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    Anyone know where the question paper could be obtained from? Teachers apparently don't get it until the next day at earliest.
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    (Original post by Hydeman)
    There was an integration-only section. The one with a = 2.3 was the definite integration, but there were three indefinite integration questions before that.
    Hey man, do you remember the 3 indefinite integral questions? or their answers? thanks
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    (Original post by Hydeman)
    There was an integration-only section. The one with a = 2.3 was the definite integration, but there were three indefinite integration questions before that.
    Hey man, do you remember the 3 indefinite integral questions? or their answers? thanks Previous years there were only two indefinites, are you sure about 3?
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    (Original post by Firestartc)
    Hey man, do you remember the 3 indefinite integral questions? or their answers? thanks
    I don't remember exactly, sorry. I vaguely remember some of the types, though: one had an 8 in the numerator (which you had to take out as a constant) with a polynomial in the denominator raised to the power of 4, I think, which had to be flipped (i.e. brought into the numerator with a minus sign on the power). The chain rule one was ln(cosx), I think, and that came out to -sinx/cosx, simplified to -tanx. The third one was 1/polynominal not raised to a power (or raised to the power of 1), so it simply integrated to 1/differential of polynominal x ln|polynomial| + c.

    Hope that helps.

    (Original post by Firestartc)
    Hey man, do you remember the 3 indefinite integral questions? or their answers? thanks Previous years there were only two indefinites, are you sure about 3?
    Fairly sure, yeah.
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    (Original post by IrrationalRoot)
    Some answers:

    1. 0.74379 to 5dp
    Multiply this by e: 2.02183 to 5dp.

    2. 30, 150, 196.6, 343.4 degrees

    146.3,326.3 degrees

    3. \dfrac{2}{3}

    4. -\frac{1}{2}\tan 3t
    -\frac{1}{8}\sec^3 3t
    -\dfrac{1}{y^3}

    5. Perimeter = Arc Length + Chord Length, arc length is easy, chord length can be found with a right-angled triangle with angle \dfrac{\theta}{2}.
    Usual stuff with iterations etc; for function use f(\theta)=\theta +2\sin\frac{\theta}{2}-4.5

    6. An easy counterexample is a=b=1, c=2, d=3

    7. x=-\dfrac{1}{3} (x=2 doesn't work)

    8. a=-3,5
    b=-\frac{2}{3}

    9. f^{-1}(x)=12-3\ln(x-8)
    Domain is [9,+\infty)

    10. To show hh(x)=x simply expand out the left.
    So it is clear that h is its own inverse and so h^{-1}(-1)=\frac{1}{9}

    That's all I can remember, please correct anything that's wrong .

    EDIT: Just realised I left out diff/int questions.

    Differentiation: \ln(\cos x) gives -\tan x,

    \tan^{-1}(\frac{x}{3}) gives \dfrac{3}{9+x^2}

    e^{6x}(3x-2)^4 gives 18xe^{6x}(3x-2)^3

    Integration: a=2.3
    Could you explain Q2? Also could there be other answers for 4bii? Could you tell me what y is
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    (Original post by Hydeman)
    I don't remember exactly, sorry. I vaguely remember some of the types, though: one had an 8 in the numerator (which you had to take out as a constant) with a polynomial in the denominator raised to the power of 4, I think, which had to be flipped (i.e. brought into the numerator with a minus sign on the power). The chain rule one was ln(cosx), I think, and that came out to -sinx/cosx, simplified to -tanx. The third one was 1/polynominal not raised to a power (or raised to the power of 1), so it simply integrated to 1/differential of polynominal x ln|polynomial| + c.

    Hope that helps.


    Fairly sure, yeah.
    The chain rule one is a differentiation one lol, the other two i remember as well. :/
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    (Original post by Maths is Life)
    Could you explain Q2? Also could there be other answers for 4bii? Could you tell me what y is
    First part of Q2 is rearranging to get an equation in \cosec\theta using \cot^2\theta+1= \csc^2 \theta, which is a standard question.
    Second part you just convert \csc\theta and \sec\theta to sines and cosines, multiply out the fractions and then you get
    \tan\theta=-\frac{2}{3}.

