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OCR (not OCR MEI) Core 1 Answers Jan 2011

OCR (not OCR MEI) Core 1 Answers Jan 2011


1. (i) Length of AB = 10 (2 marks)

(ii) Gradient of AB = -3/4 (2 marks)

(iii) They are perpendicular (3 marks)


2. p = 2 and q = 5 (3 marks)


3. (i) 8128^{\frac{1}{2}} (1 mark)

(ii) 828^{-2} (1 mark)

(iii) 8838^{\frac{8}{3}} (3 marks)


4. x=13,0,43,1x = \frac{1}{3}, 0, \frac{4}{3}, 1 (6 marks)


5. (i) Sketch of y=x3y=-x^3 (2 marks)

(ii) y=(x3)3y = -(x-3)^3 (2 marks)

(iii) Stretch scale factor 5 in y direction OR stretch scale factor 5135^{-\frac{1}{3}} in x direction (2 marks)


6. (i) 10x3+x24+1-10x^{-3} + \frac{x^{-2}}{4} + 1 (4 marks)

(ii) 30x4x3230x^{-4} - \frac{x^{-3}}{2} (2 marks)


7. (i) 4(x+32)2124 (x + \frac{3}{2})^2 - 12 (4 marks)

(ii) x=32±3x = \frac{-3}{2} \pm \sqrt 3 (4 marks)

(iii) k = -9 (3 marks)


8. (i) 4x + y - 32 = 0 (6 marks)

(ii) Midpoint = (6.5, 6) (3 marks)

(iii) x = 3 (2 marks)

(iv) x3x \leq 3 or x<3x < 3 (2 marks)


9. (i) Centre (4, 1) radius 252 \sqrt{5} (3 marks)

(ii) y=1±25y = 1 \pm 2 \sqrt 5 (3 marks)

(iii) Length of chord = 8 (3 marks)

(iv) (6, -3) (6 marks)

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Thanks for that.

Please can you tell me if you can remember what the questions were for Q7iii.
Also Q9ii.

Thanks a lot.
(edited 13 years ago)
Thanks a lot! Don't suppose you could write out the question for Q4?
Reply 3
Dam messed up on 9ii forgot to make it y=1±25y = 1 \pm 2 \sqrt 5 and left it as 1±201 \pm \sqrt 20

That'll lose me a few marks :/ cause it carries on in the next question
Original post by Paves141
Dam messed up on 9ii forgot to make it y=1±25y = 1 \pm 2 \sqrt 5 and left it as 1±201 \pm \sqrt 20

That'll lose me a few marks :/ cause it carries on in the next question


What was the question for 9ii?
Original post by Primus Pilus
Thanks a lot! Don't suppose you could write out the question for Q4?


It was the one where you had to substitute U into the equation.
Original post by Paves141
Dam messed up on 9ii forgot to make it y=1±25y = 1 \pm 2 \sqrt 5 and left it as 1±201 \pm \sqrt 20

That'll lose me a few marks :/ cause it carries on in the next question


Also it doesn't carry on to the next question because the last question was about the point of intersection and the answer was a coordinate.

And... did the question say to give it in its simplest form?
Reply 7
Think I might have full marks on this. Can't remember if I simplified 1+-sqrt(20) though!
Reply 8
Original post by The Joker666 x

Original post by The Joker666 x
Also it doesn't carry on to the next question because the last question was about the point of intersection and the answer was a coordinate.

And... did the question say to give it in its simplest form?


Simplified surd form I believe
Reply 9
hey mr m please could you tell me firstly. What was question 3) iii) and also please also for 6 i) i remember seeing. 1/4x as one of the figures to differentiate is it ok if i put (4x)^-1 because it was 1/4x so it means the whole 4x is to the power of -1? thank you
Reply 10
Are you going to put up the answers for the MEI paper ? thanks
Reply 11

Original post by Mr M
OCR (not OCR MEI) Core 1 Answers Jan 2011
x



Q8 iv) Is the 'or' part of the answer, or do you mean the examiner will accept either answer?
Original post by Coda
Simplified surd form I believe


Do you remember the question for 9ii?
Original post by Coda
Q8 iv) Is the 'or' part of the answer, or do you mean the examiner will accept either answer?


He means the examiner will accept either. Can you remind me of the question 9ii. and what it asked.
Original post by The Joker666 x
Do you remember the question for 9ii?


I think that question was about something along the lines of working out the values of 'k' from something :?
Original post by DarkPeople
I think that question was about something along the lines of working out the values of 'k' from something :?


thanks
Reply 16
Given that the tangent has the equation y = k find the intersection, or something alone those lines. Basically it was a matter of disregarding the (x-a)^2 and just writing

(k-1)^2 = 20
k-1 = +-sqrt(20)
k = 1 +- sqrt(20)
k = 1 +- 2sqrt(5)

Not sure if it asked for the simplified surd, or if I even wrote it :smile:

How do you think you did? I'm looking at around 97 - 100% I think
Isn't the answer to 7iii? 9 (not -9). This was the one where the discriminant equaled 0 and you had to find k.
Reply 18
For 7 (iii) k = 9 does it not? Not -9?

Also for 6 (i) and (ii) I got:

y=5x2+4x+xy = \frac{5}{x^2} + \frac{4}{x} + x

dydx=10x34x2+1\frac{dy}{dx} = -10x^{-3} - 4x^{-2} + 1

and

d2ydx2=30x4+8x3\frac{d^2y}{dx^2} = 30x^{-4} + 8x^{-3}

Respectively - is that not correct? :s-smilie:
(edited 13 years ago)
Original post by Coda
Given that the tangent has the equation y = k find the intersection, or something alone those lines. Basically it was a matter of disregarding the (x-a)^2 and just writing

(k-1)^2 = 20
k-1 = +-sqrt(20)
k = 1 +- sqrt(20)
k = 1 +- 2sqrt(5)

Not sure if it asked for the simplified surd, or if I even wrote it :smile:

How do you think you did? I'm looking at around 97 - 100% I think


Thanks

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