Hi,(Original post by niceguy95)
Hi everyone. This is the official thread for the following exams
Maths Pure Core 1 JUN13/MPC1
Maths Pure Core 2 JUN13/MPC2
Guys, if you want to make corrections or anything, please copy and paste the whole thing, and get the answers in the right order. I won't be online for now, most of you should be able to get most of this mark scheme done. Please make corrections etc, and at the end, I will update my own post to reflect this.
THANKS
******************************** ******************************** ******************************** ******************************** **********************
Core 1
question 2)a)ii) the question asked x√12 = 7√3  √48
x= 3/2
For Q1, p was 3, k = 10, D = (7, 4).
For the surds question, x = 1.5 and for part b the answer they were looking for was 8  (15)^0.5.
The complete the square question was 2(x+1.5)^2 + 0.5 and the length of AB was 0.5(10)^0.5.
Dividing the polynomial by x+3 gave x^2  3x + 5.
The integral was 31.5 or 63/2 and the area of the shaded region was 27.
The point furthest away from the xaxis was (5, 14) and the translation was by vector (6, 7).
The last question was 20/7 ( < or =) k (< or =) 4.
Equation of normal
X=4 Y=20 normal 4/9 So Y20=4/9(x4)
3. (Circle question)
(a) (x5)^2 + (y+7)^2 = 49
Radius was 7 so to draw the circle, you plot the middle point (5, 7) and go out 7 from the centre.
This implies a shift to the right of 6 units in x and 7 down in y. This is represented by the vector [6, 7].
Also, for question 5)b)ii), it asked you to use the coordinates that we got from part a)ii) (3,13) or something I think,
and find a minimum value putting it into 1/2√n
any done that one? And did you get 1/2√100?  DID ANYONE ELSE GET THIS BECAUSE I DID?
******************************** ******************************** ******************************** ******************************** *******************************
Core 2
Trapezium Rule Question  7.12
1.
The first term of a geometric series in 80 and the common ratio is 0.5
a) work out the third term in the series  20
b) work out the sum to infinity  160
c) work out the sum of the first 12 terms in the series to 2 decimal places  159.96
2. Work out Arc > 20 x 0.8 = 16
ii) work out area of sector > 0.5 x 20^2 x 0.8 = 160
b) work out obtuse angle ABD
Ad was 15 I think, and rad was 0.8
Other length was 20
Sin0.8/15 = SinD/20
20sin0.8 = 15sinD
SinD= 20sin0.8/15
D= sin^1 ( 20sin0.8/15) in rads
D= 1.27 rads
Since its obtuse, it's Pi 1.27
D= 1.87
3.
a) (2+Y)^3 = 8 + 12y + 6y^2 + Y^3
B) 16 +12x^4
(c) Integral = 39/2
6.
a. Write log(a)b = c in index form.  a^c = b
b. Show that only one value of x satisfies 2log(2)(x+7)  log(2)(x5) = 3
log2(x+7)2  log2(x+5)  3 = 0
→log2(x2+14x+49) = 3 + log2(x+5)
→log2(x2+14x+49) = log223 + log2(x+5)
→log2(x2+14x+49) = log2(8x+40)
→x2+14x+49 = 8x+40
→x2+6x+9 = 0
→(x+3)2
7.
a) Show p=3/2
72=p96+q
24=p24+q
48=p72
48/72= 2/3
b) Find u3
72= 2/3*96 +q
72=64 + q
Q=8
u3= 2/3*72 +q
U3= 48 +q
u3= 48 + 8
= 56
The transformation of [2 0.7] was √((x2)3 +1) 0.7
2k question k = 6/5
x = 16, 104 and 136 for the last one
also.. tan x = 1
solves to
x = 135, 315
Could you copy out the new format, and add to that, and also put your additions in bold so I know what's new.
Thanks!
You are Here:
Home
> Forums
>< Study Help
>< Maths, science and technology academic help
>< Maths
>< Maths Exams

AQA Maths Core 1+2 May 2013 Watch
Announcements

 Follow
 221
 13052013 20:44

Dedicatedstudent
 Follow
 0 followers
 0 badges
 Send a private message to Dedicatedstudent
Offline0ReputationRep: Follow
 222
 13052013 20:46
I was one of the unfortunate ones who had to endure the June 2012 paper too. I got an E ...any maths paper in comparison to that one is easier. I wouldn't say this paper was easy, I flied through questions 15 then messed up the differentiation, log, series and trig questions so there go a ton of marks but I'm hoping to get a lot of follow through marks..fingers crossed!

