Technical Mathematics Grade 12 Paper 1 Gauteng Mock 2018 Memo

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION 2018 MARKING GUIDELINES TECHNICAL MATHEMATICS ( PAPER 1) 16 pages

Downloaded from hlayiso.com 11091 / 18 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION MATHEMATICS (Paper 1) MARKING GUIDELINES QUESTION 1 1.1.1 ( )( ) factorise   (3) 1.1.2 ( )( ) Critical values: –5 and 3 critical values  –  – represented on –5 5 3 5 number line answer (3)

Downloaded from hlayiso.com 11091 / 18 1.1.3 ( x – 3)(x + 5) = 5 x 2 + 2x – 15 = 5 x 2 + 2x – 15 x 2 + 2x – 20 = 0 x 2 + 2x – 20 = 0 a = 1 ; b = 2 ; c = – 20 √ x = substitution into √ ( )(– ) correct formula = ( ) √ = √ = ( √ ) = =– √ x = – 5,58 x = 3,58 (5) x = – 5,58 or x = 3,58 1.2 – – = 1,5 1,5 (1) 1.3 x + y = 4 ……………(1) 2x 2 + y = 4 + 5x …… (2) Substitute (1) into (2): 2x 2 + y = x + y + 5x substitution 2x 2 + y – 6x – y = 0 2x 2 – 6x = 0 standard form 2x(x – 3) = 0 factorise x=0 or x =3 x – values y=4 y=1 y – values 3

Downloaded from hlayiso.com 11091 / 18 OR OR From (1) make x the subject of the formula ; i.e: x = 4 – y ……….(1.1) x=4–y Substitute (1.1) into (2): 2(4– y)2 + y = 4– y + y + 5(4 – y) substitution 2(16 – 8y + y2 ) + y = 4 + 20 – 5y 32 – 16y + 2y2 + y –24 + 5y = 0 2 –10y + 8 = 0 standard form y2 – 5y + 4 = 0 (y – 1)(y – 4) = 0 factorise y = 1 or y = 4 x – values x =3 x =0 y – values OR OR From (1) make y the subject of the formula ; i.e: y = 4 – x……….(1.1) y=4–x Substitute (1.1) into (2): 2x 2 + 4 – x = x + 4 – x + 5x substitution 2x 2 – 6x = 0 standard form 2x(x – 3) = 0 factorise x=0 or x=3 x – values y=4 y=1 y – values (6) 1.4.1 T(D + 4d) = 12,5D  LCD TD + 4dT = 12,5D 4dT = 12,5D – TD transpose answer (3) 1.4.2 ( – ) substitution d= ( ) d=2 answer (2) 4

Downloaded from hlayiso.com 11091 / 18 1.5.1 ( ) = m2 + 2mn + n2 = m2 + n2 + 2mn  m2 + n2 + 2mn = 41 + 2(9) substitution = 41 + 18 = 59 (2) 1.5.2 25 24 23 22 21 20 method 5910 1 1 1 0 1 1 1110112 5910 = 1110112 If base 2 is omitted and all else is correct then penalty of 1 is applied. Answer only: 0 (2) [27] QUESTION 2 2.1.1 x=0 x=0 (1) 2.1.2  answer (2) 2.1.3 x = 8 or x = 12 answer (1) 2.2 (a = 6 ; b = – 6 ; c = 1) ( ) ( )( ) substitution  the roots are real, irrational and unequal irrational and unequal (3) [7] 5

Downloaded from hlayiso.com 11091 / 18 QUESTION 3 3.1.1 2 log 2 + log 25 = 2 log 2 + 2 log 5 2 log 5 = 2 log 10 2 log 10 =2 2 OR OR 2 log 2 + log 25 = log 4 + log 25 log 4 = log 100 log 100 =2 2 (3) 3.1.2 ( – ) = common factor 52004 ( ) ( ) =  = 24 24 (3) 3.2.1 = 10  =5 =5 x = 54 or x = 625 (2) 6

