1987 Specimen STEP solutions thread
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
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Re: 1987 Specimen STEP solutions threadLol, 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 ...
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Re: 1987 Specimen STEP solutions threadSTEP III Q9
Let
and 
and 
. Clearly 
Note that
Hence
is closed under matrix multiplication. 
Note that
s.t. 
, which is easily demonstrated to be unique. Hence has a unique identity element, as required.
Note that the cofactor matrix of
is given by 
.
Multiplying the corresponding first row elements of and 
yields that 
. This means that 
.
Note that all nxn matrices are invertible iff their determinant is nonzero, which is the case here. Hence all matrices in are invertible.
Take an invertible nxn matrix
such that 
and 
implies that 
by the associativity of matrix multiplication and thus, by , 
. If follows that any invertible nxn matrix has a unique multiplicative inverse.
Hence it is shown that
s.t. 
, as required.
It follows that is a group under matrix multiplication, as required. .
Let
be the subset of containing the 3x3 matrices of the above form with 
.
Note that
. This means that 
and therefore is closed.
Taking
implies that there exists a unique identity in H.
Also
, hence 
, which can be shown to be unique for each element of .
It follows that
Considering the group, note that ![\mathbf{MN} = \mathbf{NM} \iff [\mathbf{MN}]_{3,1} = [\mathbf{NM}]_{3,1} \iff x_1z_2 = x_2z_1](http://www.thestudentroom.co.uk/latexrender/pictures/0d/0d74ad250154dedd0896a48ebf307c45.png "\mathbf{MN} = \mathbf{NM} \iff [\mathbf{MN}]_{3,1} = [\mathbf{NM}]_{3,1} \iff x_1z_2 = x_2z_1")
. Taking 
ensures that there exists a matrix of the form 
in s.t 
, as required(?).
[To be honest, I'm not entirely sure what they want me to say for this last part.]
In fact, I've done very little on groups so it would be nice if anyone more knowledgeable could check through this to make sure it's right.
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Re: 1987 Specimen STEP solutions threadUnfortunately I'm not one of those knowledgeable people but I don't understand your notation in the last part. What does(Original post by Farhan.Hanif93)
STEP III Q9
Let and and . Clearly
Note that
Hence is closed under matrix multiplication.
Note that s.t. , which is easily demonstrated to be unique. Hence has a unique identity element, as required.
Note that the cofactor matrix of is given by .
Multiplying the corresponding first row elements of and yields that . This means that .
Note that all nxn matrices are invertible iff their determinant is nonzero, which is the case here. Hence all matrices in are invertible.
Take an invertible nxn matrix such that and
implies that by the associativity of matrix multiplication and thus, by , . If follows that any invertible nxn matrix has a unique multiplicative inverse.
Hence it is shown that s.t. , as required.
It follows that is a group under matrix multiplication, as required. .
Let be the subset of containing the 3x3 matrices of the above form with .
Note that. This means that and therefore is closed.
Taking implies that there exists a unique identity in H.
Also , hence , which can be shown to be unique for each element of .
It follows that
Considering the group, note that . Taking ensures that there exists a matrix of the form in s.t , as required(?).
[To be honest, I'm not entirely sure what they want me to say for this last part.]
In fact, I've done very little on groups so it would be nice if anyone more knowledgeable could check through this to make sure it's right.
![[\mathbf{MN}]_{3,1}](http://www.thestudentroom.co.uk/latexrender/pictures/24/24da69f659b0f9baf4922f0f4bca043e.png "[\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... -
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Re: 1987 Specimen STEP solutions threadWhat 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:
(i.e. there's a left identity)
(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. -
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Re: 1987 Specimen STEP solutions threadIn an nxn matrix M,(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 mean?
![[\mathbf{M}]_{ij}](http://www.thestudentroom.co.uk/latexrender/pictures/aa/aa420a0977d2c155bfe6a03ea6190ba8.png "[\mathbf{M}]_{ij}")
is the ij-th entry i.e. the element in the i-th row, j-th column.
This sounds similar to what I did.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...
