Advanced Programme Maths P1 Memo 2018 hlayiso.com

Downloaded from hlayiso.com GRADE 12 EXAMINATION NOVEMBER 2018 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 © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 2 of 12 QUESTION 1 1.1 (a) x2 + x =−2 x − 2 ∴ x2 + x = −2 x − 2 or − x 2 − x = −2 x − 2 ∴ x 2 + 3x + = 2 0 or x 2 − x − = 2 0 ∴ x =−1 or − 2 or x =−1 or 2 but a check reveals x − = 1 or − 2 (b) ln x 3 + 2ln x 2 7 = ∴ ln x 3 + ln x 4 7 = ∴ ln x 7 7 = ( can go straight to the answer from here ) ∴ 7ln x 7 = ∴ ln x 1 = ∴x = e 1.2 (a) y = y 0e − kt y ∴ e − kt = y0 y ∴ −kt = ln y0 y ln y ∴k = 0 −t 0,5 y 0 ln y0 (b) k= ≈ 1,216 × 10 −4 −5700 (c) 0,9 y 0 = y 0e − kt ∴ −kt = ln0,9 ln0,9 ∴t = −k ∴ t ≈ 866 years IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 3 of 12 QUESTION 2 2.1 if 3 + 2i is a root then so is 3 − 2i so our equation is : ( x + 3 ) ( x − ( 3 + 2i ) ) ( x − ( 3 − 2i ) ) = 0 ∴ ( x + 3 ) ( ( x − 3 ) − 2i ) ( ( x − 3 ) + 2i ) = 0 ( ∴ ( x + 3 ) ( x − 3 ) − 4i 2 = 2 0 ) ( ∴ ( x + 3 ) x 2 − 6 x + 13 = 0 ) ∴ x 3 − 3 x 2 − 5 x + 39 = 0 2.2 A cubic equation will have three roots. Complex roots of polynomials with real coefficients occur in conjugate pairs so there must be at least one real root. a + bi −b − ai 2.3 × −b + ai −b − ai −ab − a 2i − b 2i − abi 2 = b2 − a2i 2 ab − ab − i ( a 2 + b 2 ) = b2 + a2 −i ( a 2 + b 2 ) = b2 + a2 = −i IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 4 of 12 QUESTION 3 = = if n 1 we have 23 − 3 5 which is clearly divisible by 5. Assume true for n = k viz. that 2= 3k − 3k 5 p where p ∈  (*) Now if n= k + 1 we have : 2( 3 k +1) − 3k +1 = 23 k +3 − 3k +1 = 23 k × 23 − 3 × 3k = 8 × 23 k − 3 × 3k from ( * ) we have 23 k 5 p = + 3k so if n k + = 1 we have ( ) = 8 ( 5 p ) + 8 3k − 3 × 3k = 5 (8p ) + 5 (3 ) k = 5 (8p + 3 ) k which is clearly divisible by 5 a so, by the Principle of Mathematical Induction we have proved the result for n ∈  IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 5 of 12 QUESTION 4 4.1 (a) (b) x =0 4.2 if f is differentiable at x = 2 it must be continuous at x = 2 so lim− f ( x ) = lim+ f ( x ) x →2 x →2 so 2a − b − 1= 4b − 2a + 5 or 4a − 5b= 6 (1) but also, lim f ' ( x ) = lim+ f ' ( x ) x → 2− x →2 = a 4b − a or = a 2b ( 2) solving (1) and ( 2 ) simultaneously = 8b − 5b 6 = so b 2 = and a 4 QUESTION 5 5.1 = sector − ∆ segment ∴ 308 = 1 2 ( ) 1 182 θ − 182 sinθ 2 ( ) ∴162θ − 162sinθ − 308 0 = 5.2 f (θ ) = 162θ − 162 sinθ − 308 162θ − 162sinθ − 308 ∴θ n +1 =θ n − 162 − 162cos θ θ = 2,49984 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 6 of 12 QUESTION 6 1  1 6.1 f ( 0 ) = , so y - int  0; 2  2   2x − 3 x − 2 2 =0 x−4 ∴ 2x 2 − 3 x − 2 =0 ∴ ( 2 x + 1)( x − 2 ) = 0  1  ∴ x - ints  - 