The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

This discussion is now closed.

Check out other Related discussions

STEP I, II, III 2000 solutions

Scroll to see replies

I believe STEP III is the only paper specifically for the FM syllabus.

As to which is easier, everyone agrees STEP III is hardest by some way. Most people reckon the AEA is also easier than STEP II and STEP I. My feeling is that the difference isn't great, and the biggest factor is probably which exam format you get on with best. There's a fair degree of personal perference: some will find AEA much easier than STEP, others will find STEP easier.
Thanks Franklin, do you think doing STEP before going to uni would improve/prepare me better for maths at uni? I only did a levels maths, but doing maths in uni this fall..

also, whats the difference between STEP, A level and University maths like?

cheers!
Reply 42
Avatar for Zhen Lin
Zhen Lin
12
johnheyes91
Zhen just curious, how do you type maths and transfer it to a pdf? Cheers!


LaTeX. The other thread probably explains better than I could.
ah, thanks!
Reply 44
Avatar for speedy_s
speedy_s
7
DFranklin
Re: Q8. When it comes to solving the recurrence relation, I think you've approached it the wrong way around. Essentially, you've derived a formula for the solution, but you haven't shown it's the only possible solution.

In fact, to answer the question you don't need to derive it. Consider

cn=α2n−1+α−(2n−1)5\displaystyle c_n = \frac{\alpha^{2n-1} + \alpha^{-(2n-1)}}{\sqrt 5}

All you need is to show c0=c1=1c_0 = c_1 = 1, and cn=3cn−1−cn−2c_n = 3c_{n-1}-c_{n-2}. Then write bn=cn−anb_n = c_n - a_n; you have b0=b1=0b_0 = b_1 = 0 and bn=3bn−1−bn−2b_n = 3b_{n-1}-b_{n-2}, so trivially b_n = 0 for all n by induction.


Sorry, but why is bn=3bn−1−bn−2b_n = 3b_{n-1}-b_{n-2}?
Reply 45
Avatar for Square
Square
13
If someone could take a look at my solution to 2000/III Q6 I would be thankful!

Spoiler

speedy_s
Sorry, but why is bn=3bn−1−bn−2b_n = 3b_{n-1}-b_{n-2}?
Because an=3an−1−an−2a_n = 3a_{n-1}-a_{n-2} and cn=3cn−1−cn−2c_n = 3c_{n-1}-c_{n-2} and bn=cn−anb_n = c_n - a_n (for all n).
Square: I'm pretty sure you've misunderstood a (small) part of the question.

When it says "explain why this equation always has a non-negative root", I think you're supposed to be looking at the equation by itself, without assuming that you know about a^2 being a root. In which case:

Spoiler



I don't think this could cost you many marks, and I'm not even sure I'm right in my interpretation here.

For the rest of it, it's very hard to comment when you've included so little working; assuming your working is right, you should be OK.

For what it's worth, this really doesn't feel like it should be a "3 sides of A4" question, but that's easy for me to say.
Reply 48
Avatar for Square
Square
13
DFranklin
Square: I'm pretty sure you've misunderstood a (small) part of the question.

When it says "explain why this equation always has a non-negative root", I think you're supposed to be looking at the equation by itself, without assuming that you know about a^2 being a root. In which case:

Spoiler



I don't think this could cost you many marks, and I'm not even sure I'm right in my interpretation here.

For the rest of it, it's very hard to comment when you've included so little working; assuming your working is right, you should be OK.

For what it's worth, this really doesn't feel like it should be a "3 sides of A4" question, but that's easy for me to say.


Aha, I see where you're getting at there.

I do have quite large writing (especially when it comes to doing maths) so maybe slightly less with averaged sized writing.

I can put in some more working if you like but some of it was quite long winded and tedious and I really could not be bothered to LaTeX it.

Maybe you have some better algebra skills than I do (which wouldn't be very hard!)
I / 7

If f(x)=ax−x31+x2 \displaystyle f(x) = ax - \frac{x^3}{1 + x^2} , show that f′(x)≥0 \displaystyle f'(x) \ge 0 ∀x \displaystyle\forall x iff
Unparseable latex formula:

\displaystyle a \ge \frac{9}{8}}



Note that f(x) is a quotient and an inequality is involved. While tempting to use something such as the product rule, by using the quotient rule the denominator will always be > 0.

