2004 a STEP thread...

Daniel Freedman

STEP II, 2004, Question 3

Spoiler

Thanks for that solution, I needed some help for the last curve.
If you multiply the x term to get:

(x^2+x)(x-2)^4

, it's much easier and quicker to differentiate using the chain rule.

Reply 61

iainfs

STEP I, 2004, Q6

The first part can be done as a coordinate geometry problem, but it ran to several pages when I did it. I found a shorter way using vectors -- I'd appreciate somebody checking it over. Second part is simple.

Spoiler

Reply 62

Zuzuzu

12

II/11

Spoiler

Reply 63

brianeverit

9

2004 STEP I question 6

\text{Let mid-points of }BC,CA,AB \text{ be }D,E,F \text{ respectively}

\text{with usual notation, mid-point of }BC \text{ is } \mathbf{d}= \dfrac{1}{2}(\mathbf{b}+ \mathbf{c})

\text{So vector equation of line }Ad \text{ is } \mathbf{r}= \mathbf{a}+ \lambda \left( \dfrac{1}{2}( \mathbf{b}+ \mathbf{c})- \mathbf{a} \right)=(1- \lambda) \mathbf{a}+ \dfrac{\lambda}{2}( \mathbf{b}+ \mathbf{c})

\text{similarly for }BE \text{ we have } \mathbf{r}=(1-\mu) \mathbf{b}+ \dfrac{\mu}{2}(\mathbf{a}+ \mathbf{c})

\text{So at point of intersection we must have }(1- \lambda) \mathbf{a}+ \dfrac{\lambda}{2}( \mathbf{b}+ \mathbf{c})=(1- \mu) \mathbf{b}+ \dfrac{\mu}{2}( \mathbf{a}+ \mathbf{c})

\text{equating coefficients of } \mathbf{a}, \mathbf{b} \text{ and } \mathbf{c} \text{ we have } (1-\lambda)= \dfrac{\mu}{2},\dfrac{\lambda}{2}=(1- \mu) \text{ and }\dfrac{\lambda}{2}= \dfrac{\mu}{2}

\text{Fron whiuch we obtain } \lambda= \mu = \dfrac{2}{3}

\text{Hence, }AD \text{ and }BE \text{ meet at the point }\left( \dfrac{1}{3}(p_1+p_2+p_3),\dfrac{1}{3}(q_1+q_2+q_3)\right)

CF \text{ wil have equation }\mathbf{r}=\mathbf{c}+ \dfrac{\nu}{2}( \mathbf{a}+ \mathbf{b}- \mathbf{c}) \text{ and taking } \nu= \dfrac{2}{3} \text{ gives the same point}

\text{gradient of }AH= \dfrac{q_2+q_3}{p_2+p_3} \text{ and gradient of }BC= \dfrac{q_3-q_2}{p_3-p_2}

\text{so, if they are perpendicular then }\left( \dfrac{q_2+q_3}{p_2+p_30} \right) \left(\dfrac{q_3-q_2}{p_3-p_2} \right)=-1

\Rightarrow (q_2+q_3)(q_3-q_2)=(p_2+p_3)(p_3-p_2) \Rightarrow q_3^2-q_2^2=p_3^2-p_2^2 \Rightarrow p_2^2+q_2^2=p_3^2+q_3^2

\text {similarly, if }BH \text{ and }AC \text{ are at right angles then }p_1^2+q_1^2=p_3^2+q_3^2

\text{ and if both } p_2^2+q_2^2=p_3^2+q_3^2 \text{ and }p_1^2+q_1^2=p_3^2+q_3^2 \text{ then clearly }p_1^2+q_1^2=p_2^2+q_2^2

\Rightarrow (q_1+q_2)(q_2-q_1)=(p_1+p_2)(p_2-p_1) \Rightarrow \left( \dfrac{q_1+q_2}{p_1+p_2}\right) \left( \dfrac{q_2-q_1}{p_2-p_1} \right)=-1 \Rightarrow CH\text{ is perpendicular to }AB

(edited 8 years ago)

Reply 64

brianeverit

9

2004 STEP I question 10

(i) \text{For }u>0,\text{ }d_1= \dfrac{v^2-u^2}{6a} \text{ and }d_2= \dfrac{v^2}{2a}

