1987 Specimen STEP solutions thread

Reply 60

Dirac Spinor

OP

10

Original post by brianeverit

There is a solutions thread for 1987 with ALL questions answered.

this is the specimen paper rather than the actual one.

Reply 61

brianeverit

9

Original post by ben-smith

this is the specimen paper rather than the actual one.

Apologies, I didn't read carefully enough.
Incidentall;y, several of these questions are included in Siklos's A.P.I.M. as follows
Paper I /6 APIM No.5
Paper II/13 APIM No.9, II/2 APIM No. 16 II/16 APIM No. 23
Paper III/13 APIM No. 24 III/14 APIM No. 26 III/11 APIM No. 31

(edited 10 years ago)

Reply 62

brianeverit

9

Specimen.I.16

\int_0^{\infty}f(t)dt=1\mathrm{\ so\ }\int_0^1 \frac{1}{2}dt+\int_1^{\infty}ke^{-2t}dt=1

i.e.

[\frac{t}{2}]_0^1+k[-\frac{1}{2}e^{-2t}]_1^{\infty}=1\Rightarrow \frac{1}{2}+\frac{k}{2}e^{-2}=1 \Rightarrow k=e^2

Unparseable latex formula:

E[T]=\int_0^1 \frac{t}{2}dt+e^2\int_1^{\infty}te^{-2t}dt =[\frac{t^2}{4}]_0^1+e^2[-\frac{t}{2}e^{-2t}]_1^{\infty}+e^2\int_1^{\infty}\frac{1}{2}e^{-2t}dt=\frac{1}{4}+\freac{1}{2}+e^2[-\frac{1}{4}e^{-2t}=\frac{1}{4}+ \frac{1}{2}+ \frac{1}{4}=1

F(t)= \int_0^t f(x)dx= \int_0^t \frac{1}{2}dx= \frac{t}{2} \mathrm{\ for\ }0\leq t<1

And

F(t)=\frac{1}{2}+\int_1^te^{2-2x}dx=1-\frac{1}{2}e^{2-2t} \mathrm{\ for\ }t>1

If

P(n)

is the probability of spending n 10 pence pieces then

P(1)=F(0.5)=\frac{1}{4},\ P(2)=F(1)-F(0.5)=\frac{1}{4}

P(3)=F(1.5)-F(1)=1-\frac{1}{2}e^{-1}-\frac{1}{2}=\frac{1}{2}(1-\frac{1}{e})

And in general

P(n)=F(n/2)-F((n-1)/2)=\frac{1}{2}[(1-\frac{1}{e^n/2}-\frac{1}{e^{(n-1)/2}}]

E(C)=10 \sum_{n=0}^\infty nP(n)=10\left[\frac{1}{4}+2\times\frac{1}{4}+ 3 \times \frac{1}{2}(1- \frac{1}{e})+4 \times \frac{1}{2}(\frac{1}{e}- \frac{1}{e^2})+5 \times \frac{1}{2}(\frac{1}{e^2}- \frac{1}{e^3})+...\right]

=10\left[\frac{1}{4}+\frac{1}{2}+\frac{3}{2}+\frac{1}{2}\left(e^{-1}+e^{-2}+e^{-3}+...\right)\right]=10\left[\frac{9}{4}+\frac{1}{2}\left( \frac{e^{-1}}{1-e^{-1}}\right)\right]

=22\frac{1}{2}+\frac{5}{2}\times\frac{1}{e-1}=22\frac{1}{2}+\frac{5}{e-1}

as required
This result assumes that all calls are multiples of 10 minutes and of course, most will not be exactly. Hence it will differ from average length of call times 20p

(edited 10 years ago)

Reply 63

brianeverit

9

Original post by ben-smith

STEP I Q2

[br]PB=\sqrt{x^2+h^2},PC=\sqrt{x^2+(d+h)^2}

Cosine rule:

