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Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrong

Please note this is not the official markscheme and could therefore contain mistakes. Use accordingly



1(a)6x+4
(b)(i)-2/3
(ii)x<-2/3
2(a)(i)u2=40
(ii)u3=28,u4=33
(b)(i)u5=35
(ii)851
3(a)3x²-13x-30=0
(b)(i)6, -3/5
(ii)reject -3/5 as log negative is invaid
4(a)A=55
(b)H=55(e^(-0.136x))+30
5(a)Show that k=12
(b)(0,-28)
6(a)-(2y+3x²)/(6y+2x)
(b)13x-11y+81=0
7(a)Show that AD//BC
(b)6.99
8(a)2√17 cos(x-1.326)
(b)9√17, 1.326
9(a)Show thow that y=3ln(x+5)
(b)3x-25y=-150ln5
10(a)(2k+3)/(x+4)+(k-3)/(x-2)
(b)(21/ln5)-6
11(a)1/9-2/27x+1/27x²
(b)0.03304
(c)6ln(17/16)-45/136
12(a)Show that dh/dt=λ/√h
(b)h^(3/2)=0.513t+1.728
(c)18.4
13(a)5000
(b)3 because |x-3|
(c){t:0≤t<3/5∪t>13/5}
14(a)sin2θ(23cos²θ-8cosθ-15)=0
(b)360°,540°,491°
15)Consider sinx>0 and cos x<0 for obtuse or any other valid arguments
(edited 8 months ago)

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Thank you. 2h was definitely not enough…literally panicking after seeing this
Reply 2
Original post by Anonymo us
1(a)6x+4
(b)-2/3
(c)x<-2/3
2(a)(i)u2=40
(ii)u3=28,u4=33
(b)(i)u5=35
(ii)851
3(a)3x²-13x-30=0
(b)(i)6, -3/5
(ii)reject -3/5 as log negative is invaid
4(a)A=55
(b)H=55(e^(-0.136x))+30
5(a)Show that k=12
(b)(0,-28)
6(a)-(2y+3x²)/(6y+2x)
(b)13x-11y+81=0
7(a)Show that AD//BC
(b)6.99 8(a)2√17 cos(x-1.326)
(b)9√17, 1.326
9(a)Show that y=6ln(x+5)
(b)3x-25y=-150ln5
10(a)(2k+3)/(x+4)+(k-3)/(x-2)
(b)(21/ln5)-6 11(a)1/9-2/27x+1/27x²
(b)0.03304
(c)6ln(17/16)-45/136
12(a)Show that dh/dt=λ/√h
(b)h^(3/2)=0.513t+1.728
(c)18.4
13(a)5000
(b)3 because |x-3|
(c){t:0≤t<3/5∪t>13/5}
14(a)sin2θ(23cos²θ-8cosθ-15)=0
(b)360°,540°,491°
15)Consider sinx>0 and cos x<0 for obtuse


For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.
Original post by AshRevise
For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.


Yes i got this. If you solve for x, there are no solutions that are obtuse, hence the contradiction!
Reply 4
Original post by AshRevise
For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.

Yea any valid method will do
Reply 5
How many marks was subscriber question (Q13)
Reply 6
same

Original post by AshRevise
For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.
Reply 7
Original post by AshRevise
For 15, I got to sin2x>0 by expanding the bracket, could be wrong though, probs multiple methods.


Same. I think you could use that and then say for obtuse x u can't pos sin2x
Reply 8
Original post by LEBB2460
How many marks was subscriber question (Q13)

10
Reply 9
Can someone make this into a document with marks ??
12c isn’t 18. You had to 5-1.44 then input that value of h. H wasn’t 5
Reply 11
is there another value for 14b of 450° as it's the sin2x graph for [360,540] ?
Original post by emma7393
is there another value for 14b of 450° as it's the sin2x graph for [360,540] ?


In the q it said x≠ 450
Reply 13
Original post by Hiim
10


Original post by anonymous747
In the q it said x≠ 450


360 540 and 490 something
Reply 14
Original post by anonymous747
In the q it said x≠ 450

omg yeah i remember now thats so annoying 😭 thanks
Original post by emma7393
is there another value for 14b of 450° as it's the sin2x graph for [360,540] ?


Original post by LBJLBJ975310
12c isn’t 18. You had to 5-1.44 then input that value of h. H wasn’t 5

No you didn't since the model takes that into account with +c so it's 18 and H=5
Original post by navbh
360 540 and 490 something


Yh 490.7
Reply 17
Someone pls make a docccc
By the way, the negative log is still valid because x^2 makes any answer positive. So, there are no invalid answers!
Reply 19
Original post by LBJLBJ975310
12c isn’t 18. You had to 5-1.44 then input that value of h. H wasn’t 5


h was 5 as it was a model for the total height not change in height. Otherwise h^3/2 would be 0 when t = 0

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