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    (Original post by TheNote)
    hey guys for question 4 I got:
    r = ((3V/(a+b))/pi)^1/2

    Is this correct?
    original equation was 1/3(r^2)pi(a+b) = V

    slowly finding out i got more and more wrong.
    can anyone clear this up for me <3
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    (Original post by TheNote)
    can anyone clear this up for me <3
    I think the equation went:

    V=1/3*pi*r2*(a+b)
    3V=pi*r2*(a+b)
    3V/pi=r2*(a+b)
    3V/(pi*(a+b))=r2
    r=(3V/(pi*(a+b))1/2


    Hope that helps.
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    (Original post by TheNote)
    i am assuming he meant brackets due to a typo
    Yeah my bad!! i did mean the brackets but forgot to type them :/
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    (Original post by TheNote)
    can anyone clear this up for me <3

    i got r = ((3V(a+b))/pi)^1/2


    1/3(r^2)pi(a+b) = V

    (r^2)pi(a+b) = 3V

    (r^2)(a+b) = 3V/pi

    (r^2) = 3V/pi/(a+b)

    (r^2) = 3V(a+b)/pi

    r = (3V(a+b)/pi)^1/2
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    (Original post by user1-4)
    does anybody the last question? We can go through it and figure out the 'actual answer.'
    See if it is 4/0 or the other one
    y=1/(x-2)

    y=(x-2)-1

    (chain rule)

    dy/dx = -(x-2)-2
    -(x-2)-2 = -1

    (x-2)-2 = 1

    1/(x-2)2=1

    (x-2)2=1

    x2-4x+3=0

    (x-3)(x-1)=0

    Gradient = -1 when x= 1 or 3

    To find the y coordinates,

    y=1/x-2
    y=1/1-2=-1

    y=-x+k is a tangent at point (1,-1)

    -1=-1+k
    k=0

    y=1/3-2
    y=1

    y=-x+k is also a tangent at the point (3,1)

    1=-3+k
    k=4

    So k=0 or 4

    Hope this helps.
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    (Original post by TheNote)
    can anyone clear this up for me <3
    This is what I got for that one I think.. Name:  DSC_0033.jpg
Views: 227
Size:  397.1 KB
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    (Original post by Themexicannn)
    i got r = ((3V(a+b))/pi)^1/2


    1/3(r^2)pi(a+b) = V

    (r^2)pi(a+b) = 3V

    (r^2)(a+b) = 3V/pi

    (r^2) = 3V/pi/(a+b)

    (r^2) = 3V(a+b)/pi

    r = (3V(a+b)/pi)^1/2

    I think you went wrong on step 3. You divide by pi(a+b) to get r^2=3V/(pi(a+b))
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    (Original post by Themexicannn)
    i got r = ((3V(a+b))/pi)^1/2


    1/3(r^2)pi(a+b) = V

    (r^2)pi(a+b) = 3V

    (r^2)(a+b) = 3V/pi

    (r^2) = 3V/pi/(a+b)

    (r^2) = 3V(a+b)/pi

    r = (3V(a+b)/pi)^1/2
    (r^2)(a+b) = 3V/pi

    (r^2) = 3V/pi/(a+b)

    ^ This is the wrong step.

    r2(a+b)=3V/
    π

    r2=(3V/π)/(a+b)

    r2=(3V/π)*(1/(a+b))

    r2=3V/(π*(a+b))

    I hope this makes sense. Basically, dividing by something is the same as multiplying by the reciprocal.
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    (Original post by user1-4)
    does anybody the last question? We can go through it and figure out the 'actual answer.'
    See if it is 4/0 or the other one
    The differentiating method works, but so does the discriminant method

    1/(x-2) = y
    y= -x + k

    1/(x - 2) = -x + k
    1=(x - 2)(-x + k)
    1= -x^2 + kx +2x -2k
    0= -x^2 +(2+k)x -2k -1

    so b^2 -4ac =0 (as for tangents, there is only one intersection)

    (2+k)^2 -4(-1)(-2k-1) =0
    (2+k)(2-k) +4(-2k-1) =0 EDIT: (2+k)(2+k) +4(-2k-1) =0
    k^2 +4k +4 -8k -4 =0
    K^2 -4k = 0
    k(k-4) = 0

    therefore k=0 and k=4

    I remember the last question of Jan11 had a similar sort of question, I worked it out using differentiation, but only the discriminant method was in the mark scheme, even though I thought both methods worked fine, that question was slightly different though, so hopefully both methods will be awarded this time might be a good idea to check that mark scheme actually!
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    (Original post by SFCMorganBT)
    This is what I got for that one I think.. Name:  DSC_0033.jpg
Views: 227
Size:  397.1 KB