f′(x)=a−3x2(1+x2)−2x(x3)(1+x2)2=a+x4−3x2(1+x2)2=(a+1)x4+(2a−3)x2+a(1+x2)2 \displaystyle f'(x) = a - \frac{3x^2(1+x^2) - 2x(x^3)}{(1 + x^2)^2} = a + \frac{x^4 - 3x^2}{(1+x^2)^2} = \frac{(a+1)x^4 + (2a-3)x^2 + a}{(1+x^2)^2}

(a+1)x4+(2a−3)x2+a≥0⇒B2≤4AC \displaystyle (a+1)x^4 + (2a-3)x^2 + a \ge 0 \Rightarrow B^2 \le 4AC

(2a−3)2≤4a(a−1)⇒a≥98 \displaystyle (2a - 3)^2 \le 4a(a-1) \Rightarrow a \ge \frac{9}{8}
II / 1

Guess a solution to 1N=1a+1b \displaystyle \frac{1}{N} = \frac{1}{a} + \frac{1}{b} you are provided with 12=13+16 \displaystyle \frac{1}{2} = \frac{1}{3} + \frac{1}{6} and also 13=14+112 \displaystyle \frac{1}{3} = \frac{1}{4} + \frac{1}{12}

The solution is clear - 1N=1N+1+1N2+N \displaystyle \frac{1}{N} = \frac{1}{N + 1} + \frac{1}{N^2 + N}

To prove this is so; 1N≡1N+1+1N2+N=N+1N(N+1)=1N \displaystyle \frac{1}{N} \equiv \frac{1}{N + 1} + \frac{1}{N^2 + N} = \frac{N+1}{N(N+1)} = \frac{1}{N}

Looking at (a−N)(b−N)=N2 \displaystyle (a-N)(b-N) = N^2 for N is a prime it is apparent that the only factors of N^2 are N2=N×N=N2×1 \displaystyle N^2 = N \times N = N^2 \times 1

This would imply that if distinct natural numbers (a,b) are sought for the equation 1N=1a+1b \frac{1}{N} = \frac{1}{a} + \frac{1}{b} for N prime then there is only one pair of solutions.

wip for 2N \displaystyle \frac{2}{N}
II / 2

p[x]=(x−a)2q[x]⇒p′[x]=(x−a)(2q[x]+(x−a)q′[x]) \displaystyle p[x] = (x-a)^2q[x] \Rightarrow p'[x] = (x-a)(2q[x] + (x-a)q'[x])

g[x]=(x−a)4q[x]⇒p′[x]=(x−a)3(4q[x]+(x−a)q′[x])∗ \displaystyle g[x] = (x-a)^4q[x] \Rightarrow p'[x] = (x-a)^3(4q[x] + (x-a)q'[x]) *

Given p(x) we differentiate to find;

The roots of the equation (x−α)3(4q[x]+(x−α)q′[x])=6x5+20x4−20x3−120x2−80x+32 (x-\alpha)^3(4q[x] + (x-\alpha)q'[x]) = 6x^5 + 20x^4 - 20x^3 - 120x^2 - 80x + 32

Inspection reults in α=−2 \displaystyle \alpha = -2 So if p[x] denotes the polynomial given then p[-2] = 0. Using the factor and remainder theorem; k = 48
II / 4

eiθ=cos(θ)+isin(θ)⇒eiθeiα=ei(θ+α)=cos(θ+α)+isin(θ+α) \displaystyle e^{i\theta} = cos(\theta) + isin(\theta) \Rightarrow e^{i\theta}e^{i\alpha} = e^{i(\theta + \alpha)} = cos(\theta + \alpha) + isin(\theta + \alpha)

(eiθ)n=eniθ=cos(nθ)+isin(nθ) \displaystyle (e^{i\theta})^n = e^{ni\theta} = cos(n\theta) + isin(n\theta)

Z1=(5−i)2(1+i)=34+14i \displaystyle Z_1 = (5-i)^2(1+i) = 34 + 14i

Z2=(5−i) \displaystyle Z_2 = (5-i)

Z3=(1+i) \displaystyle Z_3 = (1+i)

arg(Z1)=2arg(Z2)+arg(Z3)⇒arctan(717)=−2arctan(15)+arctan(1)⇒arctan(717)+2arctan(15)=π4 \displaystyle arg(Z_1) = 2arg(Z_2) + arg(Z_3) \Rightarrow arctan(\frac{7}{17}) = -2arctan(\frac{1}{5}) + arctan(1) \Rightarrow arctan(\frac{7}{17}) + 2arctan(\frac{1}{5}) = \frac{\pi}{4}

The second part is alot easier if you just have a hunch.

