Unofficial mark scheme Edexcel C2 Maths 24th May 2017

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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??!

U use the sin graph m9

I'll get A thank god

Original post by Wisdom2

Pythagoras theorem then the cosine rule

wait what

Reply 43

Thatguystein

Original post by JamesHope890

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

Original post by JamesHope890

Ngl this was one of the best papers I have ever seen

Reply 44

sophieannord98

Can someone remind me what q3 was?

Reply 45

SM45367

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??!

if you've drawn the triangles, you'd realise one value is obtuse; so one value is gonna be bigger than 90

Reply 46

Brussell53

Original post by bakersgonnabake

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?

You don't find the unshaded area you do a tingle from x=1 to x=2 and calculate area of triangle and add it to the integrated area from x=1 to x=-0.25

Reply 47

Cor395

was one of the answers -3-root3 and -3+root3

Reply 48

JamesHope890

Original post by Wolfram Alpha

How the hell did you get four solutions for the trig question??? I only got two and even then my answers were 31 degrees and 328 degrees???

When you root (3sinx -1), you get +-root2 as part of the other side. This gives two sets of two results.

Reply 49

JSC16

Original post by JamesHope890

I couldn't do the first part to the trig question, otherwise it was a nice paper :h:

Reply 50

kays1234

Original post by Hasib_332

This is waaaaavy

hola @ man

Reply 51

rojanrana

Original post by Wisdom2

-35.9375?

Nvm I got the same answer as them.

Reply 52

JamesHope890

Original post by bakersgonnabake

In order to avoid errors due to negative area, I did under the curve from -1/4 to 0, plus under the curve from 0 to 1, plus the triangle between x=1,x=2 and point A which was 1/2 (25)(1)

Reply 53

JSC16

Original post by JamesHope890

When you root (3sinx -1), you get +-root2 as part of the other side. This gives two sets of two results.

nah it was four as there were two solutions each with two solutions. And i didn't get what you got

Reply 54

Aceofall

Am I the only one who got -171585 for the sum?

edit : just realized I used the wrong a value great

(edited 6 years ago)

Reply 55

Hadleigh2k11

For the trapezium rule question I got :
(B) 27. Something
(C) I thought it said 5+ , so I subtracted 5 to et 22. Something

??

Reply 56

Wisdom2

From another user:

"You need to use cosine rule to work out the length BC, and then sine rule to find that sin(x) is approx. 0.987 (forgot exact value), and then you need to realise that a positive sin value can happen twice before 180 degrees, meaning that its either x or 180 - x.
Both are valid since (180 - x) + 50 < 180, meaning that the third angle WILL be positive."

Turns out i was wrong, apologies

Original post by ranulv

wait what

Reply 57

mandp

Easiest C2 paper in existence

Reply 58

Wolfram Alpha

Original post by JamesHope890

When you root (3sinx -1), you get +-root2 as part of the other side. This gives two sets of two results.

I did that and I for x = (root 2)/3 + 1 and (-root 2)/3 + 1 but then one of the solutions was invalid??? Since was greater then 1?????

Reply 59

Wisdom2

I see. Cheers mate!

Original post by Bobrocket

You can't use Pythagoras, it wasn't a right-angled triangle.
You need to use cosine rule to work out the length BC, and then sine rule to find that sin(x) is approx. 0.987 (forgot exact value), and then you need to realise that a positive sin value can happen twice before 180 degrees, meaning that its either x or 180 - x.
Both are valid since (180 - x) + 50 < 180, meaning that the third angle WILL be positive.

EDIT:
By working out the length of BC as 12.554 (via cos. rule), you get
16*sin(50) / 12.554 = sin(x) which is approx. 0.9763
arcsin(0.9763) is approx 77.5, with the second solution being 180 - 77.5, which is 102.5

So the actual solutions for this q. are 77.5, 102.5 (both are valid)