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OCR (not OCR MEI) Core 3 answers Jan 2011

OCR (not OCR MEI) CORE 3 answers Jan 2011


1. x=a3x=\frac{a}{3} or x=3ax=-3a (3 marks)


2. Translation 3 units in negative x direction, reflection in x axis, stretch scale factor 4 in y direction. Stationary points (-5, 24) and (-3, 0) (4 marks)


3. dAdt=45000cm2hr1\frac{dA}{dt}=45000 \, cm^2 hr^{-1} (3 marks)


4. (i) 25sin(θ+16.3)25\sin(\theta + 16.3) (3 marks)

ii) θ=12.4\theta = 12.4^\circ or θ=135\theta = 135^\circ (4 marks)


5. V=24πln5V=24 \pi \ln 5 (9 marks)


6. i) dydx=2(3x316x3)(x34x2+2)2\frac{dy}{dx}=\frac{-2(3x^3-16x-3)}{(x^3-4x^2+2)^2} and show ... (4 marks)

ii) (2.398, -1.552) (5 marks)


7. i) x=e88x=\sqrt{e^8 - 8} (3 marks)

ii) f has inverse f1(x)=exf^{-1} (x) = e^x (3 marks)

iii) 6e3\frac{6}{e^3} (3 marks)

iv) 20.3 (3 marks)


8. a) i) 2 positive U shapes and 2 negative inverted U shapes (3 marks)

ii) β=3πα\beta = 3\pi - \alpha (2 marks)

b) i)
Unparseable latex formula:

\tan 2\theta = \frac{2 \tan \theta}{1-\tan^2 \theta}}

(1 mark)

ii) 225322\frac{225}{322} (6 marks)


9. i) a) Show ... (3 marks)

b) Show and
Unparseable latex formula:

x>\frac{\ln 3}{4}}

(5 marks)

ii) y2ky \geq 2 \sqrt k (5 marks)
(edited 13 years ago)

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Any ideas on the grade boundaries? :smile:
Reply 2
I put 12pi ln25 for 5, would i lose any marks for not simplifying to your answer?
Reply 3
Original post by Mr M
OCR (not OCR MEI) answers Jan 2011


1. x=a3x=\frac{a}{3} or x=3ax=-3a (3 marks)


2. Translation 3 units in negative x direction, reflection in x axis, stretch scale factor 4 in y direction. Stationary points (-5, 24) and (-3, 0) (4 marks)


3. dAdt=45000cm2hr1\frac{dA}{dt}=45000 \, cm^2 hr^{-1} (3 marks)


4. (i) 25sin(θ+16.3)25\sin(\theta + 16.3) (3 marks)

ii) θ=12.4\theta = 12.4^\circ or θ=135\theta = 135^\circ (4 marks)


5. V=24πln5V=24 \pi \ln 5 (9 marks)


6. i) dydx=2(3x316x3)(x34x2+2)2\frac{dy}{dx}=\frac{-2(3x^3-16x-3)}{(x^3-4x^2+2)^2} and show ... (4 marks)

ii) (2.398, -1.552) (5 marks)


7. i) x=e88x=\sqrt{e^8 - 8} (3 marks)

ii) f has inverse f1(x)=exf^{-1} (x) = e^x (3 marks)

iii) 6e3\frac{6}{e^3} (3 marks)

iv) 20.2 (3 marks)


8. a) i) 2 positive U shapes and 2 negative inverted U shapes (3 marks)

ii) β=3πα\beta = 3\pi - \alpha (2 marks)

b) i)
Unparseable latex formula:

\tan 2\theta = \frac{2 \tan \theta}{1-\tan^2 \theta}}

(1 mark)

ii) 225322\frac{225}{322} (6 marks)


9. i) a) Show ... (3 marks)

b) Show and
Unparseable latex formula:

x>\frac{\ln 3}{4}}

(5 marks)

ii) y2ky \geq 2 \sqrt k (5 marks)


Did you feel it was a particularly hard paper?
Thanks for this. Seems I've done a bit better than I expected.
Reply 5
Original post by Mr M
OCR (not OCR MEI) answers Jan 2011


1. x=a3x=\frac{a}{3} or x=3ax=-3a (3 marks)


2. Translation 3 units in negative x direction, reflection in x axis, stretch scale factor 4 in y direction. Stationary points (-5, 24) and (-3, 0) (4 marks)


