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    Could someone explain to me how to do this question please? (It's from the C34 Jan 15 paper)

    f(x) = x^2 - 3x + 1
    Find the range of f(x)

    I completed the square to get (x - 3/2)^2 - 5/4 , but I'm not sure what to do after that. I looked at the mark scheme but it didn't really help!

    Thank you
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    And btw do we have to then learn the counterproof and the contradiction stuff ? some people saying no and some saying yes :/ ? anyone know for sure ?
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    is B an 1/8?
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    (Original post by EspDe97)
    Could someone explain to me how to do this question please? (It's from the C34 Jan 15 paper)

    f(x) = x^2 - 3x + 1
    Find the range of f(x)

    I completed the square to get (x - 3/2)^2 - 5/4 , but I'm not sure what to do after that. I looked at the mark scheme but it didn't really help!

    Thank you
    why not just plot the graph and find the smallest/maximum value of y?
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    Someone please give me some tips....btw is sin 40=cos50?? I know a dumb question but just forgot if that how it works...


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    https://07a69ccf283966549a9350d1a669...%20Edexcel.pdf Could somebody please help with 6c, I just don't understand any of it :-(
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    (Original post by sorarosa1998)
    Someone please give me some tips....btw is sin 40=cos50?? I know a dumb question but just forgot if that how it works...


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    That is correct
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    (Original post by RonnieSaha)
    If tan(x)=1/0 ie infinity, does x equal 90, 270 etc...

    (I ask this as I have seen it before on an exam paper but would like further confirmation?)
    Yeah that's right, you just need to give the values at the asymptote. Though if we get something like this in an exam e.g. cotx=1, you could be docked marks for writing tanx=infinity, you'd have to state that it tends towards infinity, and therefore tends towards 90/270 etc.
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    (Original post by physics_maths7)
    Attachment 425523
    as somebody else has said when the denominator=0 there is an asymptote. So create an equation and solve for x. This value of x = beta. I've posted my solution if it helps, also is there a mark scheme for this?

    For B, how did you make rootx... = 0 when its at the bottom and divided by 6.
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    June 2014 4c???
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    (Original post by sorarosa1998)
    Someone please give me some tips....btw is sin 40=cos50?? I know a dumb question but just forgot if that how it works...


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    Yes. Sin(x)=Cos(90-x) and Cos(x)=Sin(90-x)

    If you cant get your head around it, expand cos(90-x) using formula booklet and see what you get
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    (Original post by MATTTT)
    Can some on help. Normally to find the range you have to draw the graph. However this question is 3 marks and they differentiate it and get x=2.

    Is there a way that you can differentiate the function and find the range ?
    please can someone show how to do this...
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    (Original post by adorablegirl1202)
    June 2014 4c???
    Simultaneous equations using the two points on f(X). remember a cannot equal 0 though...
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    (Original post by TeeEm)
    just did this question with one of my students ... (he got eaten badly)
    Any takers
    wait nah I've made an error
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    (Original post by laurenjjj)
    https://07a69ccf283966549a9350d1a669...%20Edexcel.pdf Could somebody please help with 6c, I just don't understand any of it :-(
    you know g(x)=ln2x, so for g(x^2), it must be ln2x^2. You do the same with g(x^3)

    This gives you:
    ln2x + ln2x^2 + ln2x^3 = 6
    You can then split this up using rules of logs
    (lnx +ln2) + (lnx^2 + ln2) + (lnx^3 + ln2) =6
    Bring down the powers of x and simplify, giving
    6lnx + 3ln2 = 6
    Then just solve for x
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    when x=0 then y=1
    crosses the y axis at 1
    the parabola is a u shape (positive)
    the vertex of the u is the smallest value that y can be
    vertex ---> -5/4
    use the inequality from the question 0<=x<=4
    when f(4)=(4-3/2)^2-5/4 ---> 5 (biggest value y can be)
    -5/4<=y<=5
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    (Original post by broconomist)
    64.8 years?
    year 2044
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    (Original post by suyoof123)
    For B, how did you make rootx... = 0 when its at the bottom and divided by 6.
    Firstly, I could be wrong (I'm fairly sure I'm not but if someone would check it would be helpful).
    Secondly, I don't quite understand the question but I'll try my best to explain myself.
    any fraction with a denominator of 0 when typed into a calculator will display 'math error' because you cant divide by zero. An asymptote is caused when the denominator of a function is equal to zero so we know that we have to make the denominator=0. this creates the equation (sqrt(8))*cos(x-pi/4)=0.
    0/sqrt(8)=0 therefore cos(x-pi/4)=0. Then do an inverse cosine on both sides (I don't know the correct terminology) and you find x-pi/4=pi/2. By adding pi/4 to both sides we find x=3pi/4. Therefore beta=3pi/4 because beta is simply the value of x which causes an asymptote to be created.
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    (Original post by Gilo98)
    Simultaneous equations using the two points on f(X). remember a cannot equal 0 though...
    Thanks!!! Do i just take the mod value to be positive? i mean i did origianally but then i thought should i have done it negative as well?
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    (Original post by laurenjjj)
    https://07a69ccf283966549a9350d1a669...%20Edexcel.pdf Could somebody please help with 6c, I just don't understand any of it :-(
    ln(2x)+ln2x^2+ln2x^3=6
    ln(2x*2x^2*2x^3)=6
    ln8x^6=6
    8x^6=e6
    then you should solve for x
 
 
 
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