Unparseable latex formula:

cos\theta=\dfrac{d^2-x^2-h^2-x^2-(h+d)^2}{-2\sqrt{(x^2+h^2)(x^2+(d+h)^2)}}[br]\Rightarrow Tan\theta=\frac{\sqrt{1-cos^2\theta}}{cos\theta}[br]=\dfrac{\sqrt{(x^2+h^2)(x^2+(d+h)^2)-(x^2+hd+x^2)^2}}{x^2+hd+h^2}}[br]=\dfrac{\sqrt{x^4+x^2h^2+2x^2hd+x^2d^2+h^2x^2+h^4+2h^3d+h^2d^2-x^4-2x^2hd-2x^2h^2-h^2d^2-2h^3d-h^4}}{x^2+hd+h^2}[br]=\dfrac{xd}{x^2+hd+h^2}

Note that

max[\theta] \Rightarrow max[Tan\theta]

so we require an x such that

\frac{d}{dx}[Tan\theta]=0

. (obviously will be a max because min occurs a x=infinity).

\frac{d}{dx}[Tan\theta]=\frac{d(h^2+hd-x^2)}{(x^2+hd+h^2)^2}[br]\Rightarrow x^2=hd+h^2 \Rightarrow x=\sqrt{h(h+d)}

tan theta is found more easily by noting that theta=angle APC minus angle APB and using tangent addition theorem.

Reply 64

brianeverit

9

Original post by abra-cad-abra

how did you know that that first angle was pi/7. i just guessed it as its a 7sided shape so makes sense and looking at the answer i was working towards it seemed right

Produce EQ and use exterior angle of triangle theorem.

Reply 65

brianeverit

9

Original post by ben-smith

STEP I Q7

z=1+e^{2i\alpha}=1+cos2\alpha+isin2\alpha[br]\Rightarrow |z|=\sqrt{(1+cos2\alpha)^2+(sin2\alpha)^2}=\sqrt{2+2cos\alpha}=2cos\alpha[br]arg(z)=arctan(\dfrac{sin2\alpha}{1+cos2 \alpha})=arctan(Tan(2\alpha/2))= \alpha

\displaystyle \Sigma^{n}_{r=0} \displaystyle \binom{n}{r} sin(2r+1)=\Im[\displaystyle \Sigma \displaystyle \binom{n}{r} e^{i(2r+1)\alpha}]=\Im[e^{i\alpha} \displaystyle \Sigma \displaystyle \binom{n}{r} e^{i2r\alpha}][br]=\Im[e^{i\alpha}(1+e^i\alpha)^n]=\Im[e^{i\alpha}(2cos\alpha e^{i\alpha})^n][br]=2^n cos^n\alpha\Im[e^{i(n+1)\alpha}]=2^n cos^n\alpha sin(n+1)\alpha[br]

I don't understand your summation.
Isn't

\displaystyle \Sigma^n_{r=0} \displaystyle \binom{n}{r}e^{2ir\alpha}=(1+e^{2i\alpha})^n \mathrm{\ not\ }(1+e^{i\alpha})^n

though there seems to be a compensating error in the next line and I agree with the final solution.

(edited 10 years ago)

Reply 66

brianeverit

9

Original post by ben-smith

STEP I Q7

z=1+e^{2i\alpha}=1+cos2\alpha+isin2\alpha[br]\Rightarrow |z|=\sqrt{(1+cos2\alpha)^2+(sin2\alpha)^2}=\sqrt{2+2cos\alpha}=2cos\alpha[br]arg(z)=arctan(\dfrac{sin2\alpha}{1+cos2 \alpha})=arctan(Tan(2\alpha/2))= \alpha

\displaystyle \Sigma^{n}_{r=0} \displaystyle \binom{n}{r} sin(2r+1)=\Im[\displaystyle \Sigma \displaystyle \binom{n}{r} e^{i(2r+1)\alpha}]=\Im[e^{i\alpha} \displaystyle \Sigma \displaystyle \binom{n}{r} e^{i2r\alpha}][br]=\Im[e^{i\alpha}(1+e^i\alpha)^n]=\Im[e^{i\alpha}(2cos\alpha e^{i\alpha})^n][br]=2^n cos^n\alpha\Im[e^{i(n+1)\alpha}]=2^n cos^n\alpha sin(n+1)\alpha[br]

I don't understand your summation.
Isn't

\displaystyle \Sigma^n_{r=0} \displaystyle \binom{n}{r}e^{2ir\alpha}=(1+e^{2i\alpha})^n \mathrm{\ not\ }(1+e^{i\alpha})^n

Reply 67

brianeverit

9

Original post by bogstandardname

Another solution for an uncompleted question, STEP I Question 8 is attached

You cannot have the variable in your integral as one of the limits.

