Mathematics Grade 12 IEB Paper 1 2018 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2018 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 © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 2 of 16 SECTION A QUESTION 1 (a) T100= a + 99d a + 99 ( 7 ) 512 = a = −181 (b) (1) T=1 2 (1) + 3 ∴T= 1 5 ; T2 = 7 ; T3 = 9 ∴ Constant first difference 2 = n (2) = Sn 2 [2(5) + (n − 1)(2)] n = Sn [8 + 2n ] 2 S=n 4n + n 2 Alternate: n = Sn (a + l ) 2 n Sn = ( 5 + 2n + 3 ) 2 S= n n 2 + 4n (c) 2a = 4 ∴a = 2 3a + b = 3 ∴ 3 ( 2) + b = 3 ∴ b =−3 a+b+c = 4 ∴ 2 + ( −3) + c =4 ∴c = 5 Tn = 2n −3n + 5 2 QUESTION 2 1 2 (a) (1) = T1 108 ×   ∴T1 = 72 3 2 2 = T2 108 ×   ∴T2 = 48 3 3 2 (2) = T3 108 ×   ∴T3 = 32 3 4 2 64 = T4 108 ×   ∴T4 = 3 3 520 ∴ First 4 items add up to 3 ∴x = 4 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 3 of 16 Alternative: 2 Geometric sequence with a = 72 and r = 3 =Sn ( a r n −1 ) ; r ≠1 r −1 n 2 16  3  = 81    16  log 2   = n 3  81  n=4 ∴x =4 (b) Area 1 = 2π (21)2 Area 2 = 2π (3)2 2 3 Area 3 = 2π   7 1 Common ratio: indicating a convergent series 49 a = S ∞ ; −1< r < 1 1− r 2π (21)2 S∞ = 1 1− 49 7 203 S∞ = π ∴ S∞ ≈ 2 828,6 cm2 8 QUESTION 3 1 (a) (1) Working with: , undefined for: ( ) x − 3 x − 4 ( x + 1) 2 ( x − 3 x − 4 ) ( x + 1) = 2 0 ( x − 4 )( x + 1)( x + 1) = 0 x = 4 or x = −1 (2) x 2 − 3x − 4 ≤ 0 Critical values: 4 ; − 1 ∴ −1 ≤ x ≤ 4 (b) (1) x+4≥0 ∴ x ≥ −4 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 4 of 16 (2) x +4 −3 =x ( x + 4 ) =( x + 3 ) 2 2 x + 4 = x 2 + 6x + 9 x 2 + 5x + 5 = 0 −b ± b 2 − 4ac x= 2a x ≈ −1,4 or x ≈ −3,6 (n/v) QUESTION 4 [2(1 + h )3 ] − [2(1)3 ] (a) (1) Average Gradient = (1 + h ) − 1 2(1 + h )(1 + 2h + h 2 ) − 2 Average Gradient = h 2(1 + 2h + h 2 + h + 2h 2 + h3 ) − 2 Average Gradient = h 2(1 + 3h + 3h + h3 ) − 2 2 Average Gradient = h (2 + 6h + 6h 2 + 2h3 ) − 2 Average Gradient = h h(6 + 6h + 2h 2 ) Average Gradient = h Average Gradient = 6 + 6h + 2h 2 (2) ( f '(1)= lim 6 + 6h + 2h 2 h →0 ) f '(1) = 6 Alternate: 2( x + h )3 − 2 x 3 (2) f '( x ) = lim h →0 h 2( x + h )(x 2 + 2 xh + h 2 ) − 2 x 3 f '( x ) = lim h →0 h 2( x + 2 x h + h 2 x + x 2 h + 2 xh 2 + h3 ) − 2 x 3 3 2 f '( x ) = lim h →0 h 6 x h + 6h x + 2h 2 2 3 f '( x ) = lim h →0 h h(6 x 2 + 6hx + 2h 2 ) f '( x ) = lim h →0 h f '( x ) = 6 x 2 f '(1) = 6(1)2 ∴ f '(1) 6 = IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 5 of 16 Alternate: f ( x ) = 2x 3 f '( x ) = 6 x 2 f '(1) = 6 1 −2 (b) y = 3x −10x 5 4 dy − = −6 x −3 −2x 5 dx QUESTION 5 60  0,16  (1 + 0,11) − 500000 (1 + 0,11) 10 2 (a) = 300000  1 + 12  A  A = 1269728,917 Alternate: 5×12  16  T0 − T5 : A =300 000  1 +   100(12)  A = 664 142,0648 8  11  T6 − T13 : = A 664 142,0648  1 +   100  At the end of the 13th year: 1 530 540,473 − 500 000 2  11  T14 − T15 : = A 1 030 540,473  1 +   100  At the end of the 15th year he has: R1 269 728,917  (1 + n )n − 1 (b) F = x   i   8  (15×12)   1+ − 1   100(12)   1 270 000 = x  8     100(12)  x = R3 670,114804 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 6 of 16 QUESTION 6 (a) Y-intercept:= y 2(0) + 5 ∴ y-intercept for both graphs: (0 ; 5) For horizontal asymptote for f :substitute ( −1 ; y ) in g ( x ) 2 x = +5 ∴ g ( −1) = 2( −1) + 