The Student Room Group

Mr M's OCR (not OCR MEI) Core 2 Answers May 2012 <strong><font

Scroll to see replies

Hi Mr M, thanks for the mark scheme! For the common factors one, I got the (x-2)(x+3) etc but I said the common factors were therefore 2 and -3, would that get the marks?
Reply 101
Cant you also put (2sinx -2)(2sinx -1.5) for question 4- you get sin x= 1, 3/4?
Reply 102
Original post by Nuttyclorox
I'm not sure I can let you off for the pun. Hmm


my apologies, i'll C2 it that no more maths puns invade my posts.

Original post by Don95
Cant you also put (2sinx -2)(2sinx -1.5) for question 4- you get sin x= 1, 3/4?


yea, why not?
For question 4, I had the right method, but I factorised the equation wrongly so I got sin(x)=-3/4 and -1, not 3/4 and 1. Obviously I got the wrong answers as a result. How many marks do you think I lost?
Can anyone talk me through the very last question? I'm positive I've done the correct working but I keep getting the wrong answer! (Sat working through it now as its been bugging me :biggrin: )
Original post by jodiedoodles
Can anyone talk me through the very last question? I'm positive I've done the correct working but I keep getting the wrong answer! (Sat working through it now as its been bugging me :biggrin: )


I can if you give me the exact question (i did it in the exam but i can't remember what the question was.)
Original post by cameron95
I can if you give me the exact question (i did it in the exam but i can't remember what the question was.)


A geometric progression has first term log base 2 27 and common ratio log base 2 y

Find the exact value of y for which the sum to infinity of the geometric progression is 3

Thanks! :smile:
Reply 107
Original post by cameron95
I can if you give me the exact question (i did it in the exam but i can't remember what the question was.)


The question was
A geometric progression has first term logbase2 27 and common difference logbase2 x.

i) Find the set of values of y for which the geometric progression has a sum to infinity.
ii) Find the exact value of y for which the sum to infinity of the geometric progression is 3.
Reply 108
Original post by jodiedoodles
Can anyone talk me through the very last question? I'm positive I've done the correct working but I keep getting the wrong answer! (Sat working through it now as its been bugging me :biggrin: )


find the exact value of y for which the sum to infinity is 3

sum to infinity for a geometric series = a / (1-r)

a = log2(27)
d = log2(y)
when sum to infinity = 3

3 = [log2(27)] / [1 - log2(y)]

1 can be thought of as log2(2)

use rule loga(b) - loga(c) = loga(b/c)

3 = log2(27) / log2(2/y) = log(2/y)(27)

therefore (2/y)^3 = 27

therefore (2/y) = cubert(27) = 3

therefore 3y = 2

therefore y = 2/3
(edited 11 years ago)
Original post by jodiedoodles
Can anyone talk me through the very last question? I'm positive I've done the correct working but I keep getting the wrong answer! (Sat working through it now as its been bugging me :biggrin: )


9 (a) (i)
4th terms is a + (n-1)d where a is log2(27) and d is log2(x) and n is of course 4.
Therefore the 4th terms is log2(27)+3log2(27). Then using a few laws of logs you end up at the correct answer.

9 (a) (ii)
Log2(27x^3)=6
2^6=27x^3
(2^6)/27=x^3
Cubert((2^6)/27)=4/3

9 (b) (i)
If the geometric progression has a sum to infinity, the common ratio, r must be less than 1 and more than -1, so -1<r<1 and r=log2(y) so -1<log2(y)<1

log2(y)>-1 is same as 2^(-1)<y i.e. 1/2<y

log2(y)<1 is the same as 2^1>y i.e. 2>y

So y must be in between 1/2 and 2 i.e. 1/2<y<2

9 (b) (ii)

Sum to infinity is a/1-r where a is log2(27) and r is log2(y), so..

log2(27) / (1-log2(y)) = 3
log2(27) = 3 - 3log2(y)
3 - log2(27) = 3log2(y)
2^(3-log2(27) = y^3
y = Cubert(2^3-log2(27) = Cubert (8/27) = 2/3

That explain it??
(edited 11 years ago)
Original post by RebeccaLM
Hi Mr M, thanks for the mark scheme! For the common factors one, I got the (x-2)(x+3) etc but I said the common factors were therefore 2 and -3, would that get the marks?


