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    (Original post by alkaline.)
    can I just say I did that for 9b at first and got plus/minus 5/2
    but then I thoughts
    you can't just "say" that you can't put logs at negative so +5/2 would be the answer so I crossed it out and used b^2-4ac= 0 on the quadratic that it formed - will I still get the marks???? FMLLLLLLL
    It shouldn't have been possible to get +/- as the factorised quadratic was (2x-5)^2 = 0 which just gives x = 5/2 repeated root
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    (Original post by tajtsracc)
    Would do you guys predict an A will be? More than or less than 60?
    what do u guys think 67/75 would be in ums?
    it was a hard paper but I knew all the tricks
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    (Original post by Bosssman)
    It shouldn't have been possible to get +/- as the factorised quadratic was (2x-5)^2 = 0 which just gives x = 5/2 repeated root
    ****.

    i got 2x+5 and 2x-5 somehow......

    I er factorised it and there were negatives in the quadratic so one has to be positive and one has to be negative right?
    oh well I crossed it out though and used b2-4ac so that proves it right?

    I dk if crossed out x=5/2 though so kengko;weo[fkdpknsgkm im panicking n stressing.
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    (Original post by Parhomus)
    Mtangent was=1/2 therefore Mnormal=-2. Therefore you just write y-5=-2(x-4) or whatever. I think he just expanded and rearranged it which is fine too.
    Thank you! I get how you would get that equation from there but where was the gradient of the tangent from
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    (Original post by Bosssman)
    Wasn't it 13.8 for perimeter? Some are saying its 13.6 but pretty sure it was 13.8 if you didn't round before the final answer
    I must've spent about 10 minutes on that question because of the continuous copying out of full decimals
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    (Original post by Hbassett26)
    Thank you! I get how you would get that equation from there but where was the gradient of the tangent from
    Using the first derivative, inputting the x value of P
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    (Original post by alkaline.)
    ****.

    i got 2x+5 and 2x-5 somehow......

    I er factorised it and there were negatives in the quadratic so one has to be positive and one has to be negative right?
    oh well I crossed it out though and used b2-4ac so that proves it right?

    I dk if crossed out x=5/2 though so kengko;weo[fkdpknsgkm im panicking n stressing.
    The quadratic resulted in 4x^2 - 20x + 25 = 0
    And yeah, proves the repeated root
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    (Original post by Bosssman)
    Using the first derivative, inputting the x value of P
    Ah that's what I did! So I think the X coordinate in the mark scheme is wrong because X would have to be 4 instead of 5 in order for it to work!
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    (Original post by Hbassett26)
    Ah that's what I did! So I think the X coordinate in the mark scheme is wrong because X would have to be 4 instead of 5 in order for it to work!
    Yes, the coordinate of P was (4,5)
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    (Original post by Bosssman)
    Yes, the coordinate of P was (4,5)
    +1 OP pls fix

    Another thing - 6c is confusing the hell out of me

    y=\sqrt{x^2+9}
    onto
    y=3\sqrt{x^2+1}

    Let f(x) = \sqrt{x^2+9}

    f(x) --> 3f(x) = Stretch in y direction SF 3
    However, your mark scheme says "Stretch in x direction SF 1/3"
    For that to be the case, wouldn't the equation have to be the following:
    f(x) --> f(3x) = Your answer
    Which would lead to the "original equation" having to be:
    y=\sqrt{3x^2+1}

    My logic was that the 3 is multiplying the entirety of the sqrt function.
    For example:
    5^x --> 3 * 5^x = Stretch in y direction SF 3
    5^x --> 5^3x = Stretch in x direction SF 1/3
    This would be the case because the number 3 is only multiplying the x by 3, not the entire function (which in this case is 5^x)

    You said you proved it was x direction SF 1/3 using differentiation, could you show this please? Using methods and formulae that I'd been taught, I got to a y stretch SF 3.

