Technical Mathematics Grade 12 IEB Paper 1 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2019 TECHNICAL MATHEMATICS: PAPER I MARKING GUIDELINES Time: 3 hours 150 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 2 of 12 QUESTION 1 1.1 1.1.1 6 – x2 + x = 0 x2 – x – 6 = 0 (x – 3)(x + 2) = 0 x = 3 or x = –2 1.1.2 f(x) > 0 for x 𝜖𝜖 (–2; 3) x 𝜖𝜖  OR using number line OR –2 < x < 3 1.2 3y = 34 x OR y = x2 – 6x + 9 ……… (2) ∴y = 4x ……… (1) Subst (1) → (2): 4x = x2 – 6x + 9 0 = x2 – 10x + 9 0 = (x – 9)(x – 1) x = 9 or x = 1 Subst in (1) y = 36 or 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 OR x= 2 ± 2i 1.3.2 − x 2 + 4 x − 1 = 5 + k OR 0 = x − 4x + 6 + k 2 ∆ = ( −4)2 − 4(1)(6 + k) = 16 − 24 − 4k = –4k – 8 Fw 2 real, diff roots –4k – 8 > 0 by insp from graph –4k > 8 g(x) + k will meet f twice if k < –2 5+k<3 k < –2 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 3 of 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 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 4 of 12 QUESTION 2 4   x +2 > 0 3 ( x + 2) 4 = 3 3 4 2.1  ( x + 2 ) 4  ( = 27 ) 3 x > –2 OR 27   4   3 3 4  ( x + 2) 4  ( 27 )3 1 = ( x + 2) 3 = 4   x + 2 = 34 x + 2 = 81 x = 79 Valid 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 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 5 of 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°) OR (5 cos 233,13° ; 5 sin 233,13°) OR p = +5 cos 233,13° + 5 sin 233,13° OR p = 5 cis 233,13° OR p = 5 233,13° IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 6 of 12 QUESTION 3 3.1 3.1.1 At B: x + 2 = 0 x = –2 B is (–2; 0) 3.1.2 Roots are – 12 and – 2 Eqn is y =a ( x + 12 )( x + 2 ) Subst (–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 (by symmetry) OR f' (x) = 0 –2x + 14 = 0 y E =− ( −7 ) − 14 ( −7 ) − 24 2 x=7 =−49 + 98 − 24 = 25 xF = xE = −7 : Subst. 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) At K, y = 3: 3=x+2 1=x K is (1; 3) GF = 3 – (–5) = 8 GK = 1 – (–7) = 8 1 Area ΔGFK = GF .GK 2 1 = ×8×8 2 = 32 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 7 of 12 3.2 3.2.1 Let 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  OR − ≤ x < 0  3  3 3.3 Asymptote y = –3 i.e.: q = –3 = y a.b x − 3 Subst (0; –2): –2 = a.b0 – 3 1=a Subst (1; –1): –1 = b1 – 3 2=b i.e.: y = 2x –3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 8 of 12 QUESTION 4 12  0,072  4.1 4.1.1 1 + i eff =  1 + 12    i eff = 0,074424 …… i.e.: effective rate ≈ 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 … months i.e. 38 months 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 OR 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 Reducing balance OR diminishing balance 4.3.2 110 940 = 150 000 (1 – i)2 110 940 =1–i 150 000 i ≈ 0,14 Rate is 14% 4.3.3 A = 150 000 (1 – 0,14)7 ≈ R52 189,17 The book value of car after 7 years. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 9 of 12 QUESTION 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 i.e.: f ' ( x ) = 1 6 x + 13 1 = 6 x = −12 x = –2 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 10 of 12 5.4 5.4.1 Eqn 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 At D and 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 (Use calculator)  2 400  D is  − ;  E is ( 4; −36 )  3 27  IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 11 of 12 QUESTION 6 6.1 Draw A = 1 time is between t B and t C At B and C, 1 = –t3 + 2t2 t3 – 2t2 + 1 = 0  B C A=1 Let f (t) = t3 – 2t2 + 1 f (1) = 1 – 2 + 1 = 0 ∴ t – 1 is a factor (t – 1) (t2 – t – 1) = 0 1 ± 1+4 t = 1 or t = 2 1+ 5 t= ≈ 1,6 ( t > 0 ) 2 ∴ time is 1,6 – 1 = 0,6 hours = 36 minutes 6.2 Perimeter = 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 At max, 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

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER I – MARKING GUIDELINES Page 12 of 12 6.4 At A & B, 2x 2 − 8x + 6 = 0 x − 4x + 3 = 2 0 ( x − 3 )( ) = x − 1 0 x = 3 or x = 1 B ∴ Area ∫ 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 Area = 2 3 Total: 150 marks IEB Copyright © 2019