 Follow
 223
 13052013 20:48
(Original post by The H)
It was 51 marks for an A, when it normally hovers around 61ish 
 Follow
 224
 13052013 20:50
(Original post by Gotzz)
pretty sure it was a^c=b 
 Follow
 225
 13052013 20:51
(Original post by Dedicatedstudent)
I was one of the unfortunate ones who had to endure the June 2012 paper too. I got an E ...any maths paper in comparison to that one is easier. I wouldn't say this paper was easy, I flied through questions 15 then messed up the differentiation, log, series and trig questions so there go a ton of marks but I'm hoping to get a lot of follow through marks..fingers crossed!
Posted from TSR Mobile 
 Follow
 226
 13052013 20:52
(Original post by The H)
Have you seen June 2012? 
steviep14
 Follow
 5 followers
 10 badges
 Send a private message to steviep14
Offline10ReputationRep: Follow
 227
 13052013 20:56
Core 1
Coordinate Geometry Q1:
p = 3 (2)
k = 10 (1)
m(grad) = 3/4 (1) OE i.e. accept 0.75 or 6/8 etc.
CoOrdinates D = (7, 4) (4) M1 Normal m1xm2=1, M1 grad=4/3, M1 y=4/3x+c OE, A1 7, 4
Surds (Q2):
√48 = √16 x √3 = 4√3 (n=4) (1) CAO
Solve x√12 = 7√3  √48 > 3√3/√12 > x= 3/2 (3)
Completing the Square (Q):
2(x+1.5)^{2} + 0.5 (2) A1 2(x+1.5)^{2} OE, A1 +0.5 OE
min value = 0.5 (1) CAO
Length AB min = 0.5(10)^{0.5}. (?) (2)
Polynomial (Q)
Show (x+3) is a factor > f(3) = 0 > Is a factor (2) M1 f(3) E1 Therefore is a factor
(x+3)(x^{2}  3x + 5) (2)
Integration (Q)
Integral 31.5 or 63/2 (4)
Shaded Region Area 27 (3) M1 (Area of Trapezium = 58.5), M1 58.531.5, A1 27
Circles (Q)
(x5)^{2} + (y+7)^{2} = 49 (3) A1 (x5)^{2}, A1 (x+7)^{2}, A1 49 (accept 7^{2})
Centre: (5, 7) (2) A1 5, A1 7
r = √49 = 7 (1) CAO
The point furthest away from the xaxis (5, 14) (2) A1 5, A1 14
Translation by the vector (6, 7). (3) B1 Translation, A1 6, A1 7
Inequalities (Last Question)
Prove b^{2}4ac >= 0 and switch signs (4)
The last question was 20/7 ( <=) k (<=) 4. (4) B2 20/7<= k NMS, B2 k <= 4 NMS
******************************** ******************************** ******************************** ******************************** *******************************
Core 2
Geometric Series (Q1):
U_{3} = 20 (1) CAO
Sum to Infinity = 160 (2) M1 a/1r seen or used, A1 160
Sum of first 12 terms = 159.96 (2dp) (2) M1 Use of formula, A1 159.96
Circles and Radians: (Q)
S=r(theta) = 16cm (1) CAO
A=1/2 x r^{2} x theta = 160cm^{2} (2) CAO
Obtuse Angle = 1.87 radians (3) M1 Sine Rule M1 1.22r A1 1.87r
Integration (Q3):
Integral using Trapezium Rule: 7.12 (4)
(Unknown Q)
Integral = 39/2
Logs (Q6):
log_{(a)}b = c in index form.  a^{c} = b (1)
Show that only one value of x satisfies 2log_{(2)}(x+7)  log_{(2)}(x5) = 3
(x+3)^{2} (6) M1 1 law of logs used correctly, M1 second law of logs used correctly, M1 x^{2}+14x+49 M1 x^{2}+6x+9, A1 (x+3)^{2} E1 Only one root?
Arithmetic Series: (Q)
Show p=3/2
72=96p+q
24=24p+q
48=72p
48/72= 2/3  (Or simultaneous equations solving q then p) (4)
U_{3} = 56 (1) CAO
Transformations (Q):
Stretch SF 2 xdirection (2) B1 Stretch, A1 SF 2 XDirection
The transformation of [2 0.7] was √((x2)^{3} +1) 0.7
g(4) = 2.33 (3)
k = 6/5
x^{2} and x^{6} cancel  left with 16 + 12x^{4}. for binomial question
Trig (Last Question)
tan x = 1 > x = 135, 315 (2) A1 135, A1 315
Show identities equal one another (3)
x = 16, 104 and 136 (6) M1 (3cosx2), M1 cos x=2/3, M1 3x = 48, A3 16, 104, 136
Marks are a guide.. I have no clue, but it is an indication 
Dedicatedstudent
 Follow
 0 followers
 0 badges
 Send a private message to Dedicatedstudent
Offline0ReputationRep: Follow
 228
 13052013 20:56
(Original post by Hart1995)
I was aswell! These year12s don't know the true meaning of aqa torture hahaha. So yeah this paper.was alot easier! A lot of people from my year was resitting it
Posted from TSR Mobile 
matt258095
 Follow
 0 followers
 0 badges
 Send a private message to matt258095
Offline0ReputationRep: Follow
 229
 13052013 21:04
you people complain too much... i took june12 core2 paper and got a B