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. -
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Re: 1987 Specimen STEP solutions threadNever knew about this. Thanks.(Original post by 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:
(i.e. there's a left identity)
(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. -
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Re: 1987 Specimen STEP solutions threadCan you have a look at the last bit of question 13 please. I haven't really convinced myself there...(Original post by DFranklin)
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Re: 1987 Specimen STEP solutions threadI 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".(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... -
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Re: 1987 Specimen STEP solutions threadWhich paper are you working from?(Original post by ben-smith)
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Re: 1987 Specimen STEP solutions threadSTEP 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:
Adding the two areas we mentioned earlier we get:
Multiplying by 7 gives the required result.Last edited by ben-smith; 14-01-2012 at 01:04. -
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Re: 1987 Specimen STEP solutions threadLet the cross section of the cylinder have centre O and radius
. Let the contact force exerted by the ramp on the cylinder have magnitude R. Note that there is no reaction force exerted by the horizontal surface as the cylinder is on the point of losing contact with it. Friction must be limiting on the vertical surface. On the point of rolling:
(1)
(2)
Taking moments about O:
(3)
Subbing (3) into (1):
, as required.
Note that
, as required.
Furthermore, in order for the cylinder to be rolling up the slope, we require the clockwise moment to be greater than or equal to the anti-clockwise moment.
By (2):
We require
By the earlier result,
so it's safe to divide through by it to obtain the required result 
.
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Re: 1987 Specimen STEP solutions threadLet A have initial speed
and immediately before the collision, speed 
. Immediately after the collision, let A have speed 
and let B have speed 
.
By the conservation of energy:
, as required.
Whilst sliding from O to disc B, disc A is subject to a constant deceleration of
. Hence 
Also by conservation of momentum:
(1)
By Newton's Law of Restitution:
. (2)
(1)+(2):
, as required.
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Re: 1987 Specimen STEP solutions threadLet the string have modulus
. Whilst in equilibrium, note that 
.
Whilst the particle moves in the described horizontal circle, let the string have length
; be inclined at an angle 
to the downwards vertical and exert a tension T on the particle. It follows that the circle must have radius 
.
(1)
(2)
Note that from (2),
. Furthermore, 
i.e. In order for the particle to be moving in such a circle, the string must stretched to a longer length than what it was when hanging in equilibrium, as required.
From (1):
.
Observe that
and that 
.
Hence the required result
follows.
![max[\theta] \Rightarrow max[Tan\theta]](http://www.thestudentroom.co.uk/latexrender/pictures/f6/f6225bf8e62adfd2b530e62cd50e7673.png "max[\theta] \Rightarrow max[Tan\theta]")
