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Solomon C1, Paper H question! watch

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    I'm stuck on the following:

    5. (a) Sketch on the same diagram the graphs of y = [(x − 1)^2](x − 5) and y = 8 − 2x. [DID THIS]
    Label on your diagram the coordinates of any points where each graph meets
    the coordinate axes.
    (b) Explain how your diagram shows that there is only one solution, α, to
    the equation
    (x − 1)2(x − 5) = 8 − 2x [DID THIS and was due to the fact that they only intersected once]
    (c) State the integer, n, such that
    n < α < n + 1


    Also, on this same paper the following:
    8. (a) Prove that the sum of the first n positive integers is given by:
    ½ n(n+1)

    (b) Hence, find the sum of
    (i) the integers from 100 to 200 inclusive, [DID THIS and got 15'150]
    (ii) the integers between 300 to 600 inclusive which are divisible by 3

    PLS HELP ASAP,
    Thanks.
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    the second question can be done with the sum of an arithmetic series proof which should be in your C1 book with d = 1 & a = 1
    Or if you wan to think about it another way you pair the higest number with the lowest number and add the 1 + n = n+1
    you then conitnue with the next inner most pair 2 + n - 1 = n +1
    continuing this you end up with n/2 pairs if n is even and (n-1)/2 pairs + the middle term (n + 1)/2 term if odd
    both of which sum to n/2(n+1)

    the last part is equal to to 3(sum(intergers 100 to 200)) as
    so 300 + 303 + 306 +.... 600 = 3(100 + 101 + 102 +... + 200)
    then just use part b(i)
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    Ok, the first bit is wrong. I've attached a file to make things clearer. Im stuck on questions 3b), 5c), and 8b)ii)

    Thanks
    Attached Files
  1. File Type: doc Questions.doc (96.5 KB, 104 views)
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    Firstly I could swear this looks like C2??
    Besides if you consider that 'alpha' is the pt of intersect of the 2 graphs then u can make f(x)=0.now using that and a suitable iteration formula. Actually as it is for 1 mark maybe draw your graph a bit neater and then jst read it off the graph???OR keep ur sketch and like wat Retropmot sed use values inbetween the ones you already kno,(intersects with the coordinate axes etc.)
    FOr the last one about inclusive concerning 300-600 is the same way you approached the previous question just ur d=3 n a=300 n l=600

    Hopefully wat ive sed is helpful and correct!!
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    (Original post by Fade Into Black)
    Firstly I could swear this looks like C2??
    It's coz i took the problem i had in two threads and put it into one attachment, if you know wht i mean. And yes, it's helped me out finally.
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    On the attached sheet, on question 3c) I don't understand how, according to the mark scheme, once I have the area of the graph, and to find the volume, they multiply it by (8)^2 then x2. Now I understand the x2 bit coz it's area of cross section * length, but why (8)^2?

    PLS HELP.
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    The first part i made a stupid mistake it should be fairly obvious that its 4 if you draw the diagram well,
    Not sure what your problem with 3bii is?

    As for that 8^2 in the area question it tells you that each unit on the diagram is 8 cm so if you work out the cross sections from that you need to multiply it by 8^2 in order to get the actual area

    Hope that helps
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    But it says 1 unit on each axis represents 8cm, and on the x-axis there are 2 units and on the y-axis there's 1. So its not 1 and 1 unit, but its 2 and 1 unit. As for the first part, once i drew the diagram it made so much sense and i realized the stupid mistake i made.
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    if you have a square that 1unit by 1 unit the area is 1 unit ^2

    if 1 unit represents 8 cm the each side is 8cm then the area is 1*8^2 = 64cm^2

    if the square was 4 by 4 units then the area would be 16unit^2

    each of these units has area 8^2 so the area in cm is 16*8^2cm^2


    U've worked out the area of the cross-section in units^2 so in order to get the cross section in cm^2 you need to multiply it by 8^2. Then to get the volume you multiply its by thickness to get the volume.
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    Ah right, now I get it. So it has to the be area of the units, * by 8^2. Ok, thanks a lot, this makes sense now.
 
 
 
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