Advanced Higher Maths 2019 Unofficial Markscheme Watch

leavemeblank
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Sort of unofficial marking scheme for AH Maths 2019:

1. Differentiation (a) 6x^5cot5x - 5x^6cosec^2(5x) (b) -27x^2/(x^3-4)^2 (c) - 4
2. Matrix algebra (a) p = -3 (b) Row 1: q+16, 5, Row 2: 8 - 3q, -12, Row 3: 20 - 2q, -7) (c) Not a square matrix
3. Graphs (a) Even as symmetrical about x-axis: f(x) = f(-x) (b) reflection of negative parts across x-axis (w shape)
4. Partial fractions (a) 3 + (4x+19)/(x^2-x+12) (b) 3 + 5/(4-x) - 1/(x+3)
5. Parametric differentiation (a) 2t^2 + 7t (b) 1/2(8t^2 + 42t + 49)
6. Related rates of change: -5/3π cms^-1 (units usually required!)
7. Summation formula: (a) 3n^2 + 16n (b) 1520 - 3p^2 - 16p
8. Homogenous second order differential equation: y = 3e^-4x -3e^-7x
9. Binomial theorem general term: (a) (7Cr)(2)^7-r(-d)^r(x)^14-5r (b) d = 5
10. Implicit differentiation (a) dy/dx = y-2x/2y-x (b) k = +/- 4 ( i still think its +/- 2 but most folk say 4)
11. Proof by counterexample and contrapositive (a) n=4,n=7 or n=0, etc. (b) if n is even, n^2-2n+7 is odd. (ii) Proof
12. Number bases (application of division algorithm): 276 to the base 10 = 543 to the base 7.
13. Variables separable differential equation: V = 12 - 10e^-kt
14. Proof by induction: prove true for n=1, LHS = RHS = 1, assume true for n = k, consider n = k + 1
Summation of k+1 from n=1 r! r = (k + 1)! - 1 + (k+1)!(k+1) = (k+1)! + (k+1)!(k+1) - 1 = (k+1)![1 + k+1] - 1 = (k+1)!(k+2) - 1 = (k + 2)! - 1 = ((k+1)+1)! -1 therefore true for n = k+1. If true for n=k, also true for n=k+1 but since also true for n=1, by induction true for all positive integers, n.
15. Vectors. (a) Sub parametric equations into the equation of plane to prove. (b) Angle = 90-cos^-1(n.d/|n||d|) = 13.15 degrees
(c) Don't intersect - not sure if this is right but I did the cross product with π3 and L1 for the d vector, got parametric equations, solved for the separate variables and as they didn't satisfy both of the z co-ordinates, no intersection.
16. Applied integration: (a) 1/32(e^4 - 13) (b) π/2(e^4 - 13) units^3
17. (a) 63, -21, 7 hence r = -1/3 (b)(i) -1 < -1/3 < 1. (ii) 189/4 (47.25) (c) (i) ar^2/ar = ar/a gives quadratic when cross multiply, solve for x= -3. (ii) -7, 7, -7 (iii) S2n = 0 as (1-r^n) = 0.
18. (a) (i) a - root3ai (ii) w = 2a (cos (-π/3) + isin (-π/3) (b) (i) k=2, m = -9 (ii) Really not sure on this one, keep changing my mind but 2(cos5π/9 + isin5π/9) and 2(cos-7π/9 + isin(-7π/9).

There we go! Let me know if there's any mistakes and disagreements. This is generally the consensus on what has already been discussed above. Remember there's no chance this is 100% accurate and you will always get partial and follow through marks if you make any mistakes. DLB Maths have started uploading solutions so check them out! -
https://www.youtube.com/user/DLBmaths/videos
Last edited by leavemeblank; 3 weeks ago
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Evil Homer
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leavemeblank

Hey hope you don't mind but I moved this to a separate thread to make it easier for people to find
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leavemeblank
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(Original post by Evil Homer)
leavemeblank

Hey hope you don't mind but I moved this to a separate thread to make it easier for people to find
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KateM04
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@leavemeblank

Thank you for this!
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ionasmith.
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What is the working for Q18. I was really confused by it.
Thanks in advance
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sbneelu
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(Original post by ionasmith.)
What is the working for Q18. I was really confused by it.
Thanks in advance
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The Grislybear
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For q13 is anyone getting v=e^kt + 2 for the final answer?
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Razza2000
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For Q7b, in the brackets is it not just p in the brackets instead of (p+1)?
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ionasmith.
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Has anyone got working for Q6, 10b and 13 as well?
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The Grislybear
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Size:  112.1 KBThat’s my working for 10)b)
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Size:  114.7 KBAnd that’s the working for 13) however my answer ain’t the same as what the original posts answer so so idk with 13 tbh
(Original post by ionasmith.)
Has anyone got working for Q6, 10b and 13 as well?
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leavemeblank
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Yep, thank you. Just realised I made this mistake in the exam!
(Original post by Razza2000)
For Q7b, in the brackets is it not just p in the brackets instead of (p+1)?
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Razza2000
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Are these all your answers though? Let's just say I did waaaaaayyyyyyy worse that you :/
(Original post by leavemeblank)
Yep, thank you. Just realised I made this mistake in the exam!
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onlyvips
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I think your answer and working for 10b is wrong. Because there is no stationary point when the tangent is x=k. x=k is a vertical straight line so the gradient is undefined. Meaning dy/dx has a division by zero. So you equate the denominator of dy/dx to 0 and solve giving k= +-4
(Original post by The Grislybear)
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Size:  112.1 KBThat’s my working for 10)b)
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Size:  114.7 KBAnd that’s the working for 13) however my answer ain’t the same as what the original posts answer so so idk with 13 tbh
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onlyvips
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For 13, im think the answer is the other way around V = 12 - 10e^-kt Not, V =e^kt + 2
(Original post by leavemeblank)
Sort of unofficial marking scheme for AH Maths 2019:

