Advanced Programme Maths P1 Memo 2019 hlayiso.com

Downloaded from hlayiso.com GRADE 12 EXAMINATION NOVEMBER 2019 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 © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 2 of 15 QUESTION 1 1.1 Solve for x ∈  without using a calculator and showing all working: (a) Solve x 2 − 12 = x ∴ x 2 − 12 = x or x 2 − 12 = −x ∴ x2 − x − = 12 0 or x 2 + x − = 12 0 ∴ ( x − 4 )( x + = 3 ) 0 or ( x + 4 )( x − = 3) 0 ∴= x 4 or − 3 or − 4 or 3 A check reveals x = 4 or 3 (b) e x + 12e − x 8 = ∴ e 2 x − 8e x + 12 = 0 ( ∴ ex − 2 ex − 6 = 0 )( ) ∴e = x = 2 or e x 6 = ∴ x ln 2 or = x ln 6 ∴x =0,693 or 1,792 1.2 If z= a + bi and z= 2 23 − 6z then find all possible values of a and b. ( a + bi ) =23 − 6 ( a + bi ) 2 ∴ a 2 + 2abi + b 2i 2 = 23 − 6a − 6bi ( ) ∴ a 2 − b 2 + ( 2ab ) i = ( 23 − 6a ) − ( 6b ) i ∴ 2ab − = 6b ∴ a =−3 or b =0 ( also a 2 − b 2 =23 − 6a ) so, if a = −3 then 9 − b 2 = 42 ∴ b2 = −32 (not possible since b is real ) ∴b =0 ∴ a 2 = 23 − 6a ∴ a =−3 + 4 2 = ∴ a 2,66 or − 8,66 and = b 0 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 3 of 15 1.3 Solve f ( x ) = x 4 + x 3 − 2 x 2 + 2 x + 4 = 0 in  if it is given that f (1 − i ) = 0 if 1 − i is a root then so is 1 + i so ( x − (1 − i ) ) ( x − (1 + i ) ) is a factor so ( x − 1) − i 2 is a factor 2 ( ) so x 2 − 2 x + 2 is a factor by inspection : x 4 + x 3 − 2x 2 + 2x + 4 = ( x − 2x + 2)( x + 3 x + 2) = ( x − 2x + 2) ( x + 1)( x + 2) 2 2 2 ∴x 1 = − i or 1 + i or − 1 or − 2 QUESTION 2 Use Mathematical Induction to prove that: n ∑2 = 2 i =1 i n +1 −2 we wish to prove that : 2 + 22 + ..... + 2n = 2n +1 − 2 first , let ' s consider if n = 1 LHS = 2 and RHS = 22 − 2 = 2 so, it is true for n = 1 Assume it is true for n = k ∴ 2 + 22 + ..... + 2k= 2k +1 − 2 ( * ) adding the next term to each side gives : 2 + 22 + ..... + 2k + 2k +1= 2k +1 − 2 + 2k +1 2 2k +1 − 2 =× = 2k + 2 − 2 = 2( ) − 2 k +1 +1 but this is just with n= k + 1 so we have proved that it is true for n= k + 1 ∴ by the principle of mathematical induction the result is true for n ∈  IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 4 of 15 QUESTION 3 Determine f ' ( x ) by first principles if f ( x=) x +3 x +3+h − x +3 f ' ( x ) = lim h →0 h x +3+h − x +3 x +3+h + x +3 lim × h →0 h x +3+h + x +3 h = lim h →0 h ( x + 3 + h + x + 3) 1 = lim h →0 x +3+h + x +3 1 = 2 x +3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 5 of 15 QUESTION 4 x 2 + bx − 6 4.1 Consider the function: f ( x ) = 2x − a Determine the values of a and b if the function has a vertical asymptote at 1 x = 4 and an oblique asymptote of= y x+4 2 For a vertical asymptote at x = 4 the denominator must be zero when x = 4 so a = 8 Now, ( 2x − 8 )   1 f (x) = x + 4  + rem (where rem is a constant ) 2  ∴ f ( x ) = x 2 + 4 x − 32 + rem so, b = 4 x 2 + ax + b 4.2 Determine the values of a and b if the function f ( x ) = has a 2x − 3 stationary point at (1; 2 ) = we know that f (1) 2 and = f ' (1) 0 x 2 + ax + b 1 + a + b so = =2 or a + b =−3 2x − 3 −1 ( 2x + a )( 2x − 3 ) − 2 ( x 2 + ax + b ) ∴f '(x) = ( 2x − 3 ) 2 now f ' (1) = ( 2 + a )( −1) − 2 (1 + a + b ) = 0 substituting the value of a + b gives : −2 − a − 2 (1 − 3 ) =0 ∴ a =−2 − 2 ( −2 ) =2 and b =−5 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 6 of 15 