1987 Specimen STEP solutions thread Watch

oh_1993
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STEP I Q9

(a) - 2 - root3 < x < - root3

(b) 2root2
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Ree69
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(Original post by ben-smith)
x
STEP I Q9 (a)

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Consider the cases where  x + \frac{x-1}{x+1} = 2 and where  -(x + \frac{x-1}{x+1}) = 2 .

The first equation can be solved to get x = \pm \sqrt{3} and the second can be solved to get x = -2 \pm \sqrt{3}.

We've now found most of the critical values in the awkward equation  |x + \frac{x-1}{x+1}| &lt; 2 . Usually the inequality will only ever change direction at any given critical point. However, notice the expression on the LHS is not defined at x = -1; this must be taken into consideration too.

Thus the critical regions to consider are:

 x &lt; -2-\sqrt{3}

-2-\sqrt{3} &lt; x &lt; -\sqrt{3}

-\sqrt{3} &lt; x &lt; -1

-1 &lt; x &lt; -2+\sqrt{3}

-2+\sqrt{3} &lt; x &lt; \sqrt{3}

\sqrt{3} &lt; x

Plugging in a value in each of the critical regions will show that the only ones that are valid solutions for the equation are -2-\sqrt{3} &lt; x &lt; -\sqrt{3} and  -2+\sqrt{3} &lt; x &lt; \sqrt{3}


STEP I Q9 (b)

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Because of the modulus signs we need to seperately consider the region for which  cosx - sinx &lt; 0 and subtract this negative value in the given integral.

The integral will be easier if you notice that  cosx - sinx \equiv \sqrt{2}\ cos(\frac{\pi}{4} + x). From this it should become quite apparent that  cosx - sinx &lt; 0\ \forall\ x \in (\frac{\pi}{4}, \frac{5\pi}{4}) - i.e., if  \frac{\pi}{4} &lt; x &lt; \frac{5\pi}{4}, then  cosx - sinx &lt; 0 .

Looking at the integral, we only need to consider the range between 0 and \pi anyway.

So \displaystyle\int^\pi_0 |cosx - sinx|\ dx = \displaystyle\int^\frac{\pi}{4}_  0 \sqrt{2}\ cos(\frac{\pi}{4} + x)\ dx - \displaystyle\int^\pi_\frac{\pi}  {4} \sqrt{2}\ cos(\frac{\pi}{4} + x)\ dx = 2\sqrt{2}
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klausw
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(Original post by Farhan.Hanif93)
STEP III Q7
Solution
Setting y=kx^{\alpha}, y'= \alpha k x^{\alpha -1} , y'' =\alpha(\alpha -1)kx^{\alpha -2} in the differential equation:

\alpha (\alpha -1)kx^{\alpha} + (x-2)(\alpha kx^{\alpha} -kx ^{\alpha}) = 0
\implies kx^{\alpha} (\alpha -1)(x+\alpha -2) =0

Clearly this holds if \alpha =1 and hence there exists a solution of the form x^\alpha as required.

Setting y=xv, y'=xv'+v, y''=xv''+2v' in the DE:

\implies v'' + v' =0

\implies v=Ae^{-x}

\implies \boxed{y=Axe^{-x}}


They call that a STEP question...

\implies v=Ae^{-x}+B

\implies \boxed{y=Axe^{-x}}+Bx

Is this a right correction?
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Farhan.Hanif93
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(Original post by klausw)
\implies v=Ae^{-x}+B

\implies \boxed{y=Axe^{-x}}+Bx

Is this a right correction?
Yes, you're correct. I think I missed the dash on the v and hence why my expression for v was off.
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Stray
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STEP I - Q5

a)
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Answer: 67 '7's are written.

The key to this question is getting the correct 'last digit' written.

1-9: 9 Digits Running total = 9
10-99: 90 * 2 Digits Running total = 189

This leaves 1000-189 = 811 digits

811 / 3 = 270, remainder 1

So - we write all of the next 270 numbers, taking us to 369.

We then write the first digit of 370, and stop. So the 7 in 370 is never written.

