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# STEP I, II, III 2002 Solutions Tweet

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1. Re: STEP I, II, III 2002 Solutions
(Original post by ziedj)
Oh.. isn't that unnecessarily complicated when you have already found via sketching that the curve passes through (0,1), so c = 1? Is there a particular reason?
I suppose it is unnecessary to find 'c' in terms of 'a', since they do tell us the tangent intersects at (0,1). Perhaps he just wanted to briefly examine the general case - before we note that the tangent passes through (0,1).
2. Re: STEP I, II, III 2002 Solutions
(Original post by ziedj)
Where did that part on the end (not the coefficient of x, everything else) come from?
For what it's worth I set it out a slightly different way
Tangent is at x=a (>0) and passes through (0,1)

and

So we must have

Which quickly cancels down to give
3. Re: STEP I, II, III 2002 Solutions
My solution to question 14 STEP II
Attached Files
4. STEP II.pdf (42.7 KB, 23 views)
5. Last edited by grusini; 18-07-2010 at 01:52.
6. Re: STEP I, II, III 2002 Solutions
Completing Dadeyemi's work on STEP III Q2, last part
Attached Files
7. STEP III Q2.pdf (32.5 KB, 42 views)
8. Re: STEP I, II, III 2002 Solutions
(Original post by SimonM)
STEP III, Question 1

Spoiler:
Show

As , the volume tends to 2\pi
Sorry could you explain how you get to 2 divided by pi? I got the -(ln^2 (a) +2 ln(a) +2)/a but could not justify whether that expression would tend to 0 or 1, im thinking now probably 0 as if you sub in x = e^2 then e^5 that gets progressively smaller and smaller, so did you mean 2pi?

Also a little suggestion maybe it would have been easier to use the substitution u = ln x.
9. Re: STEP I, II, III 2002 Solutions
(Original post by Diego Granziol)
Sorry could you explain how you get to 2 divided by pi? I got the -(ln^2 (a) +2 ln(a) +2)/a but could not justify whether that expression would tend to 0 or 1, im thinking now probably 0 as if you sub in x = e^2 then e^5 that gets progressively smaller and smaller, so did you mean 2pi?

Also a little suggestion maybe it would have been easier to use the substitution u = ln x.
It's 2 pi, in latex that is 2\pi. Perhaps I forgot the tags?
10. Re: STEP I, II, III 2002 Solutions
Solution to 2002 STEP II Q6
Draw a decent diagram so you can see the lines and angles clearly.
Let l1, l2 and l3 be the lengths of lines l1, l2 and l3 from a horizontal plane to the point of intersection. Let vertical height of point of intersection above the plane be h. Then: -
l3 sin /4 = l1 sin /6 = l2 sin = h
l3 sin = l1 cos = l2
Therefore
l1 cos sin = l1 sin /6
cos sin = 1/2
l3 sin sin = l3 sin /4
sin sin = 1/
(sin sin )^2 + (cos sin )^2 = 3/4
sin = /2
= /3
Same diagram for second part but
replace with and
/4, and /6 with 2, /3 and
l3 sin 2 = l1 sin = l2 sin /3 = h
l3 sin = l1 cos = l2
l1 / l3 = tan = sin 2 / sin
tan = 2 cos
l1 sin = l1 cos sin /3
(tan ) ^ 2 + 3 (cos ) ^ 2 = 4
(tan ) ^ 4 - 3 (tan ) ^ 2 - 1= 0
(tan ) ^ 2 = (3 + )/2
(3 - )/2 is not a solution as (tan ) ^ 2
Last edited by welshenglish; 26-07-2010 at 22:12.
11. Hi Simon,
I'm working my way through the gaps in the solutions. I've put in an answer for 2002, STEP II Q6. Was your list up to date? If so, can you include this answer.

Cheers, Peter
12. Re: STEP I, II, III 2002 Solutions
Answer to Step III 2002 Q9.

For the case .
Where is the angle of the plane.

The area of the fluid = and forms a trapezium.
If you cut the trapezium into two parts, a rectangle and a square angled triangle with angle , the dimensions of the parts are:

Triangle: Width
Height

Rectangle: width
let height

but

Therefore the centre of mass of the triangle is from the side of the container and from the base. The mass of the triangle is units.

The centre of mass of the rectangle is from the side of the container and from the base. Its mass is units.

Using: Moments of whole = Sum moments of parts we can find x and y
To find x:

To find y:

For the case .
The area of the fluid = and forms a square angled triangle with angle .

The area can be found using

where b and c are the hypotenuse and base of the triangle respectively.
Also

Therefore by substitution:

The distance of the centre of mass from the side is .

