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    i'm so confused now was it pi -pi or pi/2 -pi/2 because everyone is saying different things. i thought that it was between -pi and pi but now i'm not sure...
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    (Original post by Linguist131)
    veldt127 Is there any chance you can explain / show us working for question 7b)? I thought I got the answer at the time but I seem to remember my answer being different to yours - the question was to solve
    3g(x+1) + 1 = 0
    with g(x) = arcsinx
    if I remember correctly? Thank you - and thanks so much for your answers, they're great!
    It was 3g(x+1)+pi=0
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    (Original post by Kay Fearn)
    Yeah but i just left it as it's exact value, i thought that's what you had to do when it came to stuff with pi :-(
    I think I did the same as you, I left it as x=sin(-pi/3) - 1
    I think we might lose 1 mark but I'm hoping not seeing as it's technically correct
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    (Original post by jack_98)
    do you know for 8b how many marks
    It was 6 marks.
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    (Original post by jj202)
    It was 6 marks.
    for 8b wasn't the range -pi/2 to pi/2 or something like that. I remember only being able to use two values from the 4 mentioned by OP
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    was 5i) tan(4/3) can't remember what i put and for 8b) did people use quadratic equation and get like -4 and 1 before you inverse tan to get the answer
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    (Original post by tigermoth890)
    Hi I think that is the one where you had 0=e^2x(3sintheta - 4costheta)

    Something like that.
    Then you end with tantheta = 3/4 or something.
    do you remember the question and how you solved it , thank you so much . Because i forgot my final answer

    Yay thank u so much but i forgot how i solved it
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    did it specify the normal equation in the form y= ?
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    (Original post by veldt127)
    Like I said, if people have the questions I can work through them, I've done working for all the ones I could remember
    Q8 b was this: solve the equation 6cot2x + 3tanx = cosec^2x - 1
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    (Original post by particlestudent)
    Im sure only the middle two values were allowed for 8B
    It was between -ve and +ve pi, though to get full marks I only think you need two solutions, so you should be fine lad
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    (Original post by Ashleighh37)
    does anyone know how many marks each questions is worth?
    http://www.thestudentroom.co.uk/show...847081&page=92
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    (Original post by veldt127)
    I'm not 100% sure on all of these, but I think most are correct.

    Question 1
    Spoiler:
    Show
    A. -5, 6
    B. 6
    Question 2
    Spoiler:
    Show
    A. (20-4x^2)/(x^2+5)^2
    B. >sqrt(5), <-sqrt(5)
    Question 3
    Spoiler:
    Show
    A. R= sqrt(5)
    a= 26.57°
    B. 33.0°, 273.9°
    C. 86.1°
    Question 4
    Spoiler:
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    Ai. 21
    ii. ln(2.5)
    iii. 25
    B. Show that question
    C. 1.4368, 1.4373
    D. Show the root was in [1.4365, 1.4375]
    Question 5
    Spoiler:
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    i. 0.9463
    ii. 0.5cosec(4Y)
    Question 6
    Spoiler:
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    A. A=3, B=4
    B. Normal eqn. -0.5x + 17.5
    Question 7
    Spoiler:
    Show
    A. graph of arcsin, sort of a rotated S shape. Went through (0,0), (-1,-pi/2), (1,pi/2)
    B. -1-0.5sqrt(3)
    Question 8
    Spoiler:
    Show
    A. Trig proof
    B. -2.848, -1.277, 0.294, 1.865
    Question 9
    Spoiler:
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    A. 6.740 - dose after 4 hours
    B. show that dose equalled a value after 7 hours
    C. 5ln(2 + 2e^-1)
    A = 5, B = 2
    PDF with working for 1-4 and 9 : http://www.thestudentroom.co.uk/atta...5&d=1466508775
    isn't your answer for 7 part b outside the range of what x could be? wasn't it between -1 and 1?
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    (Original post by jj202)
    Q8 b was this: solve the equation 6cot2x + 3tanx = cosec^2x - 1
    Ah right. You knew acot2x + tanx = cotx from part a

    so 3cotx = cosec^2x - 2

    cosex^2x = 1 + cot^2x

    Then it's a quadratic; cot^2x - 3cotx - 1 = 0
    cot x = 3(+/-)sqrt(13) divided by 2

    tan x is the reciprocal of that; I'm not entirely sure on the +1/-1/-2 part but it was along those lines.
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    (Original post by samkts)
    isn't your answer for 7 part b outside the range of what x could be? wasn't it between -1 and 1?
    It was actually g(x+1) so the range would be lower - move the graph 1 to the left.
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    The examiner must think I'm on drugs for one of the questions. I misread the limits as -pi/4 to pi/2 for one question, rather than pi/4 to pi/2 (the tan(-3/4) one). So I kept getting two answers and starting over. Did this about 3 times before realising.

    (Original post by PeterMJP)
    If you put 2cosec4y but did all working out correct how marks would you lose and what do people think an A will be?
    Probably just the 1.
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    What do u think the grade boundaries are
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    how many mark would i lose if i didn't put the inequalities signs correctly?
    I did however get + and - root5
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    (Original post by jwanjwan)
    i agree however i dont know about 7 part b, i cant really remember the question but i believe i left it in terms of sin but your answer looks more complete
    I got sin(-pi/3)-1 or something
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    (Original post by DomWoodcock)
    Yeah I think I remember the limits between -pi/2 and pi/2
    No, the limits were between -pi/4 and pi/2. I m sure of this cos I also thought it was between -pi/2 and pi/2 but changed it later
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    (Original post by keyadeb)
    how many mark would i lose if i didn't put the inequalities signs correctly?
    I did however get + and - root5
    Probably only one mark cos you calculated the critical values
 
 
 
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