# STEP I 2014 solutions

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#1
STEP I Solutions now complete. Well done everyone!

STEP I:
1: Solution by DFranklin
2: Solution by Tarquin Digby
3: Solution by DFranklin
4: Solution by DFranklin
5: Solution by DFranklin
6: Solution by Tarquin Digby
7: Solution by Farhan.Hanif93
8: Solution by Pascal678
9: Solution by JosepthML
10: Solution by JosephML
11: Solution by DFranklin
12: Solution by DFranklin
13: Solution by Smaug123
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#2
STEP I, Q1

(i) 3 = 2^2 - 1^2, 5 = 3^2 - 2^2, 8 = 3^2 - 1^2, 12 = 4^2 -2^2, 16 = 5^2 - 3^2

(ii) If n = 2k+1 is odd, then n = (k+1)^2 - k^2

(iii) if n = pq (p < q), then if we let a = (q - p) / 2, b = (p+q)/2, then b^2-a^2 = pq. If n = 4k, take p = 2, q = 2k to get a = k-1, b = k+1.

(iv) Suppose 4k+2 = (a^2-b^2), then (a-b)(a+b) = 4k+2. Note that since a+b = (a-b) + 2b, and 2b is even, either both (a+b), (a-b) are even or both are odd. Since 4k+2 is even, they must both be even, and so 4k+2 must be divisible by 4, which is impossible.

(v) Using (iii) we find a = (q - p) /2, b = (p+q) /2 as one solution, and factorising n = 1.(pq) we get a second solution a = (pq - 1)/2, b = (pq+1) / 2. These are clearly distinct since pq > q - p. Note that a 3rd solution would provide a new factorization of pq which is not possible since p, q are prime.

(vi) 675 = 25 * 27 = 5^2 * 3^3. Every factorization 675 into two factors p, q will have both p, q odd. So (q-p)/2 and (p+q)/2 will always both be integers. So the question is how many factorizations there are.

Factorize ; there are 3 choices for r and 4 for s givinig 12 possible factorizations. But this double counts (e.g. 1 x 675 and 675 x 1 are both counted), so we need to divide by 2 giving a final answer of 6.
2
6 years ago
#3
Might as well do the only question I could get a full solution for

STEP I 2014 question 8

part i)
Spoiler:
Show

where

The line is

i.e.

If is defined similarly then we have the same sets of coordinates and gradient but this time in terms of b

The line is

and intersect when:

so

So Thus

Thus the point of intersection is T(ab,(a-1)(b-1))

part ii)
Spoiler:
Show

If the point of intersection lies on the curve, x=ab when y=(a-1)(b-1)

Thus as b->a

So the point of intersection T lies on the curve C as LHS=RHS at this point as b->a

part iii)
Spoiler:
Show

First we find the gradient of the curve C at any point on the curve in terms of x

Compare this to the gradient of which These are very similar and as b->a the gradient of C at the general point of intersection which lies on is

Furthermore the point of intersection lies on the curve and and so satisfies for the curve for various values of x and y, thus every tangent to C is of the form for some a

0
#4
STEP I, Q4.

Choose coordinate axes so the short hand has position (0, a) and the short hand has position .
If x is the distance between the hands, then . Write S for , C for , then we have
x^2 = b^2 S^2 + a^2 - 2ab C + b^2 C^2 = b^2 + a^2 - 2abC.

Differentiate w.r.t. theta: , so . Differentiate again:

This = 0 when , or equivalently (and x not = 0).

So (a^2+b^2-2abC)C = ab(1-C^2), so (a^2+b^2)C - ab = abC^2, so abC^2 -(a^2+b^2)C + ab = 0.

So C = = = a/b or b/a. Since b > a, and c <=1, b/a is impossible, so C = a / b.

So for a max/min rate of change, we must have x^2 = b^2 + a^2 - 2ab (a/b) = b^2 - a^2 as desired. (I'm not going to worry about whether it's a max or min, because the everything "flips" for theta negative, and so we get into worrying about whether "largest" means most positive, or "biggest magnitude" etc. It's not terribly well worded).

