AQA FP3 June 2015 Unofficial Mark Scheme

I got 0.426 for 7)b)iii) but not 100% sure and for 5bii I think it was 18x^3 -27x^4

Reply 4

ArchmageRohan

8

I got 0.5something for the last one, but not sure, will need what other people got to verify...

Reply 5

Doomlar

OP

13

Original post by liamhull96

I got 0.426 for 7)b)iii) but not 100% sure and for 5bii I think it was 18x^3 -27x^4

Original post by ArchmageRohan

I got 0.5something for the last one, but not sure, will need what other people got to verify...

Thus far it seems everyone I spoke to got something different so we need someone to do working; I'll do later I'm in lessons at the moment!

Posted from TSR Mobile

Reply 6

I'll have another go now with one of my teachers

Reply 7

ScienceGeek!

What level of difficulty do people think this was compared to June 2014?

Reply 8

Original post by Doomlar

Thus far it seems everyone I spoke to got something different so we need someone to do working; I'll do later I'm in lessons at the moment!

Posted from TSR Mobile

I got 18x^3 -27x^4 for the first 2 terms of the expansion too

Reply 9

Going through again I got AB to be 0.425, using the triangle of OA and the perpendicular line from the base to A to find the height which is 3/4, and then working out the bases using pythagorous which is (3*3^0.5)/4 then using another triange of OB and the perpendicular line about O at height with B and using pythagorous again to find the small distance at the top of that triangle which is ((5+2*13^0.5)/16)^0.5 and the taking the base of the smaller away from the larger so (3*3^0.5)/4 - ((5+2*13^0.5)/16)^0.5 = 0.4254... = 0.425

(edited 8 years ago)

Reply 10

Oraeng

9

BUZZING that I got question 6!!
Couldn't do 7)b)ii) (which was 6 marks) and got something like 0.337 for iii) doubt that's right though. iii) was 3 marks.

Couldn't do the maclauren show f''(x)=0 - didn't seem right for 1 mark. Does anyone know what you were supposed to do?

(edited 8 years ago)

Reply 11

Also I got 3pi/4 for the area and I'm pretty sure that was right

Reply 12

username1138850

3

I agree with everything on here ( at least the parts I answered :| )

Reply 13

Original post by Oraeng

BUZZING that I got question 6!!
Couldn't do 7)b)ii) (which was 6 marks) and got something like 0.337 for iii) doubt that's right though. iii) was 3 marks.

Couldn't do the maclauren show f''(x)=0 - didn't seem right for 1 mark. Does anyone know what you were supposed to do?

In The differential equation, letting x=0 left f''(0) = 36sin (0) therefore f''(0) =0

Reply 14

Oraeng

9

Original post by manraj97

In The differential equation, letting x=0 left f''(0) = 36sin (0) therefore f''(0) =0

Ahh of course thank you. What about the next bit? Did you have to keep differentiating it until you got 2 values which weren't 0 or was there another way to do it?

Reply 15

masterchef97

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} + ... \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{[?]}} \\[br]\item[br]\begin{enumerate}[br]\item A(\dfrac{3}{2}, \dfrac{\pi}{6}) \hfill \textbf{\underline{[?]}} \\[br]\item Show OB = \dfrac{\sqrt{13} + 1}{4} \hfill \textbf{\underline{[?]}} \\[br]\item AB = 1.15 ??? \hfill \textbf{\underline{[?]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]\end{enumerate}[br]

i got 0.425 for the very last question and so did all my classmates....and for the question about ln((1+2x)(1-2x)) it also said state the range for which it was valid?
Also is (1/2)lnx equivalent to ln(root x)??

Reply 16

Original post by Oraeng

Ahh of course thank you. What about the next bit? Did you have to keep differentiating it until you got 2 values which weren't 0 or was there another way to do it?

I kept differentiating it till I got 2 values which wernt 0 (which were at f'''(x) and f''''(x) )

I thought this was wierd though as the differentiations were tedious, but the question only carried 3 marks

Reply 17

Oraeng

9

Original post by manraj97

I kept differentiating it till I got 2 values which wernt 0 (which were at f'''(x) and f''''(x) )

I thought this was wierd though as the differentiations were tedious, but the question only carried 3 marks

Yeah thanks, I thought they would be too time consuming if I did that so left it and moved on. Think I did pretty well on the rest of the paper though except the last two bits so shouldn't be too damaging :smile:

Reply 18

ArchmageRohan

8

Original post by liamhull96

Going through again I got AB to be 0.536, using the triangle of OA and the perpendicular line from the base to A to find the height which is 3/4, and then working out the bases using pythagorous which is (3*3^0.5)/4 then using another triange of OB and the perpendicular line about O at height with B and using pythagorous again to find the small distance at the top of that triangle which is (5+2*13^0.5)/16 and the taking the base of the smaller away from the larger so (3*3^0.5)/4 - (5+2*13^0.5)/16 = 0.505358... = 0.536

That rings a bell, that's exactly what I got, but again can't be sure. I used the cosine rule. What do you get when you use that?

(edited 8 years ago)

Reply 19