RuSSeLLMaGuiRe
 Follow
 0 followers
 1 badge
 Send a private message to RuSSeLLMaGuiRe
Offline1ReputationRep: Follow
 230
 13052013 21:07
Right, made an account for this.
I used a new calculator for this exam and you can check back through the previous 99 calculations (I cleared it for the exam obviously) and these are my answers, workings and what I think the question was (the question part references are most likely incorrect):
1. The first term of a geometric series is 80 and the common ratio is 0.5.
(a) Find the value of u3.
A: un=ar^(n1)
u3=80x0.5^(31)=20
(b) Find the sum to infinity of this series.
A: s∞=a/(1r)=80/(10.5)=160
(c) Find the sum of the first 12 terms of this series, giving your answer to two decimal places.(??)
A: sn=(a(1r^n))/(1r)
s12=(80(10.5^12))/(10.5)=159.96
2. Sector AOB has radius 20cm and angle AOB=0.8 radians.
(a) Find the arc length of the sector.
A: arc length=rθ=20cm x0.8=16cm
(b)Find the area of the sector.
A: area=0.5r^2θ=0.5x(20cm)^2x0.8=160cm^2
(On my calculator I have 0.5x16^2x0.8=102.4, I'm guessing this is a mistake?)
(c)OB meets a line from B at point D. Line BD is 15cm. Find the size of obtuse angle ODB to 3 significant figures.
A: sinA/a=sinB/b
=>sinA=asinB/b
sinODB=20cmsin0.8/15cm=0.95647...
α=1.27...
ODB=πα=π1.27...=1.87
3. (a)(i) Using binomial expansion, or otherwise, expand (2+x^2)^3.
A: 2^3+ncr(3,1) 2^2 x^2+ncr(3,2) 2 (x^2)^2 +(x^2)^3
=8+12x^2+6x^4+x^6
(ii) Hence show (2+x^2)^3+(2x^2)^3 can be written in the form p+qx^4.
A: (2+x^2)^3+(2x^2)^3
=8+8+12x^212x^2+6x^4+6x^4+x^6x^6
=16+12x^4
(b)(i) Integrate ∫ (2+x^2)^3+(2x^2)^3 dx.
A: ∫ (2+x^2)^3+(2x^2)^3 dx
=∫ (16+12x^4) dx
=16x+(12x^3)/3+c
=16x4x^3+c
(b)(ii) Hence find the value of ∫2:1 (2+x^2)^3+(2x^2)^3 dx.
A: ∫2:1 (2+x^2)^3+(2x^2)^3 dx
=[16x4x^3]2:1
=324/8(164)
=19.5
4. [Can't remember the exact order of this question. I've only got a log calculation in my calculator]
(a) Sketch the graph of y=9^x, indicating the points where the curve crosses the coordinate axis.
A: This question comes up every year, it's a exponential shaped graph with the point (0,1) clearly marked.
(b) Using logarithms, find x to three decimal places given 15=9^x.
A: 15=9^x
=> x=log9(15)=1.23
(c) The curve y=9^x is reflected in the y axis to give y=f(x). Find an expression for f(x).
A: f(x)=9^x
5.(a) Using the trapezium rule and 5 ordinates (4 strips) find an approximation for ∫2:0 √(8x^3+1) dx, giving your answer to 3 decimal places.
A: h=0.5
f(x)=√(8x^3+1)
∫2:0 √(8x^3+1) dx
≃0.5h(f(x0)+f(xn)+2(f(x1)+f(x2 )+...f(xn1)))
=0.5x0.5(1+√65+2(√2+3+2√7))
=7.12
(b) Describe fully the transformation that maps the curve y=√(8x^3+1) to y=√(x^3+1).
A: Stretch in the x direction, scale factor 2.
(c) y=√(x^3+1) is translated via vector [2, 0.7] to give curve g(x). Find g(4).
A: g(x)=√((x2)^3+1)0.7
g(4)=√((42)^3+1)0.7=2.3
...more coming once i decipher my remaining calculations...
By the way, these are my calculations and answers, please validate them before taking them as the right answers. 
I Persia I
 Follow
 9 followers
 2 badges
 Send a private message to I Persia I
Offline2ReputationRep: Follow
 231
 13052013 21:08
(Original post by dannyb2012)
U know for the 9th question of tanx=1 my teacher said do the inverse of the positive value only so i did and got 45degrees so i got 180  45 and 360 + 45, X = 135, 405 Anyone else remember this and then for the very last one i got summit like 14degrees and cant remember the other 2 values U had to factorise i got summit like (3cos(theta)2) (2cos(theta)+3) Anyone else confirm or better ? 
 Follow
 232
 13052013 21:11
(Original post by Dzintra)
I had to sit the June 12 one, got a D. Todays was honestly so much easier! 
 Follow
 233
 13052013 21:14
(Original post by solosdk)
Hi,
Could you copy out the new format, and add to that, and also put your additions in bold so I know what's new.
Thanks! 
 Follow
 234
 13052013 21:15
What was the actual question for the 2^k?

