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# AQA Maths Core 1+2 May 2013 Watch

1. (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

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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 x-axis 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 Y-20=-4/9(x-4)

3. (Circle question)
(a) (x-5)^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?

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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)(x-5) = 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 √((x-2)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
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!
2. 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 1-5 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!
3. (Original post by The H)
It was 51 marks for an A, when it normally hovers around 61ish
Oh wow.
4. (Original post by Gotzz)
pretty sure it was a^c=b
Quite possibly. It's all faded into the mists of my memory until the grades come out
5. (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 1-5 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!
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
6. (Original post by The H)
Have you seen June 2012?
I had to sit the June 12 one, got a D. Todays was honestly so much easier!
7. 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.
Co-Ordinates 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)(x2 - 3x + 5) (2)

Integration (Q-)
Integral 31.5 or 63/2 (4)
Shaded Region Area 27 (3) M1 (Area of Trapezium = 58.5), M1 58.5-31.5, A1 27

Circles (Q-)
(x-5)2 + (y+7)2 = 49 (3) A1 (x-5)2, A1 (x+7)2, A1 49 (accept 72)
Centre: (5, -7) (2) A1 5, A1 -7
r = √49 = 7 (1) CAO
The point furthest away from the x-axis (5, -14) (2) A1 5, A1 -14
Translation by the vector (6, -7). (3) B1 Translation, A1 6, A1 -7

Inequalities (Last Question)
Prove b2-4ac >= 0 and switch signs (4)
The last question was 20/7 ( <=) k (<=) 4. (4) B2 20/7<= k NMS, B2 k <= 4 NMS

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Core 2

Geometric Series (Q1):
U3 = 20 (1) CAO
Sum to Infinity = 160 (2) M1 a/1-r 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 r2 x theta = 160cm2 (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. - ac = b (1)
Show that only one value of x satisfies 2log(2)(x+7) - log(2)(x-5) = 3
(x+3)2 (6) M1 1 law of logs used correctly, M1 second law of logs used correctly, M1 x2+14x+49 M1 x2+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)
U3 = 56 (1) CAO

Transformations (Q-):
Stretch SF 2 x-direction (2) B1 Stretch, A1 SF 2 X-Direction
The transformation of [2 -0.7] was √((x-2)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 (3cosx-2), 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
8. (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
Hahaha I agree with you there! I retook core 2 again in January but I had some circumstances so I couldn't revise for it properly and ended up with another awful grade. So this is third time lucky! But I know I messed up a lot of the big questions :/ ...how do you think you did?
9. you people complain too much... i took june12 core2 paper and got a B
10. 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^(n-1)
u3=80x0.5^(3-1)=20

(b) Find the sum to infinity of this series.
A: s=a/(1-r)=80/(1-0.5)=160

(c) Find the sum of the first 12 terms of this series, giving your answer to two decimal places.(??)
A: sn=(a(1-r^n))/(1-r)
s12=(80(1-0.5^12))/(1-0.5)=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^2
θ=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=
π-α=π-1.27...=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+(2-x^-2)^3 can be written in the form p+qx^-4.
A:
(2+x^-2)^3+(2-x^-2)^3
=
8+8+12x^-2-12x^-2+6x^-4+6x^-4+x^-6-x^-6
=16+12x^-4

(b)(i) Integrate
(2+x^-2)^3+(2-x^-2)^3 dx.
A:
(2+x^-2)^3+(2-x^-2)^3 dx
=
∫ (16+12x^-4) dx
=16x+(12x^-3)/-3+c
=16x-4x^-3+c

(b)(ii) Hence find the value of
∫2:1 (2+x^-2)^3+(2-x^-2)^3 dx.
A:
∫2:1 (2+x^-2)^3+(2-x^-2)^3 dx
=[16x-4x^-3]2:1
=32-4/8-(16-4)
=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 co-ordinate 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(xn-1)))
=
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)=
√((x-2)^3+1)-0.7
g(4)=
√((4-2)^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.
11. (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 ?
You had to find the inverse tan of -1 which was -45, but that wasn't in the range so you had to add 180 to that to get 135, and add 180 to that to get 315, which were the two answers I got.
12. (Original post by Dzintra)
I had to sit the June 12 one, got a D. Todays was honestly so much easier!
Oh yeah
13. (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!
I Believe the geometric series bit to be wrong if the common ratio is 0.5 then the sum of the 12 terms is 80(1-0.5^12) / 1-0.5 = 79.48 ???? the first bit is right but not sure about this last bit you state as 2 answers surely wont be the same as if it 159.96 then it would be 80(1-0.5^12) / 0.5 only and forget about the 1 ??
14. What was the actual question for the 2^k?
15. 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
16. 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
17. 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!
18. Carry on get the answers coming guys I need them!
19. 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.
20. 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

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