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    (Original post by milliondollarcorpse)
    That's not the way to solve equations with the modulus function! What are you solving here, x in terms of a? Add one to both sides and divide through by x (|2x+a|=1/x). Then do a quick sketch of the graphs |2x+a| and 1/x and consider points of intersection.
    You can solve modulus functions like that, treat the modulus as a bracket, and expand, and after you found your solution, put a minus infront of the bracket, and solve again so you have two solutions. For this question as it involves "a" is gonna be hard to do, your method is useful by drawing the graph, but sometimes it will be hard to pin point where exactly the lines intersect so the method I stated will be required.
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    i hope you want to know the name so you can send an angry letter to the artist.
    what an abomination that was.
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    (Original post by samir12)
    You can solve modulus functions like that, treat the modulus as a bracket, and expand, and after you found your solution, put a minus infront of the bracket, and solve again so you have two solutions. For this question as it involves "a" is gonna be hard to do, your method is useful by drawing the graph, but sometimes it will be hard to pin point where exactly the lines intersect so the method I stated will be required.
    Well yeah, you can do it like that (not the way I personally do it), but he didn't include the 'putting a minus infront of the bracket' bit so I thought I'd just explain how I go about solving them. And you don't need to pin point exactly where the lines intersect, you only need to see the 'regions' they'll intersect in so you can form equations with the appropriate signs.

    (Original post by indie_couture)
    thankyou!
    I was wondering if you could help with domains/range.
    How do you know if it is all real numbers XER or whatever.
    for the range do you put in the x values of the domain or x=0?
    i can only sometimes get these right xxx
    Do you have any questions to post in particular?

    Drawing a quick sketch of the graph is always a good idea if you can't see the domain/range. For the range you would use the domain yes. However if the domain of x is simply that it's an element of the reals (or something along those lines), then you need to consider what the graph looks like. If you have any particular questions you're stuck on I'm sure myself and a few others can help.
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    (Original post by milliondollarcorpse)
    Well yeah, you can do it like that (not the way I personally do it), but he didn't include the 'putting a minus infront of the bracket' bit so I thought I'd just explain how I go about solving them. And you don't need to pin point exactly where the lines intersect, you only need to see the 'regions' they'll intersect in so you can form equations with the appropriate signs.


    Do you have any questions to post in particular?

    Drawing a quick sketch of the graph is always a good idea if you can't see the domain/range. For the range you would use the domain yes. However if the domain of x is simply that it's an element of the reals (or something along those lines), then you need to consider what the graph looks like. If you have any particular questions you're stuck on I'm sure myself and a few others can help.
    thanks. Okay well this isnt what i have been previously stuck on but:
    The functions f and g are defined by:
    f- 3x-4
    g- 2/(x+3)
    solve gf(x)=6
    so
    2/(3x-4)+3 = 6
    2= 6(3x-1)
    then this is where i differ from the m/s:
    2= 18x-6
    18x=8
    x=2.25
    but the m/s do
    2= 6(3x-1)
    2/6=3x-1
    the and 1 to 2/6 then divide by 3 to get 4/9. Why doesnt it work my way? x

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    18x=8
    x= 8/18
    x= 4/9.

    Nothing else :P You did 18/8, which is incorrect.
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    The song in the background of the new Femibion advert. I literally can't hear enough to search the lyrics
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    (Original post by indie_couture)
    how would you do this

    if f(x)= 4-x^2

    solve ff(0)
    so 4-(4-x^2)^2
    x^4-8x^2-12=0

    i dont know how i would do it.
    xxx
    Umm, couldn't we have just found f(0)=4 and then solve f(4)? :o:
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    find the range of:
    (2x-3)/(x-2) XER, x>2
    how would you do this?
    i thought you put x=2 but this would give 1, when the answer is 2?!?!? xxx
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    (Original post by The Hatman)
    Owl City - Fireflies
    Thanks dude
    I will make sure to +rep you tomorrow!!!
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    (Original post by indie_couture)
    find the range of:
    (2x-3)/(x-2) XER, x>2
    how would you do this?
    i thought you put x=2 but this would give 1, when the answer is 2?!?!? xxx
    For these questions I'd try and construct a table to help you (might not).
    so x=2 y=infinity
    x=3 y=3
    x=4 y=2.3
    then try x with a large number i.e. 100000, then y is like 2.00001 or something like that. so now you know (or already do :rolleyes: ) that the minimum value is 2 and it is tending towards infinity, so with luck, the range should be f(x)>2.

