 0 − ( −4 ) ± ( −4 ) − 4 (1)( 6 ) 2 x= 2 (1) 4 ± −8 4 ± 2 −2 =x = 2 2 x = 2 ± −2 OF x= 2 ± 2i 1.3.2 − x 2 + 4 x − 1 = 5 + k OF 0 = x − 4x + 6 + k 2 ∆ = ( −4)2 − 4(1)(6 + k) = 16 − 24 − 4k = –4k – 8 Dus 2 reële, verskillende wortels –4k – 8 > 0 Deur inspeksie van grafiek –4k > 8 g(x) + k ontmoet f twee keer k < –2 indien 5 + k < 3 k < –2 IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 3 van 12 1× 24 + 1× 23 + 0 × 22 + 1× 21 + 1× 20 1.4 1× 25 + 1× 24 + 0 × 23 + 0 × 22 + 1× 21 + 1× 20 16 + 8 + 2 + 1 = 32 + 16 + 2 + 1 27 = 51 9 = 17 ∆L 1.5 ε= L 182 − L 0,77 = L 0,77L + L = 182 182 L= 1,77 = 102,824858 …… ≈ 1,02825 × 102 IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 4 van 12 VRAAG 2 4   x +2 > 0 3 ( x + 2) 4 = 3 3 4 ( )  ( ) x + 2 4 = 27 3 27 2.1 x > –2 OF   4   3 3 4  ( x + 2) 4  ( 27 = )3 1 ( x + 2) 3 = 4   x + 2 = 34 x + 2 = 81 x = 79 Geldig 2.2 2.2.1 ( 2 3 − 3 − 2 2 )( 2 3 − 3 + 2 2 ) = ( 3 − 2 2 )( 3 + 2 2 ) =3–8 = –5 3 ⋅ 22 x +1 − 22 x −2 + 4 x 2.2.2 4 ⋅ 22 x − 3 3 ⋅ 22 x ⋅ 21 − 22 x ⋅ 2−2 + 22 x = 4 ⋅ 22 x ⋅ 2−3 = ( 22 x 3 ⋅ 21 − 2−2 + 1 ) 4. 22 x ⋅ 2−3 1 6− +1 = 4 1 4⋅ 8 27 2 = × 4 1 27 = 2 2.3 2 ( 5 − 2i ) − i ( 6i − 1) = 10 − 4i − 6i 2 + i = 10 − 4i + 6 + i = 16 − 3i IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 5 van 12 2.4 2.4.1 i –3 point R ray –4 r –4 2.4.2 r = p r 2 = ( −3 ) + ( −4 ) 2 2 = 9 + 16 = 25 r=5 −4 4 tanθ = = −3 3 θ = 180° + 53,13° = 233,13° P = (5; 233,13°) OF (5 cos 233,13° ; 5 sin 233,13°) OF p = +5 cos 233,13° + 5 sin 233,13° OF p = 5 cis 233,13° OF p = 5 233,13° IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 6 van 12 VRAAG 3 3.1 3.1.1 By B: x + 2 = 0 x = –2 B is (–2; 0) 3.1.2 Wortels is –12 en –2 Vergelyking is y =a ( x + 12 )( x + 2 ) Vervang (–13; –11): –11 = a (–1)(–11) –11 = 11a –1 = a −1( x + 12 )( x + 2 ) y= y=− x 2 − 14 x − 24 3.1.3 xE = −7 (deur simmetrie) OF f' (x) = 0 –2x + 14 = 0 y E =− ( −7 ) − 14 ( −7 ) − 24 2 x=7 =−49 + 98 − 24 = 25 xF = xE = −7 : Vervang in g: y F = –7 + 2 = –5 EF = y E − y F = 25 – (–5) = 30 3.1.4 xE = xG = −7 G is (–7; 3) By K, y = 3: 3=x+2 1=x K is (1; 3) GF = 3 – (–5) = 8 GK = 1 – (–7) = 8 1 Oppervlakte ΔGFK = GF .GK 2 1 = ×8×8 2 = 32 IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 7 van 12 3.2 3.2.1 Laat y = 0 −2 0= −3 x −2 3= x 3x = –2 2  2  x= − OR  − ;0  3  3  3.2.2  2  2 3.2.3 x ∈  − ; 0  OF − ≤ x < 0  3  3 3.3 Asimptoot y = –3 d.w.s. q = –3 = y a.b x − 3 Vervang (0; –2): –2 = a.b0 – 3 1=a Vervang (1; –1): –1 = b1 – 3 2=b d.w.s. y = 2x –3 IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 8 van 12 VRAAG 4 12  0,072  4.1 4.1.1 1 + i eff =  1 + 12    i eff = 0,074424 …… d.w.s. effektiewe koers ≈ 7,44% n  0,072  = 4.1.2 150 000 120 000  1 +  12  n 15  0,072  = 1+ 12  12  5 log 0,072    = n  1+ 12   4    n ≈ 37,3… maande d.w.s. 38 maande 48 6 12 6  0,12   0,14   0,12   0,14  4.2 = × 10  1 +6   1+  − 300 000  1 + 1+ 12   2  I 2  12   2   ≈ R4 331 715,06  6 0,12  36   0,12 12  0,14 6 OF 2 × 10  1 + − 300 000   1 + 1+   12    12   2  ≈ R4 331 715,06 0,12 0,14 i= i= 12 2 4.3 4.3.1 Verminderende saldo 4.3.2 110 940 = 150 000 (1 – i)2 110 940 =1–i 150 000 i ≈ 0,14 Koers is 14% 4.3.3 A = 150 000 (1 – 0,14)7 ≈ R52 189,17 Die boekwaarde van motor na 7 jaar. IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 9 van 12 VRAAG 5 f ( x + h) − f ( x ) 5.1 f ' ( x ) = lim h →0 h −3 ( x + h ) + 1 − ( −3 x + 1) = lim h →0 h −3 x − 3h + 1 + 3 x − 1 = lim h →0 h 3h = lim h →0 h =3 2 x 5 5.2 5.2.1 f ( = x) − x x 1 − = 2 − 5x 2 5 − 32 f '(x) = x 2 y 2 x (1 − x ) 2 5.2.2= = 2x − 4 x 2 + 2x 3 dy ∴ =2 − 8 x + 6 x 2 dx 5.3 f (x = ) 3 x 2 + 13 x f ' ( x= ) 6 x + 13 = mtan = tan 45 ° 1 d.w.s. f ' ( x ) = 1 6 x + 13 1 = 6 x = −12 x = –2 IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 10 van 12 5.4 5.4.1 Vergelyking is y =( x + 2 )( x − 1)( x − 6 ) ( = ( x + 2) x 2 − 7x + 6 ) = x 3 − 7 x 2 + 6 x + 2 x 2 − 14 x + 12 = x 3 − 5 x 2 − 8 x + 12 b = –5; c = –8; d = 12 5.4.2 f'(x) = 3 x 2 − 10 x − 8 By D en E, 3 x 2 − 10 x − 8 =0 ( 3 x + 2 )( x − 4 ) = 0 2 xD = − xE = 4 3 400 yD = y E = −36 27 (Gebruik sakrekenaar)  2 400  D is  − ;  E is ( 4; −36 )  3 27  IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 11 van 12 VRAAG 6 6.1 Teken A = 1 Tyd is tussen t B en t C By B en C, 1 = –t3 + 2t2 t3 – 2t2 + 1 = 0 B C A=1 Laat f (t) = t3 – 2t2 + 1 f (1) = 1 – 2 + 1 = 0 ∴ t – 1 is ’n faktor (t – 1)(t2 – t – 1) = 0 1 ± 1+4 t = 1 of t = 2 1+ 5 t= ≈ 1,6 ( t > 0 ) 2 ∴ tyd is 1,6 – 1 = 0,6 uur = 36 minute 6.2 Omtrek = 2πr + 2L 400 = 2πr + 2L 200 = πr + L 200 – πr = L S = πr 2 + 2r⋅L = πr 2 + 2r (200 – πr) S = 400r – πr 2 ds = 400 – 2πr dr By maksimum, 400 – 2πr = 0 200 r= m π ∫ ( 2x + 3 x − 1) dx −1 2 6.3 3x 2 + 1 = 2 ⋅ ln x + − x +c 2 +1 = 2ln x + x − x + c 3 IEB Copyright © 2019 BLAAI ASSEBLIEF OM

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT: TEGNIESE WISKUNDE: VRAESTEL I – NASIENRIGLYNE Bladsy 12 van 12 6.4 By A en B, 2x 2 − 8x + 6 = 0 x − 4x + 3 = 2 0 ( x − 3 )( ) = x − 1 0 x = 3 of x = 1 B ∴ Oppervlakte ∫f ( x ) = dx A 3 = ∫ ( 2x − 8x + 6 ) dx 2 1 3  2x 3 8x 2  =  − + 6x   3 2 1  2 ( 27 )  2  =  − 4 ( 9 ) + 6 ( 3 ) −  − 4 + 6  3  3  2 = −2 3 2 Oppervlakte = 2 3 Totaal: 150 punte IEB Copyright © 2019