# STEP II/III 2014 solutions

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#1
Papers found here. (Credit to Tarquin Digby for those [and Charles98 for his photos])

STEP II:
1: Solution by mikelbird
2: Solution by LightBlueSoldier
3: Solution by Stray
4: Solution by Khallil
5: Solution by Blazy
6: Solution by DFranklin
7: Solution by cliverlong
8: Solution by LightBlueSoldier, Solution by Stray
9: Solution by Brammer
10: Solution by JosephML
11: Solution by Brammer
12: Solution by shamika
13: Solution by DJMayes

STEP III:
1: Solution by jtSketchy
2: Solution by Brammer
3: Solution by edrraa, Solution by mikelbird
4: Solution by Farhan.Hanif93
5: Solution by metaltron
6: Solution by ctrls
7: Solution by Farhan.Hanif93
8: Solution by DFranklin
9: Solution by JosephML
10: Solution by JosephML
11: Solution by Brammer
12: Solution by DJMayes
13: Solution by DJMayes
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6 years ago
#2
II/10:

Spoiler:
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For the first part, get the two equations for and and combine them:

and , giving:

Now we wish to maximize for a given value of so we take what is known as the partial derivative of with respect to i.e. ignore and differentiate wrt to :

.

To get the max, we solve which gives . Putting this back into will give us :

.

The sketch of this is fairly straight forward with the set of points which can be reached being those with and .

Finally the maximum possible distance reached by the particle is when which gives us .
0
6 years ago
#3
II/12:

(i)

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. The denominator is just 1 - F(t). The numerator is , since the first term is tends to the derivative of F(t) (i.e. f(t)), as t tends to 0.

(ii)

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In this case, . giving immediately .

(iii)

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Note that . Therefore we want to solve . Separating the variables gives

, i.e.

. When t = a, F(t) = 0, so c = log a, and so rearranging we get . Differentiating, we have .

As a check, we can insert F(t) and f(t) into the definition of the hazard function; we get , as required.

(iv)

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We prove each implication separately. If , then , where x is just a dummy variable. This gives . Subbing this into the formula of h(t), we get h(t) = k, a constant.

On the other hand, if h(t) = k, then if we solve , which (after separating the variables and integrating to find F(t)), gives , which differentiates to the required form of f(t).

(v)

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Similar to (iii) and (iv), solve , to give . Finally, differentiating, we get

Note: this is the Weibull distribution, which is an important distribution in actuarial science to model very extreme losses. Of course, there was no reason you needed to know that to complere the question. The concept of the hazard function is also a fundamental theoretical actuarial tool, but I've never used it in practice.
0
6 years ago
#4
Some photos: II 2:

And II 8

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1
6 years ago
#5
III/9

Spoiler:
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We have
,

so differentiating gives us:

and again:

.

For the initial conditions just put in .

Now we must show that this fits with :

And so we indeed have .

Now letting and and subbing into we get:

For we must have the component of to be zero. Hence giving:

.

Similarly writing in terms of and we have:

which then gives

which simplifies to the required result. Combining the two results gives us:

We are given that . So from the definition of we have:

And so we get the required . Finally since and are both acute we must have
0
6 years ago
#6
II/4:
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I'll edit the last part into this post in a short while.
Last edited by user2020user; 1 year ago
2
6 years ago
#7
(Original post by JosephML)
III/9
.
I only got up as far as the result involving uksin - how many marks do you think I'd have got?
0
6 years ago
#8
(Original post by JosephML)
III/9
I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?
0
6 years ago
#9
III/6

Spoiler:
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By letting with the given inequality and noting that the inequality holds for where , you get,

Since Repeating the process by setting gives the required inequality.

(i) Choosing , if we can prove that for all , then the result follows. To do this, we differentiate it twice, so

Noting that for and so it follows that for this range of Since the conditions for the inequality in the original question is met and we can deduce for the given range of and the result follows.

(ii) Here you can prove each bound separately. Since for you can show that , and , we can multiply over the denominators without worrying about flipping the inequality. This gives,

So we set and prove it is positive in the given domain using the inequality again. Differentiating twice gives,

Noting again the the two functions are positive in the domain and that the result follows.

Finally for the upper bound, rearranging the inequality as,

And setting , again it is sufficient to show that for the given domain to establish the inequality. Differentiating twice,

Note that this gives the same inequality from part (i), so we can quickly deduce that for the given domain. Since we can deduce that and the result follows.

Note: Most of this is just tedious manipulation to show the initial inequalities are met, so I've omitted some of the details of showing certain functions are positive and that f(0) and f'(0) are equal to zero in the various cases. If there's an faster way to do this question that I've missed, I would be interested to hear about it.
1
6 years ago
#10
(Original post by Kasel)
I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?
Ah, you're right. I missed that bit. I'll correct the solution, thanks.

(Original post by CD315)
I only got up as far as the result involving uksin - how many marks do you think I'd have got?
I should suspect somewhere between 8 and 12 marks.
0
6 years ago
#11
(Original post by Khallil)
II/4
Do you think we would have had to prove arctan(x) + arctan(1/x)=pi/2?
There are a few short proofs but I forgot to include any in my solution.
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#12
STEP III, Q4

(i)
Note, expanding the square in the integrand of :

, as required.

Hence , which can interpreted as the area under a curve that is always above the -axis. It follows that with over the interval of integration i.e. for every s.t. . Solving:

.

