(Original post by adrienne_om)
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 
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 11062012 09:14

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 11062012 20:17
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
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.Last edited by Xero Xenith; 11062012 at 21:00. 
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 183
 11062012 21:03
(Original post by Xero Xenith)
I think the result only holds for complex numbers modulus 1. 
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 184
 12062012 11:12
(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
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. 
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 185
 12062012 14:28
(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
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.Last edited by Extricated; 12062012 at 14:38. 
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 186
 12062012 14:35
(Original post by Extricated)
Probably being a bit stupid here but how did you go from 1sqrt(1cos^2 x) to 1cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally.
(Original post by sonofdot)
Good spot, fixed 
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 12062012 15:09
(Original post by Extricated)
Probably being a bit stupid here but how did you go from 1sqrt(1cos^2 x) to 1cos2x?(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 
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 188
 22022013 18:30
(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.
m_{3} has direction ; m_{4} 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 nonzero solutions. 
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 22022013 20:03
(Original post by yukki0822)
sorry im really confused that why does the modulus of r equal to 1? 
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 17032013 21:33
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
Last edited by DJMayes; 17032013 at 21:36. 
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 191
 20052013 22:38
(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 
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 192
 29052013 23:27

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 193
 25062013 18:00
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
We check the remaining cases where as follows (draw your own diagrams):
all the best for your STEP exams tomorrow! 
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 194
 01072013 09:41
(Original post by Generic Name)
Does this graph have an oblique asymptote? 
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 13072013 14:06
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.
Could you please clarify? 
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 13072013 14:42
(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?

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 13072013 15:07
(Original post by davros)
Yes  you've answered your own question 
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 24112013 17:04
(Original post by Dadeyemi)
Some more;
Did these quite a while ago I'm afraid some may be partial solutions. 
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 27112013 01:46
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. 
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 200
 11012014 07:29
k= 1/2a x {1 +/ sqrt[(1+3a)/(1a)]}
= 1/2a x {1 +/ sqrt[1+ 4a/(1a)]}
since 4a/(1a)>0 so the sqrt >1, and k>0, hence cannot take the negative sqrt.
k = 1/2a x {1 + sqrt[1+ 4a/(1a)]}Last edited by tccheng; 11012014 at 08:35.
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Updated: November 23, 2015
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