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1987 Specimen STEP solutions thread

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Lol, all 4 mechanics questions ftw.
Does anyone else want to do the honours of finishing off III?
I might be able to bu right now I'm feeling sleepy and don't trust my answers. A second ago I thought the area of a triangle was base*height ... :colondollar:

Reply 21

Farhan.Hanif93

STEP III Q9

...

Reply 22

Dirac Spinor

Original post by Farhan.Hanif93

STEP III Q9

...

Unfortunately I'm not one of those knowledgeable people but I don't understand your notation in the last part. What does

[\mathbf{MN}]_{3,1}

mean?

What I did for the last part was a bit simple but I simply multiplied two matrices in this form and then inverted the order and equated entries. This showed that one of the entries didn't matter so if you let the other two be 0 but this other one non zero then commutativity would hold for all B. I don't know whether it's valid...

Reply 23

DFranklin

What either of you have done is fine for "justfiy your answers". Justify here means "you're not going to get marks for just writing 'yes'".

The only other thing I'll say is that there are "minimal" axioms for a group: The closure + associativity axioms are the same, but the other two become:

\exists e \in G : ea = a \forall a \in G

(i.e. there's a left identity)

\forall b \in G, \exists a\in G : ab = e

(i.e. there's a left inverse)

It can be helpful to state that G is a group if these conditions apply and then you have less to verify.

Reply 24

Farhan.Hanif93

Original post by ben-smith

Unfortunately I'm not one of those knowledgeable people but I don't understand your notation in the last part. What does

[\mathbf{MN}]_{3,1}

mean?

In an nxn matrix M,

[\mathbf{M}]_{ij}

is the ij-th entry i.e. the element in the i-th row, j-th column.

What I did for the last part was a bit simple but I simply multiplied two matrices in this form and then inverted the order and equated entries. This showed that one of the entries didn't matter so if you let the other two be 0 but this other one non zero then commutativity would hold for all B. I don't know whether it's valid...

This sounds similar to what I did.

Question did feel a bit pointless, though. I was expecting a catch at the end i.e. either H wasn't a subgroup or G wasn't commutative under matrix multiplication but that wasn't the case.

Along with the fact that a lot of people have done nothing on groups by the time they get on to STEP, I can understand why they got rid of them.

Reply 25

Farhan.Hanif93

Original post by DFranklin

\exists e \in G : ea = a \forall a \in G

(i.e. there's a left identity)

\forall b \in G, \exists a\in G : ab = e

(i.e. there's a left inverse)

It can be helpful to state that G is a group if these conditions apply and then you have less to verify.

Never knew about this. Thanks.

Reply 26

Dirac Spinor

Original post by DFranklin

...

Can you have a look at the last bit of question 13 please. I haven't really convinced myself there...

Reply 27

DFranklin

Original post by ben-smith

Can you have a look at the last bit of question 13 please. I haven't really convinced myself there...

I think I would have to do the question myself to be give a "proper answer", but I'd be surprised if the correct answer isn't something along the lines of "at this point, the y' equals the acceleration due to gravity, and it follows that the reaction (which is acting against gravity) must be 0. After this point, the hoop will leave the table".

Reply 28

Dirac Spinor

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)}

(edited 12 years ago)

Reply 29

Farhan.Hanif93

Original post by ben-smith

...

Which paper are you working from? :p:

Reply 30

Dirac Spinor

STEP I Q3

\frac{d^2y}{dx^2}+2\frac{dy}{dx}-3y=2e^x[br]\Rightarrow \frac{d^2y}{dx^2}-\frac{dy}{dx}+3(\frac{dy}{dx}-y)=2e^x[br]\Rightarrow \frac{dz}{dx}+3z=2e^x[br]\Rightarrow \frac{dz}{dx}e^{3x}+3e^{3x}z=2e^{4x}[br]\Rightarrow \frac{d}{dx}[ze^{3x}]=2e^{4x} \Rightarrow z=\frac{e^x}{2}+C_1e^{-3x}

Using the initial conditions z(0)=1 so C_1=1/2. Now, to solve for y:

\frac{dy}{dx}-y=\frac{e^x}{2}+C_1e^{-3x}[br]\Rightarrow \frac{d}{dx}[ye^{-x}]=1/2+C_1e^{-4x}[br]\Rightarrow y=\frac{xe^x}{2}-C_1e^{-3x}+C_2e^x[br]y(0)=1 \Rightarrow C_2=9/8[br]

Reply 31

Dirac Spinor

Original post by Farhan.Hanif93

Which paper are you working from? :p:

I managed to acquire the papers missing. I'll upload in the OP.

Reply 32

Dirac Spinor

STEP I Q4

I=\displaystyle \int^{\infty}_{1} \dfrac{1}{(x+1)\sqrt{x^2+2x-2}}dx

Let:

u^2=(x+1)^2-3 \Rightarrow x=\sqrt{u^2+3}-1 \Rightarrow \frac{dx}{du}=\frac{u}{\sqrt{u^2+3}}[br]\therefore I=\displaystyle \int^{\infty}_{1} \frac{u}{\sqrt{u^2+3}u\sqrt{u^2+3}}du[br]=\displaystyle \int^{\infty}_{1} \frac{1}{u^2+3}du=\frac{1}{\sqrt3}\displaystyle \int^{\infty}_{1} \frac{1}{\sqrt3}\frac{1}{(u/\sqrt3)^2+1}du=\frac{1}{\sqrt3}[arctan(u/\sqrt3)]^{\infty}_{1}[br]=\dfrac{\pi}{3\sqrt3}[br]

(edited 10 years ago)

STEP I - Q7

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]

Reply 35

Dirac Spinor

STEP I Q6
Consider the sector formed by the points A,D and E and let Q be the centre of the whole coin. We want to find the area of the segment subtended on A and the area of the triangle QDE. Let QD=QE=x and 2y=DE: