The A.G
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#1
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#1
The mark scheme has taken a different approach,I didn’t show the cubic equation however still arrived at the valid solution of 0.5.
Out of 6 marks, how much would I be rewarded? Name:  13305574-5598-423F-BB0B-E6FB1B9A1B3B.jpeg
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Last edited by The A.G; 1 month ago
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mqb2766
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#2
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#2
Hard to say how many marks you'd get / lose, but the main thing for me would be time/errors.
Parts ii) and iii) are fairly clearly related to solving iv), but you largely ignore them and repeat some of the calcuation. That will take extra time and mean extra errors could creep in.
You don't show the equation thats asked for and you don't justify the quartic c=1/2 solution, so would you be able to spot that under pressure/time constraints.

If I was doing it, Id possibly have considered the upper region and said 1/12 was the integral of
sqrt(x) - c
from c^2 to 1 and subtracting the triangle 1/2(1-c)^2. It should give the same cubic in c?

As soon as you involve the -x in the integral you'll get a quartic as the limits are c^2, so you may get penalized for that. So the cubic is really a hint to just consider the integral of the sqrt(x) and handle the triangle using a standard formula.

Edit - I dont fully follow your triangle area formulae on each side of the equation at the start.
Last edited by mqb2766; 1 month ago
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The A.G
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#3
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#3
(Original post by mqb2766)
Hard to say how many marks you'd get / lose, but the main thing for me would be time/errors.
Parts ii) and iii) are fairly clearly related to solving iv), but you largely ignore them and repeat some of the calcuation. That will take extra time and mean extra errors could creep in.
You don't show the equation thats asked for and you don't justify the quartic c=1/2 solution, so would you be able to spot that under pressure/time constraints.

If I was doing it, Id possibly have considered the upper region and said 1/12 was the integral of
sqrt(x) - c
from c^2 to 1 and subtracting the triangle 1/2(1-c)^2. It should give the same cubic in c?

As soon as you involve the -x in the integral you'll get a quartic as the limits are c^2, so you may get penalized for that. So the cubic is really a hint to just consider the integral of the sqrt(x) and handle the triangle using a standard formula.

Edit - I dont fully follow your triangle area formulae on each side of the equation at the start.
Essentially what I did was equate the two regions and rearranged to get one expression.
I’ve only one mock on the Mat,I realised that they don’t allow scrap paper which seems stupid since solutions most of the time take more than the amount of paper given.Before starting a question would it be better if I clearly indicated somewhere if certain parts may relate to each other since I tend to forget as I progress through it.
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mqb2766
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#4
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(Original post by The A.G)
Essentially what I did was equate the two regions and rearranged to get one expression.
I’ve only one mock on the Mat,I realised that they don’t allow scrap paper which seems stupid since solutions most of the time take more than the amount of paper given.Before starting a question would it be better if I clearly indicated somewhere if certain parts may relate to each other since I tend to forget as I progress through it.
I don't get that area for the triangle on your line 2 of your solution. Should it be
0.5(c-c^2)^2

Its almost worth assuming that the question parts guide you through a solution. Sometimes they won't but often the earlier parts are about building up your skills / intermediate results to use on later parts. Recognising this is part of the exam.
Last edited by mqb2766; 1 month ago
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The A.G
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#5
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#5
(Original post by mqb2766)
I don't get that area for the triangle on your line 2 of your solution. Should it be
0.5(c-c^2)^2

Its almost worth assuming that the question parts guide you through a solution. Sometimes they won't but often the earlier parts are about building up your skills / intermediate results to use on later parts. Recognising this is part of the exam.
Length of the triangle would be (c-c^2) and heights would just be c^2 as y=x so the triangle would be ((c-c^2)c^2)/2 which I then rearranged to one side,getting rid of the denominator
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mqb2766
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#6
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#6
(Original post by The A.G)
Length of the triangle would be (c-c^2) and heights would just be c^2 as y=x so the triangle would be ((c-c^2)c^2)/2 which I then rearranged to one side,getting rid of the denominator
I got both the length and height being the same c-c^2?
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The A.G
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#7
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#7
(Original post by mqb2766)
I got both the length and height being the same c-c^2?
Oh I see now, you’re correct,thanks for the help
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mqb2766
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#8
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#8
(Original post by The A.G)
Oh I see now, you’re correct,thanks for the help
This may sort your quartic - cubic problem as well
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