The Student Room Group

Scroll to see replies

Reply 1
II/10:

Spoiler

Reply 2
II/12:

(i)

Spoiler



(ii)

Spoiler



(iii)

Spoiler



(iv)

Spoiler



(v)

Spoiler

(edited 10 years ago)
Some photos: II 2:

ImageUploadedByStudent Room1403971645.464489.jpgImageUploadedByStudent Room1403971658.601121.jpgImageUploadedByStudent Room1403971668.513346.jpg

And II 8

ImageUploadedByStudent Room1403971688.538401.jpgImageUploadedByStudent Room1403971696.480341.jpg


Posted from TSR Mobile
Reply 4
III/9

Spoiler

(edited 10 years ago)
II/4:

Spoiler

I'll edit the last part into this post in a short while.
(edited 4 years ago)
Reply 6
Original post by JosephML
III/9
.


I only got up as far as the result involving uksin - how many marks do you think I'd have got?
Reply 7
Original post by JosephML
III/9


I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?
Reply 8
III/6

Spoiler

Reply 9
Original post by Kasel
I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?


Ah, you're right. I missed that bit. I'll correct the solution, thanks.

Original post by CD315
I only got up as far as the result involving uksin - how many marks do you think I'd have got?


I should suspect somewhere between 8 and 12 marks.
Reply 10
Original post by Khallil
II/4


Do you think we would have had to prove arctan(x) + arctan(1/x)=pi/2?
There are a few short proofs but I forgot to include any in my solution.
STEP III, Q4

(i)



(ii)



It's worth pointing out that a very similar question appeared in a IB tripos exam for the variational principles course.
(edited 10 years ago)
Original post by Kasel
...


I did but I left it out of the solution above for brevity. The one I wrote in the paper was very similar to this, but it looks really forced with regards to the 1-1 in the denominator.

ϕ=arctanb+arctan1b    tanϕ=tan(arctanb)+tan(arctan1b)1tan(arctanb)tan(arctan1b)=b+1b11\begin{aligned} \phi = \arctan b + \arctan \frac{1}{b} \implies \tan \phi & = \dfrac{\tan \left( \arctan b \right) + \tan \left( \arctan \frac{1}{b} \right)}{1 - \tan \left( \arctan b \right)\tan \left( \arctan \frac{1}{b} \right)} \\ & = \dfrac{b + \frac{1}{b}}{1 - 1} \end{aligned}

 tanϕ=limn0(b+1bn)    ϕ=arctan(limn0(b+1bn))=π2\therefore \ \tan \phi = \displaystyle \lim_{n \to 0} \left( \dfrac{b + \frac{1}{b}}{n} \right) \implies \phi = \arctan \left( \lim_{n \to 0} \left( \dfrac{b + \frac{1}{b}}{n} \right) \right) = \dfrac{\pi}{2}
(edited 10 years ago)
Reply 13
Original post by Khallil
...


To avoid limits you can say let y=arctan(x), then tan(y)=x
and
cot(y)=(1/x)=tan(pi/2 - y)
which gives arctan(x)+arctan(1/x)= y+pi/2-y=pi/2
Original post by Kasel
...


Ah, of course! :facepalm2:
STEP III - Q12

Spoiler

(edited 10 years ago)
Reply 16
STEP II - Q8 (LaTeX)

Spoiler



This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.
(edited 10 years ago)
Original post by Stray
STEP II - Q8 (LaTeX)

{n,a,b,cr}Z+\{n, a, b, c_r\} \mathbb{Z}+
cr,(0rn)=coefficient of xr in (a+bx)nc_r, (0 \leq r \leq n) = \texttt{coefficient of } x^r \texttt{ in } (a + bx)^n

cr=nCranrbr=n!r!(nr)!anrbrc_r = ^n\mathrm{C}_r \cdot a^{n-r} b^r= \frac{n!}{r!(n-r)!} \cdot a^{n-r} b^r

cmcr for 0rnc_m \geq c_r \texttt{ for } 0 \leq r \leq n

If cmc_m is a maximum, it must be cm+1\geq c_{m+1} and cm1\geq c_{m-1}


Taking first, cm+1c_{m+1} :

n!an(m+1)b(m+1)(m+1)!(n(m+1))!n!anmbmm!(nm)!\frac{n!a^{n-(m+1)} b^{(m+1)}}{(m+1)!(n-(m+1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}

