# STEP I, II, III 2010 Solutions

Watch
Announcements
#1
Index of STEP solutions; 2010 STEP papers.
Thanks to miml for originally running this thread.
4 solutions left to fill.

STEP I
1 – Solution by jasbirsingh
2 – Solution by gcseeeman
3 – Solution by matt2k8
4 – Solution by miml
5 – Solution by miml
6 – Solution by jj193
7 – Solution by jj193
8 – Solution by miml
9 – Solution by matt2k8
10 – Solution by matt2k8
11 – Solution by brianeverit
12 – Solution by curious_lemma
13 – Solution by matt2k8, Get me off the £\?%!^@ computer

STEP II
1 – Solution by Unbounded, Salavant
2 – Solution by Unbounded, Farhan.Hanif93
3 – Solution by Unbounded, matt2k8
4 – Solution by Unbounded, matt2k8
5 – Solution by matt2k8
6 – Solution by jj193
7 – Solution by matt2k8
8 – Solution by ziedj
9 –
10 –
11 – Solution by gcseeeman
12 – Solution by curious_lemma
13 – Solution by metaltron

STEP III
1 – Solution by Unbounded
2 – Solution by matt2k8
3 – Solution by Unbounded
4 – Solution by Unbounded; correction by Dalek1099
5 – Solution by Get me off the £\?%!^@ computer
6 – Solution by Get me off the £\?%!^@ computer
7 – Solution by matt2k8
8 – Solution by matt2k8
9 – Solution by AnonyMatt
10 – Solution by Pyoro
11 –
12 – Solution by Pyoro
13 –
2
10 years ago
#2
SHOTGUN STEP II QUESTION 4, WAS THE SEXIEST (easiest) QUESTION
0
10 years ago
#3
The second and third "here" don't open fully?
1
10 years ago
#4
STEP I - Q5
The binomial expansion required is

i)
Let x = 1,

ii)
Differentiate both sides to get

Let x=1

iii)
Integrate both sides over 0,1

and

(This is easier to see by writing out the sum explicitly)

iv)

Differentiate both sides twice

and

Let x=1,

Adding this to the result in ii) gives

STEP I - Q4

i)

ii)

STEP I - Q8
i)

So we have

Now let

Our equation becomes

And so

Now let b=3q
Our equation becomes

And so

Therefore each of a,b, c must be divisible by 3.
We can write

, ,

Our equation becomes

Therefore each of p,q,r must also be divisible by 3.
We can carry on the previous argument repeatedly wlog
Therefore the numbers a,b,c must contain infinitely many factors of 3.
And so, a=b=c=0 QED

ii)
Consider quartic residues modulo 5

(All numbers are congruent 0,1,2,3,4 modulo 5 so this is sufficient)

Therefore we must have
or
and
or

Also

We must have . Considering possible pairings of residues (0,0),(0,2),(1,0),(1,2), we must have both p and q divisible by 5.

We can write and
Our equation becomes

And so,

We can write ,,

Our equation becomes

Therefore each of a,b,c must also be divisible by 5.
We can carry on the previous argument repeatedly wlog
Therefore the numbers p,q,r must contain infinitely many factors of 5.
And so, p=q=r=0 QED
0
10 years ago
#5
(Original post by boromir9111)
The second and third "here" don't open fully?
They work for me, but I'll upload it onto mediafire.
0
10 years ago
#6
(Original post by miml)
They work for me, but I'll upload it onto mediafire.
They work now. Thanks!
0
10 years ago
#7
Can I shotgun II 1 please? Second-easiest question on paper (4 being the easiest <3) so shouldn't involve much pain to type up.
0
10 years ago
#8
I can do Step I question 1 :P
0
10 years ago
#9
I/12

Hopefully I've made it clear enough, couldn't get latex working.

Spoiler:
Show
E(X) = 1*P(X=1) + 2*P(X=2) + 3*P(X=3) + ... = sum of r*P(X=r), where r ranges over the positive integers.