Downloaded from hlayiso.com 11091 / 18 3.2.2 ( )– ( )  ( ) ( )  ( ) 5x + 2 = 3x 2 3x 2 – 5x – 2 = 0 (3x + 1)(x – 2) = 0 (3x + 1)(x – 2) = 0 –  or –  (5) NA 3.3.1 Modulus of z = | | = r = √ substitution in correct formula = √(– ) ( ) =√ answer (2) = 13 3.3.2 Argument of z = arg(z) = ( ) substitution into correct formula = ( ) Ref. angle = 67,38o Ref. angle – answer (3) 7

Downloaded from hlayiso.com 11091 / 18 3.3.3 z = 13cis112,62o z in any one of the three acceptable notations OR z = 13 (cos 112,62o + i sin 112,62o ) OR z = 13 (1) 3.3.4 (x – y) + 2i = 0 + (y – 3x)i ……………..(1) Comparing real parts of (1): x–y=0 comparing real parts Then: x = y …………(2) Comparing imaginary parts of (1): 2 = y – 3x……………(3) comparing imaginary parts Substitute (2) into (3): 2 = (y – 3y) 2 = – 2y – A x = – 1 – A y = – 1 (4) [23] 8

Downloaded from hlayiso.com 11091 / 18 QUESTION 4 B(–3 ; 3) g  g y=2 g 4.1.1 k =2 k = 2 For a: Substitute B(–3 ; 3) into g(x) = +2 substitution of i.e.: 3= point B into g 3–2= 1= a = – 3 (3) – 4.1.2 r =√ r=√ (1) 4.1.3 x –√ √ ] correct inequality signs OR correct values –√ x √ (2) 4.1.4   OR y (– ) (2; ) OR { y: } {y: 2 < y < } (2) 9

Downloaded from hlayiso.com 11091 / 18 4.2.1a y= ( ) y= (1) 4.2.1b Let y = 0: – y = 0 –x 2 + 2x + 3 = 0 x 2 – 2x – 3 = 0 factors (or substitution into (x – 3)(x + 1) = 0 formula) x = 3 ; x = 3 or x = – 1 x = – 1 (4) 4.2.1c Axis of symmetry: – = – correct method (– ) = 1 x = 1 f(1) Max. value: f(1) = – (1)2 + (1) + 1,5 = 2 y = 2 Turning point is (1 ; 2) (4) 4.2.1d y=–1 y = – 1 (1) 4.2.1e Let y = 0 , then  x = 0 (2) 10

Downloaded from hlayiso.com 11091 / 18 4.2.2 For f: y x – and y – intercept (1 ; 2) turning point shape 𝑓 For g: x x –intercepts –1 0 3 shape 𝑔 y = –1 asymptote (6) 4.2.3 See graph in 4.2.2 correct shaded area (1) [27] QUESTION 5 5.1 ( ) substitution in correct ( ) formula  Effective interest rate = 9,25% p.a. 9,25% (3) 5.2 A = P(1 – i)n 179 200 = 350 000(1 – i)3 substitution into correct formula 0,512 = (1 – i)3  = 0,512 √ –1=–i –0,2 = –i rate = 20% p.a. Annual rate of depreciation is 20% p.a. (3) 11

Downloaded from hlayiso.com 11091 / 18 5.3 For initial deposit: For A1 : i = and n = 6 A1 = 20 000(1 + )6 correct substitution = R 23 109, 142  R 23 109, 142 After second deposit: A2 = A1 + R 10 000 = R 33 109,142 After 3 months, new balance: For A3 : A3 =A2 (1 + i)n = 33109,142(1+ )1 correct substitution 33 916,17734 = R 33 916,17734 After a further 15 months, i.e. at the end of the For A4 : three (3) years: A4 = A3 (1 + i)n i = and n = 15 = 33 916,17734(1 + )15  R 38 388,36 = R 38 388,36 (7) [13] 12

