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

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Reply 20

Ozil5

Can someone please explain how to do 4ii, I know ive made a mistake somewhere but im not sure where

Reply 21

CC21231

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

Reply 22

ComputeiT

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

(edited 7 years ago)

Reply 23

AlfieH

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)

\displaystyle \frac{3 \pi}{10}

(2 marks)

(ii)

r=20.4

(3 marks)

3. (i)

\displaystyle 27 +27kx+9k^2x^2 +k^3x^3

(4 marks)

(ii)

k=\pm \sqrt{3}

(2 marks)

4. (i)

\displaystyle \log_3 \frac{x^2}{x+4}

(2 marks)

(ii)

x = 12

(4 marks)

5. (a)

\displaystyle \frac{x^4}{2} -x^3 +2x^2-6x+k

(3 marks)

(b) (i)

\displaystyle \frac{2}{a^2} - \frac{6}{a} + 4

(4 marks)

(ii) 4 (1 mark)

6. (i)

k=91

(3 marks)

(ii)

\displaystyle S_{16} = 978

(2 marks)

(iii)

N=38

(6 marks)

7. (i) Quotient

x^2-4x+3

and remainder 0 (3 marks)

(ii)

x=1

x=-1

x=3

(3 marks)

(iii) Show (2 marks)

(iv)

\displaystyle \frac{512}{15}

(4 marks)

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

(ii) Stretch parallel to the y axis scale factor

\displaystyle \frac{1}{9}

(2 marks)

(iii) Sketch showing

(0, \frac{1}{9})

as the only point of intersection (2 marks)

(iv)

x=6.73

(3 marks)

(v) 9.60 (3 marks)

9. (i)

\displaystyle \frac{2 \pi}{a}

(1 mark)

(ii)

a=5

and

k=0

(3 marks)

(iii)

\displaystyle x=\frac{\pi}{3a}

and

\displaystyle x=\frac{4 \pi}{3a}

(4 marks)

Question 9ii - the question said that k must be a positive constant and 0 isn't a positive constant?

Reply 24

username2097061

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)

\displaystyle \frac{3 \pi}{10}

(2 marks)

(ii)

r=20.4

(3 marks)

3. (i)

\displaystyle 27 +27kx+9k^2x^2 +k^3x^3

(4 marks)

(ii)

k=\pm \sqrt{3}

(2 marks)

4. (i)

\displaystyle \log_3 \frac{x^2}{x+4}

(2 marks)

(ii)

x = 12

(4 marks)

5. (a)

\displaystyle \frac{x^4}{2} -x^3 +2x^2-6x+k

(3 marks)

(b) (i)

\displaystyle \frac{2}{a^2} - \frac{6}{a} + 4

(4 marks)

(ii) 4 (1 mark)

6. (i)

k=91

(3 marks)

(ii)

\displaystyle S_{16} = 978

(2 marks)

(iii)

N=38

(6 marks)

7. (i) Quotient

x^2-4x+3

and remainder 0 (3 marks)

(ii)

x=1

x=-1

x=3

(3 marks)

(iii) Show (2 marks)

(iv)

\displaystyle \frac{512}{15}

(4 marks)

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

(ii) Stretch parallel to the y axis scale factor

\displaystyle \frac{1}{9}

(2 marks)

(iii) Sketch showing

(0, \frac{1}{9})

as the only point of intersection (2 marks)

(iv)

x=6.73

(3 marks)

(v) 9.60 (3 marks)

9. (i)

\displaystyle \frac{2 \pi}{a}

(1 mark)

(ii)

a=5

and

k=0

(3 marks)

(iii)

\displaystyle x=\frac{\pi}{3a}

and

\displaystyle x=\frac{4 \pi}{3a}

(4 marks)

Do you know what question 9 i and ii were?

