# STEP I, II, III 2002 Solutions

Watch
Announcements
Thread starter 12 years ago
#1
STEP I:
1: Solution by Unbounded
2: Solution by Unbounded
3: Solution by nota bene
4: Solution by SimonM
5: Solution by nota bene
6: Solution by Unbounded
7: Solution by sonofdot
8: Solution by Unbounded
9: Solution by cliverlong
10: Solution by Dadeyemi
11:Solution by Unbounded
12: Solution by Robbie10538
13: Solution by Unbounded
14: Solution by brianeverit

STEP II:
1: Solution by sonofdot
2: Solution by dadeyemi
3: Solution by sonofdot
4: Solution by dadeyemi
5: Solution by dadeyemi
6: Solution by welshenglish
7: Solution by Glutamic Acid
8: Solution by Glutamic Acid
9: Solution by tommm
10: Solution by tommm
11: Solution by tommm
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit

STEP III:
1: Solution by SimonM
2: Solution by Dadeyemi
3: Solution by Daniel Freedman
4: Solution by Dadeyemi
5: Solution by Dadeyemi
6: Solution by Elongar
7: Solution by Dadeyemi
8: Solution by Dadeyemi
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit

Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
0
reply
Thread starter 12 years ago
#2
0
reply
12 years ago
#3
STEP III, Question 1

a)

b)

0
reply
Thread starter 12 years ago
#4 is strictly increasing, but doesn't tend to infinity
0
reply
12 years ago
#5
STEP II 2002 Question 3

Spoiler:
Show  If k=1, then and , therefore it is true for k=1.

If k=2, then and , therefore it is true for k=2.

If k=3, then and , therefore it is true for k=3.

Assume, for some k  Therefore by induction, for all positive integers k.

Now consider and with r<s. Assume that and have a common factor p Since all Fermat numbers are odd, and cannot share a common factor greater than one, hence by contradiction, no two Fermat numbers have a common factor greater than 1.

All Fermat numbers are either prime or made up of prime factors. Since no Fermat numbers share a common factor greater than one, no Fermat numbers can have the same prime factor. Since there are infinitely many Fermat numbers, there must therefore be infinitely many prime numbers.
6
reply
12 years ago
#6
Question 1, STEP I, 2002
Spoiler:
Show
We need to find the points of intersection between the two ellipses: - ellipse 1 - ellipse 2

making y^2 the subject in the equation for ellipse 1: and substitution into the equation for ellipse 2:     Subbing back into the equation for y^2:    Therefore the coordinates of intersection are (2,1) and (2,-1)

From a sketch, we can see that this means that the centre of a circle passing through those two points must lie on the x-axis.

Letting the coordinates of the centre of the circle be (a,0): where r is the radius of the circle.

Substitute the point (2,1) in: So the equation of the circle now becomes:  0
reply
12 years ago
#7
Some attached

Spoiler:
Show
(the end of the question 2 is the reason why I should not do maths late at night :/
1
reply
12 years ago
#8
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.
3
reply
12 years ago
#9
1
reply
12 years ago
#10
3
reply
12 years ago
#11
3
reply
12 years ago
#12
II/8:

For x < 0, dy/dx = -y ==> 1/y dy/dx = -1 ==> ln y = -x +c ==> y = Ae^(-x); as y = a when x = -1, a = Ae so A = a/e; therefore for x < 0.

For > 0, dy/dx = y ==> 1/y dy/dx = 1 ==> ln y = x + c ==> y = Be^x; as y = b when x = 1, b = Be so B = b/e therefore for x > 0.

If a = b, there is no jump.

Note that e^x - 1 is positive if x > 0, and negative if x < 0. for x > 0.
Given that y = e^e when x = 1, e^e = Ae^(e - 1) so A = e;  for x < 0.

At x = 0, y = using the solution for x > 0. So for continuity, this must the same result when substituted into x < 0, so   (i) (ii) .
0
reply
12 years ago
#13
4
reply
12 years ago
#14
Question 2, STEP I, 2002
Third Part
Sketches to follow shortly
1
reply
12 years ago
#15
1
reply
12 years ago
#16
0
reply
12 years ago
#17
2002 II Question 12  probability such that more than l coins of the L coins will produce less than m heads

Estimating q: , where z is the probability of a coin producing less than m heads.
We can say that , which gives: for small z.
We then substitute to get as required.

z is the probability that less than m heads are made after M throws.  . as required.

K = 7, k = 2, L = 500, l = 4, M = 100, m = 48, p = 0.6. , which is a large z, so we are not allowed to approximate as needed.
0
reply
12 years ago
#18
0
reply
12 years ago
#19
0
reply
12 years ago
#20
0
reply
X

### Quick Reply

Write a reply...
Reply
new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Do you think receiving Teacher Assessed Grades will impact your future?

I'm worried it will negatively impact me getting into university/college (98)
39.68%
I'm worried that I’m not academically prepared for the next stage in my educational journey (27)
10.93%
I'm worried it will impact my future career (18)
7.29%
I'm worried that my grades will be seen as ‘lesser’ because I didn’t take exams (57)
23.08%
I don’t think that receiving these grades will impact my future (30)
12.15%
I think that receiving these grades will affect me in another way (let us know in the discussion!) (17)
6.88%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.