AQA Core 4 Exam Discussion 14/06/2012 !Poll, paper and unofficial MS (first post)!
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
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View Poll Results: How did you find the exam?
Pshh I could do it in my sleep! 14 5.56% Went pretty well 
29 11.51% It was okay... (maybe made silly mistakes?) 82 32.54% 
62 24.60% I died. 61 24.21% Fine I guess, couldn't careless. 4 1.59%
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Re: AQA Core 4 Exam Discussion 14/06/2012xD I wish! I'll be happy with an A after that ruckus of an exam ^_^(Original post by Dangerous Theory)
Lets be honest, you got AT LEAST 99UMS
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I'm sure you've pwned that paper though
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Re: AQA Core 4 Exam Discussion 14/06/2012What were the grade boundaries last June?(Original post by Cath-ay)
after looking through this paper again... maybe I didn't do as bad as I think I did. Still would like grade boundaries as low as last june
*PRAYS*
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Re: AQA Core 4 Exam Discussion 14/06/2012I know(Original post by SomeoneIveNeverMet)
Ah look at us making silly mistakes
sad times. I'm hoping it's no more than 1 or 2 marks (it was out of 6/7 marks I think, and that mistake was made on the very last line). Haha yeah I'm so so glad it was a nice vector question! Vectors were definitely my biggest worry.
What other subjects do you do?
Hmm, worst case, AAB, best case A*A*A ... other than that, I really can't tell
How about you?
I only have M2 left!
you?
I'm hoping that I still got my nice A :L so I don't need to cling onto S2 so desperately :L and I've still got biology, chemistry, general studies and S2 to go! And i've already had 6 exams and oh, well at least thats good!
Its annoying because I think I'm on the ABB/AAB boundary :/ I still only need BBB to go to Birmingham though, which is handy!
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Re: AQA Core 4 Exam Discussion 14/06/2012a* - 58(Original post by lollyage)
what were the grade boundaries last june?
a - 51
b - 46
c - 41
d - 36
e - 31 -
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Re: AQA Core 4 Exam Discussion 14/06/2012Don't worry, I'm sure you've got it, just work hard for S2 now(Original post by LGrosvenor101)
I know I'm hoping that I still got my nice A :L so I don't need to cling onto S2 so desperately :L and I've still got biology, chemistry, general studies and S2 to go! And i've already had 6 exams and oh, well at least thats good!
Its annoying because I think I'm on the ABB/AAB boundary :/ I still only need BBB to go to Birmingham though, which is handy!
Ah that sucks when do you finish then? (And get off TSR and revise! )
Birmingham is good, like Warwick I'm sure you'll be able to get the AAB though.
My insurance is AAA (for the masters course, I also have an ABB offer for the bachelors course (thank goodness!)).
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Re: AQA Core 4 Exam Discussion 14/06/2012Thanks for putting the answers up! I think I got similar answers!(Original post by Oromis263)
I'll do the rest when I get home. What do people think so far about the answers?
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Re: AQA Core 4 Exam Discussion 14/06/2012yeh i did that as AP=BP but everyone seems to have different answers what did you get ??(Original post by f1mad)
You had to use the fact that the modulus of both sides were equal (isosceles triangle). -
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I replied to your binomial one. I disagree with your range(Original post by Oromis263)
I'll do the rest when I get home. What do people think so far about the answers?
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Re: AQA Core 4 Exam Discussion 14/06/2012I got what someone else posted here.(Original post by rosie17)
yeh i did that as AP=BP but everyone seems to have different answers what did you get ??
*PRAYS*
![x = 2cos\theta y = 3sin2\theta
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\dfrac{dy}{d\theta} = 6cos2\theta, \dfrac{dx}{d\theta} = -2sin\theta
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\Rightarrow \dfrac{dy}{dx} = \dfrac{6cos2\theta}{-2sin\theta}
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\Rightarrow Using\ cos2\theta = 1 - 2sin^2 \theta
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\Rightarrow \dfrac{dy}{dx} = \dfrac{6 - 12sin^2 \theta}{-2sin\theta}
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\Rightarrow \dfrac{dy}{dx} = 6sin\theta - 3cosec\theta](http://www.thestudentroom.co.uk/latexrender/pictures/3f/3f26ee3f961c06c3c08d7d941a9c0bad.png "x = 2cos\theta y = 3sin2\theta
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\dfrac{dy}{d\theta} = 6cos2\theta, \dfrac{dx}{d\theta} = -2sin\theta
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\Rightarrow \dfrac{dy}{dx} = \dfrac{6cos2\theta}{-2sin\theta}
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\Rightarrow Using\ cos2\theta = 1 - 2sin^2 \theta
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\Rightarrow \dfrac{dy}{dx} = \dfrac{6 - 12sin^2 \theta}{-2sin\theta}
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\Rightarrow \dfrac{dy}{dx} = 6sin\theta - 3cosec\theta")
