1987 Specimen STEP solutions thread Watch

brianeverit
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#81
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(Original post by ben-smith)
STEP III Q1
a)let I_n denote the product up to n.
I_2=\frac{3}{4},I_3=\frac{2}{3},  I_4=\frac{5}{8}, I_5=\frac{3}{5}, I_6=\frac{7}{12},...
By inspection, these numbers seem to be of the form I_n=\frac{n+1}{2n} so let's guess that.
We have already proved this to be the case for n=2 so let us presume, for the purpose of induction, that I_k=\frac{k+1}{2k}. Now let n=k+1:
I_{k+1}=I_k*(1-\frac{1}{(k+1)^2})=\frac{k+1}{2(  k)}*\frac{(k+1)^2-1}{(k+1)^2}=\frac{k+2}{2(k+1)} So our guess is thus true by mathematical induction.
b) Consider the binomial expansion of (1+x)^n:
(1+x)^n=\displaystyle\sum_{r=0}^  n \displaystyle \binom{n}{r}x^r. Letting x=-1 gives us:
\displaystyle\sum_{r=0}^n \displaystyle \binom{n}{r}(-1)^r=0

\Rightarrow \displaystyle\sum_{r=0}^k (-1)^r\displaystyle \binom{n}{r}=-\displaystyle\sum_{r=k+1}^n(-1)^r \displaystyle \binom{n}{r}

=\displaystyle\sum_{r=k+1}^n(-1)^{r-1} \displaystyle \binom{n}{r}

=\displaystyle\sum_{r=k+1}^n(-1)^{r-1} \displaystyle \binom{n-1}{r-1}+\displaystyle\sum_{r=k+1}^n(-1)^{r-1} \displaystyle \binom{n-1}{r}
Notice that the i+1 th element in the first summation cancels with the ith element in the second summation leaving the first element in the first summation so:
\displaystyle\sum_{r=0}^k (-1)^r\displaystyle \binom{n}{r}=(-1)^k \displaystyle \binom{n-1}{k}
As required.
Isn't there a problem in (b) in that the statement
  \displaystyle\sum_{r=0}^k (-1)^r\displaystyle \binom{n}{r}=-\displaystyle\sum_{r=k+1}^n(-1)^r \displaystyle \binom{n}{r} is only true for odd values of n with k=\frac{n-1}{2}
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brianeverit
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(Original post by DFranklin)
STEP III, Q15: It's easy to convince yourself that the villages all remain connected if less than 3 roads are down (draw diagrams if you wish to illustrate the possible cases).
If 3 roads are down and they all go to the same village, that village is cut off. Otherwise, the villages remain connected. Again, draw diagrams if you wish.
If 4 roads are down, then it's clearly impossible for 2 roