EDEXCEL CORE 2 MAY 2016 - Unoffical Markscheme W/ Explained Answers

Hey all, sorry for the time delay in this one!! Best of wishes to Arsey too :frown:

I've not seen an official thread for this, although some great progress into formulating a markscheme has begun (who I will credit).
I think that the paper was deceiving in that it was actually more demanding than you'd think, but it was certainly better than C1 :smile:

Question 1:
- Show that A is 64
(Use the sum formula and it's straight forward enough)
- Sum to infinity was 256
64/0.25 = 256
- Difference between 9th and 10th: 1.602 (3dp was requested)
Be careful to do ar^8 instead of ar^9 and likewise goes for ar^9 instead of ar^10.
Question 2:
a) Complete the trapezium rule table - straight forward substitution for 1 mark.
b) 20.75 square units.
1/2 ( 7.5 + 2(17) was the calculation
c) 5.75 units I believe - I got this one wrong :frown:

(

Question 3:
a) Distance between points P and Q = Root34
P was (7,8) and Q was (10,13) = 3^2 + 5^2 = 34 then root the answer. It wanted its exact value.
b) (X - 7) ^2 + (Y-8)^2 = 34
c) 3x + 5y - 95 = 0
You had to find gradient of PQ then -1/m it as it is perpendicular to get -3/5 as the gradient. It passed through Q therefore you y-y1= m(X-x1)

Question 4: Nice, standard question.
a) Substute in -3/2 into F(X) and you got a remainder of 5 - just a calculator job.
b) Substite X=-2 into F(X) and there is 0 remainder thus a root.
c) X = -2, -2/3 or 1/2 leaving F(X) = (X+2)(3X+2)(2X-1)

Question 5: Binomial which got challenging in some areas.
a) I believe it was 16-288x + 1944x^2
b) A = 16 you had to to do this (1+kx)(16-288x+1944x^2) and the only single number with no X coefficient was 16 X 1 therefore 16.
c and d)
You had -288 + 16k = -232 Now add 288 to both sides leaving
16k = 52 therefore
K = 7/2
Substitute this value of k into the expansion and you get 936 ( I did the correct method andgot an answer in 900s but can't remember if this was what I 100% got so I may have made a stupid mistake).
B = 936
You had to remember you have the X^2 term and kx^2 term.

Question 6:
I thought the first question was nasty for 3 marks - maybe I just over complicated it. I got 8Pi/ 15 and -7pi/15 however my second answer was wrong. I believe that the agreed answers are:
8Pi/15 and -2pi/15 - I was so close :frown:

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b) I liked this one : 194.5 and 345.5 I got. You formulate a quadratic and X=2 or -1/4 . You had to disregard the 2 value as it is larger than one leaving sin^-1 of 1/4 which was 14.5. Then use cast and make sure to account for it being negative.

Question 7:
a) Standard integration
b) 24.3 square units - you had to find the X bound and I believe it was 9.

Question 8: Oh logs.
- I messed up - I got b = 5/3 - a hahahah! The correct answer is:
a-5/9
(Thanks to Mills.O)
B) -2.19 luckily I got this. Very sneaky one. 2^2x+5 - 7(2^y) =0
Let y = 2^x and you get 32y^2-7y =0
Divide by y to get 32y = 7
Therefore y=7/32
2^x = 7/32 and take logs to get the answer.

Question 9: A really discriminating question - tough.
a) The included angle is 120 degrees, from angles in an equilateral triangle are 60 degrees, then to find the included angle do 180 - 60 = 120 degrees. I did it in degrees, you may have converted to radians, either way you should be left with 120/360 X PI X x^2 Or use 1/2r^2 X 120 (in radians)
= PiX^2/3
b) Demanding proof - only 3 marks to. I got it in the end, cannot remember it fully though. Area of rectangle was 2xy. Area of triangle was 1/2absinC = 1/2x^2 X sin 60. Area of sector was previous answer. Add them up to equal 1000 (given in the question and then rearrange for y)
c) Even if you didn't get the previous proof, you could have attempted this. You had to find the arc length as well. You ended up with the arc length + 3x + 2y. Substitute the value of y in and do some manipulation.
d) Differentiate to equal zero and attempt to solve. X = 16.6 and P = 120.
e) Differntiate twice and state how it must be larger than zero. Even if your X value is wrong, which I believe mine was, you can still get an error carried forward and pick up these last two marks.

Hope this helps!
HUGE CREDIT TO - X_IDE_sidf
Insane memory from him, as well as marks and answer help. Thanks!
Also credit to Cookie_raider01 for clarifying the marks for each question! See the comments for it :smile:

Thanks to TripleSeven for clarity in P and Q coordinates!

Best of luck all - I think I've got between 59 and 70 so that's what I mean by deceiving - I think the grade boundaries will be 57-63 for an A. Sorry to speculate but a top 50 boundary for the A is more than likely.
Any corrections, thoughts or misunderstandings please say!!

(edited 7 years ago)

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Reply 1

cookie_raider01

cheers man, very very very helpful!
I lost all marks on logs and 1 or 2 marks on 9d as i only found x and didn't put it back to find P. And the one where you had to find area using integration, I got that wrong too as for x values, i got 2 and 0.
Lost about 9 or 10 marks . fml

I will post this thread's link in other forums!
Cheers!

Reply 2

Honeystone

Thank you, any chance of getting the marks next to the questions?

Reply 3

cookie_raider01

1.
A) 2 marks
B) 2 marks
C) 3 marks
2.
A) 1 mark
B) 3 marks
C) 2 marks
3.
A) 2 marks
B) 2 marks
C) 4 marks
4.
A) 2 marks
B) 2 marks
C) 4 marks
5.
A) 4 marks
B) 1
C) 2
D) 2
6.
A) 3
B) 6
7.
A) 3
B) 3
8.
A) 3
B)4
9.
A) 2
B) 3
C) 3
D) 5
E)2

Reply 4

TrueDAN

Original post by cookie_raider01

Thanks for the kind words bro, no problem!! That'd be sick if you share it, thanks man! Yeah I think I'm in a very similar position to you :smile:

I think it will be accounted for in the gradeboundaries being on the lower side though, I reckon you've solidly got an A and done really well! Don't worry about it xD

EDEXCEL CORE 2 MAY 2016 - Unoffical Markscheme W/ Explained Answers

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