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Mr M's OCR (not OCR MEI) Core 2 Answers May 2013

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Reply 20
Thanks
Original post by Majeue
I'd imagine you would get 2/4 marks.
Reply 21
Original post by swaggy
Hey does anyone remember the questions for 5i), 5ii), 6i) 6ii) and 9i) please


Oh thats right thanks :smile: do you know the questions for 6, 7 and 9?


6i) was that arithmetic progression where she used 6g then 7.8g for an experiment

6ii) was the geometric series where she had 1800g and you had to first prove that the maximum number of experiments she could do could be expressed as 1.3^N<91 and then use logarithms to find the value of N

7i) you had to show that the area between the curve y = x^3/2 - 1 and the x axis was 9.4

ii) you had to find the area between the curve, it's tangent at the point (4,7) and the x axis

9i) you had to find the remainder when 4x^3 -7x - 3 was divided by (x-2)

ii) you had to prove that (2x + 1) was a factor and express the cubic as the product of three linear factors

iii) you had to find out the possible values of x when 4cos^3x - 7cosx - 3 = 0
Reply 22
How many marks do you think I would get if worked area of small triangle and equation of tangent correct- 2 marks?? Thanks
Original post by Mr M
Definitely.




(Original post by kiraeil)
For the area under curve question would I get some marks for workings as i worked out equation of tangent and area of small triangle but got complete wrong answer?
thanks
(edited 10 years ago)
Reply 23
Mr M, i love you, take me back :colondollar:
Reply 24
For question 8 i a and b I put y=1 and y=4 instead of coordinates - will I lose the marks?
Reply 25
This went so well :biggrin: thanks for putting up your answers Mr. M, they were all the same as what i got! :biggrin:
Reply 26
Since there's not one in the OP:
Worked Solution to 8ii)
ax=4bx[br]xlog2a=2+xlog2b[br]x(log2ab)=2[br]As ab=2b=2a[br]x(log2a22)=2[br]x=2log2a1 a^x = 4b^x[br]\Rightarrow x\log_2a = 2+x\log_2b[br]\Rightarrow x(\log_2\frac{a}{b}) =2[br]\mathrm{As} \ ab=2 \Rightarrow b=\frac{2}{a}[br]\Rightarrow x(\log_2\frac{a^2}{2}) = 2[br]\Rightarrow x = \dfrac{2}{log_2a - 1}

And: For the sake of completion:
Worked solution to 7)i)
I=14x321 dx=[25x52x]14[br]I=(25(452)4)(25(152)1)[br]I=6454+35=67205=475=925I = \displaystyle\int^4_1 x^{\frac{3}{2}} - 1 \ dx = \left[\dfrac{2}{5}x^{\frac{5}{2}} - x \right]^4_1[br]\Rightarrow I = \left(\dfrac{2}{5}(4^{\frac{5}{2}}) - 4\right) - \left(\dfrac{2}{5}(1^{\frac{5}{2}}) - 1\right)[br]\Rightarrow I = \dfrac{64}{5}-4+\dfrac{3}{5} = \dfrac{67-20}{5} = \dfrac{47}{5} = 9\frac{2}{5}
(edited 10 years ago)
Reply 27
Found it harder than the c2 past papers. Hoping grade boundaries will be low.
Reply 28
Original post by Mr M
Mr M's OCR (not OCR MEI) Core 2 Answers May 2013


1. 6.39 (4 marks)


2. (i) 254 and 106 degrees (3 marks)

(ii) 71.6 and 252 degrees (3 marks)


3. (i) 64+960x+6000x264+960x+6000x^2 (3 marks)

(ii) c = - 11 (3 marks)


4. (a) 54x43x2+x+k\frac{5}{4}x^4 -3x^2+x+k (3 marks)

(b) (i) 12x2+k-12 x^{-2} + k (2 marks)

(ii) a = 2 (3 marks)


5. (i) 62.2 (4 marks)

(ii) 34.0 (4 marks)


6. (i) 963 (3 marks)

(ii) 17 (6 marks)


7. (i) Show ... (4 marks)

(ii) 3730\frac{37}{30} (5 marks)


8. (i) (a) (0, 1) (1 mark)

b) (0, 4) (1 mark)

c) Any value where a > 1 and any value where 0 < b < 1 (2 marks)

