OCR A2 maths january, C3/C4/S2 question.

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.school-portal.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 y-axis.
Show by calculation that the x-coordinate 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.school-portal.co.uk/Group...urceID=4777877

Just wondering where I've gone

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, -2--3, -4-1) getting (-3, 1, -5).
But they got the signs the other way round getting (3, -1, 5). Confused!

Anyone....?

Would have been nice if you had acknowledged my help on the earlier questions

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.... ?

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

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

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/y-1 over 2/y-2 = x.

The mark scheme says eliminating double decker fraction... not sure how it gets the answer: 2x+y=2xy+2

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

$a + \dfrac{b}{c} - \dfrac{\frac{d}{e+1}}{\frac{f+1}{g+1}}$

So $t=\frac{2}{y}-3$

So $x=\dfrac{\frac{2}{y}-1}{\frac{2}{y}-2} = \dfrac{2-y}{2-2y}$


Which gives the answer if you multiply by the denominator

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?

I have not multiplied the RHS by y

Consider

$x = \dfrac{1}{2} = \dfrac{4}{8}$

do you believe that the second fraction is actually 2x?