2 ;0  and ( 2;0 )   6.2 vertical asymptote : x = 4 2 x 2 − 3 x − 2 = ( x − 4 )( 2 x + 5 ) + R so, oblique asymptote is y = 2x + 5 2x 2 − 3 x − 2 6.3 f (x) = x−4 ( 4 x − 3 )( x − 4 ) − 1( 2x 2 − 3 x − 2 ) =∴f '(x) = 0 ( x − 4) 2 ∴ 4 x 2 − 19 x + 12 − 2 x 2 + 3 x + 2 =0 ∴ 2 x 2 − 16 x + 14 = 0 ∴ x 2 − 8x + 7 =0 ∴ ( x − 1)( x − 7 ) = 0 ∴x = 1 or 7 ∴ (1;1) and ( 7;25 ) are stationary points 6.4 f '' (1) < 0 so (1;1) is a local maximum f '' ( 7 ) > 0 so ( 7;25 ) is a local minimum IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 7 of 12 QUESTION 7 7.1 x 2 + xy + y 2 1 = dy dy ∴ 2x + y + x + 2y = 0 dx dx dy ∴ ( x + 2y ) = −2 x − y dx dy −2 x − y ∴ = dx x + 2y 7.2 At A,y = 0 ∴ x =1 dy −2 so, at A, = = −2 dx 1 ∴ y =−2 ( x − 1) ∴ y =−2 x + 2 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 8 of 12 QUESTION 8 FC 8.1 cos θ = CD 0,4 cos θ ∴ FC = ∴A D=CDF + DBEG + BCFG 1  ∴A =2  × 0,4 × 0,4 cos θ sinθ  + 0,4 × 0,4 cos θ ( DCDF =DBEG ) 2  =∴ A 0,16 sinθ cos θ + 0,16 cos θ =∴ A 0,08 sin 2θ + 0,16 cos θ =∴V 20 ( 0,08 sin 2θ + 0,16 cos θ ) =∴V 1,6 sin2θ + 3,2cosθ = 8.2 V 1,6 sin2θ + 3,2cosθ dV ∴= 3,2cos 2θ − 3,2sin=θ 0 dθ ∴ cos 2θ sin = θ π  ∴ cos 2θ = cos  − θ  2  π ∴ 2θ = −θ 2 π ∴ 3θ = 2 π ∴θ = 6 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 9 of 12 QUESTION 9 9.1 (a) sin= 3 θ sinθ × sin2 θ = sinθ (1 − cos2 θ ) qqq = sin − sin cos2 as required ∫ sin θ = dθ ∫ sinθ dθ − ∫ sinθ cos θ dθ 3 2 (b) cos3 θ = − cos θ + +c 3 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 10 of 12 9.2 METHOD 1 x ∫ 2 + x dx let u 2+ x = then x u −2 = and = du dx u−2 ∴ ∫ 1 du u2 1 1 − = ∫ u 2 − 2u 2 du 2 32 1 = u − 4u 2 + c 3 3 1 2 = (2 + x )2 − 4 (2 + x )2 + c 3 ALTERNATIVE 1 x ∫ 2 + x dx 1 ∫ x ( 2 + x ) 2 dx − = 1 (2 + x ) 2 − by parts f= x and g=' 1 g 2 (2 + x )2 =' 1 and = so f 1 1 = 2 x ( 2 + x ) 2 − ∫ 2 ( 2 + x ) 2 dx 3 1 4 (2 + x )2 = 2x ( 2 + x ) − 2 +c 3 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 11 of 12 QUESTION 10 10 3 1 10.1 Area = + + 3 2 ( 4 ) 6 42 ( ) 119 = (3) 32 10.2 It will be an over-approximation. As n gets larger the answer decreases.  10 3 1  10.3 = lim  Area + + 2  n →∞  3 2n 6n  10 = units 2 3 10 10.4 units 2 since this is simply a reflection of the shaded area in the y-axis. 3 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 12 of 12 QUESTION 11 7 = Area ∫ g ( x ) − f ( x ) dx 1 7 = ∫ f ( x ) + kx + 1 − f ( x ) dx 1 7 = ∫ kx + 1 dx 1 7  kx 2  =  + x  2 1 49k k so, + 7 − −1= 54 2 2 so, 49k + 14 − k − 2 = 108 so, 48k = 96 so k = 2 QUESTION 12 b vol = π ∫ f ( x )  dx 2 a b ∴175 = π ∫ − x 2 + 6 x + 4 dx a h  x3  ∴175 = π  − + 3x 2 + 4x   3 0  h3  ∴175 = π  − + 3h 2 + 4h   3  525 ∴ −h3 + 9h 2 + 12h − =0 π ∴x =5,28 cm Total: 200 marks IEB Copyright © 2018