Counting the 7s:

Per hundred: 10 unit digits (17, 27 etc) + 10 'tens' digits (70, 71)

= 20 per 100 for 1-300 = 60

In the range 301-369 we only hit '7's in units 7 times.

Total 7s written: 60 + 7 = 67


---

b)
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Answer: 365! ends in 89 '0's.

The question asks how many zeros 365! ends in. (Not what 365! is - luckily).

We can do this by considering digits. The only way to 'create' a 0 digit is to multiply by a 0 or by 5 and an even number. Each 5 gets 'used up' - i.e. (2 x 5 x 6) still only creates one zero.

Considering the unit digits:

The range 1-365 contains 36 numbers ending in 0.

It contains 37 numbers ending in 5, and an abundance of even numbers.

So the multiplication of 'units' should create 36+37 = 73 zeros on the end of 365 factorial.

Considering the 10s:

We have 36 'tens' digits: 3 of them are 0, 4 of them are 5s and again we have an abundance of even numbers. So these will produce a further (3+4 = 7) zeros.

Special cases:

Numbers ending 25 or 75 (25, 175 etc) multiply by 4 to give multiples of 100, which have 2 zeros each. There are 7 of these, and we've only accounted for 1 zero for each of them so far. So that's +7.

4 x 250 = 1000 - we've only accounted for 2 of these digits, so +1 for this case.

4 x 125 = 500, which gives us an extra 5 (in the hundreds place), adding one more zero.

Total zeros: 73 + 7 + 7 + 1 + 1 = 89


(I also confirmed both these results via a brute force algorithm on a computer).
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mikelbird
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surely the original thread (and all the solutions) were at

http://www.thestudentroom.co.uk/showthread.php?t=448558
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Dirac Spinor
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#47
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(Original post by mikelbird)
surely the original thread (and all the solutions) were at

http://www.thestudentroom.co.uk/showthread.php?t=448558
that's the actual paper, this is the specimen.
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mikelbird
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(Original post by ben-smith)
that's the actual paper, this is the specimen.
OK!!
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Stray
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STEP I - Q14

i)
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Using two 4-sided dice, the probabilities of scoring a total between 2 and 8 are:

2: 1/16 (1-1)
3: 2/16 (1-2, 2-1)
4: 3/16 (1-3, 3-1, 2-2)
5: 4/16 (1-4, 4-1, 2-3, 3-2)
6: 3/16 (2-4, 4-2, 3-3)
7: 2/16 (3-4, 4-3)
8: 1/16 (4-4)

i) For the first part of the question we only need P(8).

The benefit of landing an 8 is 6400p.
The cost of not landing an 8 is 400p.

To get the 'expected gain' we multiply each of these by their respective Probability, and subtract losses from gains:

expected gain = (1/16 * 6400) - ((1-1/16) * 400) = 400 - 375 = 25p.

ii)
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If we do the same for each of the other 7 possible scores, we find that betting on 5 is the best option, with an expected benefit of 437.5 pence.


(Does anyone else agree that this is a really easy question considering it's a STEP paper? Perhaps Stats was considered new fangled in those days )
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bogstandardname
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STEP 1 Q 1 does not appear to be solved so here goes.

It isn't immediately obvious where the inequalities come from.
we have 4x2+16xy+y2+24x = 0
Complete the square for y
(y+8x)2-(8x)2+4x2+24x=0
so (y+8x)2=(64-4)x2-24x
(y+8x)2=60x2-24x
LHS ≥ 0, so RHS must too be ≥ 0,
This implies 60x2-24x≥ 0, If we look at the graph (roots are 0 and 2/5), this is clearly where 0≥x and x≥2/5 as required.

For the next part complete square for x
4x2+(16y+24)x+y2=0
4(x2+(4y+6)x)+y2=0
4((x+(2y+3)2-(2y+3)2)+y2=0
4(x+(2y+3))2-4(2y+3)2+y2=0
4(x+(2y+3))2=4(2y+3)2-y2
4(x+(2y+3))2=4(4y2+12y+9)-y2
4(x+(2y+3))2=15y2+48y+36
LHS ≥ 0 so RHS must too be ≥ 0,
15y2+48y+36≥ 0
roots are -2 and -1.2
so restrictions on y are
-2≥y y≥-1.2

b is quite simple so can't be bothered, aplogies for mistakes, hard to remember what I did exactly
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Lord of the Flies
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Wow this is noticeably easier than the actual exam papers - especially I. 4. :eek:

That's all I have to say... :creep:
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bogstandardname
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Another solution for an uncompleted question, STEP I Question 8 is attached
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Sophie M-J
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(Original post by Lord of the Flies)
Wow this is noticeably easier than the actual exam papers - especially I. 4. :eek:

That's all I have to say... :creep:
Just did this question in about 5 minutes ... was feeling so happy... until I read this haha!!
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bogstandardname
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solution for STEP II Q7
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bogstandardname
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(Original post by Farhan.Hanif93)
STEP III Q7
Solution
Setting y=kx^{\alpha}, y'= \alpha k x^{\alpha -1} , y'' =\alpha(\alpha -1)kx^{\alpha -2} in the differential equation:

\alpha (\alpha -1)kx^{\alpha} + (x-2)(\alpha kx^{\alpha} -kx ^{\alpha}) = 0
\implies kx^{\alpha} (\alpha -1)(x+\alpha -2) =0

Clearly this holds if \alpha =1 and hence there exists a solution of the form x^\alpha as required.

Setting y=xv, y'=xv'+v, y''=xv''+2v' in the DE:

\implies v'' + v' =0

\implies v=Ae^{-x}

\implies \boxed{y=Axe^{-x}}


They call that a STEP question...
When you get to \implies v'' + v' =0 taking auxillary equations gives roots 0 and -1, so why isn't the solution in the form {y=Axe^{-x}+Bx}
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Farhan.Hanif93
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(Original post by bogstandardname)
When you get to \implies v'' + v' =0 taking auxillary equations gives roots 0 and -1, so why isn't the solution in the form {y=Axe^{-x}+Bx}
It is; my mistake. It seems I forgot to edit this solution last time so thanks for the reminder (see here). Turns out that solving the problem whilst typing it in LaTeX lead to a typo (I wrote v, instead of v' and accordingly missed the second term).
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DJMayes
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STEP III - Question 5:

Part i):

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Sketch a graph of  y = sinx and  y = kx . Both pass through the origin. Note that the gradient of the sine curve is less than or equal to 1 for all values of x. If  k &gt; 1 , the gradient of the line will always be greater than the sine curve, and so they cannot intersect, so there are no non-zero solutions.

For the next part, set the equation of the line equal to the power series expansion of sinx:

 (1-\epsilon ^2)x = x-\dfrac{x^3}{3!} +...

As we are dealing with small values of  x , we can discard terms beyond the cubic:

 x - \epsilon ^2 x = x - \dfrac{x^3}{3!}

Which then solves to give  x = 0 and  x = \epsilon \sqrt{6} . Notice we are only interested in non-zero solutions, and we are done.



Part ii):

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Draw a sine curve with several periods, and a line passing through them all. You should see that the first solution will pass between the maximum of the sine curve and it's second root, with the 3rd solution passing between the second maximum and the next root, and so on, with the (2n-1)th solution passing through the nth positive section of the curve, between  2(n-1)\pi and  (2n-1)\pi . This yields the first inequality described:

 (2n-2)\pi &lt; x_{2n-1} &lt; (2n-1)\pi

In a similar way, it can be seen that every even solution will occur between a root of the sine curve and its maximum (i.e. even solutions occur as the sine curve is increasing and odd as the sine curve is decreasing). This then yields the second inequality:

 2n\pi &lt; x_{2n} &lt; (2n+\frac{1}{2})\pi

For the final part, we know that no more solutions occur when:

 \lambda x = 1

 x = \dfrac{1}{\lambda}

The sine wave has period  2\pi , therefore approximately  \dfrac{1}{2\pi \lambda} cycles of the sine wave will occur. From our graph, we can see that each cycle gives two solutions, except for the first which gives only one.

