STEP 2018 Solutions Watch

Oskarrrrrrr
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#81
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(Original post by GcseLad-_-)
Could you try and estimate my mark Zacken (or anyone else can feel free to), I saw you've done it in previous years.

Number 1 I got all of it until proving the neceesary and sufficient condition.

Number 3 I did all of it apart from showing g(x)+g(2a-x)=b iff rotational symmetry about (a,b)

Number 4 I did the first two parts and around two steps of simplification in the third parts

Number 5 I did all of it until the last 'deduce' part where I made the correct substitution and changed the limits but didnt go further

Number 7 I wrote the line equations, found the interesection point but didn't simplify and so got into loads of algebra that I couldn't get out of

Number 8 I did the first part right but in the second part at the very last step I subbed back in v=y^1/2 instead of v=y^1/3 and so ended up with y1 correct but y2 as (1-e^-b/3)^2. I did show that my y1 was bigger than y2 using the proof that 1-e^-b/3<1-e^/b/2 and so the sketches were correct.

Finally, how did you find this compared to last years paper. Thanks!
Seems like you did very well ! Remember four ‘good’ answers is meant to get you a 1 And since you answered 6 that should hopefully all merge together , remember it’s really about how you answered questions and such , many people at Cambridge’s right now didn’t get 1 1 and we’re accepted when the college looked at their scripts and liked what they whats they saw. I did Step 1 last year and got a 1 and found this paper very very difficult in all honesty . Best guess isn’t that the 1 will Be 63-70, so I reckon you definitely got got it ! DONT stress just be confident and go into Thursday happy !
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Aditya33
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#82
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Here's my attempt at the first one
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gnsbrg
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#83
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I deleted my solutions because I didn’t read the show g(x)+g(2a-x) part lol how many marks do you think it will be worth?
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Zacken
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#84
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#84
(Original post by GcseLad-_-)
Could you try and estimate my mark Zacken (or anyone else can feel free to), I saw you've done it in previous years.

Number 1 I got all of it until proving the neceesary and sufficient condition.

Number 3 I did all of it apart from showing g(x)+g(2a-x)=b iff rotational symmetry about (a,b)

Number 4 I did the first two parts and around two steps of simplification in the third parts

Number 5 I did all of it until the last 'deduce' part where I made the correct substitution and changed the limits but didnt go further

Number 7 I wrote the line equations, found the interesection point but didn't simplify and so got into loads of algebra that I couldn't get out of

Number 8 I did the first part right but in the second part at the very last step I subbed back in v=y^1/2 instead of v=y^1/3 and so ended up with y1 correct but y2 as (1-e^-b/3)^2. I did show that my y1 was bigger than y2 using the proof that 1-e^-b/3<1-e^/b/2 and so the sketches were correct.

Finally, how did you find this compared to last years paper. Thanks!
I’ll try and have a look in a while. Remind me if I don’t.
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Aditya33
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(Original post by gnsbrg)
I deleted my solutions because I didn’t read the show g(x)+g(2a-x) part lol how many marks do you think it will be worth?
Same... I read it as f is said to be rather than show.... Hoping not more than 5 marks !!
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Oskarrrrrrr
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(Original post by Aditya33)
Here's my attempt at the first one
I made it (a+2)^2>4

But the rest of yours I had also
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IrrationalRoot
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#87
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I’ll post a solution to 6 in a bit
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dinglebells
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(Original post by Aditya33)
Here's my attempt at the first one
isnt beta = gamma^-1 clear from the fact that if k is a root then k^-1 is a root? why is all that analysis necessary?
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gnsbrg
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(Original post by Aditya33)
Same... I read it as f is said to be rather than show.... Hoping not more than 5 marks !!
Actually I didn’t know what a rotational symmetry of order 2 was so I was like oh so nice of them lol
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IrrationalRoot
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STEP II, Question 6

Before doing anything, it should be immediately obvious that this is a number theory problem, hence divisibility arguments are the go-to method here.

(i) Note that for n \geq 5, we have 5|n!+5 and n!+5&gt;5, so n!+5 is not prime. So we only need to check the primality of n!+5 for n=1,2,3,4. We see that the solutions are (n,p)=(2,7),(3,11),(4,29).

(ii) Remembering that this is a STEP question, the first part is almost certainly trying to give us a hint how to solve the second part; in particular we seek to again rule out sufficiently large values of n.

In fact we can rule out all n \geq 7. In the given equation, the first theorem tells us that m!&gt;(4n)!, that is, m&gt;4n, so from the second theorem we know there exists a prime p between 2n and m. Then we have p|m!, so p|1!\times 3! \times \ldots \times (2n-1)!, and since p is prime it follows that p divides one at least of the factors in one of these factorials, a contradiction since they are all less than 2n and p&gt;2n.

