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STEP I, II, III 2002 Solutions

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Original post by tommm
STEP II 2002 Q10

Spoiler



Just wondering, isn't the answer 81/40?
Original post by Extricated
Just wondering, isn't the answer 81/40?


I can't see at all where I've got my result from, must've posted that years ago!
Original post by tommm
I can't see at all where I've got my result from, must've posted that years ago!



Original post by Extricated
Just wondering, isn't the answer 81/40?


I posted a thread about this (ages ago though :tongue:)
Reply 163
Avatar for Rahul.S
Rahul.S
14
Original post by ben-smith
I posted a thread about this (ages ago though :tongue:)


ah lol

well I told extricated there was a slight error and he posted it up on the thread :colone:

I thought i would do a few questions for m2 rev :tongue:

you did that ques when you were in y12 :O
Original post by Rahul.S

you did that ques when you were in y12 :O


good times :tongue:
back before I had to worry about this university business.
Reply 165
Avatar for Rahul.S
Rahul.S
14
Original post by ben-smith
good times :tongue:
back before I had to worry about this university business.


:lol: dont tell me you have done most of the TRIPOS already -_-
Original post by z0tx
I find the solution above relies less on guessing and is closer to the techniques hinted at in the exercise than the other one posted on this thread.


I think this solution is actually more in the spirit of the question....
Step2002Paper1Question5.pdf60.7KB
Reply 167
Avatar for z0tx
z0tx
12
Original post by mikelbird
I think this solution is actually more in the spirit of the question....


Hm... Yes you're right.
Reply 168
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desijut
10
Original post by Dadeyemi
Some more;

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


on q2, the last part says "hence" what i did was spot that, it was the same as the sum above it but with r=0 which led me straight to pi/2
Original post by toasted-lion
Question 5, I'll pick up where Dadeymi left off:

So (ab2+(a−1)b+(a−1))=0 \left(ab^2 + (a-1)b +(a-1)\right)=0

  ⟹  (b+a−12a)2=1−aa−a2−2a+14a2 \implies \left( b + \frac{a-1}{2a} \right)^2 = \frac{1-a}{a} - \frac{a^2 - 2a +1}{4a^2}

]


Above from --- http://www.thestudentroom.co.uk/showthread.php?t=862415&page=4&p=19333071#post19333071

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

ab2+(a−1)b+(a−1)=0 ab^2 + (a-1)b +(a-1)=0

  ⟺  b2+a−1ab=1−aa \iff b^2 + \dfrac{a-1}{a}b = \dfrac{1 - a}{a}

Unparseable latex formula:

\iff b^2 + \dfrac{a-1}{a}b + \left(\dfrac{a - 1}{2a})^2\right = \dfrac{1 - a}{a} + \left(\dfrac{a - 1}{2a}\right)^2



  ⟺  (b+a−12a)2=1−aa+a2−2a+14a2 \iff \left( b + \dfrac{a-1}{2a} \right)^2 = \dfrac{1-a}{a} + \dfrac{a^2 - 2a +1}{4a^2}
(edited 11 years ago)
Original post by adrienne_om
Above from --- http://www.thestudentroom.co.uk/showthread.php?t=862415&page=4&p=19333071#post19333071

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

ab2+(a−1)b+(a−1)=0 ab^2 + (a-1)b +(a-1)=0

  ⟺  b2+a−1ab=1−aa \iff b^2 + \dfrac{a-1}{a}b = \dfrac{1 - a}{a}

Unparseable latex formula:

\iff b^2 + \dfrac{a-1}{a}b + \left(\dfrac{a - 1}{2a})^2\right = \dfrac{1 - a}{a} + \left(\dfrac{a - 1}{2a}\right)^2



  ⟺  (b+a−12a)2=1−aa+a2−2a+14a2 \iff \left( b + \dfrac{a-1}{2a} \right)^2 = \dfrac{1-a}{a} + \dfrac{a^2 - 2a +1}{4a^2}


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)}

Seems Paper II Q1 has a bad solution here:

http://www.thestudentroom.co.uk/showthread.php?t=862415&p=18023769#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 :tongue:

EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct. :smile:
(edited 11 years ago)
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.
Reply 173
Avatar for sonofdot
sonofdot
8
Original post by Xero Xenith
Seems Paper II Q1 has a bad solution here:

http://www.thestudentroom.co.uk/showthread.php?t=862415&p=18023769#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 :tongue:

EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct. :smile:


Good spot, fixed :smile:
Original post by Xero Xenith
Seems Paper II Q1 has a bad solution here:

http://www.thestudentroom.co.uk/showthread.php?t=862415&p=18023769#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 :tongue:

EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct. :smile:


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 :smile:
(edited 11 years ago)
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 :smile:


Seems like there's still just a minor thing on the third line (the one with the three dots) - square root sign should be 1−cos⁡22θ1- \cos^2 2 \theta not just 1−cos⁡2θ1- \cos^2 \theta. :smile:
Reply 176
Avatar for Rahul.S
Rahul.S
14
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 :smile:


nah you are being stupid here :biggrin: :lol:
Reply 177
Avatar for yukki0822
yukki0822
Original post by Glutamic Acid
II/7: (Scary scary vectors.)

??????????



sorry im really confused that why does the modulus of r equal to 1?
Reply 178
Avatar for BabyMaths
BabyMaths
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.
For II, Q10, I believe there's an error in the final part - the −292-\dfrac{29}{2} 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

(edited 11 years ago)

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