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    You need to flip the sign on 5a, you put greater than or equal to not less than or equal to
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    (Original post by Swindelz)
    That last italics bit isn't right - you need to sub in the value of x back into the original f(x) first, like matty9
    Bloody hell, so I lost the last accuracy mark :')

    Ah well, thanks!
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    (Original post by Parhomus)
    That's what I had but I also had no clue.
    Just realised I crossed out correct working **** my life.
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    (Original post by JamesNuttall)
    If sin(x) is 2/3, you can draw a right angle triangle to show that the remaining side must be root 5 due to pythagorus, so tan x = 2/root 5 and cos x = root 5 / 3, so sec x is 3/root 5 I think?
    What was the actual question
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    (Original post by irMike)
    For the stationary point, did you get sin x = 2/3? A few people I spoke to got this, and none of us had any clue how this became a root? Had to keep it as a decimal in the end
    If you draw a triangle and label the side opposite angle x as 2 and label the hypotenuse 3 then you can work out the third side by doing root of 3^(2) - 2^(2), which is root 5. So then you could find out the values of tanx and secx using this triangle and yeah I got -root5 as well.
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    (Original post by JamesNuttall)
    If sin(x) is 2/3, you can draw a right angle triangle to show that the remaining side must be root 5 due to pythagorus, so tan x = 2/root 5 and cos x = root 5 / 3, so sec x is 3/root 5 I think?
    ****, was this the one where you had to figure out the y co-ordinate afterwards? I crossed out my working for that assuming it was wrong because I had the wrong form... Could've had method marks for it. I deserve the C I'm getting
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    (Original post by matty9)
    You need to flip the sign on 5a, you put greater than or equal to not less than or equal to
    Fml I only wrote less than and forgot it was the maxima I'm so done with life. Went from 100% to horseshit.
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    (Original post by Parhomus)
    Just realised I crossed out correct working **** my life.
    Same...
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    Anybody want to join me in committing
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    (Original post by matty9)
    I redid question 5a and got the same answer as I did in the exam which was 8ln8-8

    Can someone confirm my working?

    Attachment 550557
    I believe this to be 100% correct
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    (Original post by JamesNuttall)
    y = 16x - e^(2x)
    dy/dx = 16 - 2e^(2x) which = at a stationary point.
    16 = 2e^(2x)
    8 = e^(2x)
    2x = ln(8)
    x = ln(8)/2

    d2y/dx2 = -4e^(2x)

    when x = ln(8)/2, dy/dx is negetive, therefore it's a max point.

    f(x) <= ln(8)/2
    Yeah, but you'd have to sub the x you worked out into f(x)
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    Can someone remind me what finding the area of the triangle question was. Coords and all tyvm
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    For the Area of OAB question:

    Was it a case of just making x = 0 and y = 0 to find A and B? I'm pretty certain I got A and B correct, then had absolutely no idea what to do to find the area. I tried integrating but I didn't look to me getting anywhere so I moved on because I was running out of time already by that point.
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    ok ive made corrections

    (Original post by irMike)
    For the Area of OAB question:

    Was it a case of just making x = 0 and y = 0 to find A and B? I'm pretty certain I got A and B correct, then had absolutely no idea what to do to find the area. I tried integrating but I didn't look to me getting anywhere so I moved on because I was running out of time already by that point.
    you use area of triangle = 1/2ab and bobs ur uncle
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    (Original post by beanigger)
    It would help if you could link answers to questions as i cant remember them

    Questions:

    1) a)Asked to find dy/dx [2]
    b) find dy/dx [2]
    c) find dy/dx[2]

    2) a) show alpha lied between values [2]
    b) re-arrange to get into the form x=e(ln5 / x) [3]
    c) use iteration thingy where X1 = 2, X3 =2.054 [2]
    d) simpsons rule on 5-xx which was 4.49 to 2dp [4]
    e) use d) to find approximate for xx which was 1.51 (could someone explain how to do this please) [2]

    3) a) solve x2 >|5x-6| *NEED CONFIRMED ANSWERS* [5]

    4) a) explain transformation of y=ex onto y=e2x+5 stretch by 1/2 in x-direction and translation by (2.5,0) [4]
    b) cant remember

    5) a) find range of f(x) = 16x - e2x (differentiate, find stationary point) f(x)>8ln8 - e8ln(8) (i got e8ln(8) but im not sure if it is right) [5]
    b) find gg(x) g(x) = 1/x therefore gg(x) = x [2]
    6) a) integrate ( ln3x ) / x2 by using by parts twice [4]
    b) volume of revolution, integrate ( ( ln3x ) /x2 )2 limits of 1 to 1/3pi( (ln3)2 + 2ln(3) -4 ) [7]

    7) a) integrate using substitution (5-2x)(1-4x)1/3 using u=1-4x limits of 1/2 and 1/4 you get 117/224 [7]

    8) a) use chain rule to show that (cosx)-1 is secxtanx [2]
    b) cant remeber
    c) i think it was a 6 marker for the m = 2 and n = 3

    9)a) show secx-tanx = -5 goes to secx+tanx=-0.2 [2]
    b) find the exact value of cosx cosx=-1/2.6 [3]
    c) solve the equation but x is changed into 2x-70 with range of -90 to 90 [3]

    i belive i jumbled up a few so pls correct and give CONFIRMED answers, ty.
    1 quotient rule. I thin p = -2. (Cant remember if 2 or -2)

    For 3a i got x <= 6, 1 <= x <= 2, 3 <= 3

    5a) 8ln8 -8 i think

    6a) It was only parts once I think, you used parts twice in part 2. In the second part you integrated [ln(3x)]^2/x^2 I think.
    My final answer was -(ln3)^2 - 2ln3 + 4 for B.

    7a) 177/224 not 117/224 (I think you made a typo there.)
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    sign up lads https://www.change.org/p/aqa-lower-g...iter=557227400
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    (Original post by umpa_dumpa)
    1 quotient rule. I thin p = -2. (Cant remember if 2 or -2)

    For 3a i got x <= 6, 1 <= x <= 2, 3 <= 3

    5a) 8ln8 -8 i think

    6a) It was only parts once I think, you used parts twice in part 2. In the second part you integrated [ln(3x)]^2/x^2 I think.
    My final answer was -(ln3)^2 - 2ln3 + 4 for B.

    7a) 177/224 not 117/224 (I think you made a typo there.)
    Yeah was -2

    5a defo correct (see my working)

    7a it was either haha have a feeling it was actually 177
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    (Original post by umpa_dumpa)
    1 quotient rule. I thin p = -2. (Cant remember if 2 or -2)

    For 3a i got x <= 6, 1 <= x <= 2, 3 <= 3

    5a) 8ln8 -8 i think

    6a) It was only parts once I think, you used parts twice in part 2. In the second part you integrated [ln(3x)]^2/x^2 I think.
    My final answer was -(ln3)^2 - 2ln3 + 4 for B.

    7a) 177/224 not 117/224 (I think you made a typo there.)
    thanks i made the changes xd
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    (Original post by JamesNuttall)
    If sin(x) is 2/3, you can draw a right angle triangle to show that the remaining side must be root 5 due to pythagorus, so tan x = 2/root 5 and cos x = root 5 / 3, so sec x is 3/root 5 I think?
    How many marks do you reckon one would get if he only made it to sin x=2/3?
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    fak i got like 40 if I'm lucky
 
 
 
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