\text{hence, }v^2-u^2=6ad_1 \text{ and }v^2=2ad_2 \Rightarrow 2ad_2-u^2=6ad_1 \text{ or }u^2=2ad_2-6ad_1

\text{but }u^2 \text{ must be positive so }6ad_1<2ad_2 \Rightarrow 3d_1<d_2

(ii) \text{if }u<0 \text{ then distance in negative direction before coming to rest is } \dfrac{u^2}{6a}

\text{hence, displacement from starting position after first movement is }d_1- \dfrac{u^2}{3a}

\text{so }d_1-\dfrac{u^2}{3a}= \dfrac{v^2-u^2}{6a} \Rightarrow d_1= \dfrac{u^2+v^2}{6a} \text{ and as before }d_2= \dfrac{y^2}{2a}

\text{eliminating }v^2 \text{ we have }2ad_2+u^2=6ad_1 \Rightarrow 6ad_1-2ad_2>0 \Rightarrow 3s_1>d_2

(i) \text{For }u>0,\text{ }d_1= \dfrac{v^2-u^2}{6a} \text{ and }d_2= \dfrac{v^2}{2a}

\text{hence, }v^2-u^2=6ad_1 \text{ and }v^2=2ad_2 \Rightarrow 2ad_2-u^2=6ad_1 \text{ or }u^2=2ad_2-6ad_1

\text{but }u^2 \text{ must be positive so }6ad_1<2ad_2 \Rightarrow 3d_1<d_2

(ii) \text{if }u<0 \text{ then distance in negative direction before coming to rest is } \dfrac{u^2}{6a}

\text{hence, displacement from starting position after first movement is }d_1- \dfrac{u^2}{3a}

\text{so }d_1-\dfrac{u^2}{3a}= \dfrac{v^2-u^2}{6a} \Rightarrow d_1= \dfrac{u^2+v^2}{6a} \text{ and as before }d_2= \dfrac{y^2}{2a}

\text{eliminating }v^2 \text{ we have }2ad_2+u^2=6ad_1 \Rightarrow 6ad_1-2ad_2>0 \Rightarrow 3s_1>d_2

\text{also }v^2>u^2 \Rightarrow 2ad_2>6ad_1-2ad_2 \Rightarrow 4ad_2>6ad_1 \Rightarrow d_2>3d_1 \text{ hence }d_2<3d_1<2d_2

\text{average speed of particle is } \dfrac{\text{total distance}} {\text{total time}} \text{ and total distance is }d_1+d_2 \text{ in each case}

\text{for }u>0, \text{ total time is } \dfrac{v-u}{3a}+ \dfrac{v}{a}= \dfrac{4v-u}{3a}

Unparseable latex formula:

\text{for }u<0 \text{ total time is } \dfrac{u}{3a}+\dfrac{v}{3a} + \dfrac{v}a}= \dfrac{4v+u}{3a}

\text{hence, average speeds are, for }u>0, \dfrac{3a(d_1+d_2}{4v-u} \text{ and for }u<0, \dfrac{3a(d_1+d_2)}{4v+u}

\text{Clearly } \dfrac{3a(d_1+d_2)}{4v-u}> \dfrac{3a(d_1+d_2)}{4v+u}\text { since numerator is same but denominator is smaller}

\text{i.e. Average speed is greater when }u>0

(edited 12 years ago)

Reply 65

brianeverit

9

2004 STEP I question 11

\text{Let }AB=BC=6a \text{ then }AD=2a,AP=3a \text{ and }BQ=3a

B\hat{A}C=B\hat{C}A=60^\circ

\text{If ladders do not slip then }

\text{Resolving horizontally for both ladders }F_A=F_C

\text{Resolving vertically }R_A+R_C=12W

\text{Taking moments about }A \text{ we have }

7W.2a \cos {60}+W.3a \cos {60}+4W.9a \cos{60}=R_C.12a \cos{60}

\text{i.e. }12R_C=53W \Rightarrow R_C= \dfrac{53}{12}W

\text{hence, }R_A=12W- \dfrac{53}{12}W= \dfrac{91}{12}W

\text{Taking moments about }B \text{ for }AB

Unparseable latex formula:

