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AQA FP1 June 2014 Unofficial Mark Scheme

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Reply 60

mwatters

Ah right, I see. Thanks for clearing that up!

Reply 61

Cluck

Original post by nomnomnoodles

I got 256π for 8b :/

I did too

Reply 62

Millerman

Original post by SherlockHolmes

Which other equation?

The equation you were given was:

9x^2 +18x + 16y^2 - 64y = c

I completed the square with both x and y terms, got all the number terms to the RHS and equated it with 1 as it is a translation of the original ellipse.

So where are you getting the 144 from? the only difference in our answers is that you've taken 73 from 144 and i haven't

Reply 63

nomnomnoodles

Original post by rhiam

for 8b i got 256pi
anyone else??

yes me too, but I think we both forgot n=0 :/

Reply 64

SherlockHolmes

Original post by Millerman

So where are you getting the 144 from? the only difference in our answers is that you've taken 73 from 144 and i haven't

If you look at my working in one of my previous posts, I got the equation of the translated ellipse in to the form given for the original ellipse. To do this, I divided by 9 and 16. 9 x 16 = 144.

Reply 65

Firefox23

Original post by thetodd

it was 4pi/15 +- pi/5 +(8pi/5)n

Do you know what the original question was? Was it cos(5x -pi/6) =root3/2

Reply 66

SherlockHolmes

Original post by Firefox23

Do you know what the original question was? Was it cos(5x -pi/6) =root3/2

cos(\frac{5x}{4}-\frac{\pi}{3})=\frac{\sqrt{2}}{2}

Reply 67

Firefox23

Original post by SherlockHolmes

cos(\frac{5x}{4}-\frac{\pi}{3})=\frac{\sqrt{2}}{2}

**** I thought it was root3/2. That's a whole question down the drain then, a whole 9 marks gone because I can't read

Reply 68

Maths Magnet

How many marks do you think i'd get for the matrix question out of 5 for getting the correct enlargement, and finding the correct angle, but in the wrong quadrant?

Reply 69

Hoppo7

Original post by Millerman

So where are you getting the 144 from? the only difference in our answers is that you've taken 73 from 144 and i haven't

I did {(X-a)^2 over 16} +{(y-b)^2 over 9} =1
Then expanded and simplified this and compared it to the equation given to you so RHS is 9x16 -a^2 -b^2

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Reply 70

Millerman

Original post by Hoppo7

I did {(X-a)^2 over 16} +{(y-b)^2 over 9} =1
Then expanded and simplified this and compared it to the equation given to you so RHS is 9x16 -a^2 -b^2

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Oh, i completed the square

Reply 71

Doomlar

Original post by Millerman

Oh, i completed the square

I tried to complete the square, managed to **** that up, then tried the approach of applying a and b to the original equation, and then ****ed that up too. I was so annoyed by the end of it, and desperate for a piss ^_^

Reply 72

Naomi_la

Original post by Doomlar

Will you be doing an M1 and D1 mark scheme too?

Reply 73

Doomlar

Original post by Naomi_la

Will you be doing an M1 and D1 mark scheme too?

Yeah, M1, D1, and C4 :smile:

Reply 74

Naomi_la

Original post by Doomlar

Yeah, M1, D1, and C4 :smile:

Thank you, they're so helpful! Just hope you're answers are right!!

Reply 75

Doomlar

Original post by Naomi_la

Thank you, they're so helpful! Just hope you're answers are right!!

So do I

Reply 76

HarunH1

Dont understand the last part to the last question, i got y=x + 5, y=x-5 because of the questions before, any help? ;D

Reply 77

Doomlar

Original post by HarunH1

Dont understand the last part to the last question, i got y=x + 5, y=x-5 because of the questions before, any help? ;D

You had your extreme values for k for the original ellipse, but you then had to translate the tangent lines by the column vector, if I remember correctly.

Reply 78

TLHroolz

Original post by Doomlar

Below is my unofficial mark scheme for today's FP1 paper; mark schemes for other papers I have sat are linked at the bottom. I would appreciate it if anyone could contribute their answers, and if any of my answers are wrong give corrections. Marks are in emboldened, underlined brackets, like so [x]. Corrected answers that are still being debated are simply underlined, as so x.

