STEP I 2012 discussion thread Watch

Extricated
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#41
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(Original post by In One Ear)
This was also my conclusion. May be wrong though?
Sorry didn't look at the question and didn't do it. I just assumed that that was a viable method and he was asking if using higher level techniques (i.e beyond c1-c4) was allowed. Yeah in that case it's probably wrong (again, haven't looked at q)
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8inchestall
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#42
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Can someone explain what the second paragraph of q1 means? I had no idea what the question was asking, I wrote like 'since L is changing and R is constant, m and c must be changing...'

I started with q8 and got a solution like C = (2x+y)^2/(3x+y)^3 or something, by using y=x/v (or v/x i really cannae remem) then I went to do 12345 in that order.

The graph question 2, I had the right idea for the parts ii and iii but I went braindead and drew my quartic as a cubic, because I only found 3 stationary points haha oh well.

And lastly, how do you find the co-ords of N from T in question 4? I tried doing it the usual way by finding two eqns of normal and equating, but my x-val came out to be a mass of algebra and it just seemed wrong so I didnt follow through with it.
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f1mad
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#43
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Q8.. wow, what a gift. Assuming you have done Fp3 on AQA .
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xxxxkrishyxxxx
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#44
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(Original post by Troll the Trolls)
Anyone do 4? I got t but when i tried to find n
the algebra got so messy it was unreal, like a million of each term so i left my answer as that dno if il get any marks for not simplifying
Yeah, my N coordinates were really messy too at first, but I eventually managed to simplify them to (p+q, p+q)
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Farhan.Hanif93
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#45
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I'm moving this post to here as it'll get lost at the bottom of page 2. I'm also of the opinion that this paper is one of the harder ones. I've had a chance to complete most of the questions on paper and I thought I'd LaTeX up some of the more difficult (in my opinion) ones.

Q3
(i)
Let O be the origin and let A, B and C be the points with coordinates (b,0), (b,\sin b) and (b,b) respectively.

From the sketch, note that Area_{OAB} < \displaystyle\int _0^b \sin x dx < Area_{OAC}

\iff \dfrac{1}{2}b\sin b < 1-\cos b < \dfrac{1}{2}b^2

\iff \boxed{1-\dfrac{1}{2}b^2 < \cos b < 1-\dfrac{1}{2}b\sin b}, as required.

(ii)
Consider the tangent to y=a^x at x=0 (namely y=x\ln a + 1) and the chord y=(a-1)x + 1. Let A, B, C and D be the points with coordinates (1,0), (1,1+\ln a), (1,a) and (0,1) respectively.

Note that Area_{OACD} > \displaystyle\int _0^1 a^x dx = \displaystyle\int _0^1 e^{x\ln a} dx

\iff \dfrac{a+1}{2} >  \dfrac{a-1}{\ln a}

\iff \ln a > \dfrac{2(a-1)}{a+1} (given that a-1>0)

Furthermore, observe that \displaystyle\int _0^1 a^x dx > Area_{OABD}

\iff \dfrac{a-1}{\ln a} > \dfrac{2+\ln a}{2}

\iff (\ln a)^2 +2\ln a - 2a + 2 < 0 (as a>1 \iff \ln a > 0)

\iff (\ln a + 1)^2 < 2a -1

\implies ln a < -1 + \sqrt{2a-1}

It follows that \boxed{\dfrac{2(a-1)}{a+1} < ln a < -1 + \sqrt{2a-1}}, as required.


Q6
WLOG, take AB=1 and let AD=k ( where 0<k<1) and let E be the tip of the flagpole.

From \triangle ADE and \triangle BDE respectively: \tan \alpha = \dfrac{DE}{k} \iff k = DE \cot \alpha (1)
\tan \beta  = \dfrac{DE}{1-k} \iff 1-k = DE\cot \beta (2)

Since AB is a diameter and C lies on the circular path, \triangle ACB is right angled at C. Moreover, CD is perpendicular to AB so \triangle ACD and \triangle BCD are right angled at D. By pythagorus:

AC^2 +BC^2 = 1 \iff k^2 + DC^2 + (1-k)^2 + DC^2 = 1 \iff DC = \sqrt{k(1-k)}.

Hence, by \triangle CDE: \tan \phi = \dfrac{DE}{\sqrt{k(1-k)}} \iff  k(1-k) = DE^2 \cot ^2 \phi (3)

Plugging (1) and (2) into the LHS of (3), it follows that \boxed{\cot \alpha \cot \beta = \cot ^2\phi}, as was to be shown.

To save me from the LaTeX, let \sin a = s(a) and \cos (a) = c(a).