When b = 2a, this first occurs when C = 1/2, i.e. when .
After t minutes have passed, the minute hand has moved through 6t degrees (60 minutes = 360 degrees), and the hour hand through 0.5t degrres (60 minutes = 360/12 =30 degrees). So they are apart by 60 degrees after 60/(6-0.5) = 60/5.5 minutes. Since 11 * 5.5 = 55 + 5.5 = 60.5 minutes, 60/5.5 = 11 - (0.5/5.5) minutes, which is slightly less than 11 minutes.
0
#5
STEP I, Q5.

(i) Not going to do the sketch, but if y = (x+2a)^3 -27a^2x, then dy/dx = 3(x+2a)^2 - 27a^2. When x = a, y = dy/dx = 0, so y = (x-a)^2(x-b) for some value b. Comparing constant coefficients, -a^2b = 8a^3, so b = -8 and so y >=0 for x >= -8. Note that for x > 0, y = 0 iff x = a.

(ii) Want to maximize xy^2 subject to x+2y <= 3. By the above, we know that (x+2y)^3 >= 27xy^2, so 27xy^2 <= (x+2y)^3 <= 3^3 = 27. We also know we need x = y if we want (x+2y)^3 = 27xy^2. So it is only in this case that we have 27xy^2 = (x+2y)^3, and of course we only have (x+2y)^3 = 27 when x+2y = 3. So xy^2 is maximized then x = y = 1.

(iii) (p+q+r)^3 = (p + 2(q+r)/2) ^2 >= 27 p (q+r)^2 / 4. But (q+r)^2 / 4 - qr = (q^2 + 2qr +r^2 -4qr) / 4 =(p-q)^2 / 4 > = 0. So (q+r)^2 / 4 >= qr, so 27p(q+r)^2/4 >= 27pqr.

For equality in (q+r)^2 / 4 >= qr, we need q = r. For equality in (p + 2(q+r)/2) ^2 >= 27 p (q+r)^2 / 4 we need p = (q+r)/2. Hence we have equality iff p = q = r.
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#6
STEP I, Q3.
(i) , so b^3/3 = (b^2/2)^2 = b^4 / 4. So b/4 = 1/3, so b = 4/3.

(ii) Now we have (b^3 / 3 - 1/3) = (b^2 / 2 - 1/2 )^2
So 4b^3 - 4= 3(b^2 - 1)^2
So 4(b-1)(b^2+b+1) = 3(b-1)^2(b+1)^2. Since b>1, can divide by b-1.
So 4(b^2+b+1) = 3(b-1)(b+1)^2
So 3(b^2-1)(b+1) -4b^2 -4b -4 = 0.
So 3b^3 +3b^2 -3b - 3 - 4b^2 -4b - 4 =0.
So 3b^3 -b^2 - 7b - 7 = 0 as desired.

The "show there's only one root and it lies between 2 and 3" is definitely easier by a sketch; I would draw 3b^3 and b^2+7b+7 on the same graph and it should be reasonably obvious.

(iii) In general we have
So. We are given b>a, so b-a is non-zero, s we can divide by it:
. If p = (b+a), q = (b-a), then the RHS is immediately .
The LHS is . But , and . So as desired.

If , then , so , so [latexp^2 = q^2/(3(q-1))[/latex].

Since p > 0 we must have q-1 > 0 and so b-a > 1.
Note that in fact p >=q, so we must also have 3(q-1) <= 1, so q <= 4/3, and so b-a <= 4/3.
0
6 years ago
#7
(Original post by DFranklin)
Thought we might as well create a solution thread... I'll do the indexing here as people post solutions.
Where's the paper?
1
6 years ago
#8
(Original post by BabyMaths)
Where's the paper?
Here https://www.dropbox.com/sh/yyl2eax4r...%20Q7%20Q8.JPG
2
6 years ago
#9
STEP 1 Q2

i)
Integrating by parts

Now substitute and remember to divide by .

ii)

Symmetry about the y axis due to term. Intersects x axis at and . Intersects y axis at .

iii)

Note that .

iv)

Now multiply by to get
2
6 years ago
#10
Thanks.
0
6 years ago
#11
STEP 1 Q6

i)

Suppose for some .

Then .

Hence for all by induction.

ii)

Let and .

Substitute .

Then we have:

Rearranging:

We're given , so

as required.

If then , and .

This satisfies .

