# STEP 2006 Solutions Thread

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Physics Enemy

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#61

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#61

Hey guys, I'm having a brain freeze here.

In STEP 1 Q11 Part 3, in the markscheme they say p < q < r at the start, where p, q and r are the velocities of A_(n-2), A_(n-1) and A_n.

The way it's stated so casually at the start, assumes it's something obvious. I don't get why though, and it's frustrating, as this assumption at the start will prove the contradiction needed.

DFranklin in particular, any help?

In STEP 1 Q11 Part 3, in the markscheme they say p < q < r at the start, where p, q and r are the velocities of A_(n-2), A_(n-1) and A_n.

The way it's stated so casually at the start, assumes it's something obvious. I don't get why though, and it's frustrating, as this assumption at the start will prove the contradiction needed.

DFranklin in particular, any help?

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DFranklin

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#62

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#62

I can easily justify p <= q <= r (if p > q, then obviously there will be a subsequent collision between A_n-2 and A_n-1, contradicting the assumption that all collisions have taken place, etc.)

It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).

It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).

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Physics Enemy

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#63

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#63

(Original post by

I can easily justify p <= q <= r (if p > q, then obviously there will be a subsequent collision between A_n-2 and A_n-1, contradicting the assumption that all collisions have taken place, etc.)

**DFranklin**)I can easily justify p <= q <= r (if p > q, then obviously there will be a subsequent collision between A_n-2 and A_n-1, contradicting the assumption that all collisions have taken place, etc.)

(Original post by

It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).

**The God**)It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).

Thanks so much mate,

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Farhan.Hanif93

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#64

Farhan.Hanif93

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#65

kiddey

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#66

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#66

I 2006 q1

hence

satisfies

if , then

so ,

satisfies

where

as and are prime

or

so

hence other value of n+m is , and

hence

satisfies

if , then

so ,

satisfies

where

as and are prime

or

so

hence other value of n+m is , and

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Dirac Spinor

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#67

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#67

STEP I Q14

Whilst this has been discussed earlier I haven't seen anyone post the actual solution so I'll post mine.

i) P(choosing red on rth go)=

(usually I would draw a tree diagram but here it just isn't necessary)

To find the maximum we should differentiate and equate the derivative to zero.

let p=P(choosing red on rth go):

take logs of both sides:

Now differentiate both sides:

To find stationary points, let

. Since f(n) can not be 0:

Rearranging we see that:

and so:

ii) A trick in these type of questions where the denominator and the numerator of the probabilities go down by one every time is to notice that the k+1th denominator is canceled by the kth numerator. Therefore all we care about is the first denominator and the last numerator as they are the only ones which don't cancel.

P(choosing red on rth go)=

so n should be the smallest it can be.

n+1 is greater than or equal to r so the smallest value of n is:

n=r-1

Whilst this has been discussed earlier I haven't seen anyone post the actual solution so I'll post mine.

i) P(choosing red on rth go)=

(usually I would draw a tree diagram but here it just isn't necessary)

To find the maximum we should differentiate and equate the derivative to zero.

let p=P(choosing red on rth go):

take logs of both sides:

Now differentiate both sides:

To find stationary points, let

. Since f(n) can not be 0:

Rearranging we see that:

and so:

ii) A trick in these type of questions where the denominator and the numerator of the probabilities go down by one every time is to notice that the k+1th denominator is canceled by the kth numerator. Therefore all we care about is the first denominator and the last numerator as they are the only ones which don't cancel.

P(choosing red on rth go)=

so n should be the smallest it can be.

n+1 is greater than or equal to r so the smallest value of n is:

n=r-1

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Farhan.Hanif93

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#68

brianeverit

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#69

brianeverit

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#70

brianeverit

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#71

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#72

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#73

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#76

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#78

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#79

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#79

!Phew Thank goodness that's done. I wouldn't be at all surprised if there are some errors in it. Please let me know if you spot any.

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#80

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