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8 years ago
#61
(Original post by SimonM)
...
STEP I Q14

solution
Note that , where is the PDF for the distribution of X for X>0.

Differentiating the above:

Hence, the PDF for X is given by:

(i)
Note that

(ii)
Note that

Integrating the second term by parts with :

, as required.

(iii)
Variance
Note that

Integrating the second term by parts with :

In part (ii), we had shown that , hence:

Median
Let be the median of the distribution of X, it follows that:

(By using the CDF given at the start)

(iv)

Using a substitution of for the integral:

Given that :

1
7 years ago
#62
STEP II Q7

Spoiler:
Show
i) Both P and Q perform circles in a counterclockwise direction. P about the k-axis, radius 1 in the i-j plane, Q about a different axis, radius 3

ii)

as required.

iii) Sketching this, the answer is clear.
1
7 years ago
#63
2005 STEP I question 13

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7 years ago
#64
2005 STEP II question 13

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7 years ago
#65
2005 STEP II question 14

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7 years ago
#66
2005 STEP III question 10

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7 years ago
#67

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7 years ago
#68

\times \dfrac{1}{1-\frac{w}{w+1})^2=w [/latex]

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6 years ago
#69
(Original post by sonofdot)
STEP II 2005 Question 8

[expand=First Part] for with y=1 when x=0

The RHS can be integrated by parts, with and

Substituting in x=0 and y=1 gives C=1/3, so we get as required.

For large x, we can say so

Looking at the first two terms of the expansion of gives, for large x:
Hi there how you work out the first few terms of the expansion (1+(3/x))^-1 ?
0
6 years ago
#70
(Original post by rath90)
Hi there how you work out the first few terms of the expansion (1+(3/x))^-1 ?
Standard binomial expansion
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5 years ago
#71
(Original post by Glutamic Acid)
II/4:

1st part

Substituting a^2 = bc - 1, LHS = , as required. Note, since a, b and c are positive, 1/(a+b), 1/(a+c) and 1/a will all lie between 0 and +infinity, so "implies and is implied by" implication signs can be used.

2nd part

Let p + q = "a", s = "b", t = "c" , from (*)

Let p + r = "a", u = b, v = c , from (*)

Let p = "a", b = "q" and r = "c", so , from (*), and we're done.

3rd part

We can use the result in the second part by choosing p = 7, q =1, s = 5, t = 13, r = 50, u = 25 and v = 130, and everything falls out neatly.

For part III does it matter which values you choose?
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5 years ago
#72
(Original post by Principia)
For part III does it matter which values you choose?
In order to use the previous result you must choose values so that p+q+s=13,p+q+r=21,p+r+u=87 ,p+r+v-187 and p=7
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5 years ago
#73
(Original post by brianeverit)
In order to use the previous result you must choose values so that p+q+s=13,p+q+r=21,p+r+u=87 ,p+r+v-187 and p=7
I did that, but can it be any values that meet those criteria? i.e. there are then an infinite amount of solutions.
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5 years ago
#74
(Original post by Principia)
I did that, but can it be any values that meet those criteria? i.e. there are then an infinite amount of solutions.
Yes, the solution is not unique but you must check that your values satisfy all the criteria.
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4 years ago
#75
(Original post by SimonM)
STEP III, Question 4

Spoiler:
Show

Assuming no zero terms (which they let us know)

Therefore we have the recurrence: in

(induction if that's what floats your boat...)

This proves what we want, and

Therefore

and

i) Geometric , .

ii) Periodic, period 2:

iii) Period, period 4:
Very nice approach by Simon. In (ii), probably need to exclude period =1, i.e. a=b and k=2 (c=1).
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4 years ago
#76
(Original post by SimonM)
STEP II, Question 1

Spoiler:
Show
Let

Therefore

Therefore the solutions are

We have

Therefore

Therefore

Let

Therefore comparing coefficients, we have:

Therefore

Therefore take
when comparing coefficients did you put this
Let

into

Therefore
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2 years ago
#77
An almost complete solution to STEP III Q 12:
1
1 week ago
#78
(Original post by nuodai)
Ah what the hell,
STEP I 2005 Question 1

Part (i)

For a 5-digit number to add up to 43 the digits must be either:
(1): (9, 9, 9, 9, 7) [ 5 ways of rearranging ]
(2): (9, 9, 9, 8, 8) [ 5!/3!2! = 10 ways of rearranging ]

5 + 10 = 15

Part (ii)

We can classify groupings of digits by how many 9s they have and, if no 9s. Also, since since 8*4 + 7 = 39 there can't be less than four 8s if there are no 9s.

FOUR NINES:
(9, 9, 9, 9, 3) [ 5 ways of rearranging]

THREE NINES:
(9, 9, 9, 8, 4) [ 5!/3! = 20 ways of rearranging ]
(9, 9, 9, 7, 5) [ 5!/3! = 20 ways of rearranging ]
(9, 9, 9, 6, 6) [ 5!/3!2! = 10 ways of rearranging ]

TWO NINES:
(9, 9, 8, 8, 5) [ 5!/2!2! = 30 ways of rearranging ]
(9, 9, 8, 7, 6) [ 5!/2! = 60 ways of rearranging ]
(9, 9, 7, 7, 7) [ 5!/2!3! = 10 ways of rearranging ]

ONE NINE:
(9, 8, 8, 8, 6) [5!/3! = 20 ways of rearranging ]
(9, 8, 8, 7, 7) [5!/2!2! = 30 ways of rearranging ]

NO NINES:
(8, 8, 8, 8, 7) [5 ways of rearranging ]

Doing some good ol' addition gives you 5 + 20 + 20 + 10 + 30 + 60 + 10 + 20 + 30 + 5 = 210 ways of rearranging the numbers. And no LaTeX required!
One of my students just came up with what I think is a really great way of solving part (b), which requires no splitting into cases, and in my opinion is more elegant than the board's published answer (and the answer already suggested on this thread):

Solution as follows:

The maximum sum is 45, by using 99999 (uniquely). To make 39, we must subtract a total of six from the digits of 99999, so use the symbol x to represent subtracting one from a digit, and / to represent moving on to the next digit in the number [e.g. x/x/x/x/xx would mean subtract one from the 1st, 2nd, 3rd, and 4th digits, and subtract two from the 5th digit; //xxx/x/xx would mean subtract nothing from the 1st and 2nd digits, then three from the 3rd digit, one from the 4th digit, and two from the 5th digit]. But now the problem just becomes how many ways are there of arranging 6 "x" symbols and 4 "/" symbols, i.e.10C4 = 210.
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