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# STEP 2005 Solutions Thread Watch

1. (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 :

2. 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.
3. 2005 STEP I question 13

4. 2005 STEP II question 13

5. 2005 STEP II question 14

Attached Images

6. 2005 STEP III question 10

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

8. (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 ?
9. (Original post by rath90)
Hi there how you work out the first few terms of the expansion (1+(3/x))^-1 ?
Standard binomial expansion
10. (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?
11. (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
12. (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.
13. (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.
14. (Original post by SimonM)
STEP III, Question 4

Spoiler:
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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).
15. (Original post by SimonM)
STEP II, Question 1

Spoiler:
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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
16. An almost complete solution to STEP III Q 12:

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