STEP maths I, II, III 1992 solutions

*bobo*

for some reason my computer won't let me access paper III anymore, says its damaged? :confused:

It could be your cache that's got damaged. Try clearing it then opening the paper again.

Reply 83

*bobo*

STEP II
11)
if the purse scrapes the edge of the cliff, it reaches its max point at 0.5d from edge of cliff.
vertical movement:
u=VsinA, v=0, a=-g
=> s= (VsinA)^2/(2g)

horizontl movement after max point:
v=VcosA, s=0.5d => t=d/(2VcosA)
vertical movement after max point:
s=((VsinA)^2/(2g))-h, u=0, a=g, t= d/(2VcosA)

((VsinA)^2/(2g))-h= 0.5g(d/(2VcosA))^2

=> 8V^4sin^4A- 8V^2(2gh+V^2)sin^2A + 2g(gd^2 + 8V^2h)=0
sin^2A= (8V^2(2gh+V^2) +/- (64V^4(V^4+4gh+4g^2h^2)-64gV^4(gd^2+8V^2h))^0.5)/16V^4

for the purse to be able to be thrown in the situations given, sinA must exist so:
(V^4 +4ghV^2 +4g^2h^2)>= g^2d^2 + 8ghV^2

(V^4- 4ghV^2 +4g^2h^2)>=(gd)^2
(V^2-2gh)>=gd
V^2>= g(2h +d)

Reply 84

STEP I Question 7(This question was partially completed on post #4)

g(x) = ax + b

g(0) = b

g(1) = a+b

g(n) = an+bn

ng(1) = an + bn

g(n) = ng(1) - bn + b

g(n) = ng(1) -n(g(0)) + g(0)

if g(1) and g(0) for any given value of n, g(n) is an integer.

f(x) = Ax^2 + Bx + C

f(-1) = A - B + C

f(0) = C

f(1) = A + B + C

f(n) = An^2 + Bn + C

f(n) = \frac{1}{2}n^2(f(-1) + f(1) - 2f(0)) + \frac{1}{2}n(f(1)-f(-1)) + f(0)

Therefore because f(-1),f(0) and f(1) are integers

Now for the bit that harder

f(\alpha) = A\alpha^2 + B\alpha + C

f(\alpha - 1) = A\alpha^2 - 2A\alpha + A + B\alpha - B + C

f(\alpha + 1) = A\alpha^2 + 2A\alpha + A + B\alpha + B + C

f(\alpha + n) = A\alpha^2 - 2A\alpha n + n^2 + Ba + Bn + C

Take the first term Aalpha^2

f(\alpha) - f(\alpha+1) + [f(\alpha) - f(\alpha-1)] = A

A = \frac{1}{2}[2f(\alpha) - f(\alpha+1) - f(\alpha-1)]

A\alpha^2 = \frac{1}{2}\alpha^2(2f(\alpha) - f(\alpha+1) - f(\alpha-1))

next term -2A\alpha n

-2A\alpha n = -\alpha n[2f(\alpha) - f(\alpha+1) - f(\alpha-1)]

Next two easier terms Ba + C

B\alpha + C = f(\alpha) - A\alpha^2 = f(\alpha) - \frac{1}{2}\alpha^2[2f(\alpha) - f(\alpha+1) - f(\alpha-1)]

Next term is Bn

Bn = \frac{1}{2}n[ (-f(\alpha+1) - f(\alpha-1)) ] - 2\alpha ( 2f(\alpha) - f(\alpha +1) - f(\alpha -1) )]

So!

f(\alpha + n) = A\alpha^2 - 2A\alpha n + n^2 + B\alpha + Bn + C

= \frac{1}{2}\alpha^2(2f(\alpha) - f(\alpha+1) - f(\alpha-1)) -\alpha n[2f(\alpha) - f(\alpha+1) - f(\alpha-1)] + n^2 + f(\alpha) - \frac{1}{2}\alpha^2[2f(\alpha) - f(\alpha+1) - f(\alpha-1)] + \frac{1}{2}n[ (-f(\alpha+1) - f(\alpha-1)) ] - 2\alpha ( 2f(\alpha) - f(\alpha +1) - f(\alpha -1) )]

As \alpha and all the functions involving alpha are integers therefore f(\alpha + n) must be an integer.

Id tidy that up too but i really cant be arsed at the moment.

What a bitch that was. I bet theres a much more elegant way. Ive most definitely made an error in that but the method is there. You just have to find the function of f(\alpha + n) in terms of functions that are known integers.

Reply 85

STEP I Question 8

Calling F(t) the integrated function of f(t)

Unparseable latex formula:

\frac{d}{dx} \int_a^x f(t) \hspace5 dt = \frac{d}{dx}[F(t) ]_a^x

= \frac{d}{dx} [ F(x) - F(a) ]

= \frac{d}{dx}(F(x)) = f(x)

(thanks to Billy (GE) for the cleaning up the tizwozz i got into about the third and last part)
Consider

Unparseable latex formula:

f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + \text{const}

.
You can just differentiate both sides from here and get

f'(x) = f(x) \Rightarrow f(x) = Ae^x

and plug this into the integrals in the second and third part:

Unparseable latex formula:

f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + 1

Unparseable latex formula:

Ae^x = \displaystyle \int_0^x Ae^t \hspace5 dt + 1

Ae^x = Ae^x - A + 1 \Rightarrow A = 1

Unparseable latex formula:

f(x) = \displaystyle \int_0^x f(t) \hspace5 dt

Unparseable latex formula:

Ae^x = \displaystyle \int_0^x Ae^t \hspace5 dt

Ae^x = Ae^x - A \Rightarrow A=0

Just attempting the final and 4th part now.

