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    Please could someone help me with this P4 question, I think I did the first bit and ended up with x is more than 1, which doesn't seem to work out for part (b). Please could someone help? Thanks
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    I get for part 1, x > 1 or x < -1/2

    So for part 2, sin t is x, and so you have 2 inequalities;
    sin t > 1
    sin t < -1/2

    The first is impossible, so persue the second.
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    Thanks a lot . I must have forgotten the minus half part, thanks for pointing it out, thanks .

    Would people mind if I used this thread to post any further questions, as I'm working through a few papers at the moment?

    And, would people mind if I used their answers along with mine (once I get the papers marked of course) for a website which could have markschemes for all the papers Edexcel are too stingy to provide (I can scan the papers too, at a much higher quality also).
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    I did that and got -1/2 < x, x > 1 for part a, how can you call that P4 :confused:
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    (Original post by mik1a)
    I did that and got -1/2 < x, x > 1 for part a, how can you call that P4 :confused:
    how can u judge p4 just by doing that qu,i'll give u another 1 p4 qu
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    I'm having a couple of troubles on the M3 also, if anyone would be kind enough to lend a hand, I'd me mucho gratefuls . Thanks in advance .
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    And this one.

    It was P4, I know that they're easy, inequalities, I just made a lousy error.
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    second part I/m not sure about trig inequalities but I had a go and got either

    pi/6 < t < pi/2
    -5pi/6 < t < -pi/2
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    M3 one

    I don't understand "with a force 5(AB) Newtons"

    What does this mean with AB in brackets? is it notation or 5*AB N?
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    (Original post by mik1a)
    M3 one

    I don't understand "with a force 5(AB) Newtons"

    What does this mean with AB in brackets? is it notation or 5*AB N?
    AB is the distance of B from A
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    (a) 2x > (3x+1)/(x+1)

    2x(x+1) > 3x+1 , x>-1 (assumption)
    2x^2 - x - 1 > 0
    (2x+1)(x-1) > 0

    2x+1 > 0 and x-1 < 0 (1)
    or,
    2x+1 < 0 and x-1 > 0 (2)

    (1) gives
    x>-1/2 and x<1 which satisfies the condition (x < -1)

    (2) gives
    x<-1/2 and x>1 - contradiction!

    So, solution is
    x>-1/2 and x<1
    i.e.
    -1/2 < x < 1
    ========

    (b) the above solution range satisfies -pi < x < pi
    so from part (a) we have
    -1/2 < sint < 1
    giving
    -pi/6 < t < pi/2
    ==========
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    (Original post by Fermat)
    (a) 2x > (3x+1)/(x+1)

    2x(x+1) > 3x+1 , x>-1 (assumption)
    2x^2 - x - 1 > 0
    (2x+1)(x-1) > 0

    2x+1 > 0 and x-1 < 0 (1)
    or,
    2x+1 < 0 and x-1 > 0 (2)

    (1) gives
    x>-1/2 and x<1 which satisfies the condition (x < -1)

    (2) gives
    x<-1/2 and x>1 - contradiction!

    So, solution is
    x>-1/2 and x<1
    i.e.
    -1/2 < x < 1
    ========

    (b) the above solution range satisfies -pi < x < pi
    so from part (a) we have
    -1/2 < sint < 1
    giving
    -pi/6 < t < pi/2
    ==========
    i disagree because in that interval the rectangular hyperbolic curve is greater than y=2x and besides since x+1 can be negative or positive the jump from 2x>(3x+1)/(x+1) to 2x(x+1)>3x+1 is flawed and only gives u 1 solution
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    hope dis helps the confusion!,best thing to do with these qu is to draw the graphs
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    I think IntegralAnomaly is right, multiplying both sides by the RHS denominator does remove a solution.
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    yea so by lookin at the graph the line is 'higher' than the curve in the intervals x>1 and -1<x<-1/2.
    and since it has been said by the james than sint cannot be >1,
    so -1<sint<-1/2.
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    for a full solution, you have to solve 2 quadratic inequalities:

    2x(x+1) > 3x+1 for x>-1

    2x(x+1) < 3x+1 for x<-1
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    (Original post by IntegralAnomaly)
    i disagree because in that interval the rectangular hyperbolic curve is greater than y=2x and besides since x+1 can be negative or positive the jump from 2x>(3x+1)/(x+1) to 2x(x+1)>3x+1 is flawed and only gives u 1 solution
    You're right.
    And there's another error in my post
    I guess I've been away from this stuff for too long!
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    (Original post by Fermat)
    You're right.
    And there's another error in my post
    I guess I've been away from this stuff for too long!
    aahh every1 makes mistakes....
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    Thanks guys you've really helped
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    Thanks guys, you've really helped .
 
 
 
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