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STEP 2005 Solutions Thread

(edited 8 years ago)

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Reply 1
I'm not going to type it up, but STEP III Question 7 is one of my favourite STEP questions
Reply 2
STEP III 2005 Question 7

This is one of my favourites too

We have 1uf(u)du=F(u)+c\displaystyle\int \frac{1}{u f(u)} \, du = F(u) +c

Making a substitution u=xmdu=mxm1dxu=x^m \Rightarrow du = mx^{m-1} dx

mxm1xmf(xm)dx=F(xm)+c    mxf(xm)dx=F(xm)+c\displaystyle\therefore \int \frac{mx^{m-1}}{x^m f(x^m)} \, dx = F(x^m) + c \iff \int \frac{m}{x f(x^m)} \, dx = F(x^m) + c

(i)

(ii)

STEP II 2005, Question 6

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sonofdot
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What happens when n = 2 in part ii)?
STEP II 2005, Question 2

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Reply 6
Daniel Freedman
What happens when n = 2 in part ii)?

Good point, edited
STEP I: Question 12
Part 1

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Part 2

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Reply 8
STEP III 2005 Question 14

:eek: that was a long one...

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II/4:

1st part



2nd part



3rd part

STEP III

Question 1


1st part



2nd part



Cheating with the graphs.
Reply 11
STEP III 2005 Question 6

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III/2 - with the proviso that I think I may have missed something somewhere.

Solution of Differential Equation



d/dx and d²/dx² of (x² + y²)



Graph


0 Point of minimum distance to origin with < c a^2



If c > a²

For this part, is it not far simpler to note that as the gradient of PR is 1p\frac{1}{p} and the gradient of SQ is q-q, and as pq=1pq = -1 it's clear that p=1qp = -\frac{1}{q} and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.
Reply 14
STEP III, Question 3

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Reply 15
STEP III, Question 4

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Reply 16
STEP II, Question 1

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Reply 17
STEP II, Question 3

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GHOSH gave all the solutions I have for 1 2005:Q2,3,4,7
I'll try another one
Reply 19
STEP III 2005 Question 9

Hopefully my diagram makes it clear what all the symbols I use mean

Diagram

First Part

Second Part

Third Part