The given sum is equal to (P(X=1) + P(X=2) + P(X=3) + ...) + (P(X=2) + P(X=3) + ...) + (P(X=3) + ...) + ... which simplifies to E(X).

------------------------

Notice that P(X=>4) = 1 - P(X<=3), and that P(X=1) equals zero.

For X to equal 2, we can either have MD, or DM, which has a total probability of 2pq.

For X to equal 3, we can either have MMD, or DDM, which has a total probability of p^2q + pq^2 = pq(p+q) = pq(1) = pq.

Therefore P(X=>4) = 1 - (pq + 2pq) = 1 - 3pq = 1 - 3p(1-p)
= 1 - 3p + 3p^2 = (1 - 3p +3p^2 -p^3) + p^3 = (1-p)^3 + p^3 = p^3 + q^3

------------------------

Now notice that for n > 1, for X to take some value n, we must have DDD....M (n-1 Ds and 1 M) or MMM...D (n-1 Ms and 1 D). Therefore, P(X=n) = qp^(n-1) + pq^(n-1). If we sum this, using the formula for a GP, we find that P(X=>n) = p^(n-1) + q^(n-1).

Using the formula for E(X) which we derived earlier, we see that

E(X) = SUM[p^(r-1) + q^(r-1)] - 1, where r ranges over the positive integers.

(the -1 is to take account for the fact that the formula for P(X=>n) gives P(X=>1) as 2, when it should in fact be 1).

So E(X) = 1/(1-p) + 1/(1-q) -1 = 1/p + 1/q -1 = (p+q)/pq -1 = 1/pq -1

Now, since pq = p(1-p) = p - p^2 = 0.25 - (p - 0.5)^2, we can see that pq has a maximum value of 0.25, and so 1/pq has a minimum value of 4. From this it follows that E(X) => 3.
0
10 years ago
#10
OK, here we go, II 1, LaTeX isn't fantastic but that's because I'm still learning how to use LyX. (I really hope this is right, but I used a grapher, looks right, circle osculates nicely). Question 11 on II I also remember being easy but only if you can draw a diagram... hard to do on computer. I'd tackle a III, but would have to start from scratch on questions and it's a little late.
0
10 years ago
#11
(Original post by miml)
...
STEP II - Q2
cos(3x)

as required.

sin(3x)

as required.

(i)

Let :

as required.

Note that .

(ii)
Considering the proper value of :

If

Considering the value of using Eustace's method:

Note that we're considering where :

as required.

Considering all values of for which Eustace's method gives the right answer (Let ):

.
0
10 years ago
#12
SHOTGUN STEP I - Q6 for tomorrow.
0
10 years ago
#13
(Original post by miml)
...
STEP I - Q1

Spoiler:
Show

1
10 years ago
#14
I/2

Let
Let

Stationary points when dy/dx = 0. e^x > 0 for all x, so we need the quadratic to = 0 i.e. statisfy the condition for 2 real roots:

2 distinct real roots if

This inequality holds if or as required.
i)
As above, stationary points when

Subbing in a and b gives:

or (hmm probably would have been easier to factorise)

Asymptote when x - b = 0 ==> x = 1/2 so these are on either side, as required.

Graph (in Wolfram Alpha):

ii)
Similar working for ii) leads to x = 1.5 or 3 (both on the same side of the asymptote at x = 0), and to the graph below:
Graph (in Wolfram Alpha):

II/11

[See Diagram Below]

Resolving Horizontally:

Substituting (**) into (*), cancelling R, and expanding using double angle formulae:

Factoring out sin and cos alpha e.t.c:

as required.

There may be (probably is) a better way to do this last part, but hopefully just plugging in the values given and expanding tan (45 - 30) for tan(15) is at least OK to show it. i.e.

(as alpha < 180)
Diagram(ish)

Hopefully there are no typos (or mistakes).
1
10 years ago
#15
I'm about to do II- q4
0
10 years ago
#16
II/3

i) using values for we have
From so
from
fom
from

so using the first one we actually have

and the last one factorises to

so using the first, as required. (*)

Also factorising the second one, then using the first one means that

Substituting this into (*) and some other algebra leads to

ii) substitute in then use binomial theorem gives

iii) using sum of GP formula, then some algebra gives the required sum as 1.