3. dAdt=45000cm2hr1\frac{dA}{dt}=45000 \, cm^2 hr^{-1} (3 marks)


4. (i) 25sin(θ+16.3)25\sin(\theta + 16.3) (3 marks)

ii) θ=12.4\theta = 12.4^\circ or θ=135\theta = 135^\circ (4 marks)


5. V=24πln5V=24 \pi \ln 5 (9 marks)


6. i) dydx=2(3x316x3)(x34x2+2)2\frac{dy}{dx}=\frac{-2(3x^3-16x-3)}{(x^3-4x^2+2)^2} and show ... (4 marks)

ii) (2.398, -1.552) (5 marks)


7. i) x=e88x=\sqrt{e^8 - 8} (3 marks)

ii) f has inverse f1(x)=exf^{-1} (x) = e^x (3 marks)

iii) 6e3\frac{6}{e^3} (3 marks)

iv) 20.2 (3 marks)


8. a) i) 2 positive U shapes and 2 negative inverted U shapes (3 marks)

ii) β=3πα\beta = 3\pi - \alpha (2 marks)

b) i)
Unparseable latex formula:

\tan 2\theta = \frac{2 \tan \theta}{1-\tan^2 \theta}}

(1 mark)

ii) 225322\frac{225}{322} (6 marks)


9. i) a) Show ... (3 marks)

b) Show and
Unparseable latex formula:

x>\frac{\ln 3}{4}}

(5 marks)

ii) y2ky \geq 2 \sqrt k (5 marks)



Okay Mr M some questions!
for 2) I got the stat points, the x direction translation, but didn't reflect it on the graph... how many marks would I have lost.

6ii) I got correct x value, but forgot to get y- again how many marks lost?

7iii) I found the gradient of fg(x) instead of gf(x)... would that make me loose all the marks? I got the correct fg(x) gradient?

Thanks!
Reply 6
hi for question 5.) if you left your answer as pi12ln25 would you get full marks?
also for question 6.ii) did you need to put both of those values in to the iterative fomula and if you only did one how many marks would you lose? thanks
Reply 7
How many method marks do you think i could possibly get for question 5 providing i got the wrong value for a? I got the correct integral and used the method for finding the volume correctly. 3-4?
Reply 8
Thanks so much Mr M, I did better than I imagined, looking at 56-60 :biggrin:

Any chance you could show your workings for 8b ii) ? The horrible tan one? :smile:
Reply 9
Can you explain the 9 mark question (q5) please?

Original post by Mr M
OCR (not OCR MEI) answers Jan 2011


1. x=a3x=\frac{a}{3} or x=3ax=-3a (3 marks)


2. Translation 3 units in negative x direction, reflection in x axis, stretch scale factor 4 in y direction. Stationary points (-5, 24) and (-3, 0) (4 marks)


3. dAdt=45000cm2hr1\frac{dA}{dt}=45000 \, cm^2 hr^{-1} (3 marks)


4. (i) 25sin(θ+16.3)25\sin(\theta + 16.3) (3 marks)

ii) θ=12.4\theta = 12.4^\circ or θ=135\theta = 135^\circ (4 marks)


5. V=24πln5V=24 \pi \ln 5 (9 marks)


6. i) dydx=2(3x316x3)(x34x2+2)2\frac{dy}{dx}=\frac{-2(3x^3-16x-3)}{(x^3-4x^2+2)^2} and show ... (4 marks)

ii) (2.398, -1.552) (5 marks)


7. i) x=e88x=\sqrt{e^8 - 8} (3 marks)

ii) f has inverse f1(x)=exf^{-1} (x) = e^x (3 marks)

iii) 6e3\frac{6}{e^3} (3 marks)

iv) 20.2 (3 marks)


8. a) i) 2 positive U shapes and 2 negative inverted U shapes (3 marks)

ii) β=3πα\beta = 3\pi - \alpha (2 marks)

b) i)
Unparseable latex formula:

\tan 2\theta = \frac{2 \tan \theta}{1-\tan^2 \theta}}

(1 mark)

ii) 225322\frac{225}{322} (6 marks)


9. i) a) Show ... (3 marks)

b) Show and
Unparseable latex formula:

x>\frac{\ln 3}{4}}

(5 marks)

ii) y2ky \geq 2 \sqrt k (5 marks)


how many marks would 3a = x, get for the first question? without further simplification
Oops! Anyone know how I can change the thread title to include Core 3?
Original post by kingjohno
Did you feel it was a particularly hard paper?