1. Differentiation (a) 6x^5cot5x - 5x^6cosec^2(5x) (b) -27x^2/(x^3-4)^2 (c) - 4
2. Matrix algebra (a) p = -3 (b) Row 1: q+16, 5, Row 2: 8 - 3q, -12, Row 3: 20 - 2q, -7) (c) Not a square matrix
3. Graphs (a) Even as symmetrical about x-axis: f(x) = f(-x) (b) reflection of negative parts across x-axis (w shape)
4. Partial fractions (a) 3 + (4x+19)/(x^2-x+12) (b) 3 + 5/(4-x) - 1/(x+3)
5. Parametric differentiation (a) 2t^2 + 7t (b) 1/2(8t^2 + 42t + 49)
6. Related rates of change: -5/3π cm^3s^-1 (units usually required!)
7. Summation formula: (a) 3n^2 + 16n (b) 1520 - 3p^2 - 16p
8. Homogenous second order differential equation: y = 3e^-4x -3e^-7x
9. Binomial theorem general term: (a) (7Cr)(2)^7-r(-d)^r(x)^14-5r (b) d = 5
10. Implicit differentiation (a) dy/dx = y-2x/2y-x (b) k = +/- 2
11. Proof by counterexample and contrapositive (a) n=4,n=7 or n=0, etc. (b) if n is even, n^2-2n+7 is odd. (ii) Proof
12. Number bases (application of division algorithm): 276 to the base 10 = 543 to the base 7.
13. Variables separable differential equation: V =e^kt + 2 (not V = 12 - 10e^-kt)
14. Proof by induction: prove true for n=1, LHS = RHS = 1, assume true for n = k, consider n = k + 1
Summation of k+1 from n=1 r! r = (k + 1)! - 1 + (k+1)!(k+1) = (k+1)! + (k+1)!(k+1) - 1 = (k+1)![1 + k+1] - 1 = (k+1)!(k+2) - 1 = (k + 2)! - 1 = ((k+1)+1)! -1 therefore true for n = k+1. If true for n=k, also true for n=k+1 but since also true for n=1, by induction true for all positive integers, n.
15. Vectors. (a) Sub parametric equations into the equation of plane to prove. (b) Angle = 90-cos^-1(n.d/|n||d|) = 13.15 degrees
(c) Don't intersect - not sure if this is right but I did the cross product with π3 and L1 for the d vector, got parametric equations, solved for the separate variables and as they didn't satisfy both of the z co-ordinates, no intersection.
16. Applied integration: (a) 1/32(e^4 - 13) (b) π/2(e^4 - 13) units^3
17. (a) 63, -21, 7 hence r = -1/3 (b)(i) -1 < -1/3 < 1. (ii) 189/4 (47.25) (c) (i) ar^2/ar = ar/a gives quadratic when cross multiply, solve for x= -3. (ii) -7, 7, -7 (iii) S2n = 0 as (1-r^n) = 0.
18. (a) (i) a - root3ai (ii) w = 2a (cos (-π/3) + isin (-π/3) (b) (i) k=2, m = -9 (ii) Really not sure on this one, keep changing my mind but 2(cos5π/9 + isin5π/9) and 2(cos-7π/9 + isin(-7π/9).

There we go! Let me know if there's any mistakes and disagreements. This is generally the consensus on what has already been discussed above. Remember there's no chance this is 100% accurate and you will always get partial and follow through marks if you make any mistakes. DLB Maths have started uploading solutions so check them out! -
https://www.youtube.com/user/DLBmaths/videos
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robmiester25
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I agree. You move the -1 using laws of logs so it becomes 1/(12-v) and 1/10
(Original post by onlyvips)
For 13, im think the answer is the other way around V = 12 - 10e^-kt Not, V =e^kt + 2
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leavemeblank
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These aren't my answers noo!! I just compiled this from what most people were saying on the AH Maths thread. I've dropped somewhere between 5-15 marks lol.
(Original post by Razza2000)
Are these all your answers though? Let's just say I did waaaaaayyyyyyy worse that you :/
I did think that, but seen a few other people say otherwise but I've changed it back! I realised I'd forgot to take into account the fact it was a -V that you had to take the log of when integrating so I was frantically panicking on this question in the last few mins of the exam!
(Original post by onlyvips)
For 13, im think the answer is the other way around V = 12 - 10e^-kt Not, V =e^kt + 2
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aidancolley
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2019 Advanced Higher Mathematics Worked Solutions
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aidancolley
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(Original post by aidancolley)
2019 Advanced Higher Mathematics Worked Solutions
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Jordin Petricor
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For question six the units should be cms^-1 instead of cm^-3s^-1 because the question was asking for rate of change in radius not volume.
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