The Student Room Group

Edexcel IAL Core Mathematics - C12 - WMA01 - JUNE 2016

Scroll to see replies

Original post by charlesdarwin
somebody remembers the answer for the second part of the trapezium rule question?



it was 14. something, I think 14.553
Original post by Einsteinj.n.r
it was 14. something, I think 14.553


For the part with the transformations, on the first part did u divide by 2 and the second part used trapezium rule again

Posted from TSR Mobile
till what i could remember..not in the same order though!
1. p=1.5, q=63
2. a. 14.556, b. 7.278 , cant remember
3. a. x< or equal to 4.5
b. x> 1.5 or x< -5 X< -5 and 1.5<X<4.5
4. arithmetic440066
5. centre(x-a)^2 + y2= a^2A= 25/8
6. angle sin 19.5 and -30
7. geometric0.817539
8. grapha- (8/5,4/5) b- (20/9, 10/9)
9. log Y^2 = 32/Xx= 0.5y=8
10. Q(x)a= -10b= 13
11. sectorarea = 0.654
12. some angle question pi/12 and 5pi/12
13. integration 11/64
Original post by Jjll99
don't you think so?


I found the exam easier so i think the gb may increase
Original post by jamessmith 2202
till what i could remember..not in the same order though!
1. p=1.5, q=63
2. a. 14.556, b. 7.278 , cant remember
3. a. x< or equal to 4.5
b. x> 1.5 or x< -5 X< -5 and 1.5<X<4.5
4. arithmetic440066
5. centre(x-a)^2 + y2= a^2A= 25/8
6. angle sin 19.5 and -30
7. geometric0.817539
8. grapha- (8/5,4/5) b- (20/9, 10/9)
9. log Y^2 = 32/Xx= 0.5y=8
10. Q(x)a= -10b= 13
11. sectorarea = 0.654
12. some angle question pi/12 and 5pi/12
13. integration 11/64


For 5, y=0 so it will be
centre(x-a)^2 + 9= a^2A= 25/8

For 4, I don't remember that kind of an answer

The rest seems familiar 😃😃
Posted from TSR Mobile
Original post by jamessmith 2202
till what i could remember..not in the same order though!
1. p=1.5, q=63
2. a. 14.556, b. 7.278 , cant remember
3. a. x< or equal to 4.5
b. x> 1.5 or x< -5 X< -5 and 1.5<X<4.5
4. arithmetic440066
5. centre(x-a)^2 + y2= a^2A= 25/8
6. angle sin 19.5 and -30
7. geometric0.817539
8. grapha- (8/5,4/5) b- (20/9, 10/9)
9. log Y^2 = 32/Xx= 0.5y=8
10. Q(x)a= -10b= 13
11. sectorarea = 0.654
12. some angle question pi/12 and 5pi/12
13. integration 11/64

from what i remember... all these seem right:smile:
Original post by khassan
For the part with the transformations, on the first part did u divide by 2 and the second part used trapezium rule again

Posted from TSR Mobile

it was 16+ the answer you got which was 14.553 or something
Original post by Ibrahim Rampuri
it was 16+ the answer you got which was 14.553 or something


Whaaaaat? How?

Posted from TSR Mobile
Original post by Ibrahim Rampuri
it was 16+ the answer you got which was 14.553 or something


Yup got the same!
Original post by jamessmith 2202
till what i could remember..not in the same order though!
1. p=1.5, q=63
2. a. 14.556, b. 7.278 , cant remember
3. a. x< or equal to 4.5
b. x> 1.5 or x< -5 X< -5 and 1.5<X<4.5
4. arithmetic440066
5. centre(x-a)^2 + y2= a^2A= 25/8
6. angle sin 19.5 and -30
7. geometric0.817539
8. grapha- (8/5,4/5) b- (20/9, 10/9)
9. log Y^2 = 32/Xx= 0.5y=8
10. Q(x)a= -10b= 13
11. sectorarea = 0.654
12. some angle question pi/12 and 5pi/12
13. integration 11/64


almost all seem familiar apart from 4 :/
Original post by khassan


it was integration of 2 + the original equation.. so when integrate 2 you get 2x and subtitute x with number of x values between 6 and -2 so 8 values so it is 2*8 + your original answer
Original post by pondsteps
Yup got the same!


For the first or second transformation?
For the first I just divided the previous by 2....for the second I used the trapezium rule again

Posted from TSR Mobile
Original post by khassan
For the first or second transformation?
For the first I just divided the previous by 2....for the second I used the trapezium rule again

Posted from TSR Mobile


no the 16+ thing is for the sec transformation.. and no u dont use the trapezium rule again
Original post by pondsteps
no the 16+ thing is for the sec transformation.. and no u dont use the trapezium rule again


Y 16?

Posted from TSR Mobile
Reply 114
Original post by jamessmith 2202
i think it is 11/64 cz..the area of the triangle was 0.5x(3/8)x(2-0.5)..=9/32
integrated region= x(x-1)(x-2) from 1 to 0.5= 7/64
9/32 -7/64 = 11/64


yea i think that's what i got
Reply 115
I don't know how some of you thought the paper was easy but I found it quite tough tbh 😟 Hope grade boundaries are lower.
Original post by Saad69
I got 30 and -19.5 I think

Posted from TSR Mobile


me too
Original post by Arzam45
Yea that's what I got.


i got a=10
Reply 118
Original post by Riri18
I don't know how some of you thought the paper was easy but I found it quite tough tbh 😟 Hope grade boundaries are lower.


It was definitely not easy, but it wasn't too hard either. I know many people at my school found it difficult so I'm sure the grade boundaries will be similar to the January 2016 grade boundaries, if not, lower. I guess the difficulty of the paper depends on how many past papers you've done, some people who found the exam very easy have done literally all the papers since 2007 lol
Original post by Raynedrops
Nothing, I have no idea how to work out the answer!


the centre was (a,0)
so circles eq will be (x-a)^2+(y-0)^2=a^2 because a is the radius
and in the second part substitute x with 4 and y with -3 to get a
i got it like 3.125

Quick Reply

Latest

Trending

Trending