6a.F_A \sion{60}=6aR_A\cos{60}-3aW \cos{60}-7W.4a \cos{60}

\Rightarrow 3 \sqrt3F_A=3a \dfrac{91W}{12}- \dfrac{31Wa}{12}= \dfrac{29Wa}{4}

\text{so }F_A= \dfrac{29}{12 \sqrt3}W=F_C

AB \text{ slips if }F_A> \mu R_A \text{ i.e. If }\mu< \dfrac{F_A}{R_A} =\dfrac{29}{12 \sqrt3} \times \dfrac{12}{91}= \dfrac{29}{91 \sqrt3}

\text{and }BC \text{ slips if }F_C> \mu R_C \text{ i.e. If } \mu< \dfrac{F_C}{R_C}=\dfrac{29}{12 \sqrt3} \times \dfrac{12}{53}= \dfrac{29}{53 \sqrt3}

\text{hence, }BC \text{ slips and } \mu= \dfrac{29}{53 \sqrt3}

2004 STEP I question 13

(i) Pr(X \leq0.8)=Pr \text{all three numbers are }\leq0.8)=0.8^3=0.512

\text{F}(x)==Pr(X=x) \text{ is }x^3 \text{ for }0<x<1 \text{ so density function is }3x^2 \text{ for }0<x<1 \text{ and }0 \text{ elsewhere}

\text{hence, E}[X]= \displaystyle \int_0^1 x.3x^2dx= \displaystyle \int_0^13x^3dx= \displaystyle \left[\dfrac{3}{4}x^4 \right]_0^1= \dfrac{3}{4}

(ii) \text{Null hypothesis is rejected if all }N \text{ numbers are less than }0.8

\text{hence, with }5\%\text{ significance level, if }(0.8)^N<0.05

Unparseable latex formula:

\text{ i.e. } \left( \dfrac{2^3}{10} \right)^N< \dfrac{1}{20} \Rightarrow 2^{3N+1}<10^{N-1} \approx\left(2^{10/3} \righ)^N-1=2^{\frac{10N-10}{3}}

Unparseable latex formula:

\text{so we have }2^{9N+3}<2^{10N-10} \Rightarrow9N-3<10N-10 \RightarrowN>13

\text{so smallest integer value of }N \text{ is }14

\text{if }a=0.8 \text{ then the null hypothesis is certain to be rejected, i.e. Probability of rejection }=1

\text{if }a=9 \text{ null hypothesis is rejected with probability }\left( \dfrac{8}{9} \right)^14

Reply 67

brianeverit

9

2004 STEP I question 14

\text{We consider the possible sequences of gold and lead coins}

\text{Denoting the coins by }G \text{ and }L \text{ respectively}

\text{total number of sequences of }n \text{ }G\text{'s and }2\text{ } L\text{'s is } \dfrac{(n+2)(n+1)}{2} \text{ each eqully likely}

(i) \text{First Pirate gets some gold coins as long as he draws one first i,e with Probability }\dfrac{n}{n+2}

(ii) \text{For second Pirate to get some gold. It is only necessary that the two }L \text{'s are separated }

\text{in the sequences}

\text{Number of such sequences is } \dfrac{(n+2)(n+1)}{2}-(n+1)= \dfrac{n(n+1)}{2}

\text{So }Pr \text{(second pirate gets some gold) is }\dfrac{n(n+1)}{(n+1)(n+2)}= \dfrac{n}{n+2}

(iii) \text{All three pirates get some gold if sequence begins and ends with }G \text { and the }L \text{'s are separated}

\text{Number of such sequences is } \dfrac{n(n-1)}{2}-(n-1)= \dfrac{(n-1)(n-2)}{2}

\text{So }Pr \text{(all three get some gold) }= \dfrac{(n-1)(n-2)}{(n+1)(n+2)}

Reply 68

brianeverit

9

2004 STEP II question 9

\text{Taking moments about }A \text{(first diagram) we have }

Mg(p \sin \alpha-q \cos \alpha)=mgr \cos \alpha

\text{so, dividing through by }\cos \alpha \text{ we have}

Mgp \tan \alpha=Mgq+mgr \Rightarrow \tan \alpha= \dfrac{Mg+mr}{Mp}

\text{In new position (second diagram) we have }

Mg(q \cos \beta-p \sin \beta)=mgr \cos \beta

\text{i.e. }Mgp \tan \beta=Mgq-mgr \Rightarrow \tan \beta= \dfrac{Mq-mr}{Mp}

Unparseable latex formula:

