Unofficial mark scheme Edexcel C2 Maths 24th May 2017

Random answers:

1) 243-135x+30x^2-10/3 x^3
2) 70.5 and 109.5
3) y - coordinate = 6
b) Approximation: 17.56
c) Approximation: 27.56
4) Circle: centre: (5,-3) and radius 2 units
y - coordinates of intersection: -3+sqrt(3) and -3-sqrt(3)
Tent question: 19.04 altoghether
FABC sector arc length= 6.20
Sector area: 10.84
Factor theorem with (x+3) (i.e. f(-3) = 0)
Factorise : f(x) = (x+3)(7-3x)(2x+1)
y = 1.22 (2 dp)
4 solutions to the the trigonometic question (sin x = (1+sqrt(2))/3 or sin x =
(1-sqrt(2))/ 3): 53.58, 126.42, 187.94, 352.06 (2dp)
Geometic series - common ratio gives the quadratic required
k = 11 or 9/11 (but k is not an integer) so k=9/11
sum of 10 terms: -152520 (or something like that)
4th term: -512/11
x = a(2a+1)(2a-1)
y = 10/9 b
10) differentiate and show 1 gives dy/dx = 0 (e.g. via factorising)
R = 32.52

This is all I can remember. Please keep profanity to a minimal. Reply with edits.

(edited 6 years ago)

Reply 1

yslnizzy

1

Original post by Redcoats

Random answers:

1) 243-30x+135x^2-10/3 x^3
2) 70.5 and 109.5
3) y - coordinate = 6
b) Approximation: 17.56
c) Approximation: 27.56
4) Can't remember
Tent question: 19.04 altoghether
FABC sector arc length= 6.20
Sector area: 10.84
4 solutions to the the trigonometic question (sin x = (1+sqrt(2))/3 or sin x =
(1-sqrt(2))/ 3
Geometic series - common ratio gives the quadratic required
k = 11 or 9/11 (but k is not an integer) so k=9/11
sum of 10 terms: -152520 (or something like that)
4th term: -512/11
x = a(2a+1)(2a-1)
y = 10/9 b
10) differentiate and show 1 gives dy/dx = 0 (e.g. via factorising)
R = 32.52

This is all I can remember. Please keep profanity to a minimal. Reply with edits.

Tbh mate, idk what ur talking about??!

Reply 2

bakersgonnabake

1

Original post by Redcoats

Random answers:

1) 243-30x+135x^2-10/3 x^3
2) 70.5 and 109.5
3) y - coordinate = 6
b) Approximation: 17.56
c) Approximation: 27.56
4) Can't remember
Tent question: 19.04 altoghether
FABC sector arc length= 6.20
Sector area: 10.84
4 solutions to the the trigonometic question (sin x = (1+sqrt(2))/3 or sin x =
(1-sqrt(2))/ 3
Geometic series - common ratio gives the quadratic required
k = 11 or 9/11 (but k is not an integer) so k=9/11
sum of 10 terms: -152520 (or something like that)
4th term: -512/11
x = a(2a+1)(2a-1)
y = 10/9 b
10) differentiate and show 1 gives dy/dx = 0 (e.g. via factorising)
R = 32.52

This is all I can remember. Please keep profanity to a minimal. Reply with edits.

What was your method for the last question? I just integrated to find the area under the main curve then didn't know how to work out the small unshaded part

This is waaaaavy

Heya everyone, did no one get for q2) x = 70.5 and 59.5,

Since sine rule and another angle is 180-50-70.5 = 59.5 only other possible angle. How did people get x = 109.5 out of 180 degrees??!

Reply 5

JamesHope890

1

Unofficial ms for the edexcel 2017 Core 2 AS exam

2)70.1, 109.9 (70.5 and 109.5 if you don't round until the very end)
3)a)17.56
b)27.56
4)a)6.20
b)10.84
c)19.04
5)a)(5,-3)
b)radius=2
c)-3+root2, -3-root2
6)y=1.22
7)4a^3 - a
10b/9
8)53.58, 126.42, 187.94, 352.06
9)4th term= -512/11
Sum up to 10=-152520
10)32.52

Hope this helps
-James

(edited 6 years ago)

Reply 6

Haoding

1

Do you have the answer for Q.1?

Reply 7

Edgerulezs

11

Original post by Specofranger

Heya everyone, did no one get for q2) x = 70.5 and 59.5,

Since sine rule and another angle is 180-50-70.5 = 59.5 only other possible angle. How did people get x = 109.5 out of 180 degrees??!

You found the other angle in the triangle :frown:

you were meant to find the 2 possible values of x so 180-70.5

Reply 8

username2371841

20

Original post by Specofranger

Heya everyone, did no one get for q2) x = 70.5 and 59.5,

Since sine rule and another angle is 180-50-70.5 = 59.5 only other possible angle. How did people get x = 109.5 out of 180 degrees??!

surely you do 180-70.5

Reply 9

ShakezZz

4

I'd tell you if I did that question......

Reply 10

bakersgonnabake

1

Original post by JamesHope890

Unofficial ms for the edexcel 2017 Core 2 AS exam

2)70.1, 109.9
3)a)17.56
b)27.56
4)a)6.20
b)10.84
c)19.04
5)a)(5,-3)
b)radius=2
c)-3+root2, -3-root2
6)y=1.22
7)4a^3 - a
10b/9
8)53.58, 126.42, 187.94, 352.06
9)4th term= -512/11
Sum up to 10=-152520
10)32.52

Hope this helps
-James

JAMES,

How did you calculate the area of the unshaded segment beneath the curve in Q10? I found the area under the curve via integegration, and the equation of the line AB but I know I got the wrong answer. What was your method for the final bit?

Reply 11

Strimpy

12

Agree with all of these except (what I assume) is your answer to 7i). The first log one

Pretty sure was 4a^3-a

Posted from TSR Mobile

(edited 6 years ago)

Reply 12

Richdash1

1

Original post by bakersgonnabake

What was your method for the last question? I just integrated to find the area under the main curve then didn't know how to work out the small unshaded part

You integrate to find out the area under the whole curve then to find out the area under the line AB you integrate that line using X -coordinates from 0 to the X coordinate of B. Then you just take the whole area from the shaded area and that's the answer

Reply 13

Amarante

8

Q7 could have been factorised further

Posted from TSR Mobile

Reply 14

Jaazal

1

243-135x+30x^2-10/3x^3
70.5, 109.5
17.56
27.56
6.20m
10.84m^2
19.04m^2
(5,-3) Radius = 2
y= 1.22
-3+root3, -3-root3
-512/11
53.58, 126.42, 187.94, 352.06 Integration area = 32.52

Reply 15

Harrysomers1

1

Original post by Strimpy

Agree with all of these except (what I assume) is your answer to 7i). The first log one

Pretty sure was 4a^3-a

EDIT: lol didn't read it properly yes u did get it right dw

Posted from TSR Mobile

Yeah I got that too
EDIT I didn't read your edit

(edited 6 years ago)

Reply 16

Owen_Tech_Vint

1

Some answers from today