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

1. (Original post by TrueDAN)
Yes that was it! Thanks
Great mark scheme by the way it was really helpful.
2. I think I got about 67. However, I made some stupid mistakes and I am not certain how much i will lose for them. Lowest I can get is 63 from what I can tell, but I know they give marks for working with mistakes. How many marks do you think I will lose for the following mistakes:

X-x1/y-y1 by accident, but otherwise good working

Missing the 1/12 in the differentiation part of the question.

1.601 rather than 1.602

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3. (Original post by LukeB98)
Sine is negative between 180 and 360 degrees. Thus the 300 degree one is a solution.
I checked on my calculator at the end to make sure my values gave me -1/4 and the calculator gave me +1/4. I am so pissed of now because i am retaking and i was aiming for full marks. Just to find out that knowing how to answer every question in maths means nothing if you have a (highly) mistake prone brain.
Well thats A* out the window

4. heres the working for question 10, realised that i forgot to find p when i got 2 mins left so rushed it and got it wrong
P=120.2
d2p/dx2 = 2000^-3, x = 16.6, d2p/dx2 > 0 hence minimum
5. Could you explain further how by letting 2^x =y that 2^2x+5 gives 32y^2
6. (Original post by Morg698)
Could you explain further how by letting 2^x =y that 2^2x+5 gives 32y^2
Using indices laws you can split up
2^2x+5 into 2^5 * 2^2x which then goes to 32 * (2^x)^2 so subbing y=2^x in you get 32y^2. Hope this helps
7. (Original post by Mills.O)
Using indices laws you can split up
2^2x+5 into 2^5 * 2^2x which then goes to 32 * (2^x)^2 so subbing y=2^x in you get 32y^2. Hope this helps
Oh of course. Thanks very much.
8. Can someone tell me what question 9 was?

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9. (Original post by AnIndianGuy)
Is it worth resitting for 100 UMS in C1 and C2 next year?
If you get higher than 240 UMS its not worth it because you need 240 UMS in AS and A2 (480 UMS)to get an A and the same requirements but an average of 90 UMS in C3 and C4(which are much harder ) in addition to 240 UMS to get an A*. so the best thing to do is focus on your A2.
This is from someone who is ressiting C2 and S1 because i got a B in both, but still managed to get an A (243 UMS)
i just resat C2 and i thought i got full UMS when i came out, only to find out that i dropped around 8 MARKS due to SILLY mistakes.
10. Does anyone know what the answers not being in only either graphical or numerical form meant for the logs question? Swear that's only normally in trig
11. [QUOTE=TrueDAN;65158533]Hey all, sorry for the time delay in this one!! Best of wishes to Arsey too
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

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 (

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 (
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
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 59-63 for an A. Sorry to speculate but a low 60 boundary for the A is more than likely.
Any corrections, thoughts or misunderstandings please say!![/QUOTE

Any chance you know what question 6B was?
12. With question 6b, I must've got my +ve's and -ve's the wrong way round when factorising (😑🔫) as I got -2 and +1/4, assuming I got the correct values for +1/4(rather than -1/4), how many marks do you think I'd've dropped in this question (out of the six marks)? Thanks
13. For the q7 on integration in part b where did the line cross the X axis apart from 0,0 for when u sub it in and how many marks out of 3 would u get if u used the wrong X coordinate when subbing in but still using correct method? Cheers
14. Hi guys

if i put 5y + 3x - 95 = 0, will i still get the mark even though it said put it as 'ax + by + c = 0'?
Hi guys

if i put 5y + 3x - 95 = 0, will i still get the mark even though it said put it as 'ax + by + c = 0'?
Yeah I'm pretty sure that's fine, in past mark schemes they seem to say either way
16. (Original post by SharkGirl)
Does anyone know what the answers not being in only either graphical or numerical form meant for the logs question? Swear that's only normally in trig
Yeah it's normally in trig, but I believe it means that you can't solely rely on some form of iteration formula to approximate the roots of the equation, such as the Newton-Raphson process (and others) covered in FP1 and C3. Otherwise you can just approximate the solution to a very high degree of accuracy (sometimes better than the 2 decimal places stated in the question).
17. (Original post by TrueDAN)
Hey all, sorry for the time delay in this one!! Best of wishes to Arsey too
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

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 (

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 (
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
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 59-63 for an A. Sorry to speculate but a low 60 boundary for the A is more than likely.
Any corrections, thoughts or misunderstandings please say!!
is this not 56??
18. how did you do the log question (8)? It was log3(3b+1)-log3(a-2)=-1. Then ii) was solve 2^(2x+5)-7(2^x)=0
19. what if i got 165.5 instead of 195.5 but i got the other right, how many marks do i lose? 1 or 2
20. what was the actual question for 8(a), the logs one?

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