So we only need to consider n=1,2, \ldots, 6. Note that n=1,2,3 immediately give solutions (n,m)=(1,1),(2,3),(3,6).

For n=4, 1! \times 3! \times 5! \times 7!=m!, we note that 7&lt;m&lt;11 (since 7,11 are primes, so 7 divides the LHS but 11 does not) and also the left has two factors of 5, forcing m=10. Indeed, with m=10 both sides are 720 after dividing out 7!, so we get the solution (n,m)=(4,10).

For n=5, 1! \times 3! \times \ldots \times 9!=m!, we note that again m&lt;11. But now the LHS has three factors of 5 and the RHS at most two, so there is no solution in this case.

Finally, for n=6, 1! \times 3! \times \ldots \times 11!=m!, we note that 11&lt;m&lt;13, hence m=12, which obviously gives a contradiction upon dividing by 11!, so there is no solution here either.

In conclusion, the only solutions are (n,m)=(1,1),(2,3),(3,6), (4,10).


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raergadgaerg
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(Original post by Aditya33)
Here's my attempt at the first one
I think for b=2(a-1) the double root is -1 so it factorises to (x+1)^2(x^2 + (a-2)x +1) and for b = -2(a+1) the double root is +1 so it factorises to (x-1)^2(x^2 + (a+2)x +1) which then leads to (a-2)^2>4 and (a+2)^2>4 respectively.
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JeremyC
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Question 3


3(i)

Spoiler:
Show

Differentiate using the quotient rule to get
f'(x) = -\frac{\sec^2{x}}{(1+\tan{x})^2}
=- \frac{1}{\cos^2{x}} \frac{1}{1 + 2\tan{x} + \tan^2{x}}
=-\frac{1}{\cos^2{x}+2\sin{x}\cos{  x}+\sin^2{x}}
=-\frac{1}{1+\sin{2x}} as required
where in the last line we used \cos^2{x}+\sin^2{x}=1 and \sin{2x} = 2\sin{x}\cos{x}

Since -1 \leq \sin{2x} \leq 1
0 \leq 1 + \sin{2x} \leq 2
and so \frac{1}{1+\sin{2x}} \geq \frac{1}{2}
and hence we find the range of f'(x) \leq -\frac{1}{2}

Graph of f(x): https://www.desmos.com/calculator/kyjynulcgf

3(ii)
Spoiler:
Show

I drew a little sketch to show what we need to prove, plotting the points (x, g(x)), (a, b), (2a-x, g(2a-x)) with rotational symmetry of order 2 about (a,b).

(Note that x is a-x to the left of a, and 2a-x is a-x to the right of a).

Rotational symmetry of order 2 about (a,b)
\Longleftrightarrow g(2a-x) - b = b - g(x) (clear from drawing a sketch)
\Longleftrightarrow g(x) + g(2a-x)  = 2b

When there is rotational symmetry of order 2 about the origin, we can write down the value of the integral \int_{-1}^1 g(x) dx = 0

(If they wanted us to actually prove it, when there is rotational symmetry of order 2 about the origin, the equation involving g(x) becomes g(x) + g(-x) = 0 so we have
\int_{-1}^1 g(x) dx = \int_{-1}^0 g(x) dx + \int_0^1 g(x) dx
=\int_0^1 g(-x) dx + \int_0^1 g(x) dx
= \int_0^1 g(-x) + g(x) dx
= \int_0^1 0 dx
= 0)

3(iii)
Spoiler:
Show

For the final part we look to our sketch in 3(i) to get an idea of what we should guess for (a,b), the point around which we have rotational symmetry of order 2.

To me it looked like the rotational symmetry of order 2 was about (\frac{\pi}{4}, f(\frac{\pi}{4})) = (\frac{\pi}{4}, \frac{1}{2}).
To check this, we need to use our result from part (ii).
f(x) + f(2 \times  \frac{\pi}{4} -x)
=\frac{1}{1+\tan^k{x}} + \frac{1}{1+\tan^k{\frac{\pi}{2}-x}}
=\frac{1}{1+\tan^k{x}} + \frac{1}{1+\cot^k{x}}
=\frac{1}{1+\tan^k{x}} + \frac{\tan^k{x}}{\tan^k{x}+1}
=\frac{1+\tan^k{x}}{1+\tan^k{x}}
=1 = 2 \times \frac{1}{2} = 2b