By sheer observation the complex number β=(1+i)(4−i)3(−20+i) \displaystyle \beta = (1+i)(4-i)^3(-20+i) shows the fact; 3arctan(14)+arctan(120)+arctan(11985)=π4 \displaystyle 3arctan(\frac{1}{4}) + arctan(\frac{1}{20}) + arctan(\frac{1}{1985}) = \frac{\pi}{4}
II / 8

−∫dy(y+3)12=−2∫−2xe−x2dx  ⟹  −2(y+3)12=−2e−x2+C  ⟹  y=e−2x2+2ke−x2+(k2−2) \displaystyle - \int \frac{dy}{(y+3)^\frac{1}{2}} = -2 \int -2xe^{-x^2} dx \implies -2(y+3)^{\frac{1}{2}} = -2e^{-x^2} + C \implies y = e^{-2x^2} + 2ke^{-x^2} + (k^2 - 2)

After subbing (0,6) ; (k+4)(k−2)=0 \displaystyle (k+4)(k-2) = 0

∴y=e−2x2−8e−x2+13 \displaystyle \therefore y = e^{-2x^2} - 8e^{-x^2} + 13 and case 2 y=e−2x2+4e−x2+1 \displaystyle y = e^{-2x^2} + 4e^{-x^2} + 1

There is a vlaue of k such that as x tends to ∞ \displaystyle \infty y tends to 1 \displaystyle 1

(ii) wip. Got
y2e−3x2 \displaystyle y^2e^{-3x^2} but the answer requires ye−3x2 \displaystyle ye^{-3x^2}
III/12:

P(3 or fewer | Holding a winning ticket) = P(3 or fewer AND holding a winning ticket)/P(holding a winning ticket).

3 or fewer and you win: you win and two others win; you win and one other wins; you win and no others win. You winning and other people winning is independent, so P(3 or fewer AND holding a winning ticket)/P(holding a winning ticket) = P(2 others or fewer win).

The others can be modelled as a binomial distribution, X ~ B(2n-1, 1/n)
P(X <= 2) = P(X = 2) + P(X = 1) + P(X = 0)
= (2n−12)1n2(1−1n)2n−3+(2n−11)1n(1−1n)2n−2+(1−1n)2n−1\dbinom{2n-1}{2}\dfrac{1}{n^2} \left(1 - \dfrac{1}{n}\right)^{2n-3} + \dbinom{2n-1}{1}\dfrac{1}{n} \left(1 - \dfrac{1}{n}\right)^{2n-2} + \left(1 - \dfrac{1}{n}\right)^{2n-1}

= (1−1n)2n−3[(2n−1)(2n−2)2!n2+(2n−1)n(1−1n)+(1−1n)2]\left(1 - \dfrac{1}{n}\right)^{2n-3}\left[\dfrac{(2n-1)(2n-2)}{2!n^2} + \dfrac{(2n-1)}{n}\left(1 - \dfrac{1}{n}\right) + \left(1 - \dfrac{1}{n}\right)^2 \right]

=(1−1n)2n−3[5−8n+3n2]=(1−1n)2n(5−8n+3n2)÷(1−1n)3=(1−1n)2n5n2−3nn2−2n+1=(1−1n)2n5−3/n1−2/n+1/n2= \left(1 - \dfrac{1}{n}\right)^{2n-3}\left[5 - \dfrac{8}{n} + \dfrac{3}{n^2}\right] = \left(1 - \dfrac{1}{n}\right)^{2n}(5 - \dfrac{8}{n} + \dfrac{3}{n^2}) \div \left(1 - \dfrac{1}{n}\right)^3 = \left(1 - \dfrac{1}{n}\right)^{2n} \dfrac{5n^2 - 3n}{n^2 - 2n + 1} = \left(1 - \dfrac{1}{n}\right)^{2n} \dfrac{5 - 3/n}{1 - 2/n + 1/n^2}

As n gets large, (1−1n)2n≈e−2 and 5−3/n1−2/n+1/n2≈5/1=5\left(1 - \dfrac{1}{n}\right)^{2n} \approx e^{-2} \text{ and } \dfrac{5 - 3/n}{1 - 2/n + 1/n^2} \approx 5/1 = 5 So the probability is approximately 5e−25e^{-2}.

5e−2≈5(1−2+42−86+1624−32120+64720−1285040)=4163≈235e^{-2} \approx 5(1 - 2 + \frac{4}{2} - \frac{8}{6} + \frac{16}{24} - \frac{32}{120} + \frac{64}{720} - \frac{128}{5040}) = \frac{41}{63} \approx \frac{2}{3}.

The last part is slightly dodgy, plus the arithmetic was rather tedious. I've been light on the algebra at several parts; none of it's particularly difficult.

