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Edexcel IGCSE Mathematics A - Paper 3H - 2016 - Unofficial Mark Scheme

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Reply 460
Original post by BOBQ
On the straight line graph that had to be drawn, did you have to stop at X=3


yes x=3 y=4
angle of 105 degrees and sides of 8cm and 12cm
Reply 462
Original post by WhiteX
yes x=3 y=4


Thanks!
Original post by igggy
Didnt he already have like 1000 or something


you are right
Original post by _Xenon_
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. :smile:

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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
Original post by BOBQ
The first part a) was ds/dt to find the velocity which = 6t^2 -24t + 7

and b) was 2



Yeah but for the first part I think you had to differentiate again because it's dv/ds
Question 22 Proof that answer is 188. :smile:
Reply 467
Does anyone remember the shaded rectangle with circle in the middle question?
Original post by joebarnes123
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.
Original post by Martins1
Question 22 Proof that answer is 188. :smile:


Haha I messed that one up so bad
Original post by joebarnes123
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.
Original post by sgsb0y
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.


yeah I think part a was 12t^2-24t+7
Original post by sgsb0y
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
Reply 473
Original post by sgsb0y
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

which =2
Original post by sgsb0y
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!
(edited 7 years ago)
Original post by sgsb0y
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.
Original post by robellie77
yeah! I agree with you, for velocity you only differentiate once, then for acceleration you do again!

Yeah because velocity is speed in a given direction, it won't require a double differentiation. Acceleration requires another differentiation. 👍



Posted from TSR Mobile
Original post by Martins1
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.


You quoted the wrong person, I'm agreeing with you.


Posted from TSR Mobile
Original post by sgsb0y
Yeah because velocity is speed in a given direction, it won't require a double differentiation. Acceleration requires another differentiation. 👍




yesss :smile:)
Reply 479
Original post by sgsb0y
Yeah because velocity is speed in a given direction, it won't require a double differentiation. Acceleration requires another differentiation. 👍

Yeah part A wouldn't be 2 marks if it was that complicated

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