# Mr M's OCR (not OCR MEI) Core 2 Answers May 2016Watch

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2 years ago
#21
Can someone please explain how to do 4ii, I know ive made a mistake somewhere but im not sure where
0
2 years ago
#22
(Original post by Mr M)
The trouble with squaring is that it can introduce extra incorrect solutions.

2 marks probably.
Ok thanks very much at least I should get a couple of marks
0
2 years ago
#23
(Original post by Ozil5)
Thank you, would you mind explaining how you get 12 too? I just tried it again and i got (x+3)(x-4). Am i just being really stupid here?
Using previous part:
x^2/(x+4) = 3^2 = 9
x^2=9(x+4)
x^2=9x+36
x^2-9x-36=0
(x-12)(x+3)=0
x=12 or -3

But log base 3 (-3) is undefined.
Therefore x = 12
1
2 years ago
#24
(Original post by Mr M)
Mr M's OCR (not OCR MEI) Core 2 Answers May 2016

1. (i) 10 cm (2 marks)

(ii) 5.04 cm (2 marks)

2. (i) (2 marks)

(ii) (3 marks)

3. (i) (4 marks)

(ii) (2 marks)

4. (i) (2 marks)

(ii) (4 marks)

5. (a) (3 marks)

(b) (i) (4 marks)

(ii) 4 (1 mark)

6. (i) (3 marks)

(ii) (2 marks)

(iii) (6 marks)

7. (i) Quotient and remainder 0 (3 marks)

(ii) or or (3 marks)

(iii) Show (2 marks)

(iv) (4 marks)

8. (i) Translation by 2 units in the positive x direction (2 marks)

(ii) Stretch parallel to the y axis scale factor (2 marks)

(iii) Sketch showing as the only point of intersection (2 marks)

(iv) (3 marks)

(v) 9.60 (3 marks)

9. (i) (1 mark)

(ii) and (3 marks)

(iii) and (4 marks)
Question 9ii - the question said that k must be a positive constant and 0 isn't a positive constant?
0
2 years ago
#25
(Original post by Mr M)
Mr M's OCR (not OCR MEI) Core 2 Answers May 2016

1. (i) 10 cm (2 marks)

(ii) 5.04 cm (2 marks)

2. (i) (2 marks)

(ii) (3 marks)

3. (i) (4 marks)

(ii) (2 marks)

4. (i) (2 marks)

(ii) (4 marks)

5. (a) (3 marks)

(b) (i) (4 marks)

(ii) 4 (1 mark)

6. (i) (3 marks)

(ii) (2 marks)

(iii) (6 marks)

7. (i) Quotient and remainder 0 (3 marks)

(ii) or or (3 marks)

(iii) Show (2 marks)

(iv) (4 marks)

8. (i) Translation by 2 units in the positive x direction (2 marks)

(ii) Stretch parallel to the y axis scale factor (2 marks)

(iii) Sketch showing as the only point of intersection (2 marks)

(iv) (3 marks)

(v) 9.60 (3 marks)

9. (i) (1 mark)

(ii) and (3 marks)

(iii) and (4 marks)
Do you know what question 9 i and ii were?
0
2 years ago
#26
(Original post by Mr M)
Mr M's OCR (not OCR MEI) Core 2 Answers May 2016

1. (i) 10 cm (2 marks)

(ii) 5.04 cm (2 marks)

2. (i) (2 marks)

(ii) (3 marks)

3. (i) (4 marks)

(ii) (2 marks)

4. (i) (2 marks)

(ii) (4 marks)

5. (a) (3 marks)

(b) (i) (4 marks)

(ii) 4 (1 mark)

6. (i) (3 marks)

(ii) (2 marks)

(iii) (6 marks)

7. (i) Quotient and remainder 0 (3 marks)

(ii) or or (3 marks)

(iii) Show (2 marks)

(iv) (4 marks)

8. (i) Translation by 2 units in the positive x direction (2 marks)

(ii) Stretch parallel to the y axis scale factor (2 marks)

(iii) Sketch showing as the only point of intersection (2 marks)

(iv) (3 marks)

(v) 9.60 (3 marks)

9. (i) (1 mark)

(ii) and (3 marks)

(iii) and (4 marks)
For 7iv I integrated between -1 and 1 instead of -1 and 3 do you think I'll get any marks?
0
2 years ago
#27
(Original post by Mr M)
Mr M's OCR (not OCR MEI) Core 2 Answers May 2016

1. (i) 10 cm (2 marks)