Downloaded from hlayiso.com 11091 / 18 QUESTION 6 6.1 Average Gradient = ( ) – (– ) = –(– ) ( ) – ( )– ( ) (– ) = substitution into correct formula ( ) 21 = = 33 –2 (4) = –2 6.2 ( ) (x + h) = 2(x +h) + 1  (x +h) = = 2x + 2h + 1 2x + 2h + 1 ( ) ( )– ( ) substitution into correct formula – – ( )  answer (4) = NB: incorrect formula – Breakdown: 0/4 ( ) 6.3 y =( – )      (5) NB: Incorrect notation: Max 13

Downloaded from hlayiso.com 11091 / 18 6.4 –  – Gradient of tangent = 2 (Because = –1) gradient of tangent 2x – 4 = 2 i.e: 2x – 4 = 2 2x = 6 x = 3 (4) x=3 [17] QUESTION 7 7.1 f (2) = 23 –2(2)2 – 4(2) + 8 f (2) =0 = 0 (x–2) is a factor of f(x) (2) 7.2 f (x) = x 3 – 2x 2 – 4x + 8 first factor = x 2 (x – 2) – 4(x – 2) (x – 2) = (x – 2)(x 2 – 4) quadratic factor f (x) = (x –2)2 (x + 2) (x 2 – 4) f (x) = (x – 2)(x – 2)(x + 2) (x 2 – 4) = (x – 2)(x – 2) OR f (x) = (x – 2)(ax 2 + bx – 4) = ax 3 + bx 2 – 4x – 2ax 2 – 2bx + 8 = ax 3 + (b – 2a)x 2 – 2bx – 4x + 8 x 3 – 2x 2 – 4x + 8 = ax 3 + (b – 2a)x 2 – (2b + 4)x + 8 ..(1) Comparing coefficients in (1) above: a =1 –2b – 4 = – 4 b=0 f (x) = (x – 2)(ax 2 + bx – 4) = (x – 2)(x 2 – 4) f (x) = (x – 2)2 (x + 2) (3) 14

Downloaded from hlayiso.com 11091 / 18 7.3 f (x) = x 3 – 2x 2 – 4x + 8 ( ) – –  – – =0 For turning points: ( ) i.e: (3x + 2)(x –2) = 0 (3x + 2)(x –2) = 0 x = or x = 2 both x – values f (2) = 0 (2 ; 0) f (2) = 0 f( )= ( ; ) f ( ) = (5) 7.4 y shape ( ) x – intercepts y – intercept (0 ; 8) turning points 8 x –2 0 2 (4) [14] 15

Downloaded from hlayiso.com 11091 / 18 QUESTION 8 8.1.1 Perimeter = 2l + 2b formula of perimeter 30 = 2x + 2y 30 = 2x + 2y y = 15 – x (2) 8.1.2 Area of rectangle = length breadth substitute y = 15 – x = x (15 – x) x (15 – x) or 15x – x 2 A(x) = 15x – x 2 (2) 8.1.3 ( ) –  ( ) – Maximum area: ( ) : 15 = 2x x = 7,5 and x = 7,5 y = 7,5 y = 7,5 (3) 8.1.4 Max. Area = 7,5(7,5) = 56,25m2 56,25m2 (1) 8.2.1 2 substitution into correct formula r= r = (2) 8.2.2 Area of circle = Substitute r into correct formula = ( ) = Area of circle = 71,66 m2 Area of circle = 71,66 m2 (2) 8.3 The circle. Correct choice (1) [13] 16

Downloaded from hlayiso.com 11091 / 18 QUESTION 9 9.1 ∫( )  =∫ ( )    ( ) (4) 9.2 Distance travelled = ∫ ( +5)dt ∫ ( +5)dt  =  =[ ( ) ( ) – substitution ( ) [ ( ) – =  (5) [9] TOTAL: 150 17