Also, and .

Therefore where .

, so which implies .

Hence .
1
#12
STEP I, Q12. Note that we can regard the dice as a 3 sided dice that takes values 1,2,3 with probability 1/2, 1/3, 1/6 respectively. Note also that E[|Y|^2] = E[Y^2].

So

as desired.

If k is an integer and E[X^2] is also an integer, then 6 must divide (k-1)^2, and so 6 must divide (k-1) (since both 2 and 3 must divide it). So if k is a single digit integer, we must have k = 7.

PDF of X: The "head" values for X are 6, 13 and 20 with probabilities 1/4, 1/6, 1/12 respectively.
The "tail" values for X are 6, 5, 4 with probabilities 1/4, 1/6, 1/12 respectively.
So combining these we finally have
P(X = 4) = 1/12, P(X=5) = 1/6, P(X=6) = 1/2, P(X=13) = 1/6, P(X=20) = 1/12.

For the sum of two rounds to = 25, we need either X=5, then X=20 or vice versa, with probability 2 x (1/6) x (1/12) = 1/36.
For the sum of two rounds to be > 25, we need either (6, 20), (13, 20), (20, 20), (13, 13) or (20, 6) or (20, 13).
The probability of this happening is 2 x ((1/2) x (1/12) + (1/6) x (1/12)) + (1/6)^2 + (1/12)^2 = 2 x (1/24 + 1/72) + 1/36 + 1/144 = (2 x (6+2) + 4 + 1) / 144 = 21 /144 = 7/48.

The expected return for the gambler is 7w/48+1/36 (ignoring the loss of the original stake).

The expected profit for the casino is 1 - 7w/48 - 1/36 = (144 - 21 w - 4) = (140-21w). So we require 21w < 140. If w is an integer, the greatest possible value is w = 6.

[Alternative calc for last bit: p(X>=13) is 1/6+1/12 = 1/4. So p(both tries >=13) = 1/16. If both aren't >=13, the only other possibility for getting over 25 is to get (6,20) or (20,6). This has probability 2 x (1/2) x (1/12) = 1/12. 1/16+1/12 = 7/48].
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#13
(Original post by Tarquin Digby)
Let and .
Where did these come from?

Substituting and simplifying we obtain .
I think this is too terse to be helpful, frankly. (Certainly you wouldn't get away with it in an exam).
0
6 years ago
#14
Question 9:

Resolving in the standard way we get:

So we get when , thus:

And so by symmetry we get

Now we define . So is the point at which the velocity is zero. For the sketches we consider the ratios:

So when we have the particle reaching its maximum height and hitting the ground without the velocity changing direction.

When we have the particle reaching its maximum height then moving back towards then hitting the ground.

When we have the particle moving back towards the origin before reaching its maximum height.

Finally when we have the particle stationary when . Also we have:

so the trajectory is a straight line.

The last part might be a little unclear, if people can't understand then I'll rewrite it.
0
6 years ago
#15
(Original post by DFranklin)
Thought we might as well create a solution thread... I'll do the indexing here as people post solutions.
Q13:
Finding the PDF:
Spoiler:
Show

Our PDF looks like a triangle with height and base . Thus (since the area of the triangle is 1, by "probability density function") we have - for ease, write this as . Also .

Write . Then we have , and . Eliminate : .

Then , and , so as required.

Now, the expression will be similar by symmetry. To get and a negative slope, the numerator will become . We still need ; hence by inspection.

Finding the mean:
Spoiler:
Show

The mean of the distribution is the integral of against the PDF. That is . We'll continue to write for the moment, and . We get that the mean is . Expanding these gives .

Now we have , by inspection, so we can tidy the expression up a bit:
.

Factorising: this is .

We'll consider the term first. The bracket factorises as , so the term is (after cancelling the from top and bottom).

Now the term is , so we get for that term.

Adding them up gives as required.

Finding the median:
Spoiler:
Show

For the median: there will be two cases, the first when the median is in the g(x) part of the PDF, and the second when the median is in the h(x) part. There is an obvious check: when , then the median will be .

Firstly, let's suppose the median lies leftwards of c. Then we have , for the median. The left-hand side is . so this in turn is . Solving for yields , so . (We pick the root that gives us .)