Reply 86

insparato

STEP I Question 8

Unparseable latex formula:

\frac{d}{dx} \displaystyle \int_a^x f(t) \hspace5 dt = \frac{d}{dx}[ \int f(t) ]_a^x

= \frac{d}{dx} [\int f(x) - \int f(a) ]

= \frac{d}{dx}(\int f(x)) = f(x)

Unparseable latex formula:

f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + 1

Consider

Unparseable latex formula:

f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + \text{const}

.
You can just differentiate both sides from here and get

f'(x) = f(x) \Rightarrow f(x) = Ae^x

and plug this into the integrals in the second and third part:

Unparseable latex formula:

f(x) = \displaystyle \int_0^x f(t) \hspace5 dt + 1

Unparseable latex formula:

Ae^x = \displaystyle \int_0^x Ae^t \hspace5 dt + 1

Ae^x = Ae^x - A + 1 \Rightarrow A = 1

Unparseable latex formula:

f(x) = \displaystyle \int_0^x f(t) \hspace5 dt

Unparseable latex formula:

Ae^x = \displaystyle \int_0^x Ae^t \hspace5 dt

Ae^x = Ae^x - A \Rightarrow A=0

Reply 87

I was mucking about with this yesterday but i must of got myself in a twist becaues i did get A = 1, or something of this nature but didnt know where it led to.

Yes that seems reasonable.

Argh i just realised i had a stupid idea that f(x) was on both sides on third equation which obviously doesnt make sense.

Reply 88

insparato

What a bitch that was. I bet theres a much more elegant way.There is. (I posted it earlier, actually).

Put

g(x) = f(x+\alpha)

. Then g is a quadratic, and g(-1),g(0) and g(1) are all integers. So g(n) is an integer for all n. Therefore so is

f(n+\alpha)

Reply 89

Nevermind, i thought nobody had finished that last part off as it was up as incomplete.

Reply 90

To my knowledge no one had completed it. Which post did you complete it in, David?

Reply 91

generalebriety

To my knowledge no one had completed it. Which post did you complete it in, David?

I gave a (spoilered) hint in post #40. Looking back at it, it's not actually right (sign of alpha is wrong), but it should have been enough I think.

Reply 92

Swayum

insparato

f(x) = Ax^2 + Bx + C

f(-1) = A - B + C

f(0) = C

f(1) = A + B + C

f(n) = An^2 + Bn + C

f(n) = n^2(f(-1) + f(1) - 2f(0)) + \frac{1}{2}n(f(1)-f(-1)) + f(0)

Just a slight pedantic note about A:

f(-1) + f(1) = A - B + C + A + B + C = 2A + 2C
f(-1) + f(1) - 2f(0) = 2A + 2C - 2C = 2A

So what you've actually got there is 2An^2 not An^2. You'll have to divide by 2. I'm just saying this in case in a few years time someone sees your solution and thinks their one is wrong.

Reply 93

Ooh, ok, so no one has actually completed it yet? :p:

Reply 94

generalebriety

Ooh, ok, so no one has actually completed it yet? :p:

What, you mean other than insparato's solution about 10 posts earlier and my shorter solution for the end bit? (Or are we talking about different questions? For clarity, I am talking about STEP I, Q7).

Reply 95

DFranklin

As I understood, Aaron's solution had a mistake (pointed out by DeathAwaitsU) and yours was just a hint. I don't know though, you tell me.

Reply 96

Swayum

It's solved. I was just pointing out a minor mistake and David was explaining a more elegant solution. Regardless, Insparato finished off the bit I didn't do so the question is complete (assuming his solution was correct - I don't have the strength right now to read it).

Reply 97

Ok, well, I'll leave the link up there and Aaron is free to correct his minor mistake if he wants. I shall add a link to David's post and everyone can then fight it out amongst themselves. :wink:

Reply 98

generalebriety

Ok, well, I'll leave the link up there and Aaron is free to correct his minor mistake if he wants. I shall add a link to David's post and everyone can then fight it out amongst themselves. :wink:

Yet another solution:

g(n) = g(0)+n(g(1)-g(0)), so g(n) is always an integer.

Now set

g(n) = f(n)-f(n-1) = A(2n-1)+B

. g(0) and g(1) are both integers and so g(n) is always an integer. Since f(0) is an integer it follows that f(n) is always an integer. (This is obvious, but if you want to prove it, suppose not, and let n be the integer of smallest modulus such that f(n) is not an integer. We know f(0) is an integer, so n =/= 0. If n > 0, we have f(n-1) is an integer, and so f(n) = f(n-1)+g(n) is an integer. If n < 0 we have f(n+1) is an integer and so f(n) = f(n+1)-g(n+1) is an integer).

Finally, consider

h(x) = f(x+\alpha)

. Since h is a quadratic and h(-1),h(0),h(1) are all integers, h is an integer for all n. Therefore

f(x+\alpha)

is an integer for all n.

Reply 99