QUESTION 5 Consider the function f defined as follows:  0.5 x + 4 x < −4  3 −4 ≤ x < −2  f (x) =  2 x = −2 0.5 x 2 + 1 −2 < x < 2   g ( x ) x≥2 Answer the following questions paying careful attention to the notation you use: 5.1 Determine lim f ( x ) if it exists. If not, explain why. x →−4 lim − f ( x ) 2 but = = lim+ f ( x ) 3 x →−4 x →−4 ∴ lim f ( x ) d .n.e. since they are unequal x →−4 5.2 Why is f discontinuous at x = −2 ? lim f ( = x ) 3 but f ( − = 2) 2 x →−2 ∴ discontinuous since they are unequal 5.3 What type of discontinuity occurs at x = −2 ? Removable 5.4 Determine g ( x ) if g ( x ) is a linear function and f is to be differentiable at x = 2. = we need g ( 2 ) lim− f = (x) 3 x →2 but we also need lim+ g ' ( = x ) lim− f ' ( = x ) lim = − x 2 x →2 x →2 x →2 so, g ( x= ) 2x + c but g ( 2= ) 2 ( 2 ) + c= 3 so c = −1 ∴g (x) =2x − 1 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 7 of 15 QUESTION 6 Consider the diagram below. It represents the cross-section of a semi-circular gutter with O the centre of the semi-circle. There is silt at the bottom of the gutter. The surface area of the silt CD is parallel to the surface area of the water AB. Angles in radians are as shown. If the radius of the gutter is 12 cm and the gutter is 2 m long then determine the volume of water in the gutter to the nearest litre. Remember: 1 cm3 = 1 ml and 1 litre = 1000 ml. = Area of water minor segment AB − minor segment CD = Now minor segment AB sector AOB − ∆AOB 1 2 2π 1 2 2π = 12 − 12 sin 2 3 2 3 72 3 = 48π − 2 = and minor segment CD sector OCD − ∆OCD 1 2π 1 2 π = 12 − 12 sin 2 3 2 3 72π 72 3 = − 3 2 72 3  72π 72 3  So, area of water =48π − − −  2  2   3 = 24π cm 2 = 200 × 24π cm 3 So, volume = 15079,6 cm 3 = 15 litres to the nearest litre IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 8 of 15 QUESTION 7 Below is the graph of the implicitly defined relationship: y 3 − xy =y + x 2 . Find the equation of the tangent (indicated with a dotted line) if it is known that the x-coordinate of the point of contact is 1. y 3 − xy =y + x 2 When x = 1, y 3 − y = y + 1 ∴ y 3 − 2y − 1 =0 ∴ y =−1 so, the pt . of contact is (1; −1) dy  dy  dy ∴ 3y 2 −y + x = + 2x dx  dx  dx dy dy dy ∴ 3y 2 −y −x = + 2x dx dx dx ∴ dy dx ( ) 3y 2 − x − 1 = y + 2x dy y + 2x ∴ =2 dx 3 y − x − 1 dy −1 + 2 ∴= = 1 dx 3 − 1 − 1 ∴ y − ( −1)= 1( x − 1) ∴y = x − 2 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 9 of 15 QUESTION 8 8.1 Consider the functions f ( x ) = x + a + b and g ( x ) =− x + 2 +1 f (x) = x + a + b g (x) =− x + 2 +1 (a) Determine the values of a and b a = 3 and b = −4 (b) Hence or otherwise solve: x + 3 + x + 2 ≤ 5. x +3 + x +2 ≤5 ∴ x +2 ≤5− x +3 ∴ − x + 2 ≥ −5 + x + 3 ∴− x + 2 +1≥ x + 3 − 4 ∴ −5 ≤ x ≤ 0 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 10 of 15 8.2 Given the graph y = f ( x ) , draw on your own set of axes in your Answer Book a rough sketch of y = f ( x ) . 8.3 Consider the function, f drawn below. 