 Follow
 235
 13052013 21:19
Dannyb, what I have is correct, and you must be wrong, it isn't 79.48.
Also you could easily work 3rd term out by doing 80 x 0.5 x 0.5 
steviep14
 Follow
 5 followers
 10 badges
 Send a private message to steviep14
Offline10ReputationRep: Follow
 236
 13052013 21:21
Well.. I lost 2 or 3 on the prove the identity question
lost 4 or so on the log question haha
didn't get the 2^k one...
nor did I get the translations fully...
but all in all, hopefully only lost 12 or 13... :P 
 Follow
 237
 13052013 21:24
I know I lost 1 mark for saying scale factor of 1/2 instead of 2, and up to 2 marks on the last question as I only got two solutions. Hopefully that's all though!

 Follow
 238
 13052013 21:24
Carry on get the answers coming guys I need them!

 Follow
 239
 13052013 21:25
Does anyone remember how much the transformation question was worth all together. Question 5 that is. I got g(4) as 7. 34 and was wondering how many marks I would lose.

steviep14
 Follow
 5 followers
 10 badges
 Send a private message to steviep14
Offline10ReputationRep: Follow
 240
 13052013 21:26
For the polynomial on core one...
there was a question of prove only one real root
so.. b^2  4ac < 0
(3)^2  4 x 1 x5 = 9  20
= 11
which is < 0
therefore, no real roots for the quadratic
only x = 3 is a root
Related discussions:
 AQA maths core 1, 13th May 2015
 AQA Mathematics exam core 1 and core 2 on 13th of may ...
 AQA A Level Maths Core 1  18th May 2016 [Exam Discussion ...
 AQA FP1 Further Pure 1 17th May 2013 Official Thread
 GCSE exams in May/June 2013
 maths core 1 jan 2013 complete unofficial markscheme
 AQA Alevel Mathematics Core 1 and Core 2  19th and 22nd ...
 AQA Core 4  Monday 10th June 2013 (AM)  Official Thread
 AQA Core 2 Monday 13th May
 AQA C2 Unofficial Mark Scheme 20 May 2015
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
 SherlockHolmes
 Notnek
 charco
 Mr M
 TSR Moderator
 Nirgilis
 usycool1
 Changing Skies
 James A
 rayquaza17
 RDKGames
 randdom
 davros
 Gingerbread101
 Kvothe the Arcane
 The Financier
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 Reality Check
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 LeCroissant
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Moltenmo
Updated: August 15, 2013
Share this discussion:
Tweet