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Edexcel IAL Core Mathematics - C12 - WMA01 - JUNE 2016

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There are some questions I remember but couldn't do
Any ideas on the answers?

- The question with the circle with centre (a, 3). Another coordinate on the circle is P(4, -3). Find a

- The logarithm question to convert log 2 y = 5 - log 2 x (the 2's should be small) into y^2 = k / x
- How to solve the logarithm equation simultaneously with another equation I can't remember

- How to find/prove r = 2 for the circle segment

- The answer in the Sum of n question Where a = 97 and d = -3. I think I got 56
(edited 7 years ago)
Original post by charlesdarwin
What did you get for the trapezium rule?


I got something like 14.553. I don't think I got the second part right though
Reply 62
Original post by Raynedrops
I got something like 14.553. I don't think I got the second part right though


yeah i got 14.555
Reply 63
Original post by Raynedrops
I got something like 14.553. I don't think I got the second part right though


Was the second part of that question somewhere around 132?
(not sure if it was this question)
Original post by charlesdarwin
What did you get for the trapezium rule?


i think it was 14.556?
what did u get for (A) in centre of circle
Original post by Raynedrops
There are some questions I remember but couldn't do
Any ideas on the answers?

- The question with the circle with centre (a, 3). Another coordinate on the circle is P(4, -3). Find a

- The logarithm question to convert log 2 y = 5 - log 2 x (the 2's should be small) into y^2 = k / x
- How to solve the logarithm equation simultaneously with another equation I can't remember

- How to find/prove r = 2 for the circle segment

- The answer in the Sum of n question Where a = 97 and d = -3. I think I got 56
who else didnt have enough time? :biggrin:
Original post by Arzam45
Yea that's what I got.

a was 23? I think, idont remeber
Original post by Arzam45
Was the second part of that question somewhere around 132?
(not sure if it was this question)


It was the equation in the integrate brackets but slightly different
i) original equation / 2
ii) 2 + original equation

I think you were suppose to manipulate the answer in the first part
Original post by azizmansoor
what did u get for (A) in centre of circle


Nothing, I have no idea how to work out the answer!
Reply 70
What grade boundary for an A do you guys expect?
Reply 71
Can somebody please tell me a way to find the paper we did? I think teachers get the papers straight after the exam but I'm a private candidate so that's not going to work for me. Anybody have the paper as such?
June 14 was 80 it was easier so less than that I hope

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What is the answer of a in the circle question ? did any one get it as 25/8
i calculated a and it was 25/8 not sure tho
My equation for the circle was (x-a)^2+y^2=a^2. Since the point (4,-3) was a coordinate on the circle, I plugged in the numbers into my equation in order to get a.
(4-a)^2+9=a^2.
16-8a+9=0.
a=25/8
Yup , thats exactly what i did :smile:
Original post by EUSHA
im 100% sure it was 27/64 in the last one.
it was area under line- area under curve.
calculation was like this:integ(limit 2 to .5)[-.25x+.5]-integ(limit 2 to .5)[x^3-3x^2+2x]


yup that was the answer.. I integrated from 1 to 0.5 as those were the points the curve was crossing the x-axis.. then subtracted area of the triangle from the integrated part:smile:
Reply 78
Original post by Einsteinj.n.r
yup that was the answer.. I integrated from 1 to 0.5 as those were the points the curve was crossing the x-axis.. then subtracted area of the triangle from the integrated part:smile:


Are you sure about this because i got 11/64
Original post by queenoflean
My equation for the circle was (x-a)^2+y^2=a^2. Since the point (4,-3) was a coordinate on the circle, I plugged in the numbers into my equation in order to get a.
(4-a)^2+9=a^2.
16-8a+9=0.
a=25/8


Same

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