Please help us get together an unofficial mark scheme for the Edexcel IGCSE Mathematics A HIGHER TIER exam today. If you can remember any questions/answers (preferably with question numbers) please post below. Please note that the questions which are currently empty are labelled 'not done' and will be filled in ASAP. Many thanks.
Question 1
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Question 2
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Question 3
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Question 4 (not done)
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Question 5 (not done)
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Question 6 (not done)
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Question 7 (not done)
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Question 8 (not done)
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Question 9 (not done)
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Question 10
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Question 11 (not done)
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Question 12 (not done)
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Question 13 (not done)
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Question 14 (not done)
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Question 15 (not done)
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Question 16
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Question 17 (not done)
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Question 18 (not done)
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Question 19 (not done)
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Question 20
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Question 21
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Question 22
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[Circle theorems was 63 degrees. Cube and sphere one was 1637 (?)]
If you remember the question numbers please let me know, thanks.
Cube and sphere one was 850?! Differentiation velocity one was: (a). 12t-24 (b). 2
Cube and sphere one was 850?! Differentiation velocity one was: (a). 12t-24 (b). 2
Are you sure a was that was the answer to a? I differentiated the given equation once to get a quadratic-looking equation and for part b I differentiated again to get 12t-24=0 and then I solved it to get t=2.
Cube and sphere one was 850?! Differentiation velocity one was: (a). 12t-24 (b). 2
Are you sure a was that was the answer to a? I differentiated the given equation once to get a quadratic-looking equation and for part b I differentiated again to get 12t-24=0 and then I solved it to get t=2.
Are you sure a was that was the answer to a? I differentiated the given equation once to get a quadratic-looking equation and for part b I differentiated again to get 12t-24=0 and then I solved it to get t=2.
Are you sure a was that was the answer to a? I differentiated the given equation once to get a quadratic-looking equation and for part b I differentiated again to get 12t-24=0 and then I solved it to get t=2.
I'm not entirely sure but I'm fairly confident you had to differentiate the first one a second time because you are finding velocity not speed... And yeah the second part was definitely 2
Are you sure a was that was the answer to a? I differentiated the given equation once to get a quadratic-looking equation and for part b I differentiated again to get 12t-24=0 and then I solved it to get t=2.
The first part a) said to find VELOCITY which is speed / time or ds/dt
so you differentiated the equation s = 2t^3 + 12t^2 + 7t
which = 6t^2 + 24t + 7
b) was to find ACCELERATION which = dv/dt when velocity = 0
Are you sure a was that was the answer to a? I differentiated the given equation once to get a quadratic-looking equation and for part b I differentiated again to get 12t-24=0 and then I solved it to get t=2.
no! I don't agree with you, for velocity you only differentiate once, then for acceleration you do again!
Are you sure a was that was the answer to a? I differentiated the given equation once to get a quadratic-looking equation and for part b I differentiated again to get 12t-24=0 and then I solved it to get t=2.
No, 12t-24=0 was part of the working for part b.
Part a asked you to differentiate to get 6t^2 - 24t + 7 Part b asked you to find the time when acceleration was equal to 0. Therefore, acceleration= 6t^2 - 24t + 7 differentiated to get 12t-24=a when a = 0, 12t-24=0 12t=24 t=2 therefore the time was 2 seconds.
Part a asked you to differentiate to get 6t^2 - 24t + 7 Part b asked you to find the time when acceleration was equal to 0. Therefore, acceleration= 6t^2 - 24t + 7 differentiated to get 12t-24=a when a = 0, 12t-24=0 12t=24 t=2 therefore the time was 2 seconds.
You quoted the wrong person, I'm agreeing with you.