II/4

i. Let u = a-x, so du = -dx
so

using the facts that and that x and u are both dummy variables.

adding these the 2 expressions for I gives so

First integral:
2+x-x^2 = (x+1)(2-x), so ln(2+x-x^2) = ln(x+1) + ln(2-x) = ln(x+1) + ln(1 - (x+1))

So the integral is equivalent to

so we can use the above result with a = 1, f(x) = ln(x+1) and find that the first integral has the value 1/2.

Second integral:
using compound angle formulae

and
so we can write the second integral as

so we can use the above result with and to find that the integral is equal to .

ii)
Let

Set x = 1/u, then some re-arranging, and the fact that u and x are just dummy variables to find that also,

So we can add the 2 expressions for J to find that

so
0
10 years ago
#17
(Original post by matt2k8)
I'm about to do II- q4
(Original post by ziedj)
Won't be happy...
0
10 years ago
#18
Woops didn't see that post haha
0
10 years ago
#19
(Original post by matt2k8)
Woops didn't see that post haha
You saved me a lot of typing, no worries
0
10 years ago
#20
II/12

I did this in a hurry so there might be a few mistakes!
Spoiler:
Show

Since f is a probability density function, the integral of f(x) between 0 and 1 will be equal to 1. This means that [INT(a) between 0 and k] + [INT(b) between k and 1] will equal one. Carrying out the integration and rearranging, we find that k = (1-b)/(a-b).

Now k > 0 => (1-b)/(a-b) > 0 => 1-b > 0 => b < 1
and k < 1 = (1-b)/(a-b) < 1 => 1-b < a-b => a > 1

------------------------------

E(X) = [INT(ax) between 0 and k] + [INT(bx) between k and 1]
= (ak^2)/2 + b(1 - k^2)/2

and if we substitute in our expression for k, factorise/expand/cancel factors, we arrive at E(X) = (1 - 2b + ab)/2(a-b) which is what we wanted.

------------------------------

We have have two possible scenarios: either m < k, or m > k. Consider the first: we must have [INT(a) between 0 and k] > 0.5, and so ak < 0.5. If we substitute in for k and rearrange we see that this implies that a+b > 2ab. To find m in this case we must solve

[INT(a) between 0 and m] = 0.5
i.e. am = 0.5, and so m = 1/(2a)

The second possible situation, where m > k, occurs when [INT(a) between 0 and k] < 0.5. Carrying out the integration gives the condition that a+b < 2ab. We can now solve for m as before:

[INT(a) between 0 and k] + [INT(b) between k and m] = 0.5
=> ak + bm - bk = 0.5
=> k(a-b) + bm = 0.5
=> 1 - b + bm = 0.5
=> b(m-1) = -0.5
=> m = 1 - 1/(2b)

-------------------------------

Suppose first that m = 1/(2a). Then, after some rearranging and factorising, E(X) - m = (1 - 2b + ab)/2(a-b) - 1/(2a) = (b(a-1)^2)/2(a-b) > 0. Therefore E(X) > m.

Now suppose that m = 1 - 1/(2b). Then E(X) - m = (1 - 2b + ab)/2(a-b) - 1 + 1/(2b) = (a(b-1)^2)/2(a-b) > 0, and so E(X) > m.

Thus, whatever the value of m, E(X) > m.
0
X

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 (139)
42.77%
I'm worried that I’m not academically prepared for the next stage in my educational journey (38)
11.69%
I'm worried it will impact my future career (27)
8.31%
I'm worried that my grades will be seen as ‘lesser’ because I didn’t take exams (68)
20.92%
I don’t think that receiving these grades will impact my future (34)
10.46%
I think that receiving these grades will affect me in another way (let us know in the discussion!) (19)
5.85%