# STEP II 2015 Solutions Thread

Announcements
#1
STEP II 2015

Q1) team-f-maths (attached to first post)
Q2) Gawain
Q3) Gawain
Q4) team-f-maths (attached to first post)
Q5) Gawain
Q6)
Q7) Goose Lane
Q8) Gawain
Q9) Brammer ; Alt - mikelbird
Q10) mikelbird
Q11) mikelbird
Q12) brianeverit
Q13) brianeverit

0
#2
0
#3
0
#4
Q3)Extra triangles gained adding a rod length 8:
(8,7,1-6) = 6 Triangles
(8,6,2,5) = 4 Triangles
(8,5,3-4)= 2 Triangles

So Through a similar method Therefore Trivially: and Induction:

Base: With 4 rods we can make 3 triangles, also so our base case works.

Assumption: Induction: Simplifying we get: Using the second difference formula we get: 0
#5
0
7 years ago
#6
How tough was part 2 compared to part 1. Part 3 simply follows from part 2, doesnt it?
0
7 years ago
#7
Question 6 solution:
i) sec^2 (pi/4 - x/2) = 1/cos^2(pi/4 - x/2) = 2/2cos^2(pi/4 - x/2) = 2/(1+cos(pi/2-x)) = 2/(1+sin x)

ii) take y = pi - x.
=> dy = -dx. Substituting and noticing that sin x = sin(pi - x) gives that integral( xf(sin(x) ) = integral( (pi-x)f(sin x)).
We use this fact in evaluating I = integral(x/(1+sin x)). We get 2I = integral(pi/(1+sin x)). Hence I = pi/2.integral(1/(1+sin x)).

iii) let (1+sinx)^2 = t for simplicity of writing this answer.
now J = integral ( (2x^3 - 3pix^2) / t )
Let J1 = integral(2x^3/t) and J2 = integral(-3pix^2/t).
so J = J1 + J2.

Now, lets solve J1.
J1 = integral(2x^3/t) ....................(1)
=> J1 = integral( 2(pi-x)^3/t ).......(2)
Expand the cubic term and add (1) and (2).
you will get 2J1 = integral( 2(pi^3 - 3pi^2x + 3pix^2)/t ).
So J1 = integral( (pi^3 - 3pi^2x + 3pix^2)/t ).

Now consider J = J1+J2.The 3pix^2 term gets cancelled.
so J = integral( (pi^3 - 3pi^x)/t ).
Integral 3pi^2x/t = 3pi^3/2 integral((1/t))
So J = -pi^3/2 integral( 1/t ).
=> J = pi^3.integral(sec^4z/4) for z = pi/4-x/2.

sec^4z can broken into sec^2z(1+tan^2z) and integrated to get tanz + tan^3z/3.
replace z and put the limits. you will get the answer.

The final expression is (pi^3)/4.(tan z - tan^3 z/3) for z = pi/4-x/2. thus, putting limits 0 and pi gives -2pi^3/3.
0
7 years ago
#8
It is probably slightly easier to notice: From which the conclusion follows immediately.
2
7 years ago
#9
You might as well look for Physicsmaths's post in the STEP thread and link that for the final part of Q6.
0
7 years ago
#10
For Q11 you had to do CLM and conservation of energy. Then you had to show the something about the second collision. The last part you had to show whether theta reached a particular value. That's all I remember. I think we'll be needing a paper to confirm most of the questions 0
7 years ago
#11
(Original post by Gawain)
STEP II 2015

Q1) team-f-maths (attached to first post)
Q2) Gawain
Q3) Gawain
Q4) team-f-maths (attached to first post)
Q5) Gawain
Q6)
Q7)
Q8) Gawain
Q9)
Q10)
Q11)
Q12)

The last part of Q4 isn't right here. Obviously ignore the vertical lines, and translate it (0 pi) The graph doesn't look like an arctan(x) graph for (iii).
0
7 years ago
#12
Someone gona do Q10? I can't remember it. I did not have time to finish it in the exam. I wrote down a load of crap hoping it was right last minute.

Posted from TSR Mobile
0
7 years ago
#13
Here is the solution to 6.

Did not do the first parts as they were trivial.

Posted from TSR Mobile
0
7 years ago
#14
How much would that last part be out of , d you think?
1
7 years ago
#15
(Original post by Chazatthekeys)
The last part of Q4 isn't right here. Obviously ignore the vertical lines, and translate it (0 pi) The graph doesn't look like an arctan(x) graph for (iii).
Actually the graph looks exactly like that for x<-1, translated up pi for -1<x<1, and translated up 2pi for x>1. So it all joins up at -1 and 1.

Its h(x) = arctan(x/(1-x^2)) + npi, where n varies depending on where you are
1
7 years ago
#16
(Original post by raff97)
Actually the graph looks exactly like that for x<-1, translated up pi for -1<x<1, and translated up 2pi for x>1. So it all joins up at -1 and 1.

Its h(x) = arctan(x/(1-x^2)) + npi, where n varies depending on where you are
Weird. Guess I got that wrong

Posted from TSR Mobile
0
7 years ago
#17
(Original post by physicsmaths)
Weird. Guess I got that wrong

Posted from TSR Mobile
I really want DFranklin and shamika come in and look at our solutions for STEP I and II. Have we still not got our hands on a paper? They should definitely be here for III tomorrow morning
0
7 years ago
#18
0
7 years ago
#19
Nitpick: In the answer to Q4(ii), arctan(1/2) isn't pi/4
0
6 years ago
#20
7.(i) It is clear that there are many options, simplest is perhaps choosing a circle with centre on the x-axis that intersects at (0,a) and (0,-a). Our circle has equation Subbing in for the points of intersection, we get .

If , the circle of radius r cannot intersect C at diametrically opposed points, since any two points are at most apart. And if , if there were two such points of intersection, the circles would need to overlap, which doesn't satisfy the exactly two condition.

(ii)We can see from the diagram that we have two isoceles triangles, each defined by the centre of D and the points of intersection on the circles C1 and C2. Let us call the height of these triangles and respectively. We have . Now consider the triangle defined by the centre of D, which we will give coordinates of , and the points and . Adding a perpendicular from the centre of D to the x-axis, pythagoras gives us and Subtracting (2) from (1), we obtain: Substituting for in terms of and and rearranging gives the required result for p.

Now to find q, we add the equations (1) and (2), and substitute for p, resulting in . Note that q could be either positive or negative, since there are 2 possible circles D reflected on the x-axis.

Clearly , so , and a little more rearranging gives the final result.
1
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

#### Were exams easier or harder than you expected?

Easier (21)
26.58%
As I expected (24)
30.38%
Harder (29)
36.71%
Something else (tell us in the thread) (5)
6.33%