Reply 68

brianeverit

9

Original post by Farhan.Hanif93

STEP I - Q11

You cannot assume that the friction force on the incline is

\nu R

since friction is not limiting there.

Reply 69

brianeverit

9

Original post by bogstandardname

Another solution for an uncompleted question, STEP I Question 8 is attached

Why bother using an integrating factor when the differential equation is variable separable?

Reply 70

Yung_ramanujan

Original post by brianeverit

Specimen.I.16

\int_0^{\infty}f(t)dt=1\mathrm{\ so\ }\int_0^1 \frac{1}{2}dt+\int_1^{\infty}ke^{-2t}dt=1

i.e.

[\frac{t}{2}]_0^1+k[-\frac{1}{2}e^{-2t}]_1^{\infty}=1\Rightarrow \frac{1}{2}+\frac{k}{2}e^{-2}=1 \Rightarrow k=e^2

Unparseable latex formula:

E[T]=\int_0^1 \frac{t}{2}dt+e^2\int_1^{\infty}te^{-2t}dt =[\frac{t^2}{4}]_0^1+e^2[-\frac{t}{2}e^{-2t}]_1^{\infty}+e^2\int_1^{\infty}\frac{1}{2}e^{-2t}dt=\frac{1}{4}+\freac{1}{2}+e^2[-\frac{1}{4}e^{-2t}=\frac{1}{4}+ \frac{1}{2}+ \frac{1}{4}=1

F(t)= \int_0^t f(x)dx= \int_0^t \frac{1}{2}dx= \frac{t}{2} \mathrm{\ for\ }0\leq t<1

And

F(t)=\frac{1}{2}+\int_1^te^{2-2x}dx=1-\frac{1}{2}e^{2-2t} \mathrm{\ for\ }t>1

If

P(n)

is the probability of spending n 10 pence pieces then

P(1)=F(0.5)=\frac{1}{4},\ P(2)=F(1)-F(0.5)=\frac{1}{4}

P(3)=F(1.5)-F(1)=1-\frac{1}{2}e^{-1}-\frac{1}{2}=\frac{1}{2}(1-\frac{1}{e})

And in general

P(n)=F(n/2)-F((n-1)/2)=\frac{1}{2}[(1-\frac{1}{e^n/2}-\frac{1}{e^{(n-1)/2}}]

E(C)=10 \sum_{n=0}^\infty nP(n)=10\left[\frac{1}{4}+2\times\frac{1}{4}+ 3 \times \frac{1}{2}(1- \frac{1}{e})+4 \times \frac{1}{2}(\frac{1}{e}- \frac{1}{e^2})+5 \times \frac{1}{2}(\frac{1}{e^2}- \frac{1}{e^3})+...\right]

=10\left[\frac{1}{4}+\frac{1}{2}+\frac{3}{2}+\frac{1}{2}\left(e^{-1}+e^{-2}+e^{-3}+...\right)\right]=10\left[\frac{9}{4}+\frac{1}{2}\left( \frac{e^{-1}}{1-e^{-1}}\right)\right]

=22\frac{1}{2}+\frac{5}{2}\times\frac{1}{e-1}=22\frac{1}{2}+\frac{5}{e-1}

as required
This result assumes that all calls are multiples of 10 minutes and of course, most will not be exactly. Hence it will differ from average length of call times 20p

Why is P(1) =1/4 . You have to pay 10p at beginning of the call. shouldnt it p(2)=1/4 and p(1)=1 ?

Reply 71

Yung_ramanujan

For Q15 I got E(C)=0 by symmetry and Var (C)= 8.5

Then for the context defining new random variable K as 36 independent random variables from C. So E(K)=0 and Var(K)=306.

Then would i use the normal distribution definition of outlier to determine if she should believe it to be a mistake?

