So distance PO is still sqrt(8) and distance PM is sqrt(6.25) I believe? I remember on my first run through of the paper i had root 8 and root 8.5, seem to get a different answer every time

Reply 30

ibysaiyan

Original post by Jono404

Just realised i didn't divide the Y co-ord by 2! so the midpoint is (-7/2, 4), careless mistake :smile:

Ah ! I was not wrong then xD yea... both ways of doing it =]

Reply 31

ibysaiyan

Original post by Jono404

So distance PO is still sqrt(8) and distance PM is sqrt(6.25) I believe? I remember on my first run through of the paper i had root 8 and root 8.5, seem to get a different answer every time

LOl I subbed into the calc. gave me a value of 2.5
I am bored now.Done c1,c2 and c4 revision... I find c3 to be a little tricky.

(edited 13 years ago)

Reply 32

Maria1234

oh you seems to be so kind :smile:

yes we could do!
but you said you have c3 june 2010.where did you get it?
Do you want it to send to me?Do you have c4 aswell?
thanks

Original post by ibysaiyan

Honey same here.I guess we'll have to wait a little more. I got c3 june paper but in need of others. By the way how is your preparation going ? =] We could open up a revision thread.

Reply 33

needssomehelpplease

Watcha need I did all of them last night. Any problems ?

Ok i think i might just be dumb but i can not do 1bii) 1c) for q5 i dont understand how to find 'r squared' i know theres 25 and 36 and keep thinking its root 61? Not sure how to do 6ai) and 6aii) and lastly 7aii)

:frown:

I think question 1 is really obvious but my mind is just completely blank. Any help will be appreciated. :smile:

Reply 34

ibysaiyan

Original post by needssomehelpplease

I think question 1 is really obvious but my mind is just completely blank. Any help will be appreciated. :smile:

Lets see...
1 bii to find equation AD we know AD is perpendicular to AB hence we use the gradient AB to be the reference gradient. i.e say if gradient AB is -1/2,then gradient of AD will be 2.
For part c once you got the equation. Use substitution to get the co-ordinate.
Question 5: r = square root 61 or r^2 = 61.
I had difficulty with part 6 (i) too for the you need to consider all the sides:
so you get:3x(y) +4x(y)+5x(y)+3x (4x) = 144
sort them out and you should end up with the given equation.

One way of solving part 7 ii is by finding out the determinant.If determinant < 0 then we have no real roots.

=]
EDIT: Radius is 5 as said below.Correction made by
Jono404

(edited 13 years ago)

Reply 35

ibysaiyan

Original post by Maria1234

oh you seems to be so kind :smile:

yes we could do!
but you said you have c3 june 2010.where did you get it?
Do you want it to send to me?Do you have c4 aswell?
thanks

Nope.I will send you the c3 qp as for the rest I don't.
:smile:

Reply 36

Jono404

The radius of the circle is 5 as it touches the y axis and it's centre has x-coordinate -5, hence r=5^2

Reply 37

ibysaiyan

Original post by Jono404

The radius of the circle is 5 as it touches the y axis and it's centre has x-coordinate -5, hence r=5^2

I thought it was ssqr.root 61 ?
OO i see now.. it's because the radius has crossed Y axis which means x = 0.So we ignore 36...

(edited 13 years ago)

Reply 38

qwerty54321

So sorry about the delay of copying this up.
Here is the mark scheme that i copied down from class:

1a) y=-2/3x +14/3
gradient is -2/3
b) y=-2/3x+9
ii) you'll know if you got this one right or not if you got to the right answer.. if you didn't quote me and i'll do it later :smile:

c) y=2, x=4 --> (4,2)

2a) 14-6root5
b)-11+5root5

3ai) likewise you'll know if yuo got this one right.. you must say x+3 is a factor

ii) long division correct --> (x+3)(x+5)(x-1)

b)p(2)=35
ci)p(-1) = -16
p(0) = -15
therefore p(-1) is less, meaning the lowest point on the graph is nearer to when y=-1, to y = 0

cii) draw correctly, going through -5, -3, and 1 on the x axis, and -15 at the bottom
IMPORTANT: because for part ci we found that the lowest point (the vertex) is around where y=-1, the lowest point on yourgraph CAN'T be where it crosses the y axis. the vertex must be slightly to the left of the y axis. quote me if you need help.

4a) 1/5x^5 - 4x^2 + 9x (+x)
42/5
4aii) y=9 integrate 9
9x
sub in 2
=18
18-42/5 = 48/5

b) dy/dx = 4x^3 -8
where x = 1
= -4
ii) where x=1, y=2
so equation is y=-4x+6

5a) (x+5)^2 + (y-6)^2 = 5^2
bi) 25=25
bii) gradient = -4/3
y=-4/3x-2/3
iii) m = -7/2,4
PM^2 =2^2 + (3/2)^2
pm = 5/2
PO = 2root2

sp p is closer to M

6ai)you'll know if you got this one right or not
ii)
v=6x^2y
rearrange 12=x^2 + y -> y=12/x -x
v = 6x^2(12/x-x)
v = 72x - 6x^3

bi) dv/dx = 72 - 18x^2.. when x=2 it equals 0
c) d^2v/dx^2 = -36x
sub in when x=2 = -72 <0 so a maximum

7) 2(x-5)^2 +3
no real roots - when try to solve can't squareroot a negative number

bi) you'll know if you got this right, ask me if you want me to go through it.
ii) (7k+1)(k-1) <0 (PLEASE NOTE THERE SHOULD BE A LESS THAN OR EQUALS SIGN, WHERE I HAVE JUST PUT A LESS THAN because i dont know how to do it otherwise!)
iii) -1/7 < k < 1

again, there should be a less than or equals sign, not just less than. !!

Hope this helped everyone, sorry again about the delay, and if you're stuck quote me and i'll try and help :smile:

Reply 39

qwerty54321

Original post by needssomehelpplease

Thank you for the paperrr Jono! Now don't suppose anyone has the mark scheme or knows all the answers?