Divide both sides by the common factor

n!an(m+1)bmm!(n(m+1))!\frac{n!a^{n-(m+1)}b^m}{m!(n-(m+1))!}

Giving

bm+1anm\frac{b}{m+1} \leq \frac{a}{n-m}

bnbmam+abn-bm \leq am+a

bnaam+bmbn-a \leq am+bm

bnaa+bm\frac{bn-a}{a+b} \leq m

This is almost the LHS of what we are required to show, just needs a little fiddling

bn+b(a+b)a+bm\frac{bn+b-(a+b)}{a+b} \leq m

b(n+1)a+b1m()\frac{b(n+1)}{a+b} -1 \leq m (**)

Now repeat the process for cm1c_{m-1}:

n!an(m1)b(m1)(m1)!(n(m1))!n!anmbmm!(nm)!\frac{n!a^{n-(m-1)} b^{(m-1)}}{(m-1)!(n-(m-1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}

Divide both sides by the common factor

n!anmbm1(m1)!(nm)!\frac{n!a^{n-m}b^{m-1}}{(m-1)!(n-m)!}

Giving

an(m1)bm\frac{a}{n-(m-1)} \leq \frac{b}{m}

ambn+bbmam \leq bn + b - bm

m(a+b)b(n+1)m(a+b) \leq b(n+1)

mb(n+1)a+bm \leq \frac{b(n+1)}{a+b}

Which combines with (**) to give the required solution

b(n+1)a+b1mb(n+1)a+b\frac{b(n+1)}{a+b} -1 \leq m \leq \frac{b(n+1)}{a+b}

Considering the possible values of mm

Let y=b(n+1)a+b\texttt{Let } y=\frac{b(n+1)}{a+b}

y1myy-1 \leq m \leq y

Either yy and y1y-1 are both integers, or yy and y1y-1 are non-integer values either side of an integer, so if mm is an integer, it is either yy and y1y-1 (2 values) or it lies between them (1 value).

For the next parts, we are asked to find either the value of mm where there is 1 value, or the larger value if there are 2. This means we can consider the function G(n,a,b)G(n,a,b) to produce the Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is \lfloor.

G(n,a,b)=b(n+1)a+bG(n,a,b) = \lfloor{\frac{b(n+1)}{a+b}}

i)
Unparseable latex formula:

G(9,1,3) = \lfloor{\frac{3(9+1)}{1+3}} = \lfloor\frac{30}{4} = \floor{7.5} = 7



G(9,2,3)=3(9+1)2+3=305=6G(9,2,3) = \lfloor{\frac{3(9+1)}{2+3}} = \lfloor\frac{30}{5} = 6

ii)
G(2k,a,a)=a(2k+1)a+a=2k+12=(k+0.5)=kG(2k,a,a) = \lfloor{\frac{a(2k+1)}{a+a}} = \lfloor\frac{2k+1}{2} = \lfloor{(k + 0.5)} = k

G(2k1,a,a)=a(2k1+1)a+a=2k2=kG(2k-1,a,a) = \lfloor{\frac{a(2k-1+1)}{a+a}} = \lfloor\frac{2k}{2} = k

iii)
If we fix nn and bb, the only part of GG that can alter is the denominator containing aa, so for large GG we want small aa, so GmaxG_{max} is when a=1a=1. (Zero is not a positive integer).

iv)
For fixed a=1a=1 and (unknown) fixed nn, the part of the function varying is:

bb+1()\frac{b}{b+1} (***)

This tends to 1 as b tends to infinity, but will never actually reach 1, so ()(n+1)(***)(n+1) tends towards but never reaches n+1n+1 so for fixed nn
G(n,1,b)max=n when b=G(n,1,b)_{max} =n \texttt{ when } b = \infty

---

This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.


The last part is incorrect. In particular the value of infinity does not exist.


Posted from TSR Mobile
Reply 18
STEP II, Q5: (somebody quickly check through this please though I think it should be fine)

(i)



(ii)



The only real difficulty here was spotting how to get rid of the -4 and -3.
(edited 10 years ago)
STEP III Q5

Beginning

Spoiler



Part i)

Spoiler



Part ii)

Spoiler

(edited 10 years ago)

Quick Reply