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# STEP I, II, III 2002 Solutions

Above from --- http://www.thestudentroom.co.uk/show...1#post19333071

Maybe a correction here -- I'm not sure -- but I got a positive sign in the RHS of the second equation ---

2. Seems Paper II Q1 has a bad solution here:

http://www.thestudentroom.co.uk/show...9#post18023769

No need to actually look at the question paper, Part ii) to this solution has some funky algebra!

The last few lines should be

Part Two

The jump from the third line to the fourth still isn't all that clear but at least now it seems correct

EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct.
3. (Original post by Xero Xenith)
I think the result only holds for complex numbers modulus 1.
Not even for all numbers of modulus 1. e.g. z = i, n = 2.
4. (Original post by Xero Xenith)
Seems Paper II Q1 has a bad solution here:

http://www.thestudentroom.co.uk/show...9#post18023769

No need to actually look at the question paper, Part ii) to this solution has some funky algebra!

The last few lines should be

Part Two

The jump from the third line to the fourth still isn't all that clear but at least now it seems correct

EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct.
Good spot, fixed
5. (Original post by Xero Xenith)
Seems Paper II Q1 has a bad solution here:

http://www.thestudentroom.co.uk/show...9#post18023769

No need to actually look at the question paper, Part ii) to this solution has some funky algebra!

The last few lines should be

Part Two

The jump from the third line to the fourth still isn't all that clear but at least now it seems correct

EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct.
Probably being a bit stupid here but how did you go from 1-sqrt(1-cos^2 x) to 1-cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally. Ofcourse doesn't affect final answer, but could lead to a little confusion
6. (Original post by Extricated)
Probably being a bit stupid here but how did you go from 1-sqrt(1-cos^2 x) to 1-cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally.
Good spot yourself, pretty sure you're right!

(Original post by sonofdot)
Good spot, fixed
Seems like there's still just a minor thing on the third line (the one with the three dots) - square root sign should be not just .
7. (Original post by Extricated)
Probably being a bit stupid here but how did you go from 1-sqrt(1-cos^2 x) to 1-cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally. Ofcourse doesn't affect final answer, but could lead to a little confusion
nah you are being stupid here
8. (Original post by Glutamic Acid)
II/7: (Scary scary vectors.)
??????????

Let the lines have direction vector which I'll denote by r, and without loss of generality let |r| = 1 so a^2 + b^2 + c^2 = 1;

.

b = 1 - a; c = 1 - a so
The solutions are a =1, b = 0, c = 0 and a = 1/3, b = 2/3, c = 2/3.

m3 has direction ; m4 direction .

Considering gives .

(i) A has position vector ; B .
Let P = , and Q =

AQ . BP = 0 so

. Showing that the latter is > 0; 1 - sqrt(6)/3 > 0 therefore 1 > sqrt(6)/3 therefore 1 > 6/9, which is true.

(ii)

So

The second equation gives . Substituting into the third gives . Substituting these into the first:
; multiplying out gives , so no non-zero solutions.
sorry im really confused that why does the modulus of r equal to 1?
9. (Original post by yukki0822)
sorry im really confused that why does the modulus of r equal to 1?
We're only interested in the direction.
10. For II, Q10, I believe there's an error in the final part - the should have an x attached, which makes the final answer 81/40. I also get this answer through an entirely different approach to the question:

Spoiler:
Show

Differentiate implicitly:

When T is stationary, , so

Plug this back into the original equation:

This gives maxima/minima at T = 0 and T = 3. T = 0 is obviously impossible, so T must equal 3, as required.

For the second part, use a trapezium to calculate the final speed of the second competitor, and use this to work out the acceleration:

And so the acceleration is then .

, which means that the speed in the second part is 14.5. Next, find the time when they are moving at equal speed, as this is when maximum displacement occurs:

This is greater than 0.75 so is fine. Then, calculate the distances for each:

Subtracting one from the other yields the final answer

11. (Original post by SimonM)
STEP I, Question 4

Spoiler:
Show

Therefore the equation of the tangent is

Since is a solution

This gives us the equation,

We have so the only intersections are 0 and 1.

We have

(given)

The area under the tangent is a trapezium with area

Since the graph is always above it, we get which is what we want to show.

Considering the half closest to the y axis, and revolving it around that axis we get:

and

Therefore
Does this graph have an oblique asymptote?
12. (Original post by Glutamic Acid)
II/8:

I think this should be:

13. Regarding STEP III / 2002 / Q8, i had thought that Dadeyemi gave an excellent answer, but a student of mine was seeking clarification for the first part regarding the periodicities of the argument and the tangent. I looked at the problem more closely just now, and agree that it needs to be tightened up a little. Here is my little "patch":-

Let . It has already been shown that
Hence (*)

Since , we have that
and so
Spoiler:
Show
.

Spoiler:
Show
When , , hence (*) holds with n = 0.

Spoiler:
Show
When , , hence (*) holds with n = 1.

We check the remaining cases where as follows (draw your own diagrams):-
Spoiler:
Show
When , , hence (*) holds with n = 0.

Spoiler:
Show
When , , hence (*) holds with n = 1.

all the best for your STEP exams tomorrow!
14. (Original post by Generic Name)
Does this graph have an oblique asymptote?
No. The only asymptote is the x axis.
15. (Original post by DFranklin)
I think this is your problem:
Spoiler:
Show
When , not a+b.
this was for question 3 step 1- I am having some trouble with it because i do not not know how to interpret the power
do you have to assume that the means take the positive root beacause you want it to be a function?

It is also confusing for me because after every step the possibilities are many for example the step nota bene has just made.

It could also be:

Summary if this question meant take the power of half to mean take the positive square root- i would have no problems
but i am not sure if this is so.

16. (Original post by nahomyemane778)
this was for question 3 step 1- I am having some trouble with it because i do not not know how to interpret the power
do you have to assume that the means take the positive root beacause you want it to be a function?
17. (Original post by davros)
thanks again
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.
You missed (0,1) and (0,-1) for question 4 I believe.
19. Re: 2002/III/Q8

Dadeyemi gave an excellent solution and SingaporeCantab had a point on the periodicity. Notice that the periodicity of complex numbers is 2n(pi) because of the specific signs for the real part and the imaginary part. Need extra care in using the tan function where the periodicity is n(pi), e.g cis(pi/4) is not equal to cis(pi/4 + pi) though tan(pi/4) = tan(pi/4+pi). I think the patch is not entirely correct.

My patch to the "patch": I think it is sufficient to say that (theta1 + theta2 - pi)/2 has a range (-3pi/2, pi/2). When
(theta1 + theta2 - pi)/2 is in the range (-3pi/2, -pi], we need to add 2pi (i.e. n=1) to bring it back to the domain of (-pi,pi]. Otherwise n=0.
20. (Original post by brianeverit)

II Q5 So
k= 1/2a x {1 +/- sqrt[(1+3a)/(1-a)]}
= 1/2a x {1 +/- sqrt[1+ 4a/(1-a)]}
since 4a/(1-a)>0 so the sqrt >1, and k>0, hence cannot take the negative sqrt.
k = 1/2a x {1 + sqrt[1+ 4a/(1-a)]}

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