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    I need some help with a few questions form the c3 june 2008 paper.

    Here's the link: https://eiewebvip.edexcel.org.uk/Rep...e_20080606.pdf

    2. (c) (i) Write down the maximum value of 5cosx +12sinx.

    (ii) Find the smallest positive value of x for which this maximum value occurs.


    3. Given that f(x) = 2- lx+1l

    (c) find the coordianates of the points p,q and r

    (d) solve f(x) = 0.5x

    4. (b) Find the range of f.

    Also how do you find the domain?


    Thank you for any help in advance.
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    2ci You can use your value of R to tell you how much R\cos(x-\alpha) extends above and below the x axis, so the maximum value is just R which is 13.

    ii You're solving 13\cos(x-\alpha)=13 so \cos(x-\alpha)=1. Since you want the smallest x such that x>0 then therefore you want the smallest solution to \cos(x-\alpha)=1 where x-\alpha>-\alpha. You can sketch a graph to find this solution.

    c)
    P: Think of transformations of graphs. The 'point' on y=|x| is at (0,0), so the point on y=|x+1| is at (-1,0) so then work out the location of the point on y=-|x+1| and hence the location of the point on y=2-|x+1| which is P.

    Q: Since it's on the y axis, the x coordinate is zero, and the y coordinate is f(0) which you can just calculate

    R: Solve f(x)=0. You should get two solutions. One of them will be -3 which you don't need to worry about since that's given and the other solution will be the one you're interested in. This is your x coordinate.

    Alternatively, if you don't like equations with modulus signs in, you could use the coodinates of P and Q to get the equation of the line connecting P, Q and R then once you have that equation, let y=0 and find the x coordinate.

    d) Probably the easiest way to do this is to split it into cases: when x+1>0 and when x+1<0.

    4 b 1/(3+1)=1/4. As x increases, x+1 gets bigger so 1/(x+1) gets smaller and in fact it gets very very close to zero but doesn't quite reach it. However since the domain of f is x>3 then f(x) doesn't quite reach 1/4 either. So f can take values 0<f(x)<1/4.

    As for finding the domain, if you're referring to part c then just put down the same as the range of f. When you were told that the domain of f is x>3 then this is just because the person writing the question decided that would be a nice domain to use. There's no way you could have figured this out yourself without being told.

    However if you are asked to find a domain of a function then you should put down the range of values where the function is defined. For example the domain of \sqrt{x} is x\geq0 and the domain of 1/x is x\neq0.
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    (Original post by ttoby)
    2ci You can use your value of R to tell you how much R\cos(x-\alpha) extends above and below the x axis, so the maximum value is just R which is 13.

    ii You're solving 13\cos(x-\alpha)=13 so \cos(x-\alpha)=1. Since you want the smallest x such that x>0 then therefore you want the smallest solution to \cos(x-\alpha)=1 where x-\alpha&gt;-\alpha. You can sketch a graph to find this solution.

    c)
    P: Think of transformations of graphs. The 'point' on y=|x| is at (0,0), so the point on y=|x+1| is at (-1,0) so then work out the location of the point on y=-|x+1| and hence the location of the point on y=2-|x+1| which is P.

    Q: Since it's on the y axis, the x coordinate is zero, and the y coordinate is f(0) which you can just calculate

    R: Solve f(x)=0. You should get two solutions. One of them will be -3 which you don't need to worry about since that's given and the other solution will be the one you're interested in. This is your x coordinate.

    Alternatively, if you don't like equations with modulus signs in, you could use the coodinates of P and Q to get the equation of the line connecting P, Q and R then once you have that equation, let y=0 and find the x coordinate.

    d) Probably the easiest way to do this is to split it into cases: when x+1>0 and when x+1<0.

    4 b 1/(3+1)=1/4. As x increases, x+1 gets bigger so 1/(x+1) gets smaller and in fact it gets very very close to zero but doesn't quite reach it. However since the domain of f is x>3 then f(x) doesn't quite reach 1/4 either. So f can take values 0<f(x)<1/4.

    As for finding the domain, if you're referring to part c then just put down the same as the range of f. When you were told that the domain of f is x>3 then this is just because the person writing the question decided that would be a nice domain to use. There's no way you could have figured this out yourself without being told.

    However if you are asked to find a domain of a function then you should put down the range of values where the function is defined. For example the domain of \sqrt{x} is x\geq0 and the domain of 1/x is x\neq0.
    Thank you for the help.
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    thats really helpful to what i wanted to thanks
    also, do you know how i can find edexcel modular unit 3 June 2010 paper?
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    (Original post by Jess_Holland)
    thats really helpful to what i wanted to thanks
    also, do you know how i can find edexcel modular unit 3 June 2010 paper?

    http://www.thestudentroom.co.uk/show....php?t=1459768

    This thread post number 3. I'm assuming you wanted c3.
 
 
 
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