# Edexcel P2 June 03

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#1
I sat the EDEXCEL P2 Maths exam on FRIDAY 13th JUNE 2003 and I'm abit worried about how I performed.

Does anyone have the answers or know where I can get them?
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16 years ago
#2
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16 years ago
#3
Question:

1a) Simplify

(x+3)/x

1b) x=0.2 (for logs question about a)

2a) Range of f

2<=y<=11

b) lamda = 5

3 Expansion of series (2 -px)^6

64 + Ax + 135x^2

p = 0.75
A =-144

4a) f(x+1) Shift graph to left by 1

b) modf(x) = Reflect anything under x axis above

c) F(modx) = Copy the positive x-axis onto the negative side so the y-axis becomes a line of symmetry

5a) Graphs simple
b) graphs cross once therefore only one solution
c) F(3) = -0.218
F(4) = 0.018

Change of sign over interval and continuous implies root in interval QED

d) Root is 3.921 to 3dp

6)a) prove c=4p^2 easy
b) show c=4 easy
c) integral = Pi(9+8ln2)

7a) equation of normal:
x + 2y =14
b) show n=14 easy
c) integral area:
97 - 3ln2

8ai) Prove tanx = -root3/2 east
aii Show ([email protected])/([email protected]) = [email protected] easy

b) show 180 is on solution, easy

c) Find other 2 solutions:

26.6 and 206.6 degs to 1dp

The mark scheme is as follows:

1a 2
1b 4
2a 3
2b 3
3 7
4a 3
4b 3
4c 4
5a 3
5b 1
5c 2
5d 4
6a 2
6b 2
6c 7
7a 4
7b 1
7c 7
8ai 5
aii 3
b 1
c 4

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16 years ago
#4
I think lambda should have been 1/5, as you got 3y = 15.
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16 years ago
#5
Sorry, 15y = 3.
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16 years ago
#6
no, the guy is right, lambda did equal 5.
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16 years ago
#7
Oh FFS.
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16 years ago
#8
in the range question, why is the bottom value 2? surely it should be 3!
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#9
I got 3<= y <=11 as well for 2a
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#10
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16 years ago
#11
Well thats wrong.

There are two ways to get the correct answer - One is to complete the square of the function, givin

f(x) = (x-1)^2 + 2 which is at a minimum when x=1 and therefore f = 2.

The second is to differentiate the function and set dy/dx = o, this gives 2x-2=0 => x=1 => f=2

Both of these give the minimum value of f as 2. It is not enough to simply put in the minimum value of x and expect it to give you the minimum value for f (consider the function sin(x) in the range 0<x<pi the limits both give f=0, but we know that sin(x) = 1 when x=pi/2)
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16 years ago
#12
i put 2< fx < 11 (without the equal sign), I gess thatll be about a mark down? Otherwise all the answwers seem to b right!!
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