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# how did find p4 watch

1. (Original post by 786)
The last question where it asked you to show that AB was 3a(root3)/2. Did anyone manage this? The point of intersection was at theta=60, therefore r=3a(1 -cos(theta)) i.e 3a/2. Using simple trigonometry i get AB as twice the given answer. What did i do wrong?
u can draw a triangle with vertices ABO and AB is 2rsin(theta) which gives the answer
2. (Original post by 786)
The last question where it asked you to show that AB was 3a(root3)/2. Did anyone manage this? The point of intersection was at theta=60, therefore r=3a(1 -cos(theta)) i.e 3a/2. Using simple trigonometry i get AB as twice the given answer. What did i do wrong?
AB = 2 times the y value at B (3a/2,Pi/3)
y = r Sin (-)
r=3a(1 - Cos(-))

y = 3a(1 - Cos(-))Sin(-)

Cos(Pi/3) = 0.5 and Sin(Pi/3) = Sqrt(3)/2

=============

y = 3a(1-0.5) X Sqrt(3)/2
y = 3a/2 X Sqrt(3)/2
y = 3aSqrt(3)/4

Double y to get full distance from A to B.

AB = 3aSqrt(3)/4 X 2
AB = 3aSqrt(3)/2
QED
3. A very pedantic question i know, but does anyone know if you lose a mark for omitting the cm^2 from an answer? (I did this for the last part of the last question)
4. (Original post by 786)
Why is AB 2rsin(theta)?
the y-coordinate of B is rsin(theta) and so AB is its double (A,B have the same x-coordinate)
5. (Original post by sam99)
A very pedantic question i know, but does anyone know if you lose a mark for omitting the cm^2 from an answer? (I did this for the last part of the last question)
yup
a is actually root3 cm

so the area has the unit of cm^2
6. (Original post by keisiuho)
yup
a is actually root3 cm

so the area has the unit of cm^2
I know it's cm^2, but will i lose a mark for not putting it?
7. These answers may not be entirely accurate, as i did p6 after p4 and have managed to forget half my answers to the first paper.

Q1
a) prove that SUM (r+1)(r+5) = 1/6n(2n+7)(n+7)
b) sum it from r=10 to n=40

I just used the standard results as it was only worth 3 or 4 marks.
SUM (r+1)(r+5) = SUM r^2 + 6 SUM r + 5 SUM 1
Then just put in the standard results, multiply out, and then refactorise to get the above result.

Q2
a) use linear interpolation on {1, 2}
b) use newton raphson with a=1

Just used the standard results for linear interpolation, and newton raphson is in the formula booklet.

Q3
a) so ab = 15
b) find roots of z^2 + 6z -30i (not sure bout middle term)

Substitute z = a + ib into the equation and multiply out, then equate coefficients of real and imaginary parts.

Q4
a) find solutions sets of MOD (x^2 - 2) = 2x

X < 0.74 and X > 2.74 (something like that anyway, not sure about after the decimal point)

Q5
a) find w
b) find arg w
c) find mod w
d) plot an argand diagram
e) find distance AB in surd form

Cant remember

Q6
a) find general solution
b) find co-ordinates of minimum value
c) plot a graph

P.I. = e^2x (I think)

Q7
a) Find general solution
b) find particular solution

y = e^-t(2Sint - Cost + 2)

Q8
a) Find points of intersection
b) Show the line = 3root3a/2
c)Area of shaded region
d)Area of badge given a=4.5cm

See previous post.
8. (Original post by sam99)
I know it's cm^2, but will i lose a mark for not putting it?
I really hope not, cos i cant remember if i put it or not!
9. q7 was y = e-t (2sint-cost) +2e-t
10. i thought it went quite well.

should be an A at least.
11. (Original post by rockindemon)
q7 was y = e-t (2sint-cost) +2e-t
i got 4sint ... forgot to divide by 2 !!! ...
12. hhmm, i got e-t (-2sint-cost) +2e-t, i probably forgot to get rid of a minus somewhere along the lines, how many marks you think i lose?
13. (Original post by mathematician)
i personally thought it went very well.
but it seems that everyone and i mean everyone was short of time and they all rekon they didnt do that great.
I thought it went really well, had 15 mins to spare after checking it through. Definitely better than P5.
14. (Original post by Hovick)
hhmm, i got e-t (-2sint-cost) +2e-t, i probably forgot to get rid of a minus somewhere along the lines, how many marks you think i lose?
most likely you'll loose 1 mark only but if their being picky 2 for the final wrong answer
15. (Original post by rockindemon)
q7 was y = e-t (2sint-cost) +2e-t
You could simplify that to: y = e^-t(2Sint - Cost + 2)
16. (Original post by rockindemon)
most likely you'll loose 1 mark only but if their being picky 2 for the final wrong answer
ACED!
no prob.
u know for the differential of
a^x did everyone remember the standard differential of a^xlna, for the NR question
17. (Original post by mathematician)
ACED!
no prob.
u know for the differential of
a^x did everyone remember the standard differential of a^xlna, for the NR question
was the area of the region something like
-4pi + 8 or 9 root 3
18. think it was 9root 3 - pi
19. (Original post by [_Z_])
think it was 9root 3 - pi
nah it werent, it cant of been,
your tellin me the area of the badge is 37.34.
length AB was 4.5, no way is it 37.34
my answer (9root3 - 4pi), gives an area of 9cm^2
20. yeh i concur

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