\text{Hence, }\tan(\alpha-\beta)= \dfrac{\tan\aloha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{\frac{Mq+mr}{Mp}-\frac{Mq-mr}{Mp}}{1+\left(\frac{Mq+mr}{Mp}\right)\left(\frac{Mq-mr}{Mp}\right)}

=\dfrac{Mp(2mr)}{M^2p^2+(Mq+mr)(Mq-mr)}= \dfrac{Mmrp}{M^2p^2+M^2q^2-m^2r^2}

= \dfrac{2mMrp}{M^2(p^2+q^2)-m^2r^2} \text{ as required}

(edited 12 years ago)

2004.II.9(i).jpg8.0KB

2004.iI.9(ii).jpg6.5KB

Reply 69

brianeverit

9

2004 STEP II question 13

\text{She gains £1 on next draw if she draws another white and loses £}n \text{ if she draws a red}

\text{so expected gain is }1\times Pr\text{(white drawn) }-n\times Pr\text{(red drawn) }=\dfrac{b-n-r}{b-n}- \dfrac{nr}{b-n}= \dfrac{b-n-r-nr}{b-n}

\text{This is zero if }b-n-r-nr=0 \rightarrow n= \dfrac{b-r}{r+1}

\text{with each subsequent throw, the probability of drawing a red increases}

\text{so expected gain de3creases and hence will become negative}

\text{So to maximise her expectation she should stop on the next throw}

\text{after the value found above i.e. when }n= \text{ the integer part of } \dfrac{b-r}{r+1}+1

\text{When }b=2k \text{ say, and }r=1 \text{ she should stop after drawing }\left[ \dfrac{2k-1}{2}+1 \right]=\left[k+ \dfrac{1}{2} \right]=k= \dfrac{b}{2} \text{ balls}

Pr \text{(she draws }\dfrac{b}{2} \text{ white balls and no red)}= \dfrac{b-1}{b} \times \dfrac{b-2}{b-1} \times \dots \times \dfrac{b- \frac{b}{2}}{b- \frac{b}{2}-1}= \dfrac{1}{2}

\text{all other terms cancelling. So expected gain is £}\dfrac{b}{2} \times \dfrac{1}{2}=\text{£} \dfrac{b}{4}

Unparseable latex formula:

\text{Now suppose }b=2k+1, r=1 \text{ the critical value is now }\left[\dfrac{2k+1-1}{2} +1 \right]=k+1 \text{ i.e. }\left(\dfrac{1}2}(b+1)\right)

Pr \text{(she draws} \dfrac{1}{2}(b+1) \text{ white balls) is }\dfrac{b+1}{b}\times \dots \times \dfrac{b-\frac{1}{2}b-\frac{1}{2}}{b-\frac{1}{2}b+\frac{1}{2}}=\dfrac{\frac{1}{2}b-\frac{1}{2}}{b}=\dfrac{1}{2}-\dfrac{1}{2b}

\text{so excpected winnings in this case are } \left(\dfrac{1}{2}-\dfrac{1}{2b}\right) \left( \dfrac{1}{2}b+ \dfrac{1}{2}\right)=\dfrac{(b-1)(b+1)}{4b}=\dfrac{b^2-1}{4b}

Reply 70

brianeverit

9

2004 STEP II question 14

\text{With reference to the diagram }Pr(A \cup B\cup C) \text{ consists of the total area inside the circles,}

\text{This however will include the overlap of }A\&B, B\&C \text{ and }C\&A \text{ twice so we must remove each of these overlaps once.}

\text{We have now however removed the central region completely. i.e. The overlap of }A,B \text{ and }C

\text{and so this must be added again which gives the required result}

\text{Let the three puddings be denoted }A,B \text{ and }C

Pr(A \text{ does not contain a sixpence) is }\left(\dfrac{2}{3}\right)^r \text { so }Pr \text{(it does) is }1-\left(\dfrac{2}{3}\right)^r

