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AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme] watch

1. (Original post by datpr0)
1. the marks for 1b were 6 not 5,
2. the differential equation solution was x = 1/5[(-1/2(1+t)^-1 + 5/2)^2 - 4],
3. the constant in the penultimate question was a not k, as stated in the question.
Have changed marks but I'm fairly sure it was k?

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2. (Original post by Jeff97)
i got -10/root56 root14 from doing a.b=mod a mod b costheta
Yeah, no; the scalar product came to 14 and mod a x mod b came to 2sqrt14 x sqrt 14 which equals 28.
=> cos(theta) = 14/18
3. (Original post by Alexandrite)
On a slightly related note how harsh are examiners if you forget to mark the question reference but your work/answer is clearly for that question?

I'm pretty sure I didn't make that mistake but at the same time I always have a nagging feeling maybe I did in one of my exams.
They aren't bothered really; all that they are looking for is the answers. I forgot to put references in the margin many times last year, don't worry about it.

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5. (Original post by CD223)
Have changed marks but I'm fairly sure it was k?

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Hmm.
6. Can anyone show me the detailed solution of question 5? That one with sinq, not sure about how many marks i will lose
7. time to put c3/c4 book on ebay!

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8. How many points will I lose for giving in 6a) the value of the angle in radians???

I found cos=1/2 but then calculated angle in radians :/
9. guys i got x as -1/8 but i didnt use sine, i did it in quadratic form with x.... how many marks will i lose? (the sinp Q or whatever)
10. (Original post by datpr0)
(let theta equal the acute angle)

Therefore, it came from cos(theta) = 14/28
=> cos(theta) = 1/2
=> theta = 60 degrees
Cos theta = 10/28 which is 69.1 :/
11. (Original post by majka)
How many points will I lose for giving in 6a) the value of the angle in radians???

I found cos=1/2 but then calculated angle in radians :/
I doubt you'd lose any marks for it, the mark scheme would likely say "or/condone [insert the angle in radiains]".
12. My predictions:
60 A*
55 A
50 B
etc etc
because it wasn't SOLID like we expected, but it was challenging and maybe a bit more difficult than last years but were supposedly getting 'smarter' so I think an increase of 1 isn't too unfair
13. Can anybody post predicted grade boundaries.
14. (Original post by Jed-Singh)
My predictions:
60 A*
55 A
50 B
etc etc
because it wasn't SOLID like we expected, but it was challenging and maybe a bit more difficult than last years but were supposedly getting 'smarter' so I think an increase of 1 isn't too unfair
second this.

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15. I think there by quite high the paper was quite easy to honest
16. (Original post by HennersPD)
Cos theta = 10/28 which is 69.1 :/
... I'm still pretty sure the scalar product came to 14.
17. (Original post by Jed-Singh)
My predictions:
60 A*
55 A
50 B
etc etc
because it wasn't SOLID like we expected, but it was challenging and maybe a bit more difficult than last years but were supposedly getting 'smarter' so I think an increase of 1 isn't too unfair
60 for an A* will be lovely <3 I do think that there will be less separation from the A* boundary to the C grade boundary due to the nature of this paper.
18. 65 for an A*
19. (Original post by CD223)
*** OFFICIAL AQA CORE 4 MPC4 JUNE 2015 EXAM DISCUSSION THREAD ***
UNOFFICIAL MARK SCHEME
1.
A) Express f(x) = (19x-2)/(5-x)(1+6x) in the form:
f(x) = A/(5-x) + B/(1+6x):

A=3 B=-1 (3 marks)

B) Find Integral between x = 4 and x = 0 of f(x). Show it equals pln5.

= 8/3ln5 (6 marks)

2.
A) Express 2cosx - 5sinx in the form Rcos(x+alpha), R>0, 0<alpha<pi/2
√29cos(x+1.19) (3 Marks)

B) Find value of x in interval 0<x<2pi where 2cosx - 5sinx is at a maximum.
(i). x=5.09 (2 Marks)
(ii) Solve equation 2cosx - 5sinx + 1 = 0
x=0.567, 3.34 (3 Marks)

3.
A) g(x) = 8x^3 - 12x^2 - 2x + d
Given when g(x) is divided by (2x+1) the remainder is -2, find d.
d=1 (2 Marks)

B) Given x = -0.5 is a solution of g(x)=0, express g(x) as 3 linear factors.
g(x) = (2x+1)(2x-1)(2x+3) (3 marks)

C) h(x) is given by 4x^2 - 1 / f(x). Simplify h(x), showing it is a decreasing function.

h(x) = 1/(2x+3), h'(x) = -2(2x+3)^-2 this is always negative, therefore it is a decreasing function (4 Marks)

4.
A) Find the binomial expansion of (1+5x)^(1.5):
= 1 + x - 2(x^2) (2 Marks)
B) Expand (8 + 3x)^(-2/3)
(i). 1/4 - (1/16)x + (5/256)x^2 (3 Marks)
(ii) Find an estimate for Cube root of 1/81.
0.2313 (2 Marks)

5. x = cos2t, y = sint
A) Find the gradient at t = pi/6.
dy/dx= -1/2 (3 Marks)

B) Equation of normal at t=pi/6
y=2x-1/2 (2 Marks)

C) Normal cuts curve at Q(cos2q, sinq). Set up a quadratic in sinq to find where the normal cuts the curve again. Then find the x co-ordinate of Q.
cos2q = -1/8 (5 Marks)

6. A = (3,2,10), B = (5,-2, 4)
Line l passes through A and has equation:
r = [3,2,10] + lambda[x,y,z]

A) Find angle between line l and AB.
Theta = 60 degrees (4 Marks)
B) C(15, 6, 2) (4 Marks)
C) E(11, 0, 0) and (23, 3, 0) (4 Marks)

7. y^3 + 2e^(-3x)y + x = k
Find k given point P (ln2, 0.5) lies on curve.
A) k=1/4 - ln2 (1 Mark)
B). Find the gradient at P:
dy/dx = 11/8 (6 Marks)

8. dx/dt = (4+5x)^0.5/5(1+t)^2
A) Given t=0 when x=0 find x in terms of t.
x=9/20-1/2(t+1)-1/20(1+t)^2
OR
x = 1/5[(-1/2(1+t)^-1 + 5/2)^2 - 4]

(7 marks)

B) Another pond models a circle of radius r. Rate of change of radius is inversely proportional to pond surface area. Write a differential equation to model this situation.
(i) dr/dt = k/(pir^2) (3 Marks)

(ii) The rate of change of radius is 4.5 when the radius is 1. Find the radius when the rate is 0.5.
r=3 metres (2 Marks)

Date: 9th June 2015
Time: 09:00am
Duration: 1h 30m
______________________________

______________________________

RESOURCES:
7a) i got k=1/4+ln(2)
20. (Original post by datpr0)
... I'm still pretty sure the scalar product came to 14.
It did

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