Reply 25

ellaatherz

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)

\displaystyle \frac{3 \pi}{10}

(2 marks)

(ii)

r=20.4

(3 marks)

3. (i)

\displaystyle 27 +27kx+9k^2x^2 +k^3x^3

(4 marks)

(ii)

k=\pm \sqrt{3}

(2 marks)

4. (i)

\displaystyle \log_3 \frac{x^2}{x+4}

(2 marks)

(ii)

x = 12

(4 marks)

5. (a)

\displaystyle \frac{x^4}{2} -x^3 +2x^2-6x+k

(3 marks)

(b) (i)

\displaystyle \frac{2}{a^2} - \frac{6}{a} + 4

(4 marks)

(ii) 4 (1 mark)

6. (i)

k=91

(3 marks)

(ii)

\displaystyle S_{16} = 978

(2 marks)

(iii)

N=38

(6 marks)

7. (i) Quotient

x^2-4x+3

and remainder 0 (3 marks)

(ii)

x=1

x=-1

x=3

(3 marks)

(iii) Show (2 marks)

(iv)

\displaystyle \frac{512}{15}

(4 marks)

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

(ii) Stretch parallel to the y axis scale factor

\displaystyle \frac{1}{9}

(2 marks)

(iii) Sketch showing

(0, \frac{1}{9})

as the only point of intersection (2 marks)

(iv)

x=6.73

(3 marks)

(v) 9.60 (3 marks)

9. (i)

\displaystyle \frac{2 \pi}{a}

(1 mark)

(ii)

a=5

and

k=0

(3 marks)

(iii)

\displaystyle x=\frac{\pi}{3a}

and

\displaystyle x=\frac{4 \pi}{3a}

(4 marks)

For 7iv I integrated between -1 and 1 instead of -1 and 3 do you think I'll get any marks?

Reply 26

Jamesk123

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)

\displaystyle \frac{3 \pi}{10}

(2 marks)

(ii)

r=20.4

(3 marks)

3. (i)

\displaystyle 27 +27kx+9k^2x^2 +k^3x^3

(4 marks)

(ii)

k=\pm \sqrt{3}

(2 marks)

4. (i)

\displaystyle \log_3 \frac{x^2}{x+4}

(2 marks)

(ii)

x = 12

(4 marks)

5. (a)

\displaystyle \frac{x^4}{2} -x^3 +2x^2-6x+k

(3 marks)

(b) (i)

\displaystyle \frac{2}{a^2} - \frac{6}{a} + 4

(4 marks)

(ii) 4 (1 mark)

6. (i)

k=91

(3 marks)

(ii)

\displaystyle S_{16} = 978

(2 marks)

(iii)

N=38

(6 marks)

7. (i) Quotient

x^2-4x+3

and remainder 0 (3 marks)

(ii)

x=1

x=-1

x=3

(3 marks)

(iii) Show (2 marks)

(iv)

\displaystyle \frac{512}{15}

(4 marks)

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

(ii) Stretch parallel to the y axis scale factor

\displaystyle \frac{1}{9}

(2 marks)

(iii) Sketch showing

(0, \frac{1}{9})

as the only point of intersection (2 marks)

(iv)

x=6.73

(3 marks)

(v) 9.60 (3 marks)

9. (i)

\displaystyle \frac{2 \pi}{a}

(1 mark)

(ii)

a=5

and

k=0

(3 marks)

(iii)

\displaystyle x=\frac{\pi}{3a}

and

\displaystyle x=\frac{4 \pi}{3a}

(4 marks)

I thought q 8 iii and 9 iii were 3 marks each?

Reply 27

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?

(edited 7 years ago)

Reply 28

AlfieH

Original post by Jamesk123

I thought q 8 iii and 9 iii were 3 marks each?

Nah 2 and 4.

Reply 29

Bobrey

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.

Reply 30

AlfieH

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.

Reply 31

Ozil5

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

Reply 32

Mr M

Study Forum Helper

Original post by ChrisWeatherilt

How did you get the answer to 9ii?

Wrongly as it happens. It's fixed now.

Reply 33

Mr M

Study Forum Helper

Original post by bantersaur0usrex

You were asked to solve something else though weren't you?

Reply 34

Mr M

Study Forum Helper

Original post by anon326589

How is that the answer to 9ii

Sorry my mistake. Corrected now.

Reply 35

username2097061

Original post by M99

I got a = 5/3 and used a completely different method? Are we certain that a = 5/3 is incorrect?

Reply 36

Mr M

Study Forum Helper

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.

Reply 37

Mr M

Study Forum Helper

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.

Reply 38

amygracebroad

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?

(edited 7 years ago)

Reply 39

Devkj

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