AP Maths P1 MG 2017 hlayiso.com

Downloaded from hlayiso.com GRADE 12 EXAMINATION NOVEMBER 2017 ADVANCED PROGRAMME MATHEMATICS: PAPER I MODULE 1: CALCULUS AND ALGEBRA MARKING GUIDELINES Time: 2 hours 200 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 GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 2 of 9 QUESTION 1 (ln x ) + 2ln x − 3 = 2 1.1 (a) 0 k = ln x ( k + 3 )( k − 1) = 0 ln x = −3 ln x = 1 −3 x= e = 0,0498 x= e= 2,718 (b) x+p = ln q x ≥ −p : x+p= ln q = x ln q − p x ≤ −p : −x − p = ln q x= − ln q − p 1.2 (a) (0; 3) (b) 0 = x 2 + 2x − 3 x 2 ≠ − 2x − 3 since LHS and RHS cannot both be zero simultaneously 3 9 (c)  2; 4  (Critical point is at zero of abs value term)   (d) y = x 2 − 2 x + 3 (Note that other branch does not contain tp) = 2x − 2 0 ∴ x= 1; y= 2 QUESTION 2 2.1 889 = Ae15 k 596 = Ae5 k 889 ∴ e10 k = 596 ∴k = 0,04 ∴A = 488 2.2 6000 = 488e0,04t ∴ t =62,72 ∴ year 2032 (or accept 2033) IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 3 of 9 QUESTION 3 − p ± p2 − 4 p 3.1 x= 2p p( p − 4) < 0 0 4 p=2 3.2 x = −2i is also a solution ∴ x 2 + 4 is a factor ( )( ) x 2 + 4 x 2 − 2 x + 5 = x 4 − 2 x 3 + px 2 − 8 x + 20 ∴ x =1 ± 2i and p=9 3.3 Note that i + i 2 + i 3 + i 4 = 0 Therefore: i + i + i + i + ..... + i 2016 = 2 3 4 0 Therefore answer = i QUESTION 4 Prove true for n = 2: 1 3 2 +1 3 LHS = 1 − = RHS = = 4 4 2(2) 4 Assume true for n = k:  1  1   1  k +1  1 − 4   1 − 9  ...........  1 − k 2  =      2k Prove true for n = k + 1  1  1   1  1  −  4  9  1 1 − ........... −  k2  1  1 −        ( k + 1)  2  k + 1  1  =   1 −  by assumption  2k   ( k + 1)  2  k + 1   ( k + 1) − 1  2 =    2k   ( k + 1)  2 k ( k + 2) = 2k ( k + 1) k +2 = 2(k + 1) But this is the formula with n = k + 1. Therefore, we have proved by PMI that the result is true for all natural values of n. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 4 of 9 QUESTION 5 5.1 (a) = lim f ( = x ) lim 4 4 x →1− x →1 4 = lim+ f ( = 4 x ) lim x →1 x →1 x f (1) = 4 Therefore, continuous at x = 1. Not differentiable due to sudden change of gradient. (b) lim f '( x ) = lim+ f '( x ) x → 2− x →2 −4 =a x2 ∴ a =−1 4 =− x + b x ∴b = 4 5.2 (a) 6x 2 − x − 1 = ( 3 x − 2 )( 2x + 1) + k p=3 (b) (i) f (x) = ( 3 x + 1)( 2x − 1) 2 ( 2 x − 1) Removable discontinuity at x = 0,5. Factor cancels out. 3 1 1 (ii) ∴ y= x+ ; x≠ 2 2 2 ( 3 x − 2 )(12x − 1) − ( 6 x 2 − x − 1) ( 3 ) (c) f '( x ) = (3x − 2) 2 ∴= 0 18 x 2 − 24 x + 5 i .e. ∆ > 0 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 5 of 9 QUESTION 6 6.1 Using the cosine rule: 102 = 102 + 82 − 2 (10 )( 8 ) cosOˆ ∴ Oˆ =1,159 radians (4) 1 (10 ) (1,159 ) = 57,96 units2 2 6.2 Area of sector = 2 1 Area of triangle = ( 8 )(10 ) sin1,159 = 36,656 2 Shaded area = 21,3 units2 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 6 of 9 QUESTION 7 − ( 4x + 3) 2 −1 7.1 y= 3 dy 1 ( 4 x + 3 ) 2 (4) − ∴ = dx 2 2 = ( 4x + 3)2 3 = = m 2 and n 3 2 dy 7.2 (a) cos y − sin x 0 = dx dy sin x ∴ = dx cos y π 1 = (b) x : = sin y+ 1 3 2 1 sin y = 2 π y= 6 π sin dy 3= 1 ∴ = dx cos π 6 tan x + x 2 + 1 7.3 (a) xr += 1 xr − 1 + 2x cos2 x x0 = −1 x1 = −1,31047803... x 2 = −1,227348576... x 3 = −1,181802226... x 4 = −1,172412988... (b) x 5 = −1,172093968... x 6 = −1,172093617... x 7 = −1,1720936... IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 7 of 9 QUESTION 8 8.1 4x3 + 3x 2 0 = ∴ x 2 ( 4x + 3) = 0 = g ''( x ) 12 x 2 + 6 x = 0 6 x (2 x + 1) 1 x = 0 or x = − 2 Therefore x = 0 is stationary (noting sign of gradient does not change) and 1 x = − is non-stationary. 2 8.2 = ∫ 4 x + 3 x dx 3 2 y y = x4 + x3 + C 4 = 1+ 1+ C y = x4 + x3 + 2 8.3 For a cubic, f''(x) is linear and hence f''(x) = 0 always has a solution. For a quartic, f''(x) is quadratic and hence f''(x) = 0 may not have real solutions. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 8 of 9 QUESTION 9 9.1 = (a) RHS sec 2 θ ( tan2 θ + 1) ( = sec 2 θ sec 2 θ ) = sec θ 4 ∫ sec θ .tan θ + sec θ dθ 2 2 2 (b) tan3 θ = + tanθ + C 3 ∫ sin x + cos x + 2 sin x cos x dx 2 2 9.2 (a) = ∫ 1 + sin 2x dx cos 2 x = x − +C 2 1 ( 6 ( x − 2 ) 3 x 2 − 12 x + 5 dx ) 1 ∫ 2 (b) 6 1 ( ) 3 = 3 x 2 − 12 x + 5 + C 2 9 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 9 of 9 QUESTION 10 10.1 The turning point of the graph is (2; 4). At the turning point, the rectangles change from under-approximating to over-approximating so the error cancels out to some extent. 10.2 (a) 2×2=4 (b) h( − x ) =3 2 h( x ) 3 2 ×2+ 2 = 5 units (c) 2–2=0 ( )+ 2 10.3 (a) y = a x − 2p 1 p = ( ) 0 a p4 + p1 2 4 a= − p3 p 4 ( 1 ) 2 (b) Area = ∫ − 3 x − 2p + dx 0 p p ( x − 2p ) p  3  − 4 1  = + x  p3 3 p   0 2 = 3 Total: 200 marks IEB Copyright © 2017