Contest all my answers; I really don't know how well that paper went; the last two answers are blank 'cos I blanked out...

Unparseable latex formula:

[br]\begin{enumerate}[br]\item y = 6.0996 \hfill \textbf{\underline{[5]}} \\[br]\end{enumerate}[br]

Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{1}[br]\item[br]\begin{enumerate}[br]\item \alpha + \beta = -4, \alpha\beta = 0.5 \hfill \textbf{\underline{[2]}} \\[br]\item[br]\begin{enumerate}[br]\item \alpha^2 + \beta^2 = 15 \hfill \textbf{\underline{[2]}} \\[br]\item \text{Show} \alpha^4 + \beta^4 = \dfrac{449}{2} \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\item 4x^2 - 2036x + 137 = 0 \hfill \textbf{\underline{[5]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]

Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{2}[br]\item \Sigma = 2906061 \hfill \textbf{\underline{[4]}} \\[br]\end{enumerate}[br]

Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{3}[br]\item z = 5.5+6.5i \hfill \textbf{\underline{[6]}} \\[br]\end{enumerate}[br]

Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{4}[br]\item[br]\begin{enumerate}[br]\item \text{Gradient} = h-7 \hfill \textbf{\underline{[3]}} \\[br]\item \text{as h} \rightarrow 0, \text{the gradient} \rightarrow m \text{ at } x=5, m=-7 \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]

Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{5}[br]\item[br]\begin{enumerate}[br]\item x=0, x=-2, y=0 \hfill \textbf{\underline{[2]}} \\[br]\item[br]\begin{enumerate}[br]\item y = -1 \hfill \textbf{\underline{[1]}} \\[br]\item {Sketch graph} \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\item x \leq -4, x \geq 2, -2 \textless x \textless 0 \hfill \textbf{\underline{[5]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]

Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{6}[br]\item[br]\begin{enumerate}[br]\item[br]\begin{enumerate}[br]\item \begin{array}{ccc} [br]0 & -1 \\ [br]-1 & 0 \end{array} \hfill \textbf{\underline{[1]}} \\[br]\item \begin{array}{ccc}[br]1 & 0 \\[br]0 & 7 \end{array} \hfill \textbf{\underline{[1]}} \\[br]\end{enumerate}[br]\item \begin{array}{ccc}[br]0 & -1 \\[br]-7 & 0 \end{Array} \hfill \textbf{\underline{[2]}} \\[br]\item[br]\begin{enumerate}[br]\item A^2 = 12I \hfill \textbf{\underline{[1]}} \\[br]\item \text{SF:} -2\sqrt{3}, y=tan{15}x \text{\underline{ OR }} \text{ SF: } 2\sqrt{3}, y=tan{105}x \hfill \textbf{\underline{[5]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]\end{enumerate}[br]

Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{7}[br]\item[br]\begin{enumerate}[br]\item x = \dfrac{7\pi}{15} + \dfrac{8\pi n}{5}, x = \dfrac{\pi}{15} + \dfrac{8\pi n}{5} \hfill \textbf{\underline{[5]}} \\[br]\item \text{Sum} = \dfrac{3848\pi}{15} \hfill \textbf{\underline{[4]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]

Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{8}[br]\item [br]\begin{enumerate}[br]\item \text{Sketch the ellipse, intersections at} x=\pm{4}, y=\pm{3} \hfill \textbf{\underline{[2]}} \\[br]\item \text{Show} -5 \textless k \textless 5 \hfill \textbf{\underline{[5]}} \\[br]\item a = -1, b = 2, c = \underline{71} \hfill \textbf{\underline{[5]}} \\[br]\item y=x+8, y=x-2 \hfill \textbf{\underline{[3]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]

C1: http://www.thestudentroom.co.uk/showthread.php?t=2685494
C2: http://www.thestudentroom.co.uk/showthread.php?t=2689168
C3:http://www.thestudentroom.co.uk/showthread.php?t=2709037
S1: http://www.thestudentroom.co.uk/showthread.php?t=2704980

So, I got basically everything right, apart from the summation of my general solution. Got 128pi, because I forgot to double it and add on the n=0 bit, but I did cancel the plus or minus bit and got the upper bound of the summation right (n=12?), and did use the summation equation... How many marks...?

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