\text{hence }Pr(A)=Pr(B)=Pr(C)=1-\left(\dfrac{2}{3}\right)^r \text{ and } Pr(A \cup B \cup C)=1

Pr \text{(all in one pudding)}= \left(\dfrac{1}{3}\right)^r \text{ so }Pr \text{(not all in same pudding)}=1-\left(\dfrac{1}{3}\right)^r

\text{i.e. }P(A \cap B)=P(B \cap C)=P(C \cap A)=1-\left(\dfrac{1}{3}\right)^r

Unparseable latex formula:

\text{so from result established above }1=3\left(1-\left(\dfrac{2}{3}\right)^r \right)-3\left(1-\left(\dfrac{1}{3}\right)^r+P(A \cap B \cap C)

\text {hence, }P(A \cap B \cap C)=1-3 \left(1- \left( \dfrac{2}{3} \right)^r \right)+3 \left(1- \left( \dfrac{1}{3} \right)^r \right)=1-3 \left( \dfrac{2}{3} \right)^r+3 \left( \dfrac{1}{3} \right)^r

\text{This is the probability that there is at least one sixpence in each pudding}

\text{and the complement is that at least one pudding has no sixpence}

\text{which thus has probability }3 \left( \dfrac{2}{3} \right)^r-3 \left( \dfrac{1}{3} \right)^r

\text{if this is to be less than }\dfrac{1}{3} \text{ then we require }3 \left( \dfrac{2}{3} \right)^r-3 \left( \dfrac{1}{3} \right)^r <\dfrac{1}{3} \Rightarrow 2^r-1<3^{r-2}

\text{We check integer values of }r.\text{ }r=1\Rightarrow 2^r-1=1>3^{-1}, r=2\Rightarrow 2^r-1=3>3^0

r=3\Rightarrow 2^r-1=7>3^1, r=4 \Rightarrow 2^r=-1=15>3^2, r=5\Rightarrow 2^r-1=31>3^4,r=6\Rightarrow2^r-1=63<3^4

\text{so smallest value of }r \text{ for which the inequality holds is }6

Pr\text{(2 sixpences in each pudding)}= \dfrac{^6 \text{C}_2.^4 \text{C}_2}{3^6}= \dfrac{15\times6}{3^6}=\dfrac{10}{81}

Unparseable latex formula:

Pr \text{(at least one sixpence in each)}=1-3\left( \dfrac{2}{3} \right)^r+3\left( \dfrac{1}[3} \right)^r=1-\dfrac{2^6}{3^5}+\dfrac{1}{3^5}=1- \dfrac{64+1}{243}= \dfrac{180}{243}

\text{so }Pr \text{(at least 2 sixpences in each | at least one in each)}=\dfrac{10}{81} \times \dfrac{243}{180}= \dfrac{2\times3}{36}= \dfrac{1}{6}

(edited 12 years ago)

Reply 71

brianeverit

9

2004 STEP III question 9

\text{We consider the forces acting on the hoop}

\text{If mouse runs at constant speed, Resolving tangentially }F=mg \sin 2\theta

\text{Resolving radially }N-mg \cos 2\theta= \dfrac{mv^2}{a}

\text{whilst if hoop remains stationary, taking moments about P}

F\times PM\cos\theta)=N\times PM\sin\theta \Rightarrow F\cos\theta=N\sin\theta

\text{i.e. }mga \sin 2\theta\cos \theta)=\left(mg \cos 2\theta + \dfrac{mv^2}{a} \right)a \sin \theta

\implies mga\sin2\theta\cos\theta-mga\cos2\theta\sin\theta=mv^2 \sin\theta

\implies mga \sin \theta=mv^2 \sin \theta \implies v^2=ag

\text{so, if mouse runs at a speed of } \sqrt{ag} \text{ the hoop remains in equilibrium}

Unparseable latex formula:

F \leq \mu N \implies mg \sin n\theta \leq \mu (mg \cos \theta+mg) \implies \mu \geq \dfrac{ \sdin \theta}{1+ \cos \theta}= \dfrac{2 \sin \frac{1}{2} \theta \cos \gfrac{1}{2} \theta}{2 \cos^2 \frac{1}{2} \theta}=\tan \frac{1}{2} \theta}