5 ∴ g ( −1) = 3 ∴ Horizontal asymptote of f: y = 3 a = f( x) + 3 substitute (0 ; 5) x +1 a =5 + 3 ∴a = 2 0 +1 a=2= = ; b 1 and c 3 2 5 (b) (1) X-intercept of f : 0 = + 3 ∴ x =− x +1 3 5 X-intercept of g : 0 =2 x + 5 ∴ x =− 2 5 5 (2) − ≤ x < −1 or x ≤ − 3 2 (c) (1) g ( x= ) 2x + 5 = 2y + 5 x 1 5 y= x − 2 2 x −5 (2) Point of intersection: 2 x + 5 = ∴ x =−5 2 The values of x for which g −1( x ) > g ( x ) : x < −5 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 7 of 16 SECTION B QUESTION 7 (a) x= 5 ± 2 ( ) ( ∴ x − 5 + 2  x − 5 − 2  =   )  0 x 2 − 5 x + 2 x − 5 x − 2 x + 23 = 0 x 2 − 10 x + 23 0 = (b) For real and equal roots: Quadratic must be a perfect square ∴ x 2 + ax + b = 0 (x + b ) 0 2 = x 2 + 2 bx + b = 0 ∴a = 2 b 2 ( ) a 2 ∴ b = 2   a2 ∴ b = … eq. 1 4 x 2 + bx + a = 0 (x + a ) 0 2 = x 2 + 2 ax + a = 0 ∴b = 2 a … eq. 2 Substitute eq1 in eq 2: a2 =2 a 4 3 ∴a2 = 23 2  3 3 2 ∴a2  = 23 3 ( )   ∴a = 4 and b = 4 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 8 of 16 Alternate: 2 For real and equal roots, ∆= b − 4ac= 0 For x 2 + ax + b = = a 2 − 4b 0: 0 a2 ∴ b = … eq1 4 For x 2 + bx + a = = b 2 − 4a … eq2 0: 0 Substitute eq1 in eq 2: 2  a2    − 4a 0 =  4 a 4 − 64a 0 = a(a − 64) 0 3 = =a 0 or= a 4 ∴a = 4 only and b = 4 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 9 of 16 QUESTION 8 (a) = A P (1 + i )n y= 2 y (1 + i )x x  200  = y 2 y 1+   100  y = (3)x y (b) x (c) (1) 750 = (3)x x = log3 750 x ≈ 6,03 It took approximately 6 years (2) Domain: x > 6 (accept: x ≥ 6 ) IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 10 of 16 QUESTION 9 (a) For point of inflection: Let g "( x ) = 0 = g '( x ) 3x 2 − 6x g "( x=) 6x − 6 6x − 6 = 0 ∴x= 1 g (1) −2 = and h(1) −2 = Hence, g and h intersect at x = 1 , the point of inflection. Alternate: For point of inflection: Let g "( x ) = 0 = g '( x ) 3x 2 − 6x g "( x=) 6x − 6 6x − 6 = 0 ∴x= 1 2 4 Point of intersection: x 3 − 3 x 2 = − x− 3 3 3 x − 9 x + 2x + 4 = 3 2 0 3 ± 21 x= or x = 1 3 Therefore, the graph of h does intersect the graph of g at its point of inflection. Alternate: For point of inflection: Let g "( x ) = 0 = g '( x ) 3x 2 − 6x g "( x=) 6x − 6 6x − 6 = 0 ∴x= 1 For co-ordinate of point of inflection: = (1)3 − 3(1)2 Substitute x = 1 in f (1) f (1) = −2 2 4 Substitute (1; −2 ) in y = − x− 3 3 2 4 RHS −=(1) − 3 3 RHS = −2 RHS = LHS Therefore, the graph of h does intersect the graph of g at its point of inflection. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 11 of 16 (b) (1) For stationary point of y = g '( x ) = y 3x 2 − 6x dy = 6x − 6 dx 6x − 6 = 0 ∴x = 1 Startionary point (1; −3 ) Min. value function (2) (i) Concave down for: x < 1 (ii) g='(1) 3(1)2 − 6(1) g '(1) = −3 (3) Decreasing gradient occurs for: 0 < x < 2 Maximum decreasing gradient occurs at the point of inflection. (c) The graph of g decreases for the interval: 0 < x < 2 We must shift the graph of g , 3 units to the left ∴k = 3 QUESTION 10 (a) (1) Since b > 2a , then b 2 > 4a 2 Since c < a Then b 2 > 4ac Alternate: b > 2a and b > c (b > 2a > a > c ) , hence b 2 > 4ac (2) y x From the given constraints: a , b and c are positive(+) Therefore: IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 12 of 16 Shape: min. value