![m_{tangent}\ at\ \theta=\dfrac{\pi}{6} \Rightarrow sin\dfrac{\pi}{6} = \dfrac{1}{2}
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m_{tangent} = -3 \Rightarrow m_{normal} = \dfrac{1}{3}](http://www.thestudentroom.co.uk/latexrender/pictures/80/80067818973086c1ae7ddbe2fdd932cc.png "m_{tangent}\ at\ \theta=\dfrac{\pi}{6} \Rightarrow sin\dfrac{\pi}{6} = \dfrac{1}{2}
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m_{tangent} = -3 \Rightarrow m_{normal} = \dfrac{1}{3}")
![y^2 = 9sin^2 2\theta
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Using\ sin2\theta = 2cos\theta sin\theta
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\Rightarrow y^2 = 36cos^2 \theta sin^2 \theta
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cos^2 \theta = \dfrac{x^2}{4}\ and\ sin^2 \theta = 1 - cos^2 \theta
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\Rightarrow y^2 = \dfrac{9}{4}x^2 (4-x^2)](http://www.thestudentroom.co.uk/latexrender/pictures/89/8912075303ba6120130e0448afeb9e21.png "y^2 = 9sin^2 2\theta
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Using\ sin2\theta = 2cos\theta sin\theta
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\Rightarrow y^2 = 36cos^2 \theta sin^2 \theta
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cos^2 \theta = \dfrac{x^2}{4}\ and\ sin^2 \theta = 1 - cos^2 \theta
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\Rightarrow y^2 = \dfrac{9}{4}x^2 (4-x^2)")
![9x^2 - 6xy + 4y^2 = 3
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implicit\ differentiation
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18x - 6y -6x\dfrac{dy}{dx} + 8y\dfrac{dy}{dx} = 0
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\Rightarrow \dfrac{dy}{dx} = \dfrac{6(3x-y)}{8y-6x}
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\Stationary\ at \dfrac{dy}{dx} =0
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3x-y = 0 \Rightarrow y = 3x
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\Sub\ this\ into\ original\ equation
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27x^2 = 3 \Rightarrow x = \dfrac{-1}{3}\ or \dfrac{1}{3} \Rightarrow y = 1\ or -1](http://www.thestudentroom.co.uk/latexrender/pictures/ec/ec9f60591e298e918bb9cc80d45b5e31.png "9x^2 - 6xy + 4y^2 = 3
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implicit\ differentiation
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18x - 6y -6x\dfrac{dy}{dx} + 8y\dfrac{dy}{dx} = 0
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\Rightarrow \dfrac{dy}{dx} = \dfrac{6(3x-y)}{8y-6x}
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\Stationary\ at \dfrac{dy}{dx} =0
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3x-y = 0 \Rightarrow y = 3x
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\Sub\ this\ into\ original\ equation
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27x^2 = 3 \Rightarrow x = \dfrac{-1}{3}\ or \dfrac{1}{3} \Rightarrow y = 1\ or -1")
![\dfrac{5x-6}{x(x-3)} = \dfrac{A}{x} + \dfrac{B}{x-3}
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5x-6 = A(x-3) + Bx
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Let x = 3 \Rightarrow 9 = 3B \Rightarrow B = 3
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Let x = 0 \Rightarrow -6 = -3A \Rightarrow A = 2
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\dfrac{5x-6}{x(x-3)} = \dfrac{2}{x} + \dfrac{3}{x-3}](http://www.thestudentroom.co.uk/latexrender/pictures/9d/9deb849dca4659fa555c8d5a383b0076.png "\dfrac{5x-6}{x(x-3)} = \dfrac{A}{x} + \dfrac{B}{x-3}
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5x-6 = A(x-3) + Bx
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Let x = 3 \Rightarrow 9 = 3B \Rightarrow B = 3
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Let x = 0 \Rightarrow -6 = -3A \Rightarrow A = 2
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\dfrac{5x-6}{x(x-3)} = \dfrac{2}{x} + \dfrac{3}{x-3}")
![4x^3 +5x -2 = (2x+1)(2x^2 +px + q) + r
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Using\ factor\ theorem,\ do\ f(\dfrac{-1}{2})
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4(\dfrac{-1}{2})^3 + 5\dfrac{-1}{2} -2 = -5,\ therefore\ r = -5
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Now,\ 4x^3 +5x +3 = (2x+1)(2x^2 +px + q),\ so\ by\ inspection\ p = -1 and q = 3](http://www.thestudentroom.co.uk/latexrender/pictures/67/675e8f4af16c70720de897c6c10a7b27.png "4x^3 +5x -2 = (2x+1)(2x^2 +px + q) + r
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Using\ factor\ theorem,\ do\ f(\dfrac{-1}{2})
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4(\dfrac{-1}{2})^3 + 5\dfrac{-1}{2} -2 = -5,\ therefore\ r = -5
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Now,\ 4x^3 +5x +3 = (2x+1)(2x^2 +px + q),\ so\ by\ inspection\ p = -1 and q = 3")
![Now\ divide\ the\ function\ previously\ found
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\Rightarrow \dfrac{2}{3}x^3 -\dfrac{1}{2}x^2 + 3x -\dfrac{5}{2} ln(2x+1) + c](http://www.thestudentroom.co.uk/latexrender/pictures/ac/acfd127cf7b1c9d4a1c4e3691199f2db.png "Now\ divide\ the\ function\ previously\ found
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\Rightarrow \dfrac{2}{3}x^3 -\dfrac{1}{2}x^2 + 3x -\dfrac{5}{2} ln(2x+1) + c")
