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AQA FP3 June 2015 Unofficial Mark Scheme

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Reply 20
Avatar for Doomlar
Doomlar
OP
13
Doing the last question again I got 0.425, which ive seen in this thread; I must have types numbers into my calculator wrong ****ing hell :frown:
Reply 21
Avatar for Oraeng
Oraeng
9
Original post by manraj97
It seems like I only lost 9 marks on the last 2 parts of the last question.

If so, I got 66/75 which should be an A* because I think the grade boundaries will be lower than last year


Pretty good going for FP3 :smile:
I will be ecstatic if I even get an A on it tbh - which is looking likely by some miracle :biggrin:
i think i got full marks on this. im on my way to cambridge woohoo
im going to apply through adjustment for maths
grade boundaries?
Original post by manishlfc
i think i got full marks on this. im on my way to cambridge woohoo
im going to apply through adjustment for maths


I didn't get full marks. I'm on my way to the job centre woohoo!
Could anyone tell me how to do 7 b)ii. I found the polar equation of AB and let it intersect C1.
Couldn't get the right answer.
(edited 8 years ago)
Original post by Doomlar
And so begins the exam season; below is my unofficial mark scheme for today's FP3 paper; I would appreciate it if anyone who sat the paper could contribute to the mark scheme, particularly if you think one or more of my answers are wrong. Marks are in emboldened, underlined brackets, like so [x]. Corrected answers that are still being debated are simply underlined, as so x.

Unparseable latex formula:

[br]\begin{enumerate}[br]\item[br]\begin{enumerate}[br]\item y(2.05) = 5.675 \hfill \textbf{\underline{[2]}} \\[br]\item y(2.1) = 6.67 \hfill \textbf{\underline{[3]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]




Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{1}[br]\item y = \dfrac{tan^{4}x+7}{4secx} \hfill \textbf{\underline{[9]}} \\[br]\end{enumerate}[br]




Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{2}[br]\item[br]\begin{enumerate}[br]\item[br]\begin{enumerate}[br]\item ln(1+2x) = 2x - 2x^{2} + \dfrac{8x^3}{3} - 4x^{4} + ... \hfill \textbf{\underline{[1]}} \\[br]\item ln[(1+2x)(1-2x)] = -4x^{2} - 8x^{4} + ... \text{range of validity?} \hfill \textbf{\underline{[3]}} \\[br]\end{enumerate}[br]\item \lim{x \to 0} = \dfrac{1}{24} \hfill \textbf{\underline{[4]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]




Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{3}[br]\item[br]\begin{enumerate}[br]\item \text{The interval of integration is infinite.} \hfill \textbf{\underline{[1]}} \\[br]\item I = \dfrac{1}{4} e^{-4} \hfill \textbf{\underline{[6]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]




Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{4}[br]\item[br]\begin{enumerate}[br]\item y = e^{-3x}(Ax+B) - 2cos3x \hfill \textbf{\underline{[7]}} \\[br]\item[br]\begin{enumerate}[br]\item Show f^{''}(0) = 0 \hfill \textbf{\underline{[1]}} \\[br]\item f(x) = 18x^{3} - 27x^{4} + ... \hfill \textbf{\underline{[3]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]\end{enumerate}[br]




Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{5}[br]\item [br]\begin{enumerate}[br]\item Show the substitution changes the differential equation. \hfill \textbf{\underline{[7]}} \\[br]\item y = \sqrt{x}(Acos(\dfrac{lnx}{2}) + Bsin(\dfrac{lnx}{2}) + \dfrac{(lnx)^{2}}{2} + 2lnx + 2 + \dfrac{1}{\sqrt{x}} \hfill \textbf{\underline{[10]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]




I can't remember my answers for the last question on the paper, apart from the last bit, so could people please tell me their answers, and the marks for each question if they remember them!


Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{6}[br]\item [br]\begin{enumerate}[br]\item Area = \dfrac{3\pi}{4} \hfill \textbf{\underline{[4]}} \\[br]\item[br]\begin{enumerate}[br]\item A(\dfrac{3}{2}, \dfrac{\pi}{6}) \hfill \textbf{\underline{[5]}} \\[br]\item Show OB = \dfrac{\sqrt{13} + 1}{4} \hfill \textbf{\underline{[6]}} \\[br]\item AB = \underline{0.425} \hfill \textbf{\underline{[3]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]\end{enumerate}[br]



what were the x and y values in question (2) that you'd sub in to find the arbitrary value for the equation, cause i'm sure i did it right but don't remember C=7
Original post by theconsciousindian
what were the x and y values in question (2) that you'd sub in to find the arbitrary value for the equation, cause i'm sure i did it right but don't remember C=7


Yeah I don't remember c=7 either...
Original post by theconsciousindian
what were the x and y values in question (2) that
you'd sub in to find the arbitrary value for the equation, cause i'm sure i did it right but don't remember C=7


C is 7/4, not 7.

I think y = 2 and x= pi/4.
(edited 8 years ago)
Yeah it was the third and fourth derivatives that gave non zero values
The range of validity for 3aii I'm pretty sure is -1/2 < x < 1/2
Original post by Help_Me_Alex
The range of validity for 3aii I'm pretty sure is -1/2 < x < 1/2


Was it not -1/2 < x <= 1/2 ?

The ln(1+x) expansion is valid for -1 < x <= 1, but not sure, any ideas?
Original post by ibanezmatt13
Was it not -1/2 < x <= 1/2 ?

The ln(1+x) expansion is valid for -1 < x <= 1, but not sure, any ideas?


I thought that at first but with both -2x and + 2x to be expanded does it not make
-0.5< x <= 0.5 and 0.5>x=>-0.5. As x must satisfy both doesn't it cancel to make just
-0.5<x<0.5 ?
Validity is definitely -0.5<x<0.5
Ah yeah, good point. I think you're right :smile:
Reply 35
Avatar for Doomlar
Doomlar
OP
13
I'll update the validity question now; I missed that bit in the exam haha!
Original post by onlinekute17
Could anyone tell me how to do 7 b)ii. I found the polar equation of AB and let it intersect C1.Couldn't get the right answer.


did the exact same thing, was so proud of myself then I let them equal eachother and BOOM. Dead end. Brilliant.

Hopefully we'll get a few for that mate. :smile:
Original post by Help_Me_Alex
The range of validity for 3aii I'm pretty sure is -1/2 < x < 1/2


Original post by ibanezmatt13
Was it not -1/2 < x <= 1/2 ?

The ln(1+x) expansion is valid for -1 < x <= 1, but not sure, any ideas?


Original post by Doomlar
I'll update the validity question now; I missed that bit in the exam haha!

Did it ask for the range of validity? If so I missed that.

I agree with the unofficial mark scheme for everything else and answered every question/sub question so I feel rather confident.
Reply 38
Avatar for Lau14
Lau14
11
In 6b I think you're missing the end of the first bracket (should be after Bsin(lnx/2) maybe?), and for question 7 part (a) was 5 marks and (b)(i) 4 marks from what I remember. Can't believe how fast you put this together and how much you can remember, thanks! :smile:
Original post by Stepidermis
did the exact same thing, was so proud of myself then I let them equal eachother and BOOM. Dead end. Brilliant.

Hopefully we'll get a few for that mate. :smile:


Original post by onlinekute17
Could anyone tell me how to do 7 b)ii. I found the polar equation of AB and let it intersect C1.
Couldn't get the right answer.


What sid you get for AB? Off the top of my head it was r=43sin⁡θr = \dfrac{4}{3\sin \theta}.
You set this equal to r=1+cos⁡2θr = 1 + \cos 2\theta to get a cubic in sin⁡θ\sin \theta, which you solve for sin⁡θ\sin \theta. One value is less than -1, so can be discarded, one is associated with A and the other gives you the given result.

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