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

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Original post by missypink<3
I put b as -2 at it was y=4b^x and it crossed the y-axis at (0,4) which is only possible is b is negative.


That's wrong - sorry.

Edit: I should explain why. Try letting x=12x=\frac{1}{2} for example.
(edited 10 years ago)
Original post by frankdatank
For 7ii) :
I wrote, Area = 37/30 = 1.23 (3sf)
then underlined the decimal, does this lose a mark?
Thanks in advance


That will be ok.
Reply 202
Original post by frankdatank
For 7ii) :
I wrote, Area = 37/30 = 1.23 (3sf)
then underlined the decimal, does this lose a mark?
Thanks in advance


Yeah, Mr M said as long as you state 37/30 it's fine.


Posted from TSR Mobile
67/72 roughly what ums for this paper?


Posted from TSR Mobile
Original post by Mr M
223532 - \frac{2}{3} \neq \frac{5}{3}

7ii) Find the equation of the tangent, then where the tangent meets the x axis. Now calculate the area of the triangle and subtract this from the value of the integral provided previously.


If cosx= -1 then x= 2pie/3
Cosine is negative in the 4th quadrant in the |CAST diagram. Therefore surely you would do pie + 2pie/3 in order to get the value of 5pie/3?

Also for question 7(ii).
I correctly found the equation of the tangent and then integrated the line between 4 and 0. I then subtracted this value from the value found in 7(i).
How many marks would I get?
Original post by germankid101
For the perimeter question 5(ii) I calculated the length BD using the formula 2rsin(θ/2) which gave me 12.461....
I then added this to 12.8 and 9 to give an answer of 34.3cm (3sf).
How many marks would I lose for this and why isn't this correct?



Original post by Mr M
I have no idea where that formula appeared from! Drop 2 marks.


I'm not sure whether you're interested but I suspect the formula works out the length of BC. You know for triangle BAC, and another triangle BDC with D on the circumference of the circle, angle BDC will be half angle BAC, or half theta. From the extended sine rule, the circumcircle of BCD will be the circle with centre A and radius 16 so:

BC/ sin(half theta) = 2R, so BC = 2Rsin(half theta).

However, germankid has made the mistake that D is not on the circumference of the circle.
(edited 10 years ago)
Original post by missypink<3
If cosx= -1 then x= 2pie/3
Cosine is negative in the 4th quadrant in the |CAST diagram. Therefore surely you would do pie + 2pie/3 in order to get the value of 5pie/3?

Also for question 7(ii).
I correctly found the equation of the tangent and then integrated the line between 4 and 0. I then subtracted this value from the value found in 7(i).
How many marks would I get?


Don't bother with that CAST rubbish. It is just confusing you.

cosx=cos(2πx)\cos x = \cos (2 \pi - x)

7ii) drop 3 marks
Reply 207
Hi Mr M,

Q1 - I worked out the wrong value value of h (stupid I know) but apart from that my method going forward was fine
Q7ii - I differentiated the equation but made a silly mistake and worked out m as 6 instead of 3. From then on however, my method was fine.
How many marks would I lose?
Thanks in advance,
Akash
Reply 208
Mr. M,

Unfortunately on question 3i) I expanded the first binomial incorrectly by using 5C1 and 5C2, as opposed to 6C1 and 6C2. Thus my answer was 64+800x+4000x^2, which I then used in ii) and so got c=-7.25.

How many marks do you think I would lose for this?

Thanks a lot
Original post by metaltron
I'm not sure whether you're interseted but I suspect the formula works out the length of BC. You know for triangle BAC, and another triangle BDC with D on the circumference of the circle, angle BDC will be half angle BAC, or half theta. From the extended sine rule, the circumcircle of BCD will be the circle with centre A and radius 16 so:

BC/ sin(half theta) = 2R, so BC = 2Rsin(half theta).

However, germankid has made the mistake that D is not on the circumference of the circle.


Yes that'll be it.
Original post by Akash B
Hi Mr M,

Q1 - I worked out the wrong value value of h (stupid I know) but apart from that my method going forward was fine
Q7ii - I differentiated the equation but made a silly mistake and worked out m as 6 instead of 3. From then on however, my method was fine.
How many marks would I lose?
Thanks in advance,
Akash



drop 1 and drop 2 I expect
Original post by dee1
Mr. M,

Unfortunately on question 3i) I expanded the first binomial incorrectly by using 5C1 and 5C2, as opposed to 6C1 and 6C2. Thus my answer was 64+800x+4000x^2, which I then used in ii) and so got c=-7.25.

How many marks do you think I would lose for this?

Thanks a lot


drop 2
Mr M how many marks would i get for 7ii if i worked out wrong tangent by using wrong x-value, but differentiated correctly, but showed you had to integrate by doing curve- tangent. Therefore, overall wrong area under curve

Also do you get penalised by putting to more decimal places
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 (4 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.


RE; 8i), I put y=1 and y=4 as my answers instead of (0,1) and (0,4). Will I still get the marks?
Original post by chuckynoriss
Mr M how many marks would i get for 7ii if i worked out wrong tangent by using wrong x-value, but differentiated correctly, but showed you had to integrate by doing curve- tangent. Therefore, overall wrong area under curve


curve - tangent will not necessarily produce a sensible answer

You will probably just get 1 mark for differentiating
Original post by martin_brundle
RE; 8i), I put y=1 and y=4 as my answers instead of (0,1) and (0,4). Will I still get the marks?


No I don't think so. Read the questions more carefully in future.
Original post by Mr M
drop 2


Hi Mr M,
Do you have a rough idea how many UMS I would get for 68/72?
Original post by missypink<3
Hi Mr M,
Do you have a rough idea how many UMS I would get for 68/72?


No sorry.
Hey Mr M,
you know for the logs question I did.
a^x=4b^x
xloga=2 +xlogb
xloga-xlogb=2
x(loga/b) = 2

then I got lost.
how many marks do you think I will gain out of 5?
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 (4 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.
How many marks do you think I would lose for the following mistakes:
1. h = (11 - 5) / 2 = 3, Area = 3/2 * (8/5 + 8/11 + 2 ( 8/8 ) ) = 6.49 - Did 3 ordinates instead of strips
6.i. a = 6, d = 1.8, S24 = 1/2 * 24 * ( 2(6) + 1.8 ( 24 - 1 ) ) = 640.8 - Did sum of first 24 instead of 30
8.c. a = 2, b = -2 - Did a negative value of b instead of 0 < b < 1
9.i. Remainder = 15/(x-2) - Divided remainder by the quotient rather than leaving it as the answer obtained using the factor theorem

Thanks in advance

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