C3:
Hi I'm doing the C3 June 2005 paper. Just have some questions, hoping someone could help me out.
Heres a link to the paper if anyone wants to help: http://www.schoolportal.co.uk/Group...urceID=4777881
Q8i, The diagram shows part of each of the curves y = e^1/5x and y=(3x+8)^1/3 The curves meet, as shown
in the diagram, at the point P. The region R, shaded in the diagram, is bounded by the two curves and by the yaxis.
Show by calculation that the xcoordinate of P lies between 5.2 and 5.3
So I know this has something to do with proving the sign changes.
So I put I used (3x+8)^1/3  e^1/5x. I put in when x = 5.2 and when x = 5.3.
I got 0.04 and 0.0006. So I did prove the sign changed. However looking in the mark scheme, it has a little table instead saying 0.04 and 0.0006 and then some other values before that. Heres a link to the mark scheme: http://www.schoolportal.co.uk/Group...urceID=4777877
Just wondering where I've gone wrong..
C4:
Just finished marking a C4 paper. Relatively simple vector question I'm confused with. Line L1 passes through (2, 3, 1) and (1, 2, 4).
Find an equation in the form r = a +tb.
I thought it would be
(2,3,1) and I was always taught to do say if you're finding out the equation for line AB, you would do B  A.
so I did (1 2, 23, 41) getting (3, 1, 5).
But they got the signs the other way round getting (3, 1, 5). Confused!

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 03012013 16:46

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 03012013 16:59
(Original post by 01234567)
C3:
Hi I'm doing the C3 June 2005 paper. Just have some questions, hoping someone could help me out.
Heres a link to the paper if anyone wants to help: http://www.schoolportal.co.uk/Group...urceID=4777881
Q8i, The diagram shows part of each of the curves y = e^1/5x and y=(3x+8)^1/3 The curves meet, as shown
in the diagram, at the point P. The region R, shaded in the diagram, is bounded by the two curves and by the yaxis.
Show by calculation that the xcoordinate of P lies between 5.2 and 5.3
So I know this has something to do with proving the sign changes.
So I put I used (3x+8)^1/3  e^1/5x. I put in when x = 5.2 and when x = 5.3.
I got 0.04 and 0.0006. So I did prove the sign changed. However looking in the mark scheme, it has a little table instead saying 0.04 and 0.0006 and then some other values before that. Heres a link to the mark scheme: http://www.schoolportal.co.uk/Group...urceID=4777877
Just wondering where I've gone
But I would guess they have just subtracted the graphs the other way
Yours would be fine if this is the case 
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 03012013 17:01
(Original post by 01234567)
C4:
Just finished marking a C4 paper. Relatively simple vector question I'm confused with. Line L1 passes through (2, 3, 1) and (1, 2, 4).
Find an equation in the form r = a +tb.
I thought it would be
(2,3,1) and I was always taught to do say if you're finding out the equation for line AB, you would do B  A.
so I did (1 2, 23, 41) getting (3, 1, 5).
But they got the signs the other way round getting (3, 1, 5). Confused!
Similarly you could use B instead of A as the starting point 
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 04012013 14:12
Hi i have a questions on S2 June 09 I was hoping someone could help me with. Link to the paper: http://www.ocr.org.uk/images/59560q...atistics2.pdf
Q5) In a large region of derelict land, bricks are found scattered in the earth.
Assume now that the number of bricks in 1 cubic metre of earth can be modelled by the distribution
Po(3).
ii) Find the probability that the number of bricks in 4 cubic metres of earth is between 8 and 14
inclusive.
iii) Find the size of the largest volume of earth for which the probability that no bricks are found is
at least 0.4.
ii) So because its 4 cubic meters not 1, we times Poisson by 4, so now we have Po(12).
If its inclusive between 8 and 14, that gives us 8 > X < 14, with those signs of inequality don't know how to make it typing. This means we're finding when x is less or equal to 14 and because its poisson, we always do one less than the first value, so is this x is less or equal to 7??
iii) For this i equated P(x=0) to more or equal to 0.4 ... not really sure where to go now.
I have the formula of poisson of e^l times l^0 over 0! but i dont know what to do now..
Q7iv) I have no idea how they got the new parameters.. anyone?
8ii) It is required to redesign the test so that the probability of making a Type I error is less than 0.01
when the sample mean is 77.0 s. Calculate an estimate of the smallest sample size needed, and
explain why your answer is only an estimate.
 I'm really not sure what they're asking here.. any pointers? I know that with Type I errors, its the same as the significance level but I'm not really sure how to calculate the smallest sample needed.
If anyone could help with the questions, I would really really appreciate it! 
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 04012013 16:44
Just did OCR c4 january 2007,
The points A and B have position vectors a and b relative to an origin O, where a = 4i + 3j − 2k and
b = −7i + 5j + 4k.
) Use a scalar product to ﬁnd angle OAB.
I got the answer as 137 degrees. However the answer is actually 43 degrees. Which is 180  137... why do I have to do 180  the answer.... ? 
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 04012013 19:17
C3 Jan 2007:
8 The parametric equations of a curve are x = 2t^2, y = 4t. Two points on the curve are P (2p^2, 4p) and
Q (2q^2, 4q).
(i) Show that the gradient of the normal to the curve at P is −p. [2]
(ii) Show that the gradient of the chord joining the points P and Q is
2 over p + q
(iii) The chord PQ is the normal to the curve at P. Show that p^2+ pq + 2 = 0. [2]
(iv) The normal at the point R (8, 8) meets the curve again at S. The normal at S meets the curve
again at T. Find the coordinates of T.
I've managed to do part i, ii and iii.
No idea how to do part iv.. any pointers? I don't see how point R or S comes into the question.. The mark scheme has equated (8,8) to 
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 05012013 15:06
Anyone....?