    Forgive me if I'm talking out my ass too
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    For 7b if I expanded the other bracket fully but didn't get the actual value of the coefficient how many marks would I get
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    [QUOTE=Bosssman;65156935]
    (Original post by Eisobdxhsonw)

    The answer to the log question was definitely 5/2

    The question was
    Show that log4(2x+3) + log4(2x+15) = 1 + log4(14x + 5)

    Also the trig question, it resulted in 5 + 3cosx, and so when x = pi, 3cosx = -3 so minimum value is 2
    I'm still confused as to why it was 5+3cosx and not 3+5cosx also as to why the min value was -2, cheers tho
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    And for 2c divided 5 by 0.2 and got a stretch of sf 25 in the y axis. Why isnt this correct?
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    [QUOTE=Eisobdxhsonw;65161255]
    (Original post by Bosssman)

    I'm still confused as to why it was 5+3cosx and not 3+5cosx also as to why the min value was -2, cheers tho
    the minimum value was 22 because the cos curve has gone by two units in the y direction in (3-5cosx) so the normal cos graph minimum value would be -1 but because its translated by vector (0,3) its gone from -1 to 2
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    (Original post by Myachii)
    +1 OP pls fix

    Another thing - 6c is confusing the hell out of me

    y=\sqrt{x^2+9}
    onto
    y=3\sqrt{x^2+1}

    Let f(x) = \sqrt{x^2+9}

    f(x) --> 3f(x) = Stretch in y direction SF 3
    However, your mark scheme says "Stretch in x direction SF 1/3"
    For that to be the case, wouldn't the equation have to be the following:
    f(x) --> f(3x) = Your answer
    Which would lead to the "original equation" having to be:
    y=\sqrt{3x^2+1}

    My logic was that the 3 is multiplying the entirety of the sqrt function.
    For example:
    5^x --> 3 * 5^x = Stretch in y direction SF 3
    5^x --> 5^3x = Stretch in x direction SF 1/3
    This would be the case because the number 3 is only multiplying the x by 3, not the entire function (which in this case is 5^x)

    You said you proved it was x direction SF 1/3 using differentiation, could you show this please? Using methods and formulae that I'd been taught, I got to a y stretch SF 3.

    Forgive me if I'm talking out my ass too
    The differentiation was for the normal through P translated to pass through maximum point M (different question)

    To get stretch parallel to x-axis s.f. 1/3 I did this

    Squared both sides of original equation to give y^2 = x^2 + 9
    Then I squared the new equation to give y^2 = 9(x^2 + 1)
    Expanded the bracket to give y^2 = 9x^2 + 9
    Now the only difference between the two equations is there's a 9x^2 instead of x^2
    So then you can have y^2 = (3x)^2 + 9 hence giving the transformation


    Your idea of multiplying the entire function is a misread of the equation as y = 3root(x^2 + 9) when it is y = 3root(x^2 +1)
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    (Original post by Bosssman)
    The differentiation was for the normal through P translated to pass through maximum point M (different question)

    To get stretch parallel to x-axis s.f. 1/3 I did this

    Squared both sides of original equation to give y^2 = x^2 + 9
    Then I squared the new equation to give y^2 = 9(x^2 + 1)
    Expanded the bracket to give y^2 = 9x^2 + 9
    Now the only difference between the two equations is there's a 9x^2 instead of x^2
    So then you can have y^2 = (3x)^2 + 9 hence giving the transformation


    Your idea of multiplying the entire function is a misread of the equation as y = 3root(x^2 + 9) when it is y = 3root(x^2 +1)
    Gotta love AQA
    Thanks for the clarification though, that was annoying me. Awful lot of work for 2 marks (never seen anything like it before either)
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    (Original post by IsThisReallyLife)
    QUESTION 8B:

    (Copied from first post):

    8) find the value of tan(x) tan(x) = -5/4 [2]
    b) re-arrange 16+9sin2x / 3 - 5cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]

    For part B, wouldnt x= 0, 2pi?