Note \dfrac{1}{2}c(p+q) - c(p)c(q)s^2(\frac{p+q}{2}) + s(p)s(q)c^2(\frac{p+q}{2})

 \equiv \dfrac{1}{2} [ c(p)c(q)(1-2s^2(\frac{p+q}{2})) - s(p)s(q)(1-2c^2(\frac{p+q}{2}))]

\equiv \dfrac{1}{2}[c(p)c(q)c(p+q) - s(p)s(q)c(\pi -(p+q))] (as \cos x \equiv \sin \left(\dfrac{\pi}{2} - x\right))

\equiv \dfrac{1}{2}c(p+q)[c(p)c(q) + s(p)s(q)] (since \cos (\pi - x) \equiv -\cos x)

\equiv \dfrac{1}{2}c(p+q)c(p-q)

It follows that \boxed{\cos p \cos q \sin ^2\dfrac{1}{2}(p+q) - \sin p \sin q \cos ^2 \dfrac{1}{2} (p+q) \equiv \dfrac{1}{2}\cos (p+q) - \dfrac{1}{2} \cos (p+q) \cos (p-q)} (*) holds for all p,q, as desired.

Note p, q >0 and p+q \leq \dfrac{\pi}{2} suggests that \cos (p+q) \geq 0 and also \cos x \leq 1 \implies 1-\cos (p-q) \geq 0. So from (*):

\mathrm{RHS}= \dfrac{1}{2}\cos (p+q) (1-\cos (p-q)) \geq 0.
\iff \cos p \cos q \sin ^2\dfrac{1}{2}(p+q) - \sin p \sin q \cos ^2 \dfrac{1}{2} (p+q) \geq 0
\iff \boxed{\cot p \cot q \geq \cot ^2 \dfrac{1}{2}(p+q)}, as required.

Replacing p and q with \alpha and \beta, note that:

\alpha + \beta \leq \dfrac{\pi}{2} \implies  \cot ^2 \dfrac{1}{2}(\alpha+\beta) \leq \cot \alpha \cot \beta = \cot ^2\phi
\iff \tan \phi \leq \tan \dfrac{1}{2}(\alpha + \beta) (as both \tan \phi and \tan \dfrac{1}{2}(\alpha + \beta) are positive given what we know)
\iff \boxed{\phi \leq \dfrac{1}{2}(\alpha + \beta)} since \tan x is increasing over the interval, as was to be shown.



Q7
(i)
Using the relation, t_n = x \implies (p+q -1) x = 0 and since x\not=0, we have the solution \boxed{p+q = 1; x\in \mathbb{R}/\{0\}}, as required.


(iii)
Consider the relation again:
t_{2n} = x, t_{2n+1} = y \implies \begin{Bmatrix} x=py+qx & (1) & \mathrm{if} & n &\mathrm{ even} \\ y=px+qy & (2) & \mathrm{if } & n &  \mathrm{odd}

Note (q-1) \times (1) - p \times 2: [(q-1)^2-p^2] x =0 and since x\not= 0, this can only hold if (q-1)^2 - p^2 =0 \iff \boxed{q\pm p = 1}, as was to be shown.

Note x\times (1) - y\times (2): (q-1)(x^2-y^2) = 0 and we have x^2 - y^2 = 0 \iff x=\pm y if q\not 1 AND p\not=0.

(ii)
Again, from the relation:
t_{3n} = x, t_{3n+1} = y, t_{3n+2} \implies \begin{Bmatrix} z=py+qx & (1) & \mathrm{if} & n\equiv 0\mod 3 \\ x=pz+qy & (2) & \mathrm{if } & n\equiv 1\mod 3 \\ y=px+qz & (3) & \mathrm{if } & n\equiv 2\mod 3

Note p\times (2)  + (3) = (4): (pq-1) y + (p^2 + q)z = 0

Also q\times (2) + (1) = (5):  (pq-1)z + (q^2+p)y = 0

So (pq-1) \times (4) - (p^2 + q) \times (5): [(pq-1)^2 - (p^2+q)(q^2+ p)]y = 0 and since y\not=0, this can only hold if (pq-1)^2 - (p^2+q)(q^2+p) =0 \iff \boxed{p^3 + q^3 + 3pq -1 =0} (*), as required.

Setting p=1-q in the LHS of (*) yields 0 so we can observe that:
p^3 + q^3 + 3pq -1 =0 \iff (p+q-1)(p^2 +p - pq + q + q^2 + 1) = 0
\iff \dfrac{1}{2}(p+q-1)[(p-q)^2 + (p+1)^2 + (q+1)^2] = 0
\iff p+q =1 or (p-q)^2 + (p+1)^2 + (q+1)^2 =0, as was to be shown.