    There aren't any 'other' answers for \dfrac{d^2y}{dx^2} in terms of y although I'm not sure exactly what you mean by that.
    Each line in my ms represents a different part, so the middle line for Q4 tells you what y is in terms of t.
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    (Original post by Mango88)
    i got that!!
    how is it over 8??
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    (Original post by Firestartc)
    The chain rule one is a differentiation one lol, the other two i remember as well. :/
    Ah, sorry. Misremembered. There were definitely three, though.
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    (Original post by Ceridwen101)
    how is it over 8??
    It's over 9000!!!
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    (Original post by IrrationalRoot)
    Some answers:

    1. 0.74379 to 5dp
    Multiply this by e: 2.02183 to 5dp.

    2. 30, 150, 196.6, 343.4 degrees

    146.3,326.3 degrees

    3. \dfrac{2}{3}

    4. -\frac{1}{2}\tan 3t
    -\frac{1}{8}\sec^3 3t
    -\dfrac{1}{y^3}

    5. Perimeter = Arc Length + Chord Length, arc length is easy, chord length can be found with a right-angled triangle with angle \dfrac{\theta}{2}.
    Usual stuff with iterations etc; for function use f(\theta)=\theta +2\sin\frac{\theta}{2}-4.5

    6. An easy counterexample is a=b=1, c=2, d=3

    7. x=-\dfrac{1}{3} (x=2 doesn't work)

    8. a=-3,5
    b=-\frac{2}{3}

    9. f^{-1}(x)=12-3\ln(x-8)
    Domain is [9,+\infty)

    10. To show hh(x)=x simply expand out the left.
    So it is clear that h is its own inverse and so h^{-1}(-1)=\frac{1}{9}

    That's all I can remember, please correct anything that's wrong .

    EDIT: Just realised I left out diff/int questions.

    Differentiation: \ln(\cos x) gives -\tan x,

    \tan^{-1}(\frac{x}{3}) gives \dfrac{3}{9+x^2}

    e^{6x}(3x-2)^4 gives 18xe^{6x}(3x-2)^3

    Integration: a=2.3
    I swear the grade boundaries are going to be lower... How the hell should we know that x=2 shouldnt work?!
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    (Original post by Hydeman)
    I don't remember exactly, sorry. I vaguely remember some of the types, though: one had an 8 in the numerator (which you had to take out as a constant) with a polynomial in the denominator raised to the power of 4, I think, which had to be flipped (i.e. brought into the numerator with a minus sign on the power). The chain rule one was ln(cosx), I think, and that came out to -sinx/cosx, simplified to -tanx. The third one was 1/polynominal not raised to a power (or raised to the power of 1), so it simply integrated to 1/differential of polynominal x ln|polynomial| + c.

    Hope that helps.


    Fairly sure, yeah.
    The chain rule one is a differentiation one lol, the other two i remember as well. :/
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    (Original post by Maths is Life)
    I swear the grade boundaries are going to be lower... How the hell should we know that x=2 shouldnt work?!
    Lol it was a bit mean of them but tbf candidates should be aware that mod equations often give spurious solutions. I always check my solutions to mod equations but still I was a little surprised that x=2 didn't actually work.

    The technical reason why x=2 didn't work is because when you're examining the two cases 5x+4=7x and 5x+4=7x, what you're really doing is looking at the cases x \geq -\frac{4}{5} and x \leq -\frac{4}{5} (respectively).

    Now in the second case we are supposing that x \leq -\frac{4}{5} and this produces the solution x=2. This is however a spurious solution since it contradicts x \leq -\frac{4}{5}.
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    IrrationalRoot, some questions about these:

    gives

    gives


    For the first one, would I still get the marks if I only partially simplified it? I initially got a 1/3 in the numerator and 1+(1/9)x2 in the denominator and I multiplied top and bottom by 3 instead of 9. Its the equivalent, but this would mean I left a 1/3 in the denominator...
    The second one - was this actually on the paper? I've no recollection of it. The product rule question was a different one, I feel...