    As it was 5 - 3cos(theta)

    If theta = pi then its:

    5 - (3*-1) = 8

    hence theta should be 0 or 2pi as this gives:

    5 - (3*1) =2 = Minimum value?

    Also i think the question had it as plural??

    Or am i missing something here?
    It would've ended up being 5+3cos(theta) because it was 16+9sin^2x which = 16+9(1-cos^2x)=25-9cos^2x. This is the difference of two squares so when you factorise you get (5-3cosx)(5+3cosx)/5-3cosx which =5+3cosx.
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    (Original post by beanigger)
    Unoffical Mark scheme for C2 AQA 2016

    It would help if you could correct the amount of marks in each question if it is wrong please

    Questions:

    1)a) integrate something to get -36x-1 + ax2 /2 [3]
    b) find the value of a when the answer to 1)a) is equal to 16 with limits of 3 to 1, a = - 2 [2]

    2)a) Draw the graph of (0.2)x and state where itcrosses axis (0,1)[2]
    b) use logarithms to solve (0.2)x = 4 , x = -0.861 [2]
    c) describe transformation of (0.2)x onto (5)x reflection in y axis [1]

    3)a) differentiate something to get 3x-1/2 -1 [2]
    b) find y co-ordinate of the maximum point M M(9,6) [3]
    c) equation of normal to curve at P i think P was (5,6) y=-2x+13 [2]
    d) the normal to curve at P is translated by (0,k) find value of k, k = 5.5 [3]

    4)a) show that a+10d=8 [2]
    b)sum to 2nd and 3rd term is 50, work out the 12th term of the series D=-2 and A=28, u12 = 6 [3]
    c) work out the sum of the 21st term minus sum of 3rd term = 90 [3]

    5)a) show that thetha (in radians) was 0.568 [2]
    b) work out area of triangle 19.9 [2]
    c) work out perimeter of shaded area, first piece of info given -> area of sector = area of shaded area

    you had to realise that the area of sector = area of triangle - area of sector
    re-arrange to get 2(area of sector) = 19.9 and solve for r to get 5.36?
    perimeter of shaded = (9-r) + (8-r) + 5 + (length of arc = r*theta)
    perimeter of shaded = 13.8 [6]

    6)a)trapezium rule to get 65.6 [4]
    b) describe transformation of graph y=(x2 +9)1/2 onto y = 5 + (x2 +9)1/2
    Translation by (0, 5) [2]
    c) describe transformation of graph y=(x2 +9)1/2 onto y=3(x2 +1)1/2 Stretch by scale factor 1/3 in x direction [2]

    7)a) expand (2+x)5 work out p q r, i think p was 10 q was -80 and r was 80 (correct me if im wrong) [4]
    b)find the coefficient of x10 in the expansion of (2+x)5 * (something else)7
    it was something like -1648 [5]

    8)a)find the value of tan(x) tan(x) = -5/4 [2]
    b) values of tanx in the range of 0 to 360 degrees 45, 129, 225, 309 [3]
    b) re-arrange 16+9sin2x / 3 - 5cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]

    9)re-arrange something into the form (c)1/2 / d^2 to get 3y y = 1/2 ( m + 12n) [4]
    b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4]
    Hmm for 9a I got y=1/2m-6n... I'm pretty sure you had a negative indice on the bottom so it would become positive when you bring it to the top???
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    (Original post by collegeboy66)
    And for 2c divided 5 by 0.2 and got a stretch of sf 25 in the y axis. Why isnt this correct?
    25 x 0.2^x=/= 5^x. You had to recognise that 0.2^x=1/5^x which is 5^-x.
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    (Original post by habibtii)
    yup, I most definitely failed. with dignity x)
    silly me,
    I took Edexcel C2. not this. no wonder the mark scheme looked like a different language to me.
    tbh, so did the Edexcel paper. but it's okay. I'll get there.
 
 
 
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