![\frac{d}{dx}[Tan\theta]=0](http://www.thestudentroom.co.uk/latexrender/pictures/71/71489003bdf1eb6793d6aed4620648cf.png "\frac{d}{dx}[Tan\theta]=0")
![\frac{d}{dx}[Tan\theta]=\frac{d(h^2+hd-x^2)}{(x^2+hd+h^2)^2}
\Rightarrow x^2=hd+h^2 \Rightarrow x=\sqrt{h(h+d)}](http://www.thestudentroom.co.uk/latexrender/pictures/64/64b3c18895a4fbe3a57c9d31e4fa83ae.png "\frac{d}{dx}[Tan\theta]=\frac{d(h^2+hd-x^2)}{(x^2+hd+h^2)^2}
\Rightarrow x^2=hd+h^2 \Rightarrow x=\sqrt{h(h+d)}")
![\frac{d^2y}{dx^2}+2\frac{dy}{dx}-3y=2e^x
\Rightarrow \frac{d^2y}{dx^2}-\frac{dy}{dx}+3(\frac{dy}{dx}-y)=2e^x
\Rightarrow \frac{dz}{dx}+3z=2e^x
\Rightarrow \frac{dz}{dx}e^{3x}+3e^{3x}z=2e^ {4x}
\Rightarrow \frac{d}{dx}[ze^{3x}]=2e^{4x} \Rightarrow z=\frac{e^x}{2}+C_1e^{-3x}](http://www.thestudentroom.co.uk/latexrender/pictures/27/273006db69433d76000b1683d187d8f1.png "\frac{d^2y}{dx^2}+2\frac{dy}{dx}-3y=2e^x
\Rightarrow \frac{d^2y}{dx^2}-\frac{dy}{dx}+3(\frac{dy}{dx}-y)=2e^x
\Rightarrow \frac{dz}{dx}+3z=2e^x
\Rightarrow \frac{dz}{dx}e^{3x}+3e^{3x}z=2e^ {4x}
\Rightarrow \frac{d}{dx}[ze^{3x}]=2e^{4x} \Rightarrow z=\frac{e^x}{2}+C_1e^{-3x}")
![\frac{dy}{dx}-y=\frac{e^x}{2}+C_1e^{-3x}
\Rightarrow \frac{d}{dx}[ye^{-x}]=1/2+C_1e^{-4x}
\Rightarrow y=\frac{xe^x}{2}-C_1e^{-3x}+C_2e^x
y(0)=1 \Rightarrow C_2=9/8](http://www.thestudentroom.co.uk/latexrender/pictures/de/debed4da9c5fd27bedab7157a3726852.png "\frac{dy}{dx}-y=\frac{e^x}{2}+C_1e^{-3x}
\Rightarrow \frac{d}{dx}[ye^{-x}]=1/2+C_1e^{-4x}
\Rightarrow y=\frac{xe^x}{2}-C_1e^{-3x}+C_2e^x
y(0)=1 \Rightarrow C_2=9/8")
![u^2=(x+1)^2-3 \Rightarrow x=\sqrt{u^2+3}-1 \Rightarrow \frac{dx}{du}=\frac{u}{\sqrt{u^2 +3}}
\therefore I=\displaystyle \int^{\infty}_{1} \frac{u}{(u^2+3)u\sqrt{u^2+3}}du
=\displaystyle \int^{\infty}_{1} \frac{1}{u^2+3}du=\frac{1}{\sqrt 3}\displaystyle \int^{\infty}_{1} \frac{1}{\sqrt3}\frac{1}{(u/\sqrt3)^2+1}du=\frac{1}{\sqrt3}[arctan(u/\sqrt3)]^{\infty}_{1}
=\dfrac{\pi}{3\sqrt3}](http://www.thestudentroom.co.uk/latexrender/pictures/03/03f154cf1ef635faaaf7cb53c4f29350.png "u^2=(x+1)^2-3 \Rightarrow x=\sqrt{u^2+3}-1 \Rightarrow \frac{dx}{du}=\frac{u}{\sqrt{u^2 +3}}
\therefore I=\displaystyle \int^{\infty}_{1} \frac{u}{(u^2+3)u\sqrt{u^2+3}}du
=\displaystyle \int^{\infty}_{1} \frac{1}{u^2+3}du=\frac{1}{\sqrt 3}\displaystyle \int^{\infty}_{1} \frac{1}{\sqrt3}\frac{1}{(u/\sqrt3)^2+1}du=\frac{1}{\sqrt3}[arctan(u/\sqrt3)]^{\infty}_{1}
=\dfrac{\pi}{3\sqrt3}")
![\displaystyle\sum _{r=0}^n \begin{pmatrix} n \\ r \end{pmatrix} \sin (2r+1)\alpha \right]= Im\left[e^{i\alpha}\displaystyle\sum _{r=0}^n \begin{pmatrix} n \\ r \end{pmatrix} (e^{2i\alpha})^r\right]](http://www.thestudentroom.co.uk/latexrender/pictures/78/7880d36a9827c2fc1588029e63b6701a.png "\displaystyle\sum _{r=0}^n \begin{pmatrix} n \\ r \end{pmatrix} \sin (2r+1)\alpha \right]= Im\left[e^{i\alpha}\displaystyle\sum _{r=0}^n \begin{pmatrix} n \\ r \end{pmatrix} (e^{2i\alpha})^r\right]")
![\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}]
=\Im[e^{i\alpha}(1+e^i\alpha)^n]=\Im[e^{i\alpha}(2cos\alpha e^{i\alpha})^n]
=2^n cos^n\alpha\Im[e^{i(n+1)\alpha}]=2^n cos^n\alpha sin(n+1)\alpha](http://www.thestudentroom.co.uk/latexrender/pictures/1d/1d7ec2c97b5c1210ba16d8e2dd3311f2.png "\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}]
=\Im[e^{i\alpha}(1+e^i\alpha)^n]=\Im[e^{i\alpha}(2cos\alpha e^{i\alpha})^n]
=2^n cos^n\alpha\Im[e^{i(n+1)\alpha}]=2^n cos^n\alpha sin(n+1)\alpha")