Note so that on , as desired.

(ii)
Let and observe that by the argument in . Moreover, we require in order to avoid integrating over the discontinuity of at i.e. for our integral to make sense.

Furthermore:

Setting allows us to write the integrand as the derivative of a product, provided that for the above reasons.

So, in the case :

, as desired, with the argument being valid for .

In the case :

Integrating the first term in the integrand of by parts with , we obtain:

So it is sufficient to search for a function s.t. with either or .

Seeing that the general solution of the second order differential equation is , forces us to take and observing that , it follows that is s.t. , and on .

Hence so will do the trick.

It's worth pointing out that a very similar question appeared in a IB tripos exam for the variational principles course.
1
6 years ago
#13
(Original post by Kasel)
...
I did but I left it out of the solution above for brevity. The one I wrote in the paper was very similar to this, but it looks really forced with regards to the 1-1 in the denominator.

0
6 years ago
#14
(Original post by Khallil)
...
To avoid limits you can say let y=arctan(x), then tan(y)=x
and
cot(y)=(1/x)=tan(pi/2 - y)
which gives arctan(x)+arctan(1/x)= y+pi/2-y=pi/2
0
6 years ago
#15
(Original post by Kasel)
...
Ah, of course!
0
6 years ago
#16
STEP III - Q12

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i)

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Note that is increasing. Thus:

So

ii)

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Differentiating, we deduce a p.d.f of , as required.

The mode occurs at a stationary point of the p.d.f (not at an endpoint; this is not possible as the range is infinite) so occurs when the derivative of the p.d.f is 0:

From which follows.

iii)

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Our integral is merely the integral of the p.d.f of the random variable , so equals 1. Then:

Where I is the integral mentioned above equalling one, from which our result follows.

iv):

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Using our p.d.f. X has median so Y has median

Applying ii) with f the p.d.f. given above, we find that :

Noting that , the result follows.

Then the inequality follows easily from the fact the exponential function is increasing.

0
6 years ago
#17
STEP II - Q8 (LaTeX)
Spoiler:
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If is a maximum, it must be and

Taking first, :

Divide both sides by the common factor

Giving

This is almost the LHS of what we are required to show, just needs a little fiddling

Now repeat the process for :

Divide both sides by the common factor

Giving

Which combines with (**) to give the required solution

Considering the possible values of

Either and are both integers, or and are non-integer values either side of an integer, so if is an integer, it is either and (2 values) or it lies between them (1 value).

For the next parts, we are asked to find either the value of where there is 1 value, or the larger value if there are 2. This means we can consider the function to produce the Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is .

i)

ii)

iii)
If we fix and , the only part of that can alter is the denominator containing , so for large we want small , so is when . (Zero is not a positive integer).

iv)
For fixed and (unknown) fixed , the part of the function varying is:

For positive integer values of

So
giving

LightBlueSoldier points out below that my original is wrong... of course the constraint isn't nearly as harsh as that, regardless of whether infinity is acceptable...

This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.
0
6 years ago
#18
(Original post by Stray)
STEP II - Q8 (LaTeX)

If is a maximum, it must be and

Taking first, :

Divide both sides by the common factor

Giving

This is almost the LHS of what we are required to show, just needs a little fiddling

Now repeat the process for :

Divide both sides by the common factor

Giving

Which combines with (**) to give the required solution

Considering the possible values of

Either and are both integers, or and are non-integer values either side of an integer, so if is an integer, it is either and (2 values) or it lies between them (1 value).

For the next parts, we are asked to find either the value of where there is 1 value, or the larger value if there are 2. This means we can consider the function to produce the Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is .

i)

ii)

iii)
If we fix and , the only part of that can alter is the denominator containing , so for large we want small , so is when . (Zero is not a positive integer).

iv)
For fixed and (unknown) fixed , the part of the function varying is:

This tends to 1 as b tends to infinity, but will never actually reach 1, so tends towards but never reaches so for fixed

---

This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.
The last part is incorrect. In particular the value of infinity does not exist.

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1
6 years ago
#19
STEP II, Q5: (somebody quickly check through this please though I think it should be fine)

(i)
Let , so by Product Rule.

Using this substitution, the question becomes:

Solve
.

Rearranging:

Separating the variables and integrating:

The solution curve passes through (1,1) so C = -4

Hence:

(ii)
We look for a and b such that: and in order to make the DE homogenous. and works.

Then substituting and :

Making the same substitution and rearranging:

The solution curve goes through (1,1), so C = 5. Hence:

The only real difficulty here was spotting how to get rid of the -4 and -3.
3
6 years ago
#20
STEP III Q5

Beginning

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If ABCD is a parallelogram, then . So .
If then so so ABCD is a parallelogram.

If ABCD is a parallelogram then it is a square if and only if its diagonals are perpendicular. That is .

Part i)

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Half the diagonal of a square is times the side length of a square, and the diagonal makes an angle of 45 degrees with the side. So by a spiral enlargement:

So

So

Part ii)

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By symmetry:

So

So so diagonals of quadrilateral are perpendicular.

So XYZT is a square if and only if .

That is .

Looking at the co-efficient of i (which is complex itself) which is always true.

Equating the other parts .

So XYZT is a square if and only if which is the condition for PQRS to be a parallelogram.
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