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STEP I, II, III 2002 Solutions

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Original post by SimonM
STEP I, Question 4

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Does this graph have an oblique asymptote?
Reply 181
Original post by Glutamic Acid
II/8:

y=e3+xexforx<0y = e^{3 + x - e^x} \text{for} x < 0


I think this should be:

y=e2e1+xexy = e^{2e -1 + x - e^x} instead?
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 α=arg(zw) \alpha = arg(z - w) . It has already been shown that tanα=tanx+yπ2 \tan \alpha = \tan \frac{x+y-\pi}{2}
Hence α=x+yπ2+nπ \alpha = \frac{x+y-\pi}{2} + n \pi (*)

Since x,y(π,π] x, y \in (-\pi, \pi] , we have that
x+yπ(3π,π] x+y-\pi \in (-3\pi, \pi] and so

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We check the remaining cases where tanα=0 \tan \alpha = 0 as follows (draw your own diagrams):-

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all the best for your STEP exams tomorrow!
Original post by Generic Name
Does this graph have an oblique asymptote?


No. The only asymptote is the x axis.
Original post by DFranklin
I think this is your problem:

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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 1/2 ^{1/2}
do you have to assume that the 1/2 ^{1/2} 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.

θ=0,y=a2+b2=a+b\theta = 0,\, y = \sqrt{a^2} + \sqrt{b^2} = |a| + |b|

It could also be:
θ=0,y=a2+b2=±a±b\theta = 0,\, y = \sqrt{a^2} + \sqrt{b^2} = \pm a \pm b

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.

Could you please clarify?
Reply 185
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 1/2 ^{1/2}
do you have to assume that the 1/2 ^{1/2} means take the positive root beacause you want it to be a function?



Yes - you've answered your own question :smile:
Original post by davros
Yes - you've answered your own question :smile:


thanks again
Original post by Dadeyemi
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.
Reply 188
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.
Reply 189
Original post by brianeverit
I agree with you. So we finally have \text{I agree with you. So we finally have }

Unparseable latex formula:

b=\dfrac{1}{2a}\left[1-a\pm \sqrt{(a-1)^2-4a(a-1)} \text{ or }k=\dfrac{b}{1-a}=\dfrac{1}{2a} \pm\dfrac{\sqrt{1+2a-3a^2}}{2a(1-a)}



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)]}
(edited 10 years ago)
Original post by tccheng
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)]}


Where does it say that k>0 ??
Reply 191
Original post by brianeverit
Where does it say that k>0 ??


You are right. Get mixed up.
Original post by Dadeyemi
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.


can't read themmmmmm :l
Reply 193
What do you mean by n is even??
Original post by Sen418
What do you mean by n is even??


Did you mean to quote someone in this?

Which question or post are you referring to?
There is a trivial error in the solution linked in the OP for this question (Trivial because it doesn't actually lead the user off-course), also the solution is a little tricky to read, so I decided to write it up in latex. I showed most of my working out and reasoning here, but it's not as long as it seems.

STEP 2002 III, Q5

This is quite a nice question, need to be careful as there is a large amount (but not tremendous) of algebra.

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Original post by SimonM
STEP III, Question 1

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Shouldn't it tend to 2pi and not 2/pi?
Original post by Brubeckian
Shouldn't it tend to 2pi and not 2/pi?


Yes, note he wrote \pi not /pi. The former indicates the LaTeX way of writing the pi symbol, and the latter a fraction (2/pi). 2\pi in LaTex gives 2π2\pi.

Edit: Also this is such a gift question this one.
Original post by SamKeene
Yes, note he wrote \pi not /pi. The former indicates the LaTeX way of writing the pi symbol, and the latter a fraction (2/pi). 2\pi in LaTex gives 2π2\pi.


Oh, my bad :P
Thanks!
For Q4 in III, I know there is an algebraic method to show what they want but is it cool if I say let x = a, and y=b.

Then I'd draw the cubics y=a3+a2y=a^3+a^2 and y=b3b2y=b^3-b^2 and show that for every corresponding value on the curves, the x value on the a curve is always less than the x value on the b curve?

Edit: I can show the second part really easily from the curves as well.
(edited 8 years ago)

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