Original post by brianeverit

?

Reply 72

brianeverit

9

Original post by Yung_ramanujan

Why is P(1) =1/4 . You have to pay 10p at beginning of the call. shouldnt it p(2)=1/4 and p(1)=1 ?

P(1) is the probability that the call lasts no more than half a minute.

Reply 73

brianeverit

9

Specimen Paper I. Questions 10 and 15.
These seem to be the only two without solutions so here are my attempts.

Specimen I.10.pdf26.9KB

Specimen I.15.pdf30.4KB

Reply 74

brianeverit

9

Paper II
Question 1

\frac{3x^2+2(a+b)x+ab}{x^3+(a+b)x^2+abx}=\frac{3x^2+2(a+b)x+ab}{x(x+a)(x+b)}=\frac{A}{x}+\frac{B}{x+a}+\frac{C}{x+b}

Unparseable latex formula:

[\impliesA(x+a)(x+b)+Bx(x+b)+Cx(x+a)=3x^2+2(a+b)x+ab

x=0 \implies abA=ab \implies A=1

x=-a \implies -aB(b-a)=3a^2-2a^2-2ab+ab \implies B=\frac{a^2-ab}{a^2-ab}=1

Comparing coefficients of

x^2,\ A+B+C=3\implies C=1

So

\frac{3x^2+2(a+b)x+ab}{x^3+(a+b)x^2+abx}=\frac{1}{x}+\frac{1}{x+a}+\frac{1}{x+b}

There are two special cases to consider,

a=b \mathrm{\ and\ }a+b=0

a=b \implies \mathrm{f}(x)=\frac{3x^2+4ax+a^2}{x^3+2ax^2+a^2x}=\frac{(3x+a)(x+a)}{x(x+a)^2}=\frac{1}{x}+\frac{2}{x+a}

(Cover up rule)

a+b=0 \implies b=-a \implies \mathrm{f}(x)=\frac{3x^2-a^2}{x^3-a^2x}=\frac{3x^2-a^2}{x(x+a)(x-a)} =\frac{1}{x}+\frac{1}{x+a}+\frac{1}{x-a}

If f

(x)=1 \mathrm{\ with\ }x,a,b, \mathrm{\ all\ positive\ integers\ then}

Case 1.

Unparseable latex formula:

a\neqb,

without loss of generality we may assume

a<b \mathrm{\ so\ }x+b>x+a>x

Hence,

\frac{1}{x}>\frac{1}{x+a}>\frac{1}{x+b} \implies \frac{1}{x}>\frac{1}{3} \implies x<3 \mathrm{\ i.e.\ }x=2

since we cannot have

x=1

So

\frac{1}{2+a}+\frac{1}{2+b}= \frac{1}{2}

with

a,b

integers, so

a=1,b=4

Case 2.

a=b \implies \frac{1}{x}+\frac{2}{x+a}=1

Again we must have

x=2 \mathrm{\ so\ } \frac{2}{2+a}=\frac{1}{2} \implies a=2

So the only possible positive integer solutions are

x=2,a=1,b=4 \mathrm{\ and\ } x=2,a=b=2

(edited 10 years ago)

Reply 75

brianeverit

9

Question 2

\frac{y-z}{b-c}=\frac{y-z}{(y^2-xz)-(z^2-xy)}=\frac{y-z}{(y^2-z^2)+x(y-z)}=\frac{1}{x+y+z}

Similarly, by symmetry,

Unparseable latex formula:

\frac{z-x}{c-a}=\frac{x-y}{a-b}=\frtac{1}{x+y+z}

(y-z)^2+(z-x)^2+(x-y)^2=\frac{(b-c)^2+(c-a)^2+(a-b)^2}{(x+y+z)^2}

Unparseable latex formula:

\implies (x+y+z)^2=\frac{2(a^2+b^2+c^2)-2(ab+bc+ca)}[2(x^2+y^2+z^2)-2(xy+yz+zx)}=\frac{a^2+b^2+c^2-bc-ca-ab}{(x^2-yz)+(y^2-zx)+(z^2-xy)}=\frac{a^2+b^2+c^2-bc-ca-ab}{a+b+c}=\Delta^2