It was certainly the hardest OCR paper I have done this session out of C1, C2, C3, C4, FP1 and M2.
Original post by J DOT A
Okay Mr M some questions!
for 2) I got the stat points, the x direction translation, but didn't reflect it on the graph... how many marks would I have lost.

6ii) I got correct x value, but forgot to get y- again how many marks lost?

7iii) I found the gradient of fg(x) instead of gf(x)... would that make me loose all the marks? I got the correct fg(x) gradient?

Thanks!


drop 1 , drop 1, drop all 3
Original post by Dunney101
hi for question 5.) if you left your answer as pi12ln25 would you get full marks?
also for question 6.ii) did you need to put both of those values in to the iterative fomula and if you only did one how many marks would you lose? thanks


question 5 is fine and I think it asked for the coordinate in 6ii - I will check if I can work out how to bring the paper back up on my screen!
Original post by Mr M


iv) 20.2 (3 marks))

Are you sure about that one? I got 20.3, and so did everybody else who posted their answer on t'other thread.
Also, what raw mark would you say would be necessary for full UMS?
Reply 16
Original post by Mr M
OCR (not OCR MEI) CORE 3 answers Jan 2011


1. x=a3x=\frac{a}{3} or x=3ax=-3a (3 marks)


2. Translation 3 units in negative x direction, reflection in x axis, stretch scale factor 4 in y direction. Stationary points (-5, 24) and (-3, 0) (4 marks)


3. dAdt=45000cm2hr1\frac{dA}{dt}=45000 \, cm^2 hr^{-1} (3 marks)


4. (i) 25sin(θ+16.3)25\sin(\theta + 16.3) (3 marks)

ii) θ=12.4\theta = 12.4^\circ or θ=135\theta = 135^\circ (4 marks)


5. V=24πln5V=24 \pi \ln 5 (9 marks)


6. i) dydx=2(3x316x3)(x34x2+2)2\frac{dy}{dx}=\frac{-2(3x^3-16x-3)}{(x^3-4x^2+2)^2} and show ... (4 marks)

ii) (2.398, -1.552) (5 marks)


7. i) x=e88x=\sqrt{e^8 - 8} (3 marks)

ii) f has inverse f1(x)=exf^{-1} (x) = e^x (3 marks)

iii) 6e3\frac{6}{e^3} (3 marks)

iv) 20.2 (3 marks)


8. a) i) 2 positive U shapes and 2 negative inverted U shapes (3 marks)

ii) β=3πα\beta = 3\pi - \alpha (2 marks)

b) i)
Unparseable latex formula:

\tan 2\theta = \frac{2 \tan \theta}{1-\tan^2 \theta}}

(1 mark)

ii) 225322\frac{225}{322} (6 marks)


9. i) a) Show ... (3 marks)

b) Show and
Unparseable latex formula:

x>\frac{\ln 3}{4}}

(5 marks)

ii) y2ky \geq 2 \sqrt k (5 marks)


for the range in 9 b ii i got the right answer but wrote f(x).. rather than g(x).. do you think i would lose any marks? thanks!
Original post by Revolution is my Name
Are you sure about that one? I got 20.3, and so did everybody else who posted their answer on t'other thread.
Also, what raw mark would you say would be necessary for full UMS?


I used Autograph to do it using Simpson's Rule with 4 strips - it is possible that Autograph can't round properly! I will check it by hand after I have something to eat.
Reply 18
Original post by Mr M
drop 1 , drop 1, drop all 3


Ahh okay thanks! stupid stupid mistake.
Okay, now for the killer question.... I used tan2(theta)=2tanA/1-Tan^2A
Therefore, I did Tan4a=2tan2A/1-tan^2A.
I got the correct tanx and cotx values, however I accidently subbed in tanx=1/4 in my tan4a equation, getting so I ended up with 2*2*1/4 divided by 1-2*1/4^2
and got something like 5/something
How many marks would I have lost for that question?
Reply 19
Hi Mr M.
Thanks a lot for the answers, for question 8)a)i), if you only drew 1 positive U and 1 negtaive U, how many marks do you think I would lose?

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