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 05012013 15:11
(Original post by 01234567)
Anyone....? 
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 05012013 15:29
(Original post by TenOfThem)
Would have been nice if you had acknowledged my help on the earlier questions
Oh sorry  I usually do I was so busy trying to get the paper done.
Belated thanks. 
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 05012013 15:31
(Original post by 01234567)
Just did OCR c4 january 2007,
The points A and B have position vectors a and b relative to an origin O, where a = 4i + 3j − 2k and
b = −7i + 5j + 4k.
) Use a scalar product to ﬁnd angle OAB.
I got the answer as 137 degrees. However the answer is actually 43 degrees. Which is 180  137... why do I have to do 180  the answer.... ?
AO and AB 
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 05012013 15:34
(Original post by 01234567)
C3 Jan 2007:
8 The parametric equations of a curve are x = 2t^2, y = 4t. Two points on the curve are P (2p^2, 4p) and
Q (2q^2, 4q).
(i) Show that the gradient of the normal to the curve at P is −p. [2]
(ii) Show that the gradient of the chord joining the points P and Q is
2 over p + q
(iii) The chord PQ is the normal to the curve at P. Show that p^2+ pq + 2 = 0. [2]
(iv) The normal at the point R (8, 8) meets the curve again at S. The normal at S meets the curve
again at T. Find the coordinates of T.
I've managed to do part i, ii and iii.
No idea how to do part iv.. any pointers? I don't see how point R or S comes into the question.. The mark scheme has equated (8,8) to
Solve that with the curve and you have S
Then you have a new p so that you can find a new line
Solve that with the curve and you have T 
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 05012013 16:33
(Original post by TenOfThem)
For R, you know that p=2, so you have the first line
Solve that with the curve and you have S
Then you have a new p so that you can find a new line
Solve that with the curve and you have T
Thanks. 
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 05012013 16:43
The parametric equations of a curve are x = t + 2/t + 1
, y = 2/t + 3
Find the cartesian equation of the curve, giving your answer in a form not involving fractions.
I've done it right up with y = 2/y  3. I've then put this back into the x equation.
Got it right into the mark scheme up till 2/y1 over 2/y2 = x.
The mark scheme says eliminating double decker fraction... not sure how it gets the answer: 2x+y=2xy+2 
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 05012013 16:47
(Original post by 01234567)
The parametric equations of a curve are x = t + 2/t + 1
, y = 2/t + 3
Find the cartesian equation of the curve, giving your answer in a form not involving fractions.
I've done it right up with y = 2/y  3. I've then put this back into the x equation.
Got it right into the mark scheme up till 2/y1 over 2/y2 = x.
The mark scheme says eliminating double decker fraction... not sure how it gets the answer: 2x+y=2xy+2
if you quote me you can use this as an example to copy

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 05012013 17:00
(Original post by TenOfThem)
Any chance you could use latex as I am not finding the prose easy to read
if you quote me you can use this as an example to copy
Q7ii.
Cheers. 
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 05012013 17:04

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 05012013 17:25

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 05012013 17:30
(Original post by 01234567)
Because you times the y side by y, don't you have to times the x side by y as well, therefore equating it to xy?
Consider
do you believe that the second fraction is actually 2x? 
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 05012013 20:05
c3 Jan 2010.
f(x) = 2 − (√3x + 1)^1/3
Find the set of values of x for which f(x) = f(x).
I equated fx to itself and to fx.. Is this the right way to do the question?
THe mark scheme has the answer down as x is less or equal to 7.
From equating fx = fx i didn't get anything, but from equating fx = fx i got x = 7... how does this then become x is less or equal to 7? is it because you have to look at it from the graph?
5)ii) Find an expression for the second derivative and hence ﬁnd the value of
the second derivative at the stationary point.
I got the right answer up to a point in the mark scheme.
To find out the second derivative at the stationary point is this the same as just equating dy/dx = 0?
From the mark scheme, they appear to have substituted x = 0 to find out the stationary point from the equation... so, is this different for the second derivative? 
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 05012013 20:06
(Original post by TenOfThem)
I have not multiplied the RHS by y
Consider
do you believe that the second fraction is actually 2x?
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