Thus the equation  g(x) + g(2a-x) = 2b is satified, and so we have rotational symmetry of order 2 about (\frac{\pi}{4}, \frac{1}{2})

The limits of the final integral are \frac{\pi}{4} \pm \frac{\pi}{12}. As we have rotational symmetry of order 2 about (\frac{\pi}{4}, \frac{1}{2}), the area above and below the line y=\frac{1}{2} will cancel out, and so
\int_{\frac{1}{6}\pi}^{\frac{1}{  3}\pi} \frac{1}{1+\tan^k{x}}dx = (\frac{\pi}{3}-\frac{\pi}{6})\times \frac{1}{2} = \frac{\pi}{12}
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Gregorius
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STEP II, Question 13

Perhaps I'm suffering mid-afternoon dopiness, but this question looks like a giveaway. Simply write down the recurrence relations:

A(n+1) = (1/2)A(n) + (1/4)B(n) + (1/4)D(n)
B(n+1) = (1/4)A(n) + (1/2)B(n) + (1/4)C(n)
C(n+1) = (1/4)B(n) + (1/2)C(n) + (1/4)D(n)
D(n+1) = (1/4)A(n) + (1/4)C(n) + (1/2)D(n)

Notice that A(0) = 1, B(0) = C(0) = D(0) = 0.

Then B(n+1) + D(n+1) = 1/2 and therefore B(n) = D(n) = 1/4 by symmetry with the starting condition

and then C(n) = (1/4)(B(n)+D(n)) + (1/2) C(n) = (1/8) + (1/2)C(n-1)

which is a nice simple recurrence relation with solution C(n) = (1/2^(n+2))(2^n - 1)
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raergadgaerg
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Hey can someone predict what you think I will get in the paper, here is what I completed:

Q1: Full solution so should be 18-20
Q2: Full solution but I used u=x1 and v = w = x2 in (i) so I'm not sure if it is valid or not
Q4: Did (i) and (ii) but couldn't get far with (iii)
Q5: Wrote down the expansion of 1/(1+x) and e^(-ax) and that's it
Q6: Did part (i) then gave up
Q8: Solved both the DEs but didn't realise the condition alpha = beta in the graph sketch so probably won't get any marks there
Q9: Couldn't even find H so probably 1 or 2 method marks

I'm hoping for around 70+ but from what I can tell it will be a bit more like 65.
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Aditya33
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How
(Original post by JeremyC)
Question 3


3(i)

Spoiler:
Show

Differentiate using the quotient rule to get
f'(x) = -\frac{\sec^2{x}}{(1+\tan{x})^2}
=- \frac{1}{\cos^2{x}} \frac{1}{1 + 2\tan{x} + \tan^2{x}}
=-\frac{1}{\cos^2{x}+2\sin{x}\cos{  x}+\sin^2{x}}
=-\frac{1}{1+\sin{2x}} as required
where in the last line we used \cos^2{x}+\sin^2{x}=1 and \sin{2x} = 2\sin{x}\cos{x}

Since -1 \leq \sin{2x} \leq 1
0 \leq 1 + \sin{2x} \leq 2
and so \frac{1}{1+\sin{2x}} \geq \frac{1}{2}
and hence we find the range of f'(x) \leq -\frac{1}{2}

Graph of f(x): https://www.desmos.com/calculator/kyjynulcgf

3(ii)
Spoiler:
Show

I drew a little sketch to show what we need to prove, plotting the points (x, g(x)), (a, b), (2a-x, g(2a-x)) with rotational symmetry of order 2 about (a,b).

(Note that x is a-x to the left of a, and 2a-x is a-x to the right of a).

Rotational symmetry of order 2 about (a,b)
\Longleftrightarrow g(2a-x) - b = b - g(x) (clear from drawing a sketch)
\Longleftrightarrow g(x) + g(2a-x)  = 2b

When there is rotational symmetry of order 2 about the origin, we can write down the value of the integral \int_{-1}^1 g(x) dx = 0

(If they wanted us to actually prove it, when there is rotational symmetry of order 2 about the origin, the equation involving g(x) becomes g(x) + g(-x) = 0 so we have
\int_{-1}^1 g(x) dx = \int_{-1}^0 g(x) dx + \int_0^1 g(x) dx
=\int_0^1 g(-x) dx + \int_0^1 g(x) dx
= \int_0^1 g(-x) + g(x) dx
= \int_0^1 0 dx
= 0)

3(iii)
Spoiler:
Show

For the final part we look to our sketch in 3(i) to get an idea of what we should guess for (a,b), the point around which we have rotational symmetry of order 2.