Unparseable latex formula:

\text{so mouse runs without slipping while } \thewtya \leq 2 \arctan \mu

\text{initially, hoop rotates in opposite sense to the mouse's motion round the circle}

(edited 11 years ago)

STEP2004.III.9.jpg16.5KB

Reply 72

brianeverit

9

2004 STEP III question 12
Last question from 2004

\text{Probability that player }i \text{ has correct shirt is } \dfrac{1}{m} \text{ and probability of wrong shirt is } \dfrac{m-1}{m}

\text{Hence, E}[C_1]=0 \times \dfrac{m-1}{m}+1 \times \dfrac{1}{m}= \dfrac{1}{m}

\text {Var}[C_1]=0^2 \times \dfrac{m-1}{m}+ 1^2 \times \dfrac{1}{m}-\left( \dfrac{1}{m} \right)^2=\dfrac{m-1}{m^2}

\text{Cov}[C_1C_2]= \text{E}[C_1C_2]-\text{E}[C_1] \text{E}[C_2]

\text{Now }C_1C_2=1 \text{ if }C_1=1 \text{ and }C_2=1 \text{ which has probability } \dfrac{1}{m} \times \dfrac{1}{m-1}= \dfrac{1}{m(m-1)}

\text{and probability that }C_1C_2=0 \text{ is }1- \dfrac{1}{m(m-1)}= \dfrac{m^2-m-1}{m(m-1)}

\text{so Cov}[C_1 C_2]= \dfrac{1}{(m(m-1)}- \left( \dfrac{1}{m} \right)^2= \dfrac{m-(m-1)}{m^2(m-1)}= \dfrac{1}{m^2(m-1)}

\text{E}[N]= \text{E}[C_1+C_2+ \dots +C_m]=m \times \text{E}[C_i]=m \times \dfrac{1}{m}=1

Unparseable latex formula:

\text{Var}[N]= \displaystyle \sum_{i=1}^m \text{Var}[C_i]+2 \sum_{i \neq j} \text{Cov}[C_iC_j]= \sum_{i=1}^m \left( \dfrac{1}{m}- \dfrac{1}{m^2} \right)+2 \sum_{i \neq j} \left( \dfrac{1}{m^2(m-1)} \right)}

=m \left( \dfrac{1}{m}- \dfrac{1}{m^2} \right)+2 \dfrac{m(m-1)}{2} \times \left( \dfrac{1}{m^2(m-1)} \right)=1- n\dfrac{1}{m}+ \dfrac{1}{m}=1

\text{A normal distribution is the limit of a large number of identical but independent}

\text{Bernoulli trials. Here we are not likely to have a very large number and the trials are not}

\text{independent so a normal distribution is not likely to be appropriate}

\text{However, since E}[N]= \text{Var}[N] \text{ a Poisson may well be reasonable}

\text{Possible arrangements of 4 numbers, none of which are in the correct position are as follows:}

(2,1,4,3),(2,4,1,3),(2,3,4,1),(3,1,4,2),(3,4,1,2),(3,4,2,1),(4,1,2,3),(4,3,1,2),(4,3,2,1)

Unparseable latex formula:

\text{so, probability that all shirts are wrong is } \dfrac{9}{4!}= \dfrac{9}245}= \dfrac{3}{8}

\text{using Po}(1) \text{ as an approximation we have }P(0)= \text{e}^{-1} \approx \dfrac{100}{272}

\text{so relative error is } \dfrac{\frac{3}{8}- \frac{100}{272}}{\frac{3}{8}}= \dfrac{102-100}{102}= \dfrac{2}{102} \approx 0.02 \text{ i.e. approximately }2\%

Reply 73

Farhan.Hanif93

18

Original post by Oh I Really Don't Care

TEP III, Question 5

Spoiler

Xero Xenith

...

As Xero pointed out in the STEP thread, there's a mistake with this solution (in particular, part (i)). It should go something more like the following:

STEP III Q5

SimonM

...

DFranklin

...

Could either of you add this to the first page? Thanks.

Reply 74

adrienne_om

Original post by Daniel Freedman

STEP II 2004, Question 7

Spoiler

Can someone please fix the fourth and third line from the bottom? I think they're typos. Thanks.

