Turn on thread page Beta
    • Thread Starter
    Offline

    3
    ReputationRep:
    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.

    

\begin{enumerate}

\item

\begin{enumerate}

\item y(2.05) = 5.675 \hfill \textbf{\underline{[2]}} \\

\item y(2.1) = 6.67 \hfill \textbf{\underline{[3]}} \\

\end{enumerate}

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{1}

\item y = \dfrac{tan^{4}x+7}{4secx} \hfill \textbf{\underline{[9]}} \\

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{2}

\item

\begin{enumerate}

\item

\begin{enumerate}

\item ln(1+2x) = 2x - 2x^{2} + \dfrac{8x^3}{3} - 4x^{4} + ... \hfill \textbf{\underline{[1]}} \\

\item ln[(1+2x)(1-2x)] = -4x^{2} - 8x^{4} + ... \text{valid for} \dfrac{-1}{2} < x < \dfrac{1}{2} \hfill \textbf{\underline{[3]}} \\

\end{enumerate}

\item \lim{x \to 0} = \dfrac{1}{24} \hfill \textbf{\underline{[4]}} \\

\end{enumerate}

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{3}

\item

\begin{enumerate}

\item \text{The interval of integration is infinite.} \hfill \textbf{\underline{[1]}} \\

\item I = \dfrac{1}{4} e^{-4} \hfill \textbf{\underline{[6]}} \\

\end{enumerate}

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{4}

\item

\begin{enumerate}

\item y = e^{-3x}(Ax+B) - 2cos3x \hfill \textbf{\underline{[7]}} \\

\item

\begin{enumerate}

\item Show f^{''}(0) = 0 \hfill \textbf{\underline{[1]}} \\

\item f(x) = 18x^{3} - 27x^{4} + ... \hfill \textbf{\underline{[3]}} \\

\end{enumerate}

\end{enumerate}

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{5}

\item 

\begin{enumerate}

\item Show the substitution changes the differential equation. \hfill \textbf{\underline{[7]}} \\

\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]}} \\

\end{enumerate}

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{6}

\item 

\begin{enumerate}

\item Area = \dfrac{3\pi}{4} \hfill \textbf{\underline{[5]}} \\

\item

\begin{enumerate}

\item A(\dfrac{3}{2}, \dfrac{\pi}{6}) \hfill \textbf{\underline{[4]}} \\

\item Show OB = \dfrac{\sqrt{13} + 1}{4} \hfill \textbf{\underline{[6]}} \\

\item AB = \underline{0.425} \hfill \textbf{\underline{[3]}} \\

\end{enumerate}

\end{enumerate}

\end{enumerate}
    Offline

    1
    ReputationRep:
    Thanks
    Offline

    1
    ReputationRep:
    7(b)I was coordinate of A(3/2, pi/6)
    Offline

    1
    ReputationRep:
    I got 0.426 for 7)b)iii) but not 100% sure and for 5bii I think it was 18x^3 -27x^4
    Offline

    8
    ReputationRep:
    I got 0.5something for the last one, but not sure, will need what other people got to verify...
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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
    Offline

    1
    ReputationRep:
    I'll have another go now with one of my teachers
    Offline

    1
    ReputationRep:
    What level of difficulty do people think this was compared to June 2014?
    Offline

    4
    ReputationRep:
    (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
    Offline

    1
    ReputationRep:
    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
    Offline

    2
    ReputationRep:
    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?
    Offline

    1
    ReputationRep:
    Also I got 3pi/4 for the area and I'm pretty sure that was right
    Offline

    3
    ReputationRep:
    I agree with everything on here ( at least the parts I answered :| )
    Offline

    4
    ReputationRep:
    (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
    Offline

    2
    ReputationRep:
    (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?
    Offline

    0
    ReputationRep:
    (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.

    

\begin{enumerate}

\item

\begin{enumerate}

\item y(2.05) = 5.675 \hfill \textbf{\underline{[2]}} \\

\item y(2.1) = 6.67 \hfill \textbf{\underline{[3]}} \\

\end{enumerate}

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{1}

\item y = \dfrac{tan^{4}x+7}{4secx} \hfill \textbf{\underline{[9]}} \\

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{2}

\item

\begin{enumerate}

\item

\begin{enumerate}

\item ln(1+2x) = 2x - 2x^{2} + \dfrac{8x^3}{3} - 4x^{4} + ... \hfill \textbf{\underline{[1]}} \\

\item ln[(1+2x)(1-2x)] = -4x^{2} - 8x^{4} + ... \hfill \textbf{\underline{[3]}} \\

\end{enumerate}

\item \lim{x \to 0} = \dfrac{1}{24} \hfill \textbf{\underline{[4]}} \\

\end{enumerate}

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{3}

\item

\begin{enumerate}

\item \text{The interval of integration is infinite.} \hfill \textbf{\underline{[1]}} \\

\item I = \dfrac{1}{4} e^{-4} \hfill \textbf{\underline{[6]}} \\

\end{enumerate}

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{4}

\item

\begin{enumerate}

\item y = e^{-3x}(Ax+B) - 2cos3x \hfill \textbf{\underline{[7]}} \\

\item

\begin{enumerate}

\item Show f^{''}(0) = 0 \hfill \textbf{\underline{[1]}} \\

\item f(x) = 18x^{3} - 27x^{4} + ... \hfill \textbf{\underline{[3]}} \\

\end{enumerate}

\end{enumerate}

\end{enumerate}


    

\begin{enumerate}

\setcounter{enumi}{5}

\item 

\begin{enumerate}

\item Show the substitution changes the differential equation. \hfill \textbf{\underline{[7]}} \\

\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]}} \\

\end{enumerate}

\end{enumerate}


    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!


    

\begin{enumerate}

\setcounter{enumi}{6}

\item 

\begin{enumerate}

\item Area = \dfrac{3\pi}{4} ??? \hfill \textbf{\underline{[?]}} \\

\item

\begin{enumerate}

\item A(\dfrac{3}{2}, \dfrac{\pi}{6}) \hfill \textbf{\underline{[?]}} \\

\item Show OB = \dfrac{\sqrt{13} + 1}{4} \hfill \textbf{\underline{[?]}} \\

\item AB = 1.15 ??? \hfill \textbf{\underline{[?]}} \\

\end{enumerate}

\end{enumerate}

\end{enumerate}
    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)??
    Offline

    4
    ReputationRep:
    (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
    Offline

    2
    ReputationRep:
    (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
    Offline

    8
    ReputationRep:
    (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?
    Offline

    4
    ReputationRep:
    (Original post by Oraeng)
    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
    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
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 5, 2016

2,694

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.