Hence,

x+y+z=\Delta

as required.

y-z=\frac{b-c}{\Delta}\implies y=z+\frac{b-c}{\delta}

and similarly

x=z+\frac{a-c}{\delta}

Substituting in

x+y+z=\delta

gives

3z+\frac{a-c}{\delta}+\frac{b-c}{\delta}=\delta \implies z=\frac{\Delta^2-(a-c)-(b-c)}{3\Delta}=\frac{\frac{a^2+b^2+c^2-bc-ca-ab}{a+b+c}-a-b+2c}{3\Delta}

=\frac{(a^2+b^2+c^2-bc-ca-ab)-(a+b+c)(a+b-2c)}{3\Delta(a+b+c)}=\frac{3c^2-3ab}{3\Delta(a+b+c)}=\frac{c^2-ab}{\Delta(a+b+c)}

And, by symmetry,

x=\frac{a^2-bc}{\Delta(a+b+c)} \mathrm{\ and\ }y=\frac{b^2-ca}{\Delta(a+b+c)}

(edited 10 years ago)

Reply 76

brianeverit

9

Question 3
To prove

Unparseable latex formula:

(\cos\theta+i\sin\theta)^n=\cosn\theta+i\sin n\theta

Statement is obviously true for

n=1

so assume true for

n=k

So

(\cos\theta+i\sin\theta)^k=\cos k\theta+i \sin k\theta

hence

(\cos \theta+i\sin \theta)^{k+1}=(\cos k\theta+i\sin k\theta)(\cos\theta+i \sin\theta)

Unparseable latex formula:

=(\cos k\theta \cos \theta-\sin k\theta \sin\theta)+i(\cos k\theta \sin\theta+\sin k\thewta \cos \theta)

=\cos(k+1)\theta+i\sin(k+1)\theta

so true for

n=k\implies \mathrm{\ true\ for\ }n=k+1

and hence true for all integer values of

n

by induction.
Let

z=\cos\theta+i \sin\theta

then

Unparseable latex formula:

x^3\cos3y+2x^2 \cos2y+2x\cosy=\mathrm{\ Re}((xz)^3+2(xz)^2+2xz) \mathrm{\ and\ }x^3\sin3y+2x^2\sin2y+2x\siny= \mathrm{\ Im}((xz)^3+2(xz)^2+2xz)

\mathrm{Hence\ }((xz)^3+2(xz)^2+2xz=-1\implies ((xz)^3+2(xz)^2+2xz+1=0 \implies (xz+1)(x^2z^2-xz+1)=0 \implies xz=1 or \frac{1}{2}(1\pm\sqrt3)

Clearly we must have

|x|=1 \implies x=\pm1,z=-1 \mathrm{\ or\ }\frac{1}{2}\pm\frac{\sqrt3}{2}

x=1,z=-1\implies y=(2k+1)\pi

x=1,z=\frac{1}{2}\pm\frac{\sqrt3}{2} \implies y=(2k\pi\pm\frac{1}{3})\pi

x=-1,z=1\implies y=(2k+1)\pi

x=-1,z=-(\frac{1}{2}\pm\frac{\sqrt3}{2} \implies y=(2k\pm\frac{2}{3})\pi

\mathrm{So\ solutions\ are\ } x=1,y=(2k+1)\pi, x=1,y=(2k\pm\frac{1}{3})\pi,

x=-1,y=2k\pi,\ x=-1,y=(2k\pm\frac{2}{3})\pi.

(edited 10 years ago)

Reply 77

brianeverit

9

Paper II
Question 10
10.(i) If A is a toad then he is telling the truth and he is a frog - contradiction, so he is lying and hence must be a frog, and since he said B was a frog then this is also a lie and so B is a toad.
(ii) If C is a frog then his statement is a lie and would therefore be a contradiction. So C must be a toad and since he is telling the truth, D must be a frog.
(iii) If E is a toad then both G and H are toads but they cannot both be telling the truth. So E is a frog and G and H are not both toads which means that G is telling a lie and hence is a frog whilst H is telling the truth and hence is a toad.
(iv) I, J and K cannot all be frogs as otherwise I would be telling the truth. So I is a frog
Either J or K must be a toad. If K is a toad and J a frog or J a toad and K a frog then J is telling the truth, hence J must be the toad and K a frog