function Y-int. : + Axis of Symm.: x = negative value b 2 − 4ac > 0 ∴ roots are real and unequal y (b) (1) g x 1 Horizontal asymptote: y = − 2 Vertical asymptote: x = −2 1 (2) p≥− 2 QUESTION 11 2 1 (a) (1) P(both letters are C)= × 6 5 1 = 15 2 4 4 2 (2) P(only one letter is C)  ×  +  ×  6 5 6 5 8 = 15 6! (b) = 360 2! (c) 4! = 24 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 13 of 16 QUESTION 12 Let the number of missiles required for firing be n. P(all will miss)= (1 − 0,9)n ∴ P(all will miss) = 0,1n P(at least 1 will hit) = 1 − 0,1n We require: 1 − 0,1n > 0,97 When n = 1, 1 − 0,11 = 0,9 When n = 2, 1 − 0,12 = 0,99 When n = 3, 1 − 0,13 = 0,999 Therefore, at least 2 missiles should be fired. Therefore, Lulu was correct. Alternate: Let the number of missiles required for firing be n. P(all will miss)= (1 − 0,9)n ∴ P(all will miss) = 0,1n Let: 1 − 0,1n = 0,97 0,03 = 0,1n log0,1 0,03 = n n ≈ 1,5 Therefore, at least 2 missiles need to be fired to ensure at least a 0,97 chance of hitting the target. Lulu was correct Alternate: First missile fired: P(a hit) = 0,9 Second missile fired: P(a hit) = 0,9 + MH = 0,9 + (0,1 x 0,9) = 0,99 Third missile fired: P(hit) = 0,9 + MH + MMH = 0,9 + (0,1 x 0,9) + (0,1 x 0,1 x 0,9) = 0,999 Hence Lulu was correct IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 14 of 16 Alternate: 2 Missiles fired: P(a hit) = 1 – P(no hit) = 1 - P(MM) = 1 - (0,1 x 0,1) = 0,99 3 missiles fired: P(a hit) = 1 – P(no hit) = 1 - P(MMM) = 1 - (0,1 x 0,1 x 0,1) = 0,999 Hence Lulu was correct. Alternate: P(M1 U M2) = P(M1) + P(M2) – P(M1 ∩ M2) = 0,9 + 0,9 – (0,9 x 0,9) = 0,99 Similarly, if 3 missiles are fired: P(M1 U M2 U M3) = 0,999 Hence Lulu was correct IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 15 of 16 QUESTION 13 3 y= − x +3 2 1 Area ∆OMN = b.h 2 1  3  Area ∆OMN = x  − x + 3 2  2  3 3 Area ∆OMN − =x 2 + x 4 2 dA For max. value x1 : =0 dx 3 3 0 − = x1 + 2 2 ∴ x1 =1 3 f ( x ) = rx 2 + bx + c where r = − 4 3 − x + 3 …. By inspection, b = 3 From: f '( x ) = 2 3 f (x) =− x 2 + 3x + c 4 Stationary point ( x;5 ) X-Intercept if f ’(x) represents x-coordinate of the Stationary Point ∴ Stationary Point (2;5) 3 Substitute (2;5) in f ( x ) = − x 2 + 3x + c 4 3 5 −=(2)2 + 3(2) + c 4 c=2 For value of x2 that give max. distance (S) between f and f': 3 2  3  S =− x + 3x + 2 −  − x + 3  4  2  3 2 9 S= − x + x −1 4 2 dS =0 dx 3 9 − x2 + = 0 2 2 x2 = 3 They differ. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 16 of 16 Alternate: 3 y= − x +3 2 1 Area ∆OMN = b.h 2 1  3  Area ∆OMN = x  − x + 3 2  2  3 2 3 Area ∆OMN − =x + x 4 2 dA For max. value x1 : =0 dx 3 3 0 − = x1 + 2 2 ∴ x1 =1 3 f ( x ) = rx 2 + bx + c where r = − 4 3 − x + 3 …. By inspection, b = 3 From: f '( x ) = 2 3 f (x) =− x 2 + 3x + c 4 Stationary point ( x;5 ) X-Intercept if f ’(x) represents x-coordinate of the Stationary Point ∴ Stationary Point (2;5) 3 Substitute (2;5) in f ( x ) = − x 2 + 3x + c 4 3 2 5 −=(2) + 3(2) + c 4 c=2 For value of x2 that give max. distance (S) between f and f': 3 2  3  S( x ) =− x + 3x + 2 −  − x + 3  4  2  3 9 S( x ) =− x2 + x − 1 4 2 S(1) = 2,75 and S(2) = 5 , Hence maximum distance is not at x = 1 Total: 150 marks IEB Copyright © 2018