(ii) Show ... (5 marks)


9. (i) 15 (2 marks)

(ii) Show f(-0.5)=0 and f(x)=(2x+1)(2x3)(x+1)f(x)=(2x+1)(2x-3)(x+1) (6 marks)

(iii) θ=2π3\theta = \frac{2\pi}{3} and θ=4π3\theta =\frac{4\pi}{3} and θ=π\theta=\pi (4 marks)


I'll be around a bit this afternoon and then part of this evening to answer questions. Please post in this thread rather than sending me private messages.
Heeeey for (4b) I never got this (i) (2 marks)
I got -6x^-4+k can u explain what I did rong
Reply 29
For 7ii) I forgot to put the exact value so I put 1.23 (3sf) instead of 37/30. How many marks will I lose? :s
Reply 30
For 8 i) i said it crossed y axis at x= 0 therefore...

But i didn't state the coordinates specifically ie. In the format (x,y). Would i still get the mark ?
does anyone remember what question 7ii was ?
Reply 32
Anyone know how the factor theorem question was worth 6 marks? Seemed a bit excessive to me...
Reply 33
Original post by kelvinbeyioku
does anyone remember what question 7ii was ?


Finding the area between the tangent, the x-axis and the curve. :smile:
(edited 10 years ago)
Reply 34
Original post by geekD96
Heeeey for (4b) I never got this (i) (2 marks)
I got -6x^-4+k can u explain what I did rong


When integrating, you're supposed to add 1 to the index and then divide the coefficient by that new number. You got the index and took 1. -3 + 1 = -2, so you had to divide 24 by -2 not -4.
Original post by milesw97
Anyone know how the factor theorem question was worth 6 marks? Seemed a bit excessive to me...


Factor theorem was worth 2 marks.

The one where you had to factorise the polynomial had 6 marks.
Reply 36
Original post by Mr M
Mr M's OCR (not OCR MEI) Core 2 Answers May 2013


1. 6.39 (4 marks)


2. (i) 254 and 106 degrees (3 marks)

(ii) 71.6 and 252 degrees (3 marks)


3. (i) 64+960x+6000x264+960x+6000x^2 (3 marks)

(ii) c = - 11 (3 marks)


4. (a) 54x43x2+x+k\frac{5}{4}x^4 -3x^2+x+k (3 marks)

(b) (i) 12x2+k-12 x^{-2} + k (2 marks)

(ii) a = 2 (3 marks)


5. (i) 62.2 (4 marks)

(ii) 34.0 (4 marks)


6. (i) 963 (3 marks)

(ii) 17 (6 marks)


7. (i) Show ... (4 marks)

(ii) 3730\frac{37}{30} (5 marks)


8. (i) (a) (0, 1) (1 mark)

b) (0, 4) (1 mark)

c) Any value where a > 1 and any value where 0 < b < 1 (2 marks)

(ii) Show ... (5 marks)


9. (i) 15 (2 marks)

(ii) Show f(-0.5)=0 and f(x)=(2x+1)(2x3)(x+1)f(x)=(2x+1)(2x-3)(x+1) (6 marks)

(iii) θ=2π3\theta = \frac{2\pi}{3} and θ=4π3\theta =\frac{4\pi}{3} and θ=π\theta=\pi (4 marks)


I'll be around a bit this afternoon and then part of this evening to answer questions. Please post in this thread rather than sending me private messages.


I got 59-62 marks do you think this will be an A? Everyone is saying it was easy, could you hazard a guess at the grade boundaries? :smile:

and also for 9ii) I got wrote (2x-3)(x+1) without the previous answer but then I wrote so x= -1/2 +2/3 or -1 will I lose a mark for having not written (2x+1)(2X-3)(X+1) ???
For 4b ii) evaluated it as lim n -> infinity 12/n = 0 therefore it is
(0) - (-12/a) = 3 therefore a = 4, as I forgot the square sign on the a and left it as a = 4 how many marks would I lose for this?
Reply 38
Original post by mcavengers
For 7ii) I forgot to put the exact value so I put 1.23 (3sf) instead of 37/30. How many marks will I lose? :s


Probably 1, but you'll have to ask MrM to be sure :smile:

Spoiler

Reply 39
Have you got a copy of the paper?


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