Reply 78

brianeverit

9

Question 11
Let A,B denote the particles projected from A and B re3spectively. Since they meet when both are at their greatest height then their maximum heights must be equal.
Max height for A is

\frac{V^2\sin^245^o}{2g} \mathrm{\ and\ for\ B\ }\frac{U^2\sin^230^o}{2g}

So

[br]]\frac{V^2\sin^245^o}{2g} =\frac{U^2\sin^230^o}{2g}\implies\frac{V^2}{4g}=\frac{U^2}{8g} \implies U^2=2V^2

as required.
Time to reach maximum height is given by

\frac{v\sin\theta}{g}

so A has traveled a distance

\frac{V^2\sin45^o \cos45^o}{g}=\frac{V^2}{2g} \mathrm{\ and\ B\ has\ traveled\ }\frac{U^2\sin30^o\cos30^o}{g}= \frac{U^2\sqrt3}{4g}= \frac{V^2\sqrt3}{2g}

hence

Unparseable latex formula:

a=\frac{V^2}{2g}+\frac{V^2\sqrt}{2g}=\frac{V^2}{2g}(1+\sqrt3)

as required.
If particles coalesce on impact then, by conservation of momentum, velocity of combined particle after impact in directionAB,

V' \mathrm{\ is\ given\ by\ }2mV'=mV\cos45^o-mU\cos30^o\implies V'=\frac{V}{2\sqrt2}(1-\sqrt3).

So particle hits ground with vertical velocity

\frac{V}{\sqrt2}

and horizontal velocity

\frac{V}{2\sqrt2}(\sqrt3-1)

so angle to the horizontal is

Unparseable latex formula:

\phi=\tan^{-1}\Left( \frac{V}{\sqrt2} \times \frac{2\sqrt2}{V(\sqrt3-L1)}\Right)=\tan^{-1}\frac{2}{\sqrt3-1}=\tan^{-1}(\sqrt3+1)

as required.

Reply 79

brianeverit

9

Question 14
MI of disc}

= \frac{Ma^2}{2}

and force acting on it is

-Mak(\omega_1-\omega_2)

at a distance of

\frac{a}{2}

from the axis
So

\frac{Ma^2}{2} \frac{d\omega_1}{dt}=-Mak(\omega_1-\omega_2)\frac{a}{2}\implies \frac{d\omega_1}{dt}=-k(\omega_1-\omega_2)

Similarly, the particle has MI

\frac{1}{2} \left(\frac{a}{2}\right)^2

and is acted on by a force

Mak(\omega_1-\omega_2)

So

\frac{Ma^2}{8}\frac{d\omega_2}{dt}=Mak(\omega_1-\omega_2)\implies\frac{d\omega_2}{dt}=4k(\omega_1-\omega_2)

Hence

\frac{d\omega_2}{dt}=-4\frac{d\omega_1}{dt}\implies \omega_2=4\omega_1+c

\omega_1=\Omega \mathrm{\ and\ }\omega_2=0 \mathrm{\ at\ }t=0\implies c=4\Omega

So

4\omega_1+\omega_2=4\Omega

as required.

\frac{d\omega_1}{dt}=-k(\omega_1-\omega_2)=-k\omega_1+4k(\Omega-\omega_1)=4k\Omega-5\omega_1

Separating the variables and integrating we have

\int\frac{d\omega_1}{5\omega_1-4\Omega}=-\int kdt

\implies \frac{1}{5}\ln(5\omega_1-4\Omega)=-kt+A

\omega_1=\Omega \mathrm{\ at\ }t=0 \implies A=\frac{1}{5}\ln\Omega \mathrm{\ so\ }\ln\frac{5\omega_1-4\Omega}{\Omega}=-5kt[br] \implies 5\omega_1-4\Omega=\Omega e^{-5kt} \implies \omega_1=\frac{\Omega}{5}(4+e^{-5kt}

1987 Specimen STEP solutions thread

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