 \Rightarrow N = 2\dfrac{1}{2\pi \lambda} - 1

 N = \dfrac{1}{\pi \lambda} - 1



And that's all of the STEP III questions for the specimen completed.
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Genesis2703
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(Original post by ben-smith)
STEP I Q3
\frac{d^2y}{dx^2}+2\frac{dy}{dx}-3y=2e^x

\Rightarrow \frac{d^2y}{dx^2}-\frac{dy}{dx}+3(\frac{dy}{dx}-y)=2e^x

\Rightarrow \frac{dz}{dx}+3z=2e^x

\Rightarrow \frac{dz}{dx}e^{3x}+3e^{3x}z=2e^  {4x}

\Rightarrow \frac{d}{dx}[ze^{3x}]=2e^{4x} \Rightarrow z=\frac{e^x}{2}+C_1e^{-3x}
Using the initial conditions z(0)=1 so C_1=1/2. Now, to solve for y:
\frac{dy}{dx}-y=\frac{e^x}{2}+C_1e^{-3x}

\Rightarrow \frac{d}{dx}[ye^{-x}]=1/2+C_1e^{-4x}

\Rightarrow y=\frac{xe^x}{2}-C_1e^{-3x}+C_2e^x

y(0)=1 \Rightarrow C_2=9/8
On the penultimate line (where you have y = ....) surely it should be a -1/4 * C1 lots of e^x, I know you changed it to a negative sign which is fine, but in my final answer I have 1/8 instead of 1/2 because you divide by the 4 when integrating?
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abra-cad-abra
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(Original post by ben-smith)
STEP I Q6
Consider the sector formed by the points A,D and E and let Q be the centre of the whole coin. We want to find the area of the segment subtended on A and the area of the triangle QDE. Let QD=QE=x and 2y=DE:
Sin(\pi/14)=y/a,Sin(\pi/7)=y/x

\Rightarrow x=a\dfrac{sin(\pi/14)}{sin(\pi/7)}
Adding the two areas we mentioned earlier we get:


a^2/2(\pi/7-sin(\pi/7))+a^2/2 \dfrac{sin^2(\pi/14)sin(2\pi/7)}{sin^2(\pi/7)}

=a^2/2(\frac{\pi}{7}-sin(\pi/7)+\dfrac{sin^2(\pi/14)2sin(\pi/7)cos(\pi/7)}{sin^2(\pi/7)}

=a^2/2(\frac{\pi}{7}+\frac{2sin^2(\pi/7)cos(\pi/7)-sin^2(\pi/7)}{sin(\pi/7)})

=a^2/2(\frac{\pi}{7}+\frac{cos(\pi/7)-1}{sin(\pi/7)})



=a^2/2(\frac{\pi}{7}-Tan(\pi/14)

Multiplying by 7 gives the required result.
how did you know that that first angle was pi/7. i just guessed it as its a 7sided shape so makes sense and looking at the answer i was working towards it seemed right
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brianeverit
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(Original post by ben-smith)
Found this paper the other day and realised there wasn't a solutions thread up and running so I figured it should be done. Unfortunately, I've only been able to locate the STEP III paper so if anyone could locate the others it would be awesome.
EDIT: Got them. Courtesy of Professor Siklos himself!

STEP I
1.
2.Solution by ben-smith
3.Solution by ben-smith
4.Solution by ben-smith
5.
6.Solution by ben-smith
7.Solution by Farhan.Hanif93
8.
9.
10.
11.Solution by Farhan.Hanif93
12.Solution by Farhan.Hanif93
13.Solution by Farhan.Hanif93
14.
15.
16.
STEP II
-
STEP III
1.Solution by ben-smith.
2.Solution by DFranklin
3.Solution by ben-smith
4.Solution by ben-smith
5.
6.Solution by Farhan.Hanif93
7.Solution by Farhan.Hanif93
8.Solution by DFranklin
9.Solution by Farhan.Hanif93
10.Solution by Kennethx
11.solution by ben-smith
12.Solution by ben-smith
13.Solution by ben-smith
14.Solution by ben-smith
15.Solution by DFranklin
16.Solution by Dfranklin
There is a solutions thread for 1987 with ALL questions answered.
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