- Human selections of numbers won't be uniform, and some will be more popular than others. So given that I have a winning ticket, it's likelier that more others will have a winning ticket so this probability will decrease. (Not sure if this is what they want or even correct.)

Consider the other people playing: X ~ B(2N - 1, 1/N) E(X) = "np" = 1/N*(2N-1) = 2 - N-1. So in total, given that you have a winning ticket, the expectation is 3 - N-1.
III/13:

If the first die to show a six does so on the rth roll then there'll (r-1) rolls where all dice fail to show a six = (qr−1)n=qrn−n(q^{r-1})^n = q^{rn - n}

On the rth roll: P(1 or more shows a six) = 1 - P(none shows a six) = 1 - q^n. Overall probability = qrn−n(1−qn)=qrn−n−qrn=qrn(q−n−1)q^{rn-n}(1 - q^n) = q^{rn - n} - q^{rn} = q^{rn}(q^{-n}-1), as required.

Let G(t) denote the probability generating function. r can take values 1, 2, 3, ... so G(t)=qn(q−n−1)t+q2n(q−n−1)t2+...\text{G}(t) = q^n(q^{-n}-1)t + q^{2n}(q^{-n} - 1)t^2 + ... Geometric series, a = (1−qn)t(1-q^n)t, r = qntq^nt. Sum to infinity = (1−qn)t1−qnt\dfrac{(1-q^n)t}{1-q^nt}.

Unparseable latex formula:

\text{G'}(t) = (1-q^n)\left[\dfrac{tq^n}{(1-q^nt}^2} + \dfrac{1}{1-q^nt} \right] \text{ and } E(R) = G(1) = \dfrac{q^n}{1-q^n} + 1


When n = 2, p = 1/6, q = 5/6 so E(R) = 25/361−25/36+1=2511+1=3611\dfrac{25/36}{1 - 25/36} + 1 = \frac{25}{11} + 1 = \frac{36}{11}.

P(S=r)=P(S≤r)−P(S≤r−1)P(S = r) = P(S \le r) - P(S \le r - 1).
Less than or equal to r for one die is 1 - q^r; not n failures. So for n-dice: (1−qr)n(1-q^r)^n, and similarly for r - 1: (1−qr−1)n(1-q^{r-1})^n. Overall probability: (1−qr)n−(1−qr−1)n(1-q^r)^n - (1-q^{r-1})^n

(1−qr)2−(1−qr−1)2=q2r−q2r−2+2qr−1−2qr(1-q^r)^2 - (1-q^{r-1})^2 = q^{2r} - q^{2r-2} + 2q^{r-1} - 2q^r r can take values 1, 2, 3, ...

=> G(t)=∑r=1∞(q2r−q2r−2+2qr−1−2qr)tr=∑r=1∞[trq2r−trq2r−2+2trqr−1−2qrtr]\displaystyle \text{G}(t) = \sum_{r=1}^{\infty}(q^{2r} - q^{2r-2} + 2q^{r-1} - 2q^r)t^r = \sum_{r=1}^{\infty} \left[ t^rq^{2r} - t^rq^{2r-2} + 2t^rq^{r-1} - 2q^rt^r \right]

=tq21−tq2−t1−tq2+2t1−tq−2qt1−qt=tq2−t1−tq2+2t−2qt1−qt= \dfrac{tq^2}{1-tq^2} - \dfrac{t}{1-tq^2} + \dfrac{2t}{1-tq} - \dfrac{2qt}{1-qt} = \dfrac{tq^2 - t}{1 - tq^2} + \dfrac{2t-2qt}{1-qt}.

G’(1)=E(S)=21−q=2p\text{G'}(1) = \text{E}(S) = \dfrac{2}{1-q} = \dfrac{2}{p} = 12 when p = 1/6. (I haven't LaTeXed the derivative of G(t), it's too late and it's just tedious.)
... many of the questions for 2000 are from 1987 and 1990 papers ...
:lolwut:
Reply 58
Avatar for SimonM
SimonM
OP
18
STEP I, Question 6

Not a proper solution I can't be bothered to upload sketches

Spoiler



Edit: Completed solution by etothepiiplusone

First part:

Spoiler



Sketch 1:

Spoiler



Sketch 2:

Spoiler



Final part:

Spoiler

(edited 5 years ago)
Reply 59
Avatar for SimonM
SimonM
OP
18
STEP I, Question 12

(i)

Spoiler


(ii)

Spoiler


(iii)

Spoiler

Related discussions

Latest

Trending

Trending

Articles for you