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AQA A2 Mathematics MPC3 Core 3 - Wednesday 15th June 2016 [Official Thread] watch

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    (Original post by MahuduElec)
    I remember doing a question where if you considered both ranges, you would incorrectly reject one answer. Ever since then, I only look at the function closest to the (x) and have never incorrectly rejected/accepted a value.

    Think of it like this, for fg(x) if the value of x does not satisfy g(x)'s domain, you won't have a value for g(x). And so you won't be able to input a value into f(x), even if the value satisfies f(x)'s domain. As what fg(x) is essentially doing is inputting the value from g(x), into the function f(x). So if your value from g(x) isn't valid/ doesn't work/ math error, you can't input that into f(x) and expect a valid value.
    if value of g(x) gave a value outside the domain of f(x)?
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    (Original post by Ano123)
    Multiply by sec x + tan x.
    In a proof you can only work on the LHS in order to prove the right hand side..
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    (Original post by liemluji)
    oops sorry it's supposed to be cos x
    Name:  Screen Shot 2016-06-14 at 18.00.10.png
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Size:  20.2 KBHere you go
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    (Original post by C0balt)
    well so they are defined by their domains which will give the range too so yeah I think they both have to be satisfied right
    The range defines the values f(x) can take, not the value you can obtain for x. You can have say 1. x=5, for f(x), but the function may not allow 2. f(x) =5. First part is governed by the domain, second by the range. Never look at the range of f(x) to see if x values are valid.

    This is what I know from C3, don't quote me as I've never been taught this, it's just something I've been following.
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    Good luck guys!! Tomorrow is the big day...
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    Say it was g(f(x)) and the value of x you found satisfies the domain of f(x). But what if the value you get for f(x) at that value does not satisfy the domain of g(x)? Shouldn't it be rejected?
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    (Original post by Ano123)
    Attachment 550055
    Try this integral.
    lol i wouldn't know where to start if you gave me a substution then i could do it
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    (Original post by MahuduElec)
    The range defines the values f(x) can take, not the value you can obtain for x. You can have say 1. x=5, for f(x), but the function may not allow 2. f(x) =5. First part is governed by the domain, second by the range. Never look at the range of f(x) to see if x values are valid.

    This is what I know from C3, don't quote me as I've never been taught this, it's just something I've been following.
    Yes, if you have f( g(x) ) then the domain of g is fed to g(x), the output (the range) is fed into f. So if you are finding x for checking fg(x) = a, then valid values of x would be in the domain of g (as long as the range of g is in the domain of f)
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    (Original post by xs4)
    to say thank you would simply underwhelm my appreciation.
    hey, whats the hardest core 3 paper you've come across?
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    The A* boundary has never dipped below 62 for C3 as far as I can remember. Why are none of the papers challenging? Do you think this year they'll pull all the stops and finally give us a difficult paper?
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    (Original post by Guls)
    hey, whats the hardest core 3 paper you've come across?
    I don't think one exists tbh. The June 2013 one with all the graph sketching was a bit iffy, but not difficult - just awkward.
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    (Original post by C0balt)
    if value of g(x) gave a value outside the domain of f(x)?
    The yes you would reject that x value. But only because the value generated by g(x) is out of the domain of f(x). Not because the original x value is.

    Say domain of f(x) is 2<x<7, your x value is 10, g(x) gives 9. You must reject x=10 for fg(x).
    Another function, h(x) gives 5 when x=10, now you accept x=10 for fh(x).

    This is more to do with the range of g/h(x) being outside the domain of f(x). I've never seen that come up on AQA c3, it'd be worth a lot of marks if it did come up though.
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    (Original post by Bunmi_ra)
    i know the reason a function may not have an inverse is because it has one-to-one mapping, but what does one-to-one mapping mean??
    and how do you work out the range of a function if its domain is all real values of x??
    1) a one to one means for every x value there is only one y value
    2)easiest option plug the equation in to your calc
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    Can someone kindly explain to me the transformation rules? I get really confused when it comes to x or y axis...
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    (Original post by MahuduElec)
    The yes you would reject that x value. But only because the value generated by g(x) is out of the domain of f(x). Not because the original x value is.

    Say domain of f(x) is 2<x<7, your x value is 10, g(x) gives 9. You must reject x=10 for fg(x).
    Another function, h(x) gives 5 when x=10, now you accept x=10 for fh(x).

    This is more to do with the range of g/h(x) being outside the domain of f(x). I've never seen that come up on AQA c3, it'd be worth a lot of marks if it did come up though.
    Alright thank you
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    (Original post by tanyapotter)
    I don't think one exists tbh. The June 2013 one with all the graph sketching was a bit iffy, but not difficult - just awkward.
    I found that awkward aswell. thanks.
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    (Original post by bartbarrow)
    Yes, if you have f( g(x) ) then the domain of g is fed to g(x), the output (the range) is fed into f. So if you are finding x for checking fg(x) = a, then valid values of x would be in the domain of g (as long as the range of g is in the domain of f)
    Fully agree with that. Most likely in C3 the range of g is going to be within the domain of f, unless they want to be sly *******s.
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    (Original post by Language student)
    Can someone kindly explain to me the transformation rules? I get really confused when it comes to x or y axis...
    For when it's two transformations, one altering x one altering y, the order does not matter.

    When both alter y, the order follows BIDMAS, so stretches before translations.

    When both alter x, the order follows Inverse BIDMAS, so translations before stretches.

    For reflections, i dont think the order matters, and I've never seen a rotation one come up in any paper.

    Posted from TSR Mobile
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    Anyone else need an A* in this? I'm really hoping it's not a piss easy paper to the point where if i mess up 1 small Q it costs me a **** tonne of UMS.
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    (Original post by Language student)
    Can someone kindly explain to me the transformation rules? I get really confused when it comes to x or y axis...
    You mean with stretches?

    The rules are:

    Stretch in x axis scale factor a, replace x with x/a
    => Therefore, a stretch (in x) of sf 1/a, x with be replaced with ax

    Same again with y except a => y/a etc


    If you are combining translation/stretch then just think about it logically:

    Example 1
    "Describe a sequence of transformations to translate y=lnx to y=4ln(x-e)" from Jan 12

    There will be a stretch and translation (hopefully you can spot that). It can be done in any order but depending on what you choose, the numbers will (sometimes) be different

    I recommend you do a translation first because I think it is less hassle

    If a translation of (a, b) replaces x with x-a and y with y-b, then we want a translation (e, 0)

    With just that transformation we have y=ln(x-e)

    Now with y=ln(x-e) we just need to transform it to y=4ln(x-e) or, rewritten, y/4=ln(x-e). With the rules above you can see that the stretch would be in the y axis and we need to replace y with y/4 so the scale factor will just be 4.

    In that question the order they are done in does not matter

    Example 2
    Describe a sequence of transformations that map f(x) => f(2x+2)

    You could approach it two ways:

    Method 1
    f(x) => f(x-2) => f(2x+2)
    so translation (-2, 0), stretch sf 1/2 x axis

    Method 2
    f(x) => f(2x) => f(2(x+1)) or f(2x+2)
    so stretch sf 1/2 x axis, translation (-1, 0)

    Notice how when you are translating it is x that is being replaced, so you would need to remember that it says 2x not x.
 
 
 
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