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# OCR MEI AS Mathematics Core 1 13/05/2013 Watch

1. (Original post by PreppyNinja)
You know the g(x)=+15 or something? How did you do that? It's the only question I left
The question before you have the equation 2x3 + 9x2 - 2x - 24
So -24 + 15 = -9
2. (Original post by Daniel__)
Anyone remember the equations of the two lines in question 2, section A? (intersection question).
Dont remember the equations but i got 7/10 and 1/10
3. (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!

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4. I think there will be a A1 marking point for the + on that question.
5. (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.
6. (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? 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
7. 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?
8. (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.
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
9. (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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10. for the rearanging question i got r=sqrt3v/pia+pib
did i do it wrong? i didnt put the pi in the bracket

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11. (Original post by Rubyturner94)
for the rearanging question i got r=sqrt3v/pia+pib
did i do it wrong? i didnt put the pi in the bracket

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no, thats right you just distributed pi on the denominator
12. (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 Hope you're not quite so worried!
13. for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
14. (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
15. for the evaluate 0.2^-2 was it okay to leave the answer as a fraction? i.e 1/0.04?
16. (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 There definitely wasn't a 13!
17. (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?

(1/5)^-2

(5/1)^2

5^2 = 25

since 1/0.04 = 25 then you should be okay even if youe method is a bit of a weird way round it.
18. (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?

(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.
19. (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
20. (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

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