I did that too. Fingers crossed it is okay!
Mr M's OCR (not OCR MEI) Core 2 Answers May 2012





5. (a) (i) u2=12u_2 =\frac{1}{2} and u3=4u_3 = 4 (2 marks)

(ii) Cycles periodically between 4 and 1/2 (1 mark)


For part I wrote ii) Oscillating sequence.

Would this get the mark?
Reply 112
Original post by Mr M
9bi) 1 mark

9bii Not sure what you have done really

7 no marks for using a decimal


Log to the base2 of 27 over 1 - log to the base 2 of y = log to the base 2 of 64

made this to 27/(2/y)=64.

So continued as you should, but had 64 instead of 8. Stupid error I know. But just asking if id get method marks.
Original post by saj1994
But for Q7i) i got 5/sqrt(29) which if rationalize the fraction i would have got 5sqrt(29)/29 and for 7ii) i got the answer u got but in positive.


Oh I see - I thought you were talking about 7b)
Reply 114
Mr M - what working was needed for 9bii, 9aii and 8ii? I haven't got the right answers, so looking for some hope in the method marks. Just wanna know where the question wanted us to go
(edited 11 years ago)
Original post by IShouldBeRevising_
Holy **** how did you get that lol


Lol just took it with me after I sat the paper today, was fine our teacher said we were welcome to :smile:
Mr M, for Question 7(a), for part (i) I just did Inverse Tan of 2/5 and then found the cosine of that, and put it down in decimal (correct but not the exact value). Did the same thing for part (ii) but found the obtuse angle of course. So both my answers were in decimal and not exact form. How many marks will I drop for that??? Thanks
Reply 117
Original post by Mr M
Oh I see - I thought you were talking about 7b)

so how many marks would i get for 7i) and 7ii) and also in 7b there is a right angle at C so u can u pythagoras rule and SOH CAH TOA
Original post by tahseen1995
9 (a) (i)
4th terms is a + (n-1)d where a is log2(27) and d is log2(x) and n is of course 4.
Therefore the 4th terms is log2(27)+3log2(27). Then using a few laws of logs you end up at the correct answer.

9 (a) (ii)
Log2(27x^3)=6
2^6=27x^3
(2^6)/27=x^3
Cubert((2^6)/27)=4/3

9 (b) (i)
If the geometric progression has a sum to infinity, the common ratio, r must be less than 1 and more than -1, so -1<r<1 and r=log2(y) so -1<log2(y)<1

log2(y)>-1 is same as 2^(-1)<y i.e. 1/2<y

log2(y)<1 is the same as 2^1>y i.e. 2>y

So y must be in between 1/2 and 2 i.e. 1/2<y<2

9 (b) (ii)

Sum to infinity is a/1-r where a is log2(27) and r is log2(y), so..

log2(27) / (1-log2(y)) = 3
log2(27) = 3 - 3log2(y)
3 - log2(27) = 3log2(y)
2^(3-log2(27) = y^3
y = Cubert(2^3-log2(27) = Cubert (8/27) = 2/3

That explain it??


Yeah thanks :smile: Got that exact working until the very last line so hopefully only lose one mark, max 2! No idea what i done, i got to 2^(3-logbase2(27)) = y^3, attempted to cube root it to get y and my calculator said syntax error! So i done 2^3 / 2^-logbase2(27) = y^3, so y^3 = 216, so y = 6 :frown: Not quite sure what I was doing wrong to get syntax error :s-smilie:
Reply 119
Original post by tahseen1995
Mr M, for Question 7(a), for part (i) I just did Inverse Tan of 2/5 and then found the cosine of that, and put it down in decimal (correct but not the exact value). Did the same thing for part (ii) but found the obtuse angle of course. So both my answers were in decimal and not exact form. How many marks will I drop for that??? Thanks


i did the exact same

Quick Reply

Latest