    You write the π symbol in a really neat way.
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    (Original post by LRJ)
    The differentiating method works, but so does the discriminant method

    1/(x-2) = y
    y= -x + k

    1/(x - 2) = -x + k
    1=(x - 2)(-x + k)
    1= -x^2 + kx +2x -2k
    0= -x^2 +(2+k)x -2k -1

    so b^2 -4ac =0 (as for tangents, there is only one intersection)

    (2+k)^2 -4(-1)(-2k-1) =0
    (2+k)(2-k) +4(-2k-1) =0
    Omg, I actually got all this, untill this point
    then somehow ended up with x = +-(-1)^1/2
    .. which can't be right because you cant root a negative (in real terms)

    How many marks do you reckon I picked up then?
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    (Original post by LRJ)
    The differentiating method works, but so does the discriminant method

    1/(x-2) = y
    y= -x + k

    1/(x - 2) = -x + k
    1=(x - 2)(-x + k)
    1= -x^2 + kx +2x -2k
    0= -x^2 +(2+k)x -2k -1

    so b^2 -4ac =0 (as for tangents, there is only one intersection)

    (2+k)^2 -4(-1)(-2k-1) =0
    (2+k)(2-k) +4(-2k-1) =0
    k^2 +4k +4 -8k -4 =0
    K^2 -4k = 0
    k(k-4) = 0

    therefore k=0 and k=4

    I remember the last question of Jan11 had a similar sort of question, I worked it out using differentiation, but only the discriminant method was in the mark scheme, even though I thought both methods worked fine, that question was slightly different though, so hopefully both methods will be awarded this time might be a good idea to check that mark scheme actually!
    Damn. That actually works. Spent ~5 minutes trying to do that but for some reason I couldn't get it to work so I used the differentiation method. Way to go for using that method. I assume that's the method OCR want their candidates to use as chain rule isn't on the C1 syllabus. Do you think I'll still receive full marks for this question even though I used differentiation? I got the same answer (k=0 or 4).

    Thanks
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    (Original post by Deceived)
    You write the π symbol in a really neat way.
    Oh, cheers ahaha
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    dayummm I got -9 for the min value of y for some reason...unless I misread my writing
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    i put pia +pib
    is that wrong?

    Posted from TSR Mobile
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    Are well done mate. I did this method but instead of putting (2+k)(2-k) I put (2-k)(2-k) which got my final awnser to
    K^2-12k
    K(K-12) therfore K=0 or 12
    (Original post by LRJ)
    The differentiating method works, but so does the discriminant method

    1/(x-2) = y
    y= -x + k

    1/(x - 2) = -x + k
    1=(x - 2)(-x + k)
    1= -x^2 + kx +2x -2k
    0= -x^2 +(2+k)x -2k -1

    so b^2 -4ac =0 (as for tangents, there is only one intersection)

    (2+k)^2 -4(-1)(-2k-1) =0
    (2+k)(2-k) +4(-2k-1) =0
    k^2 +4k +4 -8k -4 =0
    K^2 -4k = 0
    k(k-4) = 0

    therefore k=0 and k=4

    I remember the last question of Jan11 had a similar sort of question, I worked it out using differentiation, but only the discriminant method was in the mark scheme, even though I thought both methods worked fine, that question was slightly different though, so hopefully both methods will be awarded this time might be a good idea to check that mark scheme actually!
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    (Original post by Rubyturner94)
    i put pia +pib
    is that wrong?

    Posted from TSR Mobile
    hey, which question for?
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    (Original post by Rubyturner94)
    i put pia +pib
    is that wrong?

    Posted from TSR Mobile
    If you're talking about the denominator of the re-arranging question, then no. All you did was expand the brackets which is equal to if you didn't expand the brackets. Same answer. You should receive full marks. That is providing you did the rest of the question correctly.
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    (Original post by SFCMorganBT)
    hey, which question for?
    for rearranging tk make r the subject, in the denomenator everyone put pi(a+b) i put pia+pib

    Posted from TSR Mobile
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    Its the same thing Stop worrying!
    (Original post by Rubyturner94)
    for rearranging tk make r the subject, in the denomenator everyone put pi(a+b) i put pia+pib

    Posted from TSR Mobile
 
 
 
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