Mathematics Grade 12 IEB Paper 1 2017 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2017 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 © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 2 of 14 SECTION A QUESTION 1 (a) (1) ( x − 1)2 = 2(1 − x ) ( x − 1)2 = −2( x − 1) ( x − 1) + 2( x − 1) = 2 0 ( x − 1)( x − 1 + 2) =0 ( x − 1)( x + 1) = 0 x =1 x = −1 252 x (2) 5− x.5 x −2 = 5 4x 5 5 − x.5 x −2 = 1 5 5 − x + x −2 = 54 x −1 −2= 4 x − 1 1 x= − 4 (b) ( x + 1)2 < 9 Alternative ∴ x 2 + 2x + 1 < 9 ( x + 1)2 < 9 ∴ x 2 + 2x − 8 < 0 x 2 + 2x − 8 < 0 ∴ ( x + 4)( x − 2) < 0 Sketch: y = x 2 + 2 x − 8 Critical Values: –4; 2 x − int : x = −4; x = 2 Solution: {x : −4 < x < 2} Solution: { x : −4 < x < 2} Alternative ( x + 1)2 < 9 ∴ −3 < x + 1 < 3 ∴ −4 < x < 2 (c) ( x − 2)( x + 4) = 0 x + 2x − 8 = 2 0 ∴b = 2 and c = −8 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 3 of 14 −4 (d) (1) x −= 2 − 4 let x − 2 =y x −2 4 y=− − 4 LCD: y y y =−4 − 4 y 2 ∴ y 2 + 4y + 4 = 0 (2) ( y + 2)2 0 = Alternative ∴ y =−2 y 2 + 4y + 4 = 0 Roots are real and equal. ∴∆ = 42 − 4(1)(4) ∴∆ =0 ∴ Roots are real and equal. Alternative −4 x −= 2 −4 x −2 ∴ ( x − 2)2 =−4 − 4( x − 2) ∴ x 2 − 4x + 4 =−4 − 4 x + 8 ∴x =2 0 ∴ Roots are real and equal. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 4 of 14 QUESTION 2 (a) (1) (2) E = {HTT , THT , TTH } 3 ∴ P(2 tails and 1 head) = 8 (b) (1) P (A ∩ B) = 0 (2) (i) You cannot pick a R2 and a R5 coin at the same time. (ii) P(either a R5 or a R2) = P(A or B) = P(A) + P(B) mutually exclusive = 0,36 + 0,47 = 0,83 (c) (1) (2) P(exactly one machine is stamping R5 coins) = 0,3 + 0,2 = 0,5 ∴ 50% IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 5 of 14 QUESTION 3 (a) 480 163 ÷ 0,502 = R956 500 (b) R956 500 × 5% = R47 825 (c) = R956 500 + R47 825 Cost of machinery including import charges = R1 004 325 A P (1 + i ) n = n  9,5  = 1004 325 225 450  1 +   100  n 1004 325  219  = 225 450  200   1004 325  log 219   =n  200   225 450    n = 16,46171594 ∴ n ≈ 16,46 ∴ approx. 17 years (d) (1) Loan required: R1 004 325 − R225 450 R778 = 875 1 − (1 + i )− n  P = x   i    12  ( −4×12)  1 −  1 +   778 875 = x   1200    12   1200    x = R20 510,76607 ∴x = R20 510,77 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 6 of 14 (2) Outstanding Balance= A − F 24  12  =A 778 875  1 +   1200  A = 988 964,5744 A ≈ 988 964,57  12  24   1 +  − 1 F = 20 510,76607   1200    12   1200    F = 553 246,4277 F ≈ 553 246,43 = Outstanding balance 988 964,5744 − 553 246,4277 = R435 718,1467 ≈ R435 718,15 NB: If A and F are rounded to the nearest cent, consider Outstanding= balance 988 964,57 − 553 246,43 = R435 718,14 Alternative   12   −24 1 −  1 +    1200   Outstanding balance = 20 510,76607  12   1200    = R435 718,1466 ≈ R435 718,15 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 7 of 14 QUESTION 4 (a) (1) Constant second difference (2) Tn = an 2 + bn + c T1 = a + b + c = 1 T2 = 4a + 2b + c = 3 T3 = 9a + 3b + c = 6 ∴ 3a + b = 2 and 5a + b = 3 Substitute b= 2 − 3a into 5a + b = 3 ∴ 5a + (2 − 3a ) =3 2a = 1 1 a= 2 1 b= and c = 0 2 1 2 1 ∴Tn= n + n 2 2 (b) T3 = 52 cm T7 = 78 cm T3 =a + 2d =52 T7 =a + 6d =78 1 4d = 26 ∴ d= 6 2 ∴a =39 cm  1 = 39 + 42  6  T43  2 T43 = 312 cm IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 8 of 14 QUESTION 5 (a) (1) f ( x ) = x 2 − 6x + 9 2 xh + h 2 − 6h f '( x ) = lim h →0 h h(2 x + h − 6) f '( x ) = lim h →0 h f '( x=) 2x − 6 (2) f '( −3) = 2( −3) −6 =−12 1 (b) y = π x −1 +3x 3 dy = dx IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 9 of 14 SECTION B QUESTION 6 (a) (1) = Domain x ∈ ; x ≠ 3 (2) Range = y ∈ y ; y ≠ −3 (3) (i) 5 units (ii) 5 units  1 (b) (1) y = a.b x substitute  0;   4 1 = a.b0 4 1 a= 4 1  9 y = b x substitute  2;  4  4 9 1 2 = b 4 4 b2 = 9 ∴ b =±3 but b > 0 ∴ b =3 (2) f: shape intercept domain f –1 shape intercept range 1  (3) Range =  ; ∞  4  1 x (4) f (x) = .3 4 1 y For f −1 : x= .3 ; y ≥ 0 4 4 x = 3y 1 y = log3 (4 x ) for x ≥ 4 (5) See graph in Question 6 (b) (2) above. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 10 of 14 QUESTION 7 (a) f ( x ) = x 2 + 6 x + (3)2 + 5 − (3)2 f ( x ) =( x + 3)2 − 4 ∴ T.P.( −3; −4) (b) (1) x 2 + 6 x + 5 =− x − 5 x 2 + 7 x + 10 = 0 x = −2 or x = −5 A( −5;0) and B( −2; −3) (2) Horizontal shift: ∴ −5 < t < −2 (c) (1) Length MN =( − x − 5) − ( x 2 + 6 x + 5) Length MN =− x − 5 − x 2 − 6 x − 5 For max. length: Let Dx = 0 −2 x − 7 =0 7 x= − 2 2  7  7 ∴ Max. length MN =−  −  − 7  −  − 10  2  2 9 ∴ Max. length MN = units i.e. 2,25 units 4 9 (2) Vertical shift: ∴ k > 4 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 11 of 14 QUESTION 8 3 9 27 (a) (1) ;− ; ; ... 2 2 2 ∴ r =−3 and series is geometric however, series is not convergent since r < −1. x − 3 12 − x (2) = x +3 x −3 ( x − 3)2 = (12 − x )( x + 3) x 2 − 6 x + 9= 12 x + 36 − x 2 − 3 x 2 x 2 − 15 x − 27 = 0 3 x = 9 or x ≠ − 2 1 1 1 (b) = = S4 7 ; = S5 15 and S6 31 2 2 2 T= 5 S5 − S4 T5 = 8 = S6 − S5 T6 T6 = 16 = T5 = ar 4 8 = T6 = ar 5 16 T6 = r= 2 T5 1 ∴a = 2 1 n (2 − 1) Sn = 2 2 −1 1 = 2n−1 − 2 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 12 of 14 QUESTION 9 (a) f (x) = − x 3 + bx 2 + cx − 3 f (1) =−(1)3 + b(1)2 + c (1) − 3 =4 b+c = 8 f '( x ) = −3 x 2 + 2bx + c f ''( x ) = −6 x + 2b  1  1 f ''   = −6   + 2b = 1 2 2 b=2 ∴c = 6 (b) For concave up: f ''( x ) > 0 −6 x + 4 > 0 2 x< 3 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 13 of 14 QUESTION 10 340 340 − = 3 LCD: x ( x + 2) x x+2 340( x + 2) − 340 x= 3 x( x + 2) 3 x 2 + 6 x − 680 =0 = x 14,09 or x ≠ −16,09 Therefore Time = Alternative Let original time taken be represented by y. ∴ xy = 340 ... eq. 1 ( x + 2)( y − 3) =340 ... eq. 2 340 From eq. 1 y= x  340  ∴ ( x + 2)  − 3 =340  x  ∴ 3 x 2 + 6 x − 680 = 0 = ∴ x 14,09 or x ≠ −16,09 Therefore Time = IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 14 of 14 QUESTION 11 (a) (1) When x = −2 and x = 0 (2) 1 3 3 (b) y= x + x +3 5 4 dy 3 2 3 = x + substitute x = 0 dx 15 4 dy 3 = dx 4 3 Equation of tangent: y = x + c where c 3 = 4 3 Equation of tangent: = y x +3 4 For point of intersection between tangent 3 and line BC, substitute= x 2 into=y x +3 4 1  1 ∴y = 4 ∴ Pt  2;4  2  2 1 1 Area of Busi's region =  5 + 3 ×2 2 2 1 =8 units2 2 1 1 Area of Khanya's region =  3 + 4 ×2 2 2 1 =7 units2 2 Therefore Busi's region is larger. Total: 150 marks IEB Copyright © 2017