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OCR Core 2 Maths: Wednesday 25th May 2016

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Original post by milliex97
I got this too


Yes, it is
Original post by h3rmit
I used -1 and 3 as my limits, the stationary point of 1 was a maximum and didn't touch the x axis from inspection

I think I got 6.73, though I may have answers mixed up

Yes I used them thank god! Aah that explains my confusion as I knew there were three roots from the previous question but was confused that only two touched the x axis.
I got 6.73 as the answer to something else I'm pretty sure but the integration question wanted it in exact form, I got 512/15
For the second last part I got k=1 and a=2.5, is this correct? I basically worked out that half the period was the first two points added together, then went on from there to find a.
Original post by milliex97
Yes I used them thank god! Aah that explains my confusion as I knew there were three roots from the previous question but was confused that only two touched the x axis.
I got 6.73 as the answer to something else I'm pretty sure but the integration question wanted it in exact form, I got 512/15


I too got the answer but at worse you would get -1 for having the correct decimal to 3sf.
Also yeah N=38 as you can't have a decimal term number but if N=37 then the sum of the arithmetic is less than 1200 (the sum to infinity of the geometric progression). You would only loose one in that case too I believe. The question I found the worst where the first 3 parts of 8. Probably got par i) and iii) correct but I had a total mind blank on ii). I don't know if it's because I did a lot more prep for this exam but I found the C2 paper much easier than the C1.
What was the answer to part 8ii ????? That stretch transformation of y=3^x-2
unofficial mark scheme?
Original post by milliex97
What was the answer to part 8ii ????? That stretch transformation of y=3^x-2


3^x-2 is the same as (3^x)(3^-2) as you add the powers so I think it's a stretch SF 1/9 parallel to y axis.
The answer is + or - root3. Did it after with a maths teacher. Your expansion should of given the coefficient of x^2 to equal 9k^2. When the exam stated this eqauls the constant, they meant the constant of the expasion (the number without any xs, 27) as this stays the same no matter what. So you equate 9k^2 to 27 to get the answer. This also makes sence as the question stated give exact answers.
Original post by M12345689
3^x-2 is the same as (3^x)(3^-2) as you add the powers so I think it's a stretch SF 1/9 parallel to y axis.


Now you put it like that, why didn't I get it😱
Original post by black1blade
For the second last part I got k=1 and a=2.5, is this correct? I basically worked out that half the period was the first two points added together, then went on from there to find a.


With a period of 2pi/a I got a=5/3 and k= root3/2 I think.
Original post by gsdfghgfhfg
What did you get for Q9 A AND K
A=3/5
K=√3/2 ?????? (Root 3, Divided by, 2)


I think thats what i got actually
Original post by Sourestdeeds
With a period of 2pi/a I got a=5/3 and k= root3/2 I think.


Think I did some for the rearranging wrong but hopefully I would have got 1 or 2. Actually I know what happened, didn't add the two pis together when I was working out half the period from the points- douh. That is only 1 actual mistake so hopefully only -1 :P.
1i) 10
ii ) 5.04
2i) 3/10π
ii)20.4
3i)27+27k+9k^2x^2+k^3x^3
ii) I didn't understand
4i) log3(x^2/x+4)
ii) x=12 and -3 but only 12
5a) 1/2x^4 -x^3+2x^2 -6x+c
bi) 2/a^2 +6/a +4
bii) sum to infinity=4
6i) k=91
ii) sum if the first 16 terms = 978
iii) N=38
7i) (x^2 -4x+3)(x+1)
ii) (x-3)(x-1)(x+1)
iii) proof
iv)512/15
8i) translation along the x axis by +2
ii) stretch in the y axis by 1/9 or 3^-2
iii)0,1/9
iv)x=6.73
v) 9.598 (can't remember how many Sig fig I put it to)
9i) 2π/a
ii)a=5/3 k=(√3)/2
iii) (π/3a) and (4π/3a)
What was question 1 exactly. Also, what was question 6 I & ii. Thnaks
Reply 75
Avatar for h3rmit
h3rmit
15
Original post by Shariqameen
1i) 10
ii ) 5.04
2i) 3/10π
ii)20.4
3i)27+27k+9k^2x^2+k^3x^3
ii) I didn't understand
4i) log3(x^2/x+4)
ii) x=12 and -3 but only 12
5a) 1/2x^4 -x^3+2x^2 -6x+c
bi) 2/a^2 +6/a +4
bii) sum to infinity=4
6i) k=91
ii) sum if the first 16 terms = 978
iii) N=38
7i) (x^2 -4x+3)(x+1)
ii) (x-3)(x-1)(x+1)
iii) proof
iv)512/15
8i) translation along the x axis by +2
ii) stretch in the y axis by 1/9 or 3^-2
iii)0,1/9
iv)x=6.73
v) 9.598 (can't remember how many Sig fig I put it to)
9i) 2π/a
ii)a=5/3 k=(√3)/2
iii) (π/3a) and (4π/3a)


Examiners only allow in the y axis for reflections, and -3 works too if the multiplying two is used a power before logging
Original post by jacklfc99
unofficial mark scheme?



1) 1+1=2
2) 6x5-32
3)21-50=47
cant remember the rest
Original post by zzxxDash53xxzz
1) 1+1=2
2) 6x5-32
3)21-50=47
cant remember the rest


Hope this was some what helpful
Original post by groceryboi
What was question 1 exactly. Also, what was question 6 I & ii. Thnaks

1) i think it was based off knowing the factors of addition something allong the lines of 1+1
Original post by h3rmit
Examiners only allow in the y axis for reflections, and -3 works too if the multiplying two is used a power before logging


You are probally right but when i tried -3 on my calculator it gave me an error and based on my experiance from past papers they only like the single answer that also works on the calculator.And I am not sure about the stretch because that is the first thing that came to my head just hope that its right or i get some marks 😞😞

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