    (experiencing a bit of deja vu, i think i did this exact question for someone else lol)

    hope that helps.
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    (Original post by indie_couture)
    find the range of:
    (2x-3)/(x-2) XER, x>2
    how would you do this?
    i thought you put x=2 but this would give 1, when the answer is 2?!?!? xxx
    (Original post by Superman_Jr)
    For these questions I'd try and construct a table to help you (might not).
    so x=2 y=infinity
    x=3 y=3
    x=4 y=2.3
    then try x with a large number i.e. 100000, then y is like 2.00001 or something like that. so now you know (or already do :rolleyes: ) that the minimum value is 2 and it is tending towards infinity, so with luck, the range should be f(x)>2.

    (experiencing a bit of deja vu, i think i did this exact question for someone else lol)

    hope that helps.
    :p: Me! Except the top part was different! I THINK to find the range, you can use the denominator (bottom part) and set it not equal to 0 and solve the equation x - 2 =/=0 therefore, f(x) >2 is the range.

    Please can somebody confirm?

    Much appreciated. Thanks.
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    Wow. I worked June 2008's paper and I got 60/75 O_O

    Silly mistakes and I just didn't know some things!

    Can someone please simplify this:

    y - 8 = 16 ( x - 0.5ln(2-1) )

    (this is Q1 of June 2008 by the way).

    Thanks.
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    ^^
    I was wondering the same for the June 2008 paper
    I got to the same point but couldnt simplify it to what the mark scheme had..
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    I kind of get near
    8ln2 (as in MS) = 5.545

    y-8 = 16(x - 0.5ln(2-1) )
    y-8 = 16x - 8ln(2) + 8
    y=16x + 16 - 8ln(2)

    Thats theo nly way i can get the answer, although Im not sure if thats how you are meant to do it.
    I multiplied 16 x ln2 then times that by -.5 = 8ln 2 (5.455)
    then 16 x - (-1) = +8
    then re-arrange?
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    (Original post by Doughboy)
    Wow. I worked June 2008's paper and I got 60/75 O_O

    Silly mistakes and I just didn't know some things!

    Can someone please simplify this:

    y - 8 = 16 ( x - 0.5ln(2-1) )

    (this is Q1 of June 2008 by the way).

    Thanks.
    You've got the bracket in the wrong place. It's this...

    y-8=16(x-0.5(ln2-1))

    y-8=16x-8ln2+8
    y=16x-8ln2+16
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    what the..
    Show that f(x) = 0 has a root (alpha) between x = 1.4 and x = 1.45
    f( x ) = x^3 - 2x - 6

    I sub 1.4 into x.. but i get -6?
    what am i doing wrong


    EDIT NVM : The copy of the paper I have had atyop and missed the 3
    it should have been
    3x^3 - 2x -6


    Also found June 2008 good :P
    58/75 - and ive only done 1 day revision
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    (Original post by Noble.)
    You've got the bracket in the wrong place. It's this...

    y-8=16(x-0.5(ln2-1))

    y-8=16x-8ln2+8
    y=16x-8ln2+16
    Thanks.

    Just a little reminder to everyone:

    (Original post by Exam Report June 2008, Math)
    These papers are marked online and, if a pencil is used in drawing sketches of graphs, a
    sufficiently soft pencil (HB) should be used and it should be noted that coloured inks do not come
    up well and may be invisible. A number of candidates gave answers which went outside the area
    on the pages which is designated for answers and this caused the examiners some
    difficulties.
    Also, how the flap do you find P?

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    I believe (as its the only way i could figure it out)
    That because the two lines are meeting, P (x co-ord) is the midpoint of the two lines x co-ordinates
    so once you have found R
    its the midpoint between -3 and R
    then you can sub this into the formula and get the y coordinate.

    I got a bad feeling this exam will be hard, because going back to jan 2008 the papers since (june 2008 - june 2009) have not been as difficult imo
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    (Original post by Doughboy)
    Thanks.

    Just a little reminder to everyone:



    Also, how the flap do you find P?

    Is the answer (-1, 2) for P?
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    (Original post by Doughboy)
    Also, how the flap do you find P?

    considering f(x)

    what is the smallest value of a modulus function?
    and therefore what is the maximum value of f(x)?
    what value of x would result in this value?
 
 
 
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