(ii) 5.04 cm (2 marks)

2. (i) (2 marks)

(ii) (3 marks)

3. (i) (4 marks)

(ii) (2 marks)

4. (i) (2 marks)

(ii) (4 marks)

5. (a) (3 marks)

(b) (i) (4 marks)

(ii) 4 (1 mark)

6. (i) (3 marks)

(ii) (2 marks)

(iii) (6 marks)

7. (i) Quotient and remainder 0 (3 marks)

(ii) or or (3 marks)

(iii) Show (2 marks)

(iv) (4 marks)

8. (i) Translation by 2 units in the positive x direction (2 marks)

(ii) Stretch parallel to the y axis scale factor (2 marks)

(iii) Sketch showing as the only point of intersection (2 marks)

(iv) (3 marks)

(v) 9.60 (3 marks)

9. (i) (1 mark)

(ii) and (3 marks)

(iii) and (4 marks)
I thought q 8 iii and 9 iii were 3 marks each?
0
2 years ago
#28
I'm almost certainly wrong, but how comee this doesn't work:

Sin(api/5) =sin(a*2pi/5)

Thus (double angle theorem)

Sin (api/5) =2*sin(api/5)*cos(api/5)

Thus 0.5 = cos(api/5)

Thus pi/3 = api/5

Thus 5/3 = a (which is smaller than 5 so is surely the first solution if it is right?
0
2 years ago
#29
(Original post by Jamesk123)
I thought q 8 iii and 9 iii were 3 marks each?
Nah 2 and 4.
0
2 years ago
#30
How come when you substitute in 4pie/3a into sin(ax) = root(3)cos(ax) they are not equal - it gives -ve value on the cos side.
0
2 years ago
#31
(Original post by M99)
I'm almost certainly wrong, but how comee this doesn't work:

Sin(api/5) =2sin(api/5)

Thus (double angle theorem)

Sin (api/5) =2*sin(api/5)*cos(api/5)

Thus 0.5 = cos(api/5)

Thus pi/3 = api/5

Thus 5/3 = a (which is smaller than 5 so is surely the first solution if it is right?
This is correct.
0
2 years ago
#32
(Original post by ComputeiT)
Using previous part:
x^2/(x+4) = 3^2 = 9
x^2=9(x+4)
x^2=9x+36
x^2-9x-36=0
(x-12)(x+3)=0
x=12 or -3
Thank you, I think i got that down in the exam but just realised I made a mistake whilst doing it again just now
0
#33
(Original post by ChrisWeatherilt)
How did you get the answer to 9ii?
Wrongly as it happens. It's fixed now.
0
#34
(Original post by bantersaur0usrex)
Attachment 538249
You were asked to solve something else though weren't you?
0
#35
(Original post by anon326589)
How is that the answer to 9ii
Sorry my mistake. Corrected now.
0
2 years ago
#36
(Original post by M99)
I'm almost certainly wrong, but how comee this doesn't work:

Sin(api/5) =sin(a*2pi/5)

Thus (double angle theorem)

Sin (api/5) =2*sin(api/5)*cos(api/5)

Thus 0.5 = cos(api/5)

Thus pi/3 = api/5

Thus 5/3 = a (which is smaller than 5 so is surely the first solution if it is right?
I got a = 5/3 and used a completely different method? Are we certain that a = 5/3 is incorrect?
0
#37
(Original post by Gogregg)
How did we get a= 5 and k=0 for 9ii ? I think i got a = 5/3 (i think) and k = (√3)/2 or something similar?
That's correct. I didn't read the question properly. Sorry.
1
#38
(Original post by Willdabeast666)
I made a quadratic out of question 5 bi) and then factorised it, but earlier in my working out had written down the correct answer, do i lose any marks?
They might ignore subsequent working.
0
2 years ago
#39
Thanks for this!
For question 9i if I've written 2pi x 1/a do I still get the mark or did I need it to be simplified?
How many marks do you think I would get for 9iii. if I rearranged to get tan (ax) = root3 and then did tan^-1 of root3 to get ax equals 60?
And do you think I would only lose 1 mark for 6iii. if I left N as 37.3?
Finally, how many marks do you think I would lose for writing kx^2 and kx^3 in 3i. instead of writing k^2x^2 and k^3x^3?
0
2 years ago
#40
Sir for question 3ii would i get any marks for saying that k=1/9 and 0. As well as that for 9i would i get the mark for putting the period as 0<ax<2pia
0
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