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# OCR Maths Core 2 - May 25th 2016

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1. Did anyone get question 9 in the last bit ?
2. I didn't in the exam but i know how to now.. you had to differentiate P and set it equal to 0 to find the x value, then sub that into the P equation to get the minimum perimeter. For e) you differentiate your first differential and sub in your x value, then this will be a positive number showing that is it minimum
3. wrong exam mate, I think ur talking about the edexcel c2 exam
4. What did you get for Q9 A AND K
A=3/5
K=√3 / 2 ??????
5. yh i got that buy I dont know if its right, what did u get for the last question
6. (Original post by A1998lebanna)
Did anyone get question 9 in the last bit ?
I think I got pi/3a and 4pi/3a but I can't remember for certain.
7. (Original post by Hiling99)
I think I got pi/3a and 4pi/3a but I can't remember for certain.
I got the same . I just completely guessed.
8. Am i the only one who really struggled with that exam? 😂 lol well ive failed maths
9. Hiya guys,

Im an online tutor and just read over the paper and Q9 seemed to be a rough question if you didnt quite swat up on your trigonometry haha! Anyways your answers are both correct for the 2 parts of Q9 from what I see : )! Great Job haha!
10. 1pi/3a and 4pi/3a

1. 10
5.04

2. 3Pi/10
20.4

3. 27 + 27kx + 9k^2x^2 + k^3x^3
K= (root 3). K = -(root 3)

4. Log3 (X^2/X + 4)
X = -3. X = 12

5. (X^4/2) - X^3 + 2X^2 - 6X + C
(4a^2 - 6a + 2)a^2
4

6. K = 91
562
N = 38

7. Q = X^2 - 4X + 3. R = 0
X = -1. X = 1. X = 3
Stationary points are given by (dy/dx) = 0
512/15

8. Translation by +2 along X axis
Stretch by 1/9 in Y direction
Intersection with Y axis at (0,1/9)
6.73
9.60

9. Period is 2Pi/a
a = 5/3. K = (root 3)/2
Pi/3a. 4Pi/3a
12. Isn't A 3/5? AS then this scale factor is applied, it then turns in 5/3?

i.e A=3/5 so when applied The values becomes sin(5/3 x) and so 5/3 mulitplied by PI/5 gives you pi/3 which is the correct answer?

Because K=Root3/2

so A must be the inverse of the values?
13. Sorry you missread isnt A 3/5?
14. Youre wrong. the answer is 5/3.
15. (Original post by gsdfghgfhfg)
What did you get for Q9 A AND K
A=3/5
K=√3 / 2 ??????
I got a = 5/3 ?? A lot of my friends got that as well
16. (Original post by _.lukeg._)
I got a = 5/3 ?? A lot of my friends got that as well
How did you arrive at your answer, i am struggling to work it out.
18. (Original post by Stephen65golf)
How did you arrive at your answer, i am struggling to work it out.
I'm not going to remember now am I hahaha
19. (Original post by Stephen65golf)
i). y = sin (ax). This is a stretch by 1/a in the x direction. Therefore the period is 2Pi/a

ii). x = Pi/5 and x = 2Pi/5. Therefore sin(aPi/5) = k and sin(a2Pi/5) = k
1/2(2Pi/5 - Pi/5) = Pi/10 Pi/5 + Pi/10 = 3Pi/10
(3Pi a)/10 maps to Pi/2 because of the stretch in the x direction so a = 10/6 = 5/3. (It's much easier to show on a diagram! The way you have to think of it is that there is a 'solution' either side of Pi/2).
Sin((5/3)(Pi/5)) = Root3/2. And Sin((5/3)(2Pi/5)) = Root3/2

iii). Sin(ax) = Root3Cos(ax). So Tan(ax) = Root3. Therefore ax = Pi/3

X = Pi/3a, 4Pi/3a (x is positive but gives a negative solution), 7Pi/3a

Hope this helps.
20. (Original post by intue14)
i). y = sin (ax). This is a stretch by 1/a in the x direction. Therefore the period is 2Pi/a

ii). x = Pi/5 and x = 2Pi/5. Therefore sin(aPi/5) = k and sin(a2Pi/5) = k
1/2(2Pi/5 - Pi/5) = Pi/10 Pi/5 + Pi/10 = 3Pi/10
(3Pi a)/10 maps to Pi/2 because of the stretch in the x direction so a = 10/6 = 5/3. (It's much easier to show on a diagram! The way you have to think of it is that there is a 'solution' either side of Pi/2).
Sin((5/3)(Pi/5)) = Root3/2. And Sin((5/3)(2Pi/5)) = Root3/2

iii). Sin(ax) = Root3Cos(ax). So Tan(ax) = Root3. Therefore ax = Pi/3

X = Pi/3a, 4Pi/3a (x is positive but gives a negative solution), 7Pi/3a

Hope this helps.
I would like to say yes it makes perfect sense but sorry I dont really get it. Thnx

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