    Edit: Are you sure the second one isn't an integration question? I seem to remember integrating something like that...
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    your answer to the differential involving e^x is wrong, you need to use the product rule
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    (Original post by Hydeman)
    IrrationalRoot, some questions about these:

    gives

    gives


    For the first one, would I still get the marks if I only partially simplified it? I initially got a 1/3 in the numerator and 1+(1/9)x2 in the denominator and I multiplied top and bottom by 3 instead of 9. Its the equivalent, but this would mean I left a 1/3 in the denominator...
    The second one - was this actually on the paper? I've no recollection of it. The product rule question was a different one, I feel...
    Hmm, not sure exactly how strict the marking is. You'd definitely only lose one mark at most, but I don't know whether they'd take a mark off. I think I'd lean towards yes since I recall there being a fair few marks for these, but I can't remember what the exact marks where.

    With the second one I'm pretty sure this was on the paper? I certainly did that derivative anyway, unless I misread the question or something .
    What do you think the product rule Q was?
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    (Original post by IrrationalRoot)
    Some answers:

    1. 0.74379 to 5dp
    Multiply this by e: 2.02183 to 5dp.

    2. 30, 150, 196.6, 343.4 degrees

    146.3,326.3 degrees

    3. \dfrac{2}{3}

    4. -\frac{1}{2}\tan 3t
    -\frac{1}{8}\sec^3 3t
    -\dfrac{1}{y^3}

    5. Perimeter = Arc Length + Chord Length, arc length is easy, chord length can be found with a right-angled triangle with angle \dfrac{\theta}{2}.
    Usual stuff with iterations etc; for function use f(\theta)=\theta +2\sin\frac{\theta}{2}-4.5

    6. An easy counterexample is a=b=1, c=2, d=3

    7. x=-\dfrac{1}{3} (x=2 doesn't work)

    8. a=-3,5
    b=-\frac{2}{3}

    9. f^{-1}(x)=12-3\ln(x-8)
    Domain is [9,+\infty)

    10. To show hh(x)=x simply expand out the left.
    So it is clear that h is its own inverse and so h^{-1}(-1)=\frac{1}{9}

    That's all I can remember, please correct anything that's wrong .

    EDIT: Just realised I left out diff/int questions.

    Differentiation: \ln(\cos x) gives -\tan x,

    \tan^{-1}(\frac{x}{3}) gives \dfrac{3}{9+x^2}

    e^{6x}(3x-2)^4 gives 18xe^{6x}(3x-2)^3

    Integration: a=2.3
    Are you sure about your answer for inverse tan of x/3. Could there be a different definition for 'simplest form'?
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    (Original post by Maths is Life)
    Are you sure about your answer for inverse tan of x/3. Could there be a different definition for 'simplest form'?
    I mean it really doesn't get simpler than what I put, but like I said earlier I'm not sure whether they'd penalise fractions in the numerator or denominator, but if I had to choose I'd say they would.
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    (Original post by IrrationalRoot)
    Hmm, not sure exactly how strict the marking is. You'd definitely only lose one mark at most, but I don't know whether they'd take a mark off. I think I'd lean towards yes since I recall there being a fair few marks for these, but I can't remember what the exact marks where.
    I hope the boundaries are lower than last year. :sigh:

    With the second one I'm pretty sure this was on the paper? I certainly did that derivative anyway, unless I misread the question or something .
    What do you think the product rule Q was?
    I remember a product rule question having e6x as one of the functions, but I think the polynomial next to it is from an integration question where it was in the denominator and and you had to flip it (raise it to the negative power), and then divide by the product of the differential of and the new power, which was -3. I distinctly remember that.
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    (Original post by Hydeman)
    I hope the boundaries are lower than last year. :sigh:



    I remember a product rule question having e6x as one of the functions, but I think the polynomial next to it is from an integration question where it was in the denominator and and you had to flip it (raise it to the negative power), and then divide by the product of the differential of and the new power, which was -3. I distinctly remember that.
    Dw you'll be fine, forget about C3 and pick up a C4 paper . What's your offer btw? Need an A* in Maths?

    You might be right, I was struggling a bit to remember that question. Only thing is I swear that in the derivative there was some nice cancellation and I got 18x on the outside, and the only polynomial I could get to fit this (I worked backwards to get the Q from my working) was 3x-2.
 
 
 
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