OCR MEI C2 MAY 18th

Reply 180

lazershus

10

Original post by steven hawking

Did i? i might of, tbh not sure.
We can find out if anyone can remember the coordinate of the tangent, anyone?

yeah i thought it was (1, something (-4 i think)),i could be wrong! so one of us got it the wrong way round :/

Reply 181

jonny7bell

1

What did everyone get for the drawing curve question? I got x=0,+sqrt5 and -sqrt 5, y=0

Reply 182

Chrisgame

Did anybody work out how to prove the last question on the paper, with Ratio's of

2^(n-2) : 3^(n-2)

I know that a=10, and a=15, and R= 0.4, and R=0.6, but I couldn't find a way of getting the ratio from the ak = ar^(k-1) formulae.

Otherwise, challenging but fun paper, I reckon I may have got 68/72, with a slip up on the log question of making y in terms of log x, which was a rookie error.

Reply 183

steven hawking

Original post by jonny7bell

What did everyone get for the drawing curve question? I got x=0,+sqrt5 and -sqrt 5, y=0

sounds like what i got

Reply 184

lazershus

10

Original post by steven hawking

Actually i think you are wrong because if you sub -2 into the equation (the one that was given, the combined one)i think : x^3-5x +2=0
You dont get 0

ohhh, i think you're right :frown:

Upload the paper.

Original post by Chrisgame

Did anybody work out how to prove the last question on the paper, with Ratio's of

2^(n-2) : 3^(n-2)

I know that a=10, and a=15, and R= 0.4, and R=0.6, but I couldn't find a way of getting the ratio from the ak = ar^(k-1) formulae.

Otherwise, challenging but fun paper, I reckon I may have got 68/72, with a slip up on the log question of making y in terms of log x, which was a rookie error.

look back a few pages, the answer is there. (not explained by me though :[ )

Reply 187

ILM16

1

Original post by arcus

I cant quite remember the exact values,
but it was like 6=a*r, so 6/a=r, substitute that back into the sum to infinity equation, ended up with like a^2-25a+150=0, factorise that to give (a-10)(a-15)

Slightly different to the method I used, I did it the way below:

Original post by steven hawking

I didnt actually show a=10 at the beginning i just went straight ahead to use simulataneous equations and find out the values, of both the r's and a's.

That's what stumped me, as I thought you couldn't just use it to prove it?

In the end I worked backwards.

I showed a = 6/r

So a/(1-r) = 25 (sum to infinity)

a = 25 - 25r

6/r = 25 - 25r

6 = 25r - 25r^2

25r^2 - 25r + 6 = 0

Quadratic formulae gave me both 0.6 and 0.4

Put them back into a = 6/r, giving both a values.

Surely that'll get the marks, though it's the other way around? Getting the a's from the r's.

Reply 188

JRW1995

Original post by steven hawking

For finding the point where the curve crosses the tangent question, did anyone get x=2
?
As you had to solve: x^3 -5x +2=0
Then use the fact that you know it touches at the point of tangent ( i think at x=-1)
So you know (x+1)^2 is a factor of the above equation, then solve to x-2=0

.. x=2 ?????

Ring any bells ?

Thought it was x^3-3x+2=0

Reply 189

lazershus

10

Original post by JRW1995

Thought it was x^3-3x+2=0

that must be it! so i was right, i think, so it was x= -2 right?! and x=1 was repeated as it's a tangent.

Reply 190

steven hawking

Original post by JRW1995

Thought it was x^3-3x+2=0

yep you're right

Reply 191

tneduts46

Original post by lazershus

that must be it! so i was right, i think, so it was x= -2 right?! and x=1 was repeated as it's a tangent.

yeah, thats what i got

Reply 192

JRW1995

Original post by lazershus

that must be it! so i was right, i think, so it was x= -2 right?! and x=1 was repeated as it's a tangent.

Yeah that's what I got.

Completely flopped this paper though, want an A And unless grade boundries are low I did NOT get that. Resit it is :frown:

Reply 193

chris_gloyne

what did people get for the trapezium rule question? i got 0.68m^2 and then for volume times by 50 so 34m^3?

Reply 194

jonny7bell

1

Original post by JRW1995

Yeah that's what I got.

Completely flopped this paper though, want an A And unless grade boundries are low I did NOT get that. Resit it is :frown:

yeah i think thats what got because (x-1)(x+2)(x-1) is what it factored into so x=1,-2

Reply 195

darbyj

2

Original post by lazershus

ohhh, i think you're right :frown:

I think lazershus is right;

the cubic was y = x³ - 5x, the tangent was y = - 2x - 2

so x³ - 3x + 2 = 0

repeated root of (x - 1) as it is a tangent so;

(x - 1)(x - 1)(x + 2) = 0

so x = -2