Therefore

Therefore:

To find y:

Where d is the side opposite the angle theta.

By substitution

The distance of the centre of mass from the base is .

Therefore

For the case when and
and
Therefore the case where the fluid forms a triangle.

Also
Where is the angle between the centre of mass and the base of the container. and and are the distances of the centre of mass from the side and base of the container.

Container will topple when

In the case when

Therefore the container will topple.
Last edited by odp04y; 28-07-2010 at 15:22.
13. Re: STEP I, II, III 2002 Solutions
Well done odp04y,

You've obviously got a better command of LaTex than me.
14. Re: STEP I, II, III 2002 Solutions
Here's an alternative for the last part of Q5

Let the roots be a,b,c,d. We know that

abcd=576=2^6 * 3^2
(a+1)(b+1)(c+1)(d+1)=1+22+172... =3^3 * 7^2
(a-1)(b-1)(c-1)(d-1)=1-22+172-552+576=7 * 5^2.

All the roots are negative (Since all the signs in the equation are "+".)

If the "+1" equation has a factor of 7^2 and the "-1" has a factor 5^2, 6 must be a duplicated root. Also since the "-1" has a factor of 7, 8 must be a root. It is now easy to deduce that the other root is 2.

So the roots are -2, -6 (twice) and -8.

By the fundamental theorem of algebra, this is the only solution set.
15. Re: STEP I, II, III 2002 Solutions
STEP III Q14

Spoiler:
Show

Var[X + Y] = E[(X+Y)^2] - E^2[X+Y] = E[X^2 + 2XY + Y^2] - E^2[X+Y]
= E[X^2] + 2E[XY] + E[Y^2] - (E[X] + E[Y])^2
= E[X^2] + 2E[XY] + E[Y^2] - E[X]^2 - 2E[X]E[Y] - E[Y]^2
= E[X^2] - E[X]^2 + E[Y^2] - E[Y]^2 + 2(E[XY] - E[X]E[Y])
= Var[X] + Var[Y] + 2Cov[X,Y]
as required.

i) Var[W + L + D] = E[(W + L + D)^2] - E^2[W + L + D]

E[W+L+D] = m => E^2[W+L+D] = m^2
E[(W+L+D)^2] = E[W^2 + L^2 + D^2 + 2(WL+WD+LD)]
= 3(m/3)^2 + 2.3(m/3)^2 = m^2/3 + 2m^2/3 = m^2

=> Var[W+L+D] = m^2 - m^2 = 0

ii) Consider W:
E[W] = m/3
Var[W] = E[W^2] - E^2[W] =

E[W^2] = 1*P(W=1) + 4*(P(W=2)) + ... + n^2(P(W=n)) for n = 1, 2, ..., m
= 1*(m/3)(1-m/3)^m-1 + 4*(m/3)(m/3)(1-m/3)^m-2 + ... + n^2(m/3)^n(1-m/3)^m-n + ... + m^2(m/3)^m
=

Very much unfinished, and I don't think I've done part i) correctly...
16. Re: STEP I, II, III 2002 Solutions
OK, I know this is an old thread, but I would be forever grateful if someone could clarify a tiny bit of the solution for II/7:

(Original post by Glutamic Acid)
II/7:

So

The second equation gives . Substituting into the third gives . Substituting these into the first:
; multiplying out gives , so no non-zero solutions.
I have the same three equations. From the second equation, I get the same value of . However, when I substitute it into the third equation, I get , the negative of what Glutamic Acid got (which is the answer that works, putting into gives a false equation). I'm probably just being an idiot, but I'm tearing my hair out over this, can someone put me out of my misery?
17. Re: STEP I, II, III 2002 Solutions
(Original post by Asboob)
I have the same three equations. From the second equation, I get the same value of . However, when I substitute it into the third equation, I get , the negative of what Glutamic Acid got (which is the answer that works, putting into gives a false equation). I'm probably just being an idiot, but I'm tearing my hair out over this, can someone put me out of my misery?
I suspect it's just a typo on his part, and he meant to put since he's used that value in the following line (having multiplied top and bottom by -1).
18. Re: STEP I, II, III 2002 Solutions
2002 STEP I question 14

19. Re: STEP I, II, III 2002 Solutions
2002 STEP II question 12

20. Re: STEP I, II, III 2002 Solutions
2002 STEP question 13

21. Re: STEP I, II, III 2002 Solutions
2002 STEP II question 14

22. Re: STEP I, II, III 2002 Solutions
2002 STEP III question 9

Attached Thumbnails

23. Re: STEP I, II, III 2002 Solutions
2002 STERP III question 10

Attached Thumbnails

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