2 6 Given that ∫ f ( x ) dx −38,7 = and ∫ f ( x ) dx 74,7 = determine: 0 2 6 (a) ∫ f ( x ) dx − 0 = 38,7 + 74,7 = 36 6 (b) ∫ f ( x ) dx = 38,7 + 74,7 = 113,4 0 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 11 of 15 QUESTION 9 Consider the function f ( x )= x ln ( x ) − x 2 + 4, x > 0 9.1 Given that f is continuous at every value in its domain, justify why f has at least one root on the interval x ∈ [1; 5] f (1) = − 5 f ( 5 ) = 2,6 since f (1) < 0 and f ( 5 ) > 0 and f is continuous on [1; 5] f must cross the x − axis at least once on [1; 5] so there is at least one root on [1; 5] 9.2 Use Newton-Raphson iteration to find this root. You should: • use an initial guess of 1 • show the iterative formula you use • show your first two approximations • give your answer to 5 d.p. f ( x )= x ln ( x ) − x 2 + 4  1 1 1 ( ) − ∴ f ' ( x ) = ln ( x ) + x   − x 2 + 4 2 ( 2 x ) x 2 1 ( ) − ∴f '(x = ) ln ( x ) + 1 − x x 2 + 4 2 f ( xn ) xn += xn − f ' ( xn ) 1 xn ln ( xn ) − xn 2 + 4 = xn − 1 ( ) − ln ( xn ) + 1 − xn xn + 42 2 x0 = 1 x1 = 5,045085... x2 = 3,423819... x ≈ 3,23903 ( to 5 d .p.) IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 12 of 15 QUESTION 10 Consider the diagram below where rectangles are being formed in the first quadrant. The bottom left corner is on the origin while the top right corner is on the curve y = e − x Find, to 3 decimal places, the maximum area which can be achieved in this way. A = xe − x dA ∴ = e − x − xe − x = 0 dx ∴ e − x (1 − x ) = 0 ∴x = 1 ∴ Amax =e −1 =0,368 units 2 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 13 of 15 QUESTION 11 11.1 Robyn is using rectangles to estimate the shaded area. Calculate her percentage error to one decimal place. f ( x= ) x2 + 1 Robyn ' s estimate is (1× 5 ) + (1× 10 ) + (1× 17 ) + (1× 26 ) = 58 units 2 5 136 1 exact answer = ∫ x 2 + 1 dx = or 45 units 2 1 3 3 58 percentage error is × 100 = 27,9% 1 45 3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 14 of 15 3 x 2 + 11x − 5 11.2 (a) Resolve into partial fractions. x3 + 3x 2 − 4 3 x 2 + 11x − 5 3 x 2 + 11x − 5 = ( x 3 + 3 x 2 − 4 ( x − 1) x 2 + 4 x + 4 ) 3 x 2 + 11x − 5 = ( x − 1)( x + 2 ) 2 A B C = + + x − 1 x + 2 ( x + 2 )2 A = 1 by cover up method ( ) so 1 x 2 + 4 x + 4 + B ( x − 1)( x + 2 ) + C ( x − 1)= 3 x 2 + 11x − 5 adding x 2 + 4 x + 4 + Bx 2 + Bx − 2B + Cx − C = 3 x 2 + 11x − 5 = (1 + B ) x 2 + ( 4 + B + C ) x + ( 4 − 2B − C ) = so = , B 2 and C 5 1 2 5 = + + x − 1 x + 2 ( x + 2 )2 3 x 2 + 11x − 5 (b) Hence, or otherwise, determine ∫ dx x3 + 3x 2 − 4 3 x 2 + 11x − 5 ∫ x 3 + 3 x 2 − 4 dx 1 2 5 = ∫ + + dx x − 1 x + 2 ( x + 2 )2 5 = ln x − 1 + 2ln x + 2 − +c ( x + 2) 11.3 Determine ∫ xe 2x dx = let f ( x ) x then = f '(x) 1 1 2x = let g ' ( x ) e 2 x then = g (x) e 2 1 2x 1 2x then ∫ xe= 2x dx xe − ∫ e dx 2 2 1 2x 1 2x = xe − e + c 2 4 11.4 Determine ∫ cosec 2 x cot 2 x dx cot 3 x = +c − 3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER I – MARKING GUIDELINES Page 15 of 15 QUESTION 12 Consider the diagram below. f ( x ) = ax 2 b 160 The area bounded by the function f, the x-axis, the lines x = 0 and x = b is units 2 . 3 When this area is rotated around the x-axis the resulting volume is 1280π units 3 . Determine the values of a and b. b 160 ∫ ax dx = 3 2 0 b  ax 3  160 ∴  =  3 0 3 ab3 160 ∴ −0 = 3 3 ∴ ab = 3 160 (1) b π ∫ ( ax 2 ) dx = 1280π 2 0 b  a2 x 5  ∴ =  1280  5 0 a 2b5 ∴ = 1280 5 6400 ( 2 ) ∴ a 2b5 = now , squaring both sides of equation (1) gives a 2b 6 = 25600 ( 3 ) dividing equation ( 3 ) by equation ( 2 ) gives b = 4 5 substituting this value into (1) gives a = 2 Total: 200 marks IEB Copyright © 2019