To me it looked like the rotational symmetry of order 2 was about (\frac{\pi}{4}, f(\frac{\pi}{4})) = (\frac{\pi}{4}, \frac{1}{2}).
To check this, we need to use our result from part (ii).
f(x) + f(2 \times  \frac{\pi}{4} -x)
=\frac{1}{1+\tan^k{x}} + \frac{1}{1+\tan^k{\frac{\pi}{2}-x}}
=\frac{1}{1+\tan^k{x}} + \frac{1}{1+\cot^k{x}}
=\frac{1}{1+\tan^k{x}} + \frac{\tan^k{x}}{\tan^k{x}+1}
=\frac{1+\tan^k{x}}{1+\tan^k{x}}
=1 = 2 \times \frac{1}{2} = 2b

Thus the equation  g(x) + g(2a-x) = 2b is satified, and so we have rotational symmetry of order 2 about (\frac{\pi}{4}, \frac{1}{2})

The limits of the final integral are \frac{\pi}{4} \pm \frac{\pi}{12}. As we have rotational symmetry of order 2 about (\frac{\pi}{4}, \frac{1}{2}), the area above and below the line y=\frac{1}{2} will cancel out, and so
\int_{\frac{1}{6}\pi}^{\frac{1}{  3}\pi} \frac{1}{1+\tan^k{x}}dx = (\frac{\pi}{3}-\frac{\pi}{6})\times \frac{1}{2} = \frac{\pi}{12}
If you were to guess , how much would the part where you have to establish necessary and sufficient conditions for the function to have rotational symmetry of order 2 be worth ? Surely not more than 5 marks , right ?
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Nebrox
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Step II Q.13

I don't know enough Latex to write this up properly, so if anyone wants to do so, feel free.

i) After the whistle is blown once, since we're given that the parcel starts with child A, applying the given probabilities:

A(1)=1/2
B(1)=1/4
C(1)=0
D(1)=1/4

We can also see that:

A(2) = 1/2A(1) + 1/4B(1) + 1/4D(1) = 1/4 + 1/16 + 1/16 = 3/8
B(2) = 1/4A(1) + 1/2B(1) + 1/4C(1) = 1/8 + 1/8 + 0 = 1/4
C(2) = 1/4B(1) + 1/2C(1) + 1/4D(1) = 1/16 + 0 + 1/16 = 1/8
D(2) = 1/4A(1) + 1/4C(1) + 1/2D(1) = 1/8 + 0 + 1/8 = 1/4

ii) We can get the following general formulae for B(n+1) and D(n+1):

B(n+1) = 1/4A(n) + 1/2B(n) + 1/4C(n)
D(n+1) = 1/4A(n) + 1/4C(n) + 1/2D(n)

And so:

B(n+1) + D(n+1) = 1/4A(n) + 1/2B(n) + 1/4C(n) + 1/4A(n) + 1/4C(n) + 1/2D(n) = 1/2(A(n) + B(n) + C(n) + D(n)) = 1/2 (since A(n) + B(n) + C(n) + D(n) = 1 for all n)

Given the parcel started at A, and B and D lie either side of A, by symmetry we see that B(n) = D(n) and so:

B(n) = D(n) = 1/4

Also note similar recurrence relations for A and C:

A(n+1) = 1/2A(n) + 1/4B(n) + 1/4D(n) = 1/2A(n) + 1/8
C(n+1) = 1/4B(n) + 1/2C(n) + 1/4D(n) = 1/2C(n) + 1/8

Which are standard recurrence relations, which can be solved to get:

A(n) = 1/4 + (1/2)^n+1
C(n) = 1/4 - (1/2)^n+1
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Aditya33
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(Original post by raergadgaerg)
I think for b=2(a-1) the double root is -1 so it factorises to (x+1)^2(x^2 + (a-2)x +1) and for b = -2(a+1) the double root is +1 so it factorises to (x-1)^2(x^2 + (a+2)x +1) which then leads to (a-2)^2>4 and (a+2)^2>4 respectively.
Quite right. I will correct it soon. Apologies for the same.
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Cai Jianlin
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My answer for question 8, it is a bit different
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Cai Jianlin
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Name:  A8BD5635-16DB-4434-968A-028EA7324202.jpg.jpeg
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Size:  25.5 KBName:  BE2CAE0A-34D0-4A58-AF4E-8748F3C520DE.jpg.jpeg
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Size:  26.4 KB it is rather easy to find the equations of t. However, I failed to do the last cuz I wrongly reflected the graph along x- axis, not sure how many marks I can have
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Cai Jianlin
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Lol I used a very silly method to do it but I could do it correctly
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