STEP maths I, II, III 1990 solutions

Attempt at STEP III, Q8.

paper I/4 & 7
Comments and /or corrections welcomed.

Can anyone help me with I Question 7. I don't understand the validity conditions. When I solve INT 1/u du = INT 2/(x + 1) dx i get...

ln|u| = 2ln|x + 1| + C.

Im not sure why this isn't valid for x < -1 but I confess to not really knowing much about why we really use mod signs on logs so thats clearly the key to me not understanding. Any help is much appreciated

1990 Paper II numbers 12,13,14,16

Attempt at STEP III, Q8.

STEP I, Question 7

$\dfrac{dy}{dx} +Py=Q$

Let $y = uv$

So $\displaystyle uv' + vu' + Puv = Q \Rightarrow v' + v \left (\frac{u'}{u} + P \right) = \frac{Q}{u}$

Therefore, we can express $v'$ in terms of $Q, u, x$ if $P+\dfrac{u'}{u} = 0$

That is $\boxed{u'+Pu = 0}$.

Solving

$\displaystyle \frac{dy}{dx} - \frac{2y}{x+1} = (x+1)^{\frac{5}{2}}$

Let $y = uv$, to find a suitable u, solve $u'+Pu =0$, that is $\frac{du}{dx} -\frac{2u}{x+1} = 0$

Some separating variables gives $u = A(x+1)^2$

We must now solve $v' = \frac{(x+1)^{\frac{5}{2}}}{A(x+1)^2}$, that is $v' = \frac{1}{A}(x+1)^{\frac{1}{2}}$, so $v = \frac{2}{3A}(x+1)^{\frac{3}{2}}+D$

Therefore $y = uv = \frac{2}{3}(x+1)^{\frac{7}{2}}+F(x+1)^2$

$y(0) = \frac{2}{3}+G = 1 \Rightarrow G = \frac{1}{3}$

Therefore $\boxed{y = \frac{1}{3} \Big (2(x+1)^{\frac{7}{2}}+(x+1)^{2}\Big)}$

Valid for all $x>-1$ (I think)

STEP I
10) i)total resistive forces= 50+2(v-3)= 2(22 +v) where v is pos. in direction of turning point from starting point.
using P=Fv, where F=resistive forces due to no acceleration
300=2v(22+v)
2v^2 + 44v- 300=0
v=0.25 (in forward journey), v=1.25(backward journey) to 2dp
using v=s/t:
total time for race= 5000/0.25 + 5000/1.25
= 24000 seconds (supposed to answer to nearest second but don't have a calc on me)


ii) total energy expended using P=W/t
W=300x24000=7200000

x(1)t(1)+x(2)t(2)= 7200000

x(1)= v(1)(50+2(v(1)-3))= v(1)(44 + 2v)
t(1)= 5000/x(1)

x(2)= v(2)(50+2(v(2) +3))=v(2)(56+2v(2))
t(2)=5000/v(2)

=> 5000(44+2v(1))+5000(56+2v(2))=7200000

=> 100+ 2v(1) + 2v(2)= 1440
=> v(1) +v(2)=670 so is independant of x(1) and x(2) values.

I just had a quick go at I/14, and have not checked my working and reasoning, so likely loads of mistakes. (Also I'm not 100% sure what the last part wants)

With Juggins strategy to draw trice and then stop there are eight possible ways of not ending up with a score of 0, WWW, RRR, WRR (in any order) and WWR (in any order).
P(WWW)=1/8
P(RRR)=(3/10)^3=27/1000
P(WRR in any order)=$3\times\frac{9}{200}$
P(WWR in any order)=$3\times\frac{3}{40}$

$\frac{1}{8}+\frac{27}{1000}+\frac{9}{40}+\frac{27}{200}=\frac{64}{125}$
So 1-(64/125)=61/125, which is the probability he ends up with zero points in total.

The average score is $0\times\frac{61}{125}+3\frac{1}{8}+4\frac{9}{40}+5\frac{27}{200}+6\frac{27}{2000}=\frac{264}{125}$

Muggins with his obtained N points will have a probability of 1/2 to obtain N+1 points, probability 3/10 to obtain N+2 points and probability 1/5 to end up at zero points. Therefore the average score is $\frac{1}{2}(N+1)+\frac{3}{10}(N+2)=\frac{4}{5}N+ \frac{11}{10}$

The greatest possible average final score can be obtained by applying the above formula after each try and see if the average increases, and in that case continue - until the formula suggests the average will not increase by taking part in one more draw.

Edit: As David points out in post #86 the following should probably be added at the end for a complete solution :