[Jack:)]

Hey, this is what i got for C1 MEI today.

any discussion/corrections are welcome!

1. y=-2x+7 and (0,7) and (7/2,0)

2.[b] = SQRT(3a/2c)

3. i) 25
    ii) 4/9

4. (x-3)/(x+2)

5. i) 30
   ii) 8/11

6. i) 10
   ii) -720 (x^3)

7. -SQRT(5) < K < SQRT(5)

8. b= -4 c= -8

9. 6n+9 Show that it is always odd and n must be a multiple of 3 for (6n+9)/9

10.i) SQRT(20) = SQRT(20) QED
    ii)1/3 * -3 = -1 QED
    iii) Midpoint AC: (2,2)   Found they intersect at (2,2) and not (5,3) 

11.i) 2x^3 - 5x^2 - 23x - 10
    ii) Cubic graph drawn correct way up passing through -2, -0.5 and 5
    iii) (0,-18)
    iv) (0,-40)

12.i) (4,1) (2.6,-2.6) (0.4,-0.4) give or take 0.1 ish
     ii)Showed
     iii) x= (3+SQRT(5))/2 and (3-SQRT(5))/2 

[Axion]

Mate your question 8 is wrong. It was b = -4, c = -8.

[Jack:)]

(Original post by Axion)
Mate your question 8 is wrong. It was b = -4, c = -8.

Yes, sorry i forgot to make the second equation equal to 85

[Axion]

(Original post by Jack:))
Yes, sorry i forgot to make the second equation equal to 85

lol you remember all of that?

[Jack:)]

(Original post by Axion)
lol you remember all of that?

Yep, it’s a shame my pre-exam memory is not that good!
Oh well, it’s only 3 marks tops and hopefully i get a nice examiner and i get a mark for knowing to use simultaneous equations.

[Axion]

(Original post by Jack:))
Yep, it’s a shame my pre-exam memory is not that good!
Oh well, it’s only 3 marks tops and hopefully i get a nice examiner and i get a mark for knowing to use simultaneous equations.

Don't worry I know I've lost about 7-8 marks already from stupid mistakes

[Jack:)]

(Original post by Axion)
Don't worry I know I've lost about 7-8 marks already from stupid mistakes

Thats the biggest problem with C1, its so easy to make mistakes, i am thinking that the grade boundries will be quite high tho, 60 for an A

[dayofs]

(Original post by Jack:))
Thats the biggest problem with C1, its so easy to make mistakes, i am thinking that the grade boundries will be quite high tho, 60 for an A

For the first one, do we need to write the coordinates of the point where it crosses the axis? I didn't see that -3 from me .

[Georgina :)]

HA just got so worried because I didn't recognise ANY answers.. I don't do MEI!

[thesaintx]

So much love for this. Thank you ever so much

[steven hawking]

How can you remember all this, geeeeeeeeeeeeeeeeeeeee

[tehforum]

(Original post by steven hawking)
How can you remember all this, geeeeeeeeeeeeeeeeeeeee

Looks at username...

Lol

[steven hawking]

(Original post by tehforum)
Looks at username...

Lol

What about it?

[Stephybob]

I have an issue with your question 11, if you take the original factorised form of (x+1/2)(x+2)(x-5) then the y intercept would be -5 not -10. I put the equations 2x^3 -5x^2 -23x -10 and the expanded form of the factorised equation into autograph and they aren't the same. So to get from the factorised equation to one with a coefficient of 2, you can't just double everything or I believe you get a false y intercept (but no idea how you actually get there).

So if I'm correct, the answer for the y intercept of f(x) -8 is (0, -13) and the answer to f(x-3) is -20?

[Stephybob]

Oh and 11ii was to draw the graph of f(x)

[thesaintx]

(Original post by Stephybob)
I have an issue with your question 11, if you take the original factorised form of (x+1/2)(x+2)(x-5) then the y intercept would be -5 not -10. I put the equations 2x^3 -5x^2 -23x -10 and the expanded form of the factorised equation into autograph and they aren't the same. So to get from the factorised equation to one with a coefficient of 2, you can't just double everything or I believe you get a false y intercept (but no idea how you actually get there).

So if I'm correct, the answer for the y intercept of f(x) -8 is (0, -13) and the answer to f(x-3) is -20?

If i'm not mistaken they asked us to put the original factorised form into the form which starts 2x^3 and then to solve from there, so I think the answer's -40 as a result..

Hope it helps

[cloudhang]

grade boundaries for an A?
how did everyone find it

[Jack:)]

(Original post by Stephybob)
Oh and 11ii was to draw the graph of f(x)

Ty will add that in,

[Stephybob]

(Original post by thesaintx)
If i'm not mistaken they asked us to put the original factorised form into the form which starts 2x^3 and then to solve from there, so I think the answer's -40 as a result..

Hope it helps

Hmm you're probably right but seeing as both are f(x) then it shouldn't matter which one you use if they equal the same thing? As both should give the same y intercept? Unless they wanted us to find g(x) or something which has the same roots as f(x) but a different intercept?

[Jack:)]

(Original post by cloudhang)
grade boundaries for an A?
how did everyone find it

Judging on previous papers my geuss would be 63 Max and 58 Min but personally im thinking around 61. Of course it is hard to tell