Reply 75

desijut

10

Original post by Daniel Freedman

STEP II, 2004, Question 3

Spoiler

Is (2,0) not a point of inflexion?

Reply 76

DFranklin

18

Original post by desijut

Is (2,0) not a point of inflexion?

No, it's definitely a minimum.

You're probably thinking that if f''(a) = 0 then x = a is a point of inflection. But strictly speaking, you need f''(x) to change sign at x = a, which it doesn't, here.

[For the specific function here, note that x(x+1) > 0 for x > 0, and obviously (x-2)^4 >=0 with equality iff x = 2. So in fact x(x+1)(x-2)^2 >= 0 with equality iff x = 2, and so it's a maximum.

I would say that the solution you've quoted doesn't really explain (to my mind) why it's a maximum, which may explain some of your confusion].

Reply 77

desijut

10

Original post by DFranklin

No, it's definitely a minimum.

You're probably thinking that if f''(a) = 0 then x = a is a point of inflection. But strictly speaking, you need f''(x) to change sign at x = a, which it doesn't, here.

[For the specific function here, note that x(x+1) > 0 for x > 0, and obviously (x-2)^4 >=0 with equality iff x = 2. So in fact x(x+1)(x-2)^2 >= 0 with equality iff x = 2, and so it's a maximum.

I would say that the solution you've quoted doesn't really explain (to my mind) why it's a maximum, which may explain some of your confusion].

Ok thanks, and yes, no reason was provided for the minimum and the way it was written said it was somehow implied, but it makes sense now.

Reply 78

snaptwig

Original post by Aurel-Aqua

2004 II, Question 12

Making the comparison:

h>m

\iff \sqrt{\frac{1}{a}\ln\left(\frac{2}{1+e^{-a}}\right)} > \frac{1}{2a}

\iff 2\ln\left(\frac{2}{1+e^{-a}}\right) > 1

\iff \frac{2}{1+e^{-a}} > e^{1/2}

\iff 2e^{-1/2}-1>e^{-a}

\iff ln(2e^{-1/2}-1)>-a

\iff -ln(2e^{-1/2}-1)<a

, as required.

Step II, Question 12.

There seems to be a substantial breakdown in my understanding- how is the jump made from

\iff \sqrt{\frac{1}{a}\ln\left(\frac{2}{1+e^{-a}}\right)} > \frac{1}{2a}

to

\iff 2\ln\left(\frac{2}{1+e^{-a}}\right) > 1

It seems as though the LHS is squared, to give a coefficient of

\frac{1}{a}

While the RHS has remained unaltered...

Clearly this method works as it generates the required inequality, but why does this not multiply out to

\iff \frac{1}{a} \ln\left(\frac{2}{1+e^{-a}}\right) > \frac{1}{4a^2}

? Which would also make the solution much harder to reach...

(edited 11 years ago)

Reply 79

snaptwig

Original post by Aurel-Aqua

2004 II, Question 12

Finding the mode:
To get the mode, we differentiate to find the maximum point (since

f(0) = 0

and

f(x)>0

for other

x

:

f'(m) = 0 = ke^{-am^2}(1-axm^2) \iff 2am^2 = 1

. In case

a<\frac{1}{2}

,

m^2

will be less than 1, so this case has no mode (since it would be outside the range

0\leq m \leq 1

. In case

a > \frac{1}{2}

,

\boxed{m=\frac{1}{2a}}

.

EDIT- in fact, going back to part b) the finding of the mode,

2am^2 = 1

leads to

\boxed{m=\frac{1}{2a}}

where it should instead lead to

\boxed{m=\frac{1}{\sqrt{2a}}}

If this is the case, then this solves the issue with the comparison, as

Unparseable latex formula:

\iff \sqrt{\frac{1}{a}\ln\left(\frac{ 2}{1+e^{-a}}\right)} > \frac{1}{\sqrt{2a}}

will multiply out as desired.

Also-

m^2

will be greater than 1 if

a<\frac{1}{2}

, which also does not give a modal value. But if

m^2

was less than 1, it would give a modal value.

Can someone confirm/correct what i've suggested in these two posts?

(edited 11 years ago)

2004 a STEP thread...

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