Now let's suppose the median lies rightwards of c. We then have , for the median. The left-hand side is . Hence and so , picking the root that gives us . This becomes . (Notice that our check-case of works for both of these.)

Finally we need the conditions under which the median lies leftwards or rightwards of c. By symmetry it is clear that the median is leftwards of c iff the slope of g is less than the negative-slope of x, so that the plot of the PDF is "more shallow" on the left-hand side. That is the only way that the left-hand side of the diagram can contain more than half the area. This happens precisely when , or when , or when , or when , or when .
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#16
(Original post by JosephML)
Question 9:

Resolving in the standard way we get:

Finally when we have the particle stationary when .
Note that if , then , and then . So the path is actually a straight line.

(I don't know that they expected this, but to be honest I think they probably did).
1
6 years ago
#17
Guys - in Q9, I correctly found T_H and T_L and identified what T meant (x vel becoming zero) - however, I misinterpreted the graphs (probably as I was in a rush) and forgot that x would eventually change direction. So I drew just a sudden drop (don't ask..). I also understood what the first 2 inequalities meant, and explained (i.e. x vel becoming zero before landing) - but the whole idea I'm getting at here is that I didn't spot that it would change direction. How many marks am I looking at for this?

Also - how many marks would you estimate the introductory part to Q13 to be worth?

Thanks!
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#18
(Original post by CD315)
Guys - in Q9, I correctly found T_H and T_L and identified what T meant (x vel becoming zero) - however, I misinterpreted the graphs (probably as I was in a rush) and forgot that x would eventually change direction. So I drew just a sudden drop (don't ask..). I also understood what the first 2 inequalities meant, and explained (i.e. x vel becoming zero before landing) - but the whole idea I'm getting at here is that I didn't spot that it would change direction. How many marks am I looking at for this?
I'm not sure on this one, to be honest. There's very little content in the question, which makes me think the sketches were probably a significant amount of the marks. But on the other hand, unless you mess up, the sketches aren't really much work either. If I had to guess I'd think you're probably looking at between 5 and 12 marks for the graphs + explanation. With about 8 being my best absolute guess.

Also - how many marks would you estimate the introductory part to Q13 to be worth?
Just the "show that"? 3-4 marks at most.
0
6 years ago
#19
(Original post by DFranklin)
I'm not sure on this one, to be honest. There's very little content in the question, which makes me think the sketches were probably a significant amount of the marks. But on the other hand, unless you mess up, the sketches aren't really much work either. If I had to guess I'd think you're probably looking at between 5 and 12 marks for the graphs + explanation. With about 8 being my best absolute guess.
Yes, there's surprisingly little content in it, so I was confused as to how the marks would have been distributed. I'll take 8!

Just the "show that"? 3-4 marks at most.[/QUOTE]

Yeah, the show that + finding a similar expression for h(x).
0
#20
STEP I, Q11. (Interesting question, this!).

Disclaimer: I honestly can't remember the last pulley question I did, which means it's probably 20+ years ago. So methods etc. may not be standard.

For the first part, let T be the tension in the string. The acceleration of M is g - T/M, and the accel of m is g - T/m. For the string not to stretch, these must be equal.
So g - T/M = T/m - g, so Mmg - Tm = TM -Mmg, so 2Mmg = T(M+m), so T = 2Mmg/(M+m).
Then the acceleration of M = g - 2mg/(M+m) = g(M-m)/(M+m) as required.
The force on the pulley is 2T = 4Mmg/(M+m).

For the 2nd part. Label the M1, M2 pulley P, and let the tension it its string be T. Let the tension in the other string be S. For legibility, write A for the acceleration a2.

Again, the acceleration of M is g - T/M and the acceleration of m is g-T/m. But relative to the pulley, the accelerations are (g+A) - T/M and (g+A) -T/m. So (replacing g in the earlier calculation with g+A) we end up with a
new value for the tension .

Now, since the pulley is weightless, the net force on it must be zero, so we have 2T = S, so . Now since the acceleration of M is A, we know the net force on M is MA.

So we must have Mg - S = MA. So . So so as required.

Now suppose .
This is true iff so
iff
iff . So iff iff , iff [/latex]4 \mu = m[/latex],

But iff iff iff .
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