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OCR MEI AS Mathematics Core 1 13/05/2013

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Original post by Magenta96
oh wow I think I've slipped up on that too, can't believe I didn't realise the significance of r > 0 in the question!

Always read the question!!!


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Reply 181
I think there will be a A1 marking point for the + on that question.
Reply 182
Original post by jojo19apr
For the last question I got k=0 and k=4. I got that because y= - x + k, it had a gradient of -1 so the points on the curve must have done too. Then all I could think to do was differentiate (which as a retaker I can do but I know most of you may not have seen that yet), to find the x coordinates of where the gradient was -1 on the graph. Both seemed to match up graphically, so I hoped it worked ok!


Oh my God. I'm so relieved to see that someone else did that.

y=1/(x-2)

y=(x-2)-1

(chain rule)

dy/dx = -(x-2)-2
-(x-2)-2 = -1

(x-2)-2 = 1

1/(x-2)2=1

(x-2)2=1

x2-4x+3=0

(x-3)(x-1)=0

Gradient = -1 when x= 1 or 3

To find the y coordinates,

y=1/x-2
y=1/1-2=-1

y=-x+k is a tangent at point (1,-1)

-1=-1+k
k=0

y=1/3-2
y=1

y=-x+k is also a tangent at the point (3,1)

1=-3+k
k=4

So k=0 or 4

Hope this helps.
Reply 183
Original post by janicee
What did people get on the circle question? I couldn't find out where the circle intersected the y axis so I used the formula and got 2+root11 and 2-root11. Did I get this completely wrong? :frown: Gah I'm scared.


I did both ways, the formula and by setting y and x to 0 according to which intercept I was trying to find. I got the same as you with both methods, I only did both as I doubted myself with the formula thinking I may have gone wrong!!! So reassuringly, we got the same :biggrin:
You know the second last question in question 12? Did anybody else use the quadratic formula to solve the question to find the X coordinate?
Reply 185
Original post by Deceived
Oh my God. I'm so relieved to see that someone else did that.

y=1/(x-2)

y=(x-2)-1

(chain rule)

dy/dx = -(x-2)-2
-(x-2)-2 = -1

(x-2)-2 = 1

1/(x-2)2=1

(x-2)2=1

x2-4x+3=0

(x-3)(x-1)=0

Gradient = -1 when x= 1 or 3

To find the y coordinates,

y=1/x-2
y=1/1-2=-1

y=-x+k is a tangent at point (1,-1)

-1=-1+k
k=0

y=1/3-2
y=1

y=-x+k is also a tangent at the point (3,1)

1=-3+k
k=4

So k=0 or 4

Hope this helps.


:biggrin: Everyone came out with different answers and I was so sure of 4 and 0 because both points had a gradient of -1 and I couldn't see it could be any different! Differentiation for once saved me on this :biggrin:
Original post by jojo19apr
They won't know grade boundaries until they have all of the final results, then they adjust the boundaries to have a certain number of candidates walking away each grade. Hence for a hard paper they are lowered and vice versa, this seemed a reasonably fair paper so the boundaries may be slightly higher. But the general guide is 80% = A, 70% = B, 60% = C, 50% = D, 40% = E and 40% > U ........ Hope that gave you a vague idea!


I get that - but what do you all think it will be?


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for the rearanging question i got r=sqrt3v/pia+pib
did i do it wrong? :frown: i didnt put the pi in the bracket

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(edited 10 years ago)
Reply 188
Original post by Rubyturner94
for the rearanging question i got r=sqrt3v/pia+pib
did i do it wrong? :frown: i didnt put the pi in the bracket

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no, thats right you just distributed pi on the denominator
Reply 189
Original post by janicee
Ogod I failed. Well thanks for replying! I think I did okay on the other questions, fingers crossed!


Don't worry, there was no answer of (x-3)^2+(y-2)^2=20, that was the equation of the circle we were given! We didn't have to use complete the square on this either, it was substituting x and y in for 0 according to which intercept you were finding :biggrin: Hope you're not quite so worried!
Reply 190
for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
Reply 191
Original post by janicee
Do you mean 3(x-2)^2 -7? Because that's what I got


Same, it shouldn't be 3x^2 in front, just 3.
It also said in the form a(x+b)^2+c, so it can't be right
Reply 192
for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
Reply 193
Original post by janicee
Oh okay I failed. I'm such an idiot. Omg someone shoot me please :'(


The last question was 12, the one with k in :smile: There definitely wasn't a 13!
Reply 194
Original post by Soho32
for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?


the answer was 25;

(1/5)^-2

(5/1)^2

5^2 = 25


since 1/0.04 = 25 then you should be okay :smile: even if youe method is a bit of a weird way round it.
Reply 195
Original post by Soho32
for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?


I think you had to give the answer as 25.

(0.2)-2

=(1/5)-2

=1/(1/5)2

=1/(1/25)

=25

You may be completely fine but there's a possibility you could lose an A1 mark.
Original post by Soho32
for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?

i got 25 which is the same as 1/0.04 :smile:
Original post by Soho32
for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?


I did exactly this and for some reason didn't see that 1/0.04=25.
Really hope I don't lose a mark for it
Reply 198
Original post by krishkmistry
I get that - but what do you all think it will be?


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80% = 57-58

but I think that question 13 (iii) was awkward for a lot of people, except for the few that spotted the correct way to do it.
i've heard from a few of the higher graders in my year that thye found this a big hard at points so i'm going to say 55-53.
Reply 199
Original post by Deceived
For the very last question, did people get k=0 or k=4?

Thanks


Both :biggrin: Checked a few ways and these are the only 2 numbers that work

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