By plugging p=1-q, into the equations (1), (2) and (3) and attempting to solve, you end up with \boxed{x=y=z}. Note also that (p-q)^2 + (p+1)^2 + (q+1)^2 =0 \iff p=q=-1 as squares are non-negative. By plugging these into each of (1), (2) and (3), they all reduce to \boxed{x+y+z=0}[/latex], as required.



If you're looking for any other specific solutions, just mention it and I'm sure someone will be able to post one up.
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In One Ear
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#46
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(Original post by f1mad)
Q8.. wow, what a gift. Assuming you have done Fp3 on AQA .
Why was it a gift? I thought integrating factor can't work for it? You are forced with the substitution they give you (and it so happens that the same sub works for the last equation).
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Farhan.Hanif93
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#47
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(Original post by 8inchestall)
And lastly, how do you find the co-ords of N from T in question 4? I tried doing it the usual way by finding two eqns of normal and equating, but my x-val came out to be a mass of algebra and it just seemed wrong so I didnt follow through with it.
If you used the condition given from the start, it turns out that P(p,q) and Q(q,p), which makes the algebra much neater.
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xxxxkrishyxxxx
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(Original post by Farhan.Hanif93)
If you used the condition given from the start, it turns out that P(p,q) and Q(q,p), which makes the algebra much neater.
Did you get N as (p+q, p+q)? Please say you did!
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8inchestall
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(Original post by Farhan.Hanif93)
If you used the condition given from the start, it turns out that P(p,q) and Q(q,p), which makes the algebra much neater.
So there isn't a trick as such? Using the gradient of the normal and equate 2 new lines was the way to go then ahh
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Wahrheit
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#50
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(Original post by In One Ear)
Why was it a gift? I thought integrating factor can't work for it? You are forced with the substitution they give you (and it so happens that the same sub works for the last equation).
FP3 does give you practice using substitutions to solve differential equations though.
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jukebox123
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#51
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This was a very hard paper indeed, think I got around 50-60 marks
Do you lot reckon this will be enough for a 2?
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Farhan.Hanif93
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#52
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(Original post by xxxxkrishyxxxx)
Did you get N as (p+q, p+q)? Please say you did!
Yes.
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In One Ear
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#53
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(Original post by Wahrheit)
FP3 does give you practice using substitutions to solve differential equations though.
True. Actually the bookwork they have at the end of the fp3 text book with the v=e^t sub is actually much harder than the sub in this question. The difficulty in this question was integrating once you'd transformed the equation and then getting it in terms of v (imo).
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MHRed
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#54
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I felt before the exam I'll do well to get a 2 (and having messed up my fp3, and now no longer expecting A* in FM, considerably less pressure was on this exam).

I didn't complete 1 question fully, however made it to near the end of 3questions, about half way with 1 and two fragmentary Qs. Time'll tell!

(Original post by actuarialmaestro:p)
****.. **** ****.... think I have probably failed it... only 1 complete solution..2 partial solutions and other fragmentary answers..... really worried now...
What's/where's your offer?
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Thing8
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#55
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are results out same day as other a levels?
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f1mad
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#56
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(Original post by In One Ear)
Why was it a gift? I thought integrating factor can't work for it? You are forced with the substitution they give you (and it so happens that the same sub works for the last equation).
Nope, no integrating factor. What I was implying is, one would be comfortable and spot what to use.

xv*dv/dx = 2-2v^2

v/(2-2v^2) dv/dx = 1/x

I v/(2-2v^2) dv = I 1/x

-1/4ln(2-2v^2) = 1/x +K

Do a load of re-arranging, sub in what v was and you'll end up with:

x^2(y^2-x^2)= C
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Z1G
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#57
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I did really badly on this exam but at-least my offer doesn't depend on it.

I thought I'd done question one but I worked it out for distance squared then forgot to square root it at the end! So get cubed, instead of ^3/2

Then for question 2 I went completely blank. Tried to do the first part using the quadratic formula instead of the graph I'd just sketched! Then I put answers to the other two parts, but they were educated guesses and so probably wrong.

Question 3 I did the first section of, then question 4 I did the first two parts but couldn't do the last. I also did the first two parts for question 7, and the first part of question 8 correctly. I used the right substition and got an answer for the last part of question 8 but it looked too 'messy' to be correct.

Oh well at-least if other people found it difficult I might get a higher grade than a U! Got to keep my chin up for STEP II tomorrow now haha, it's so depressing coming away from a bad paper like on monday :/
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MHRed
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The question on integrating the cos2x ln (cos x) or whatever it was...

I did the first two parts easily, but how do you do the last bit?!
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DFranklin
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#59
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For Q1, last part, I get m=\sqrt[3]{\frac{b}{a}} and the distance (a^{2/3}+b^{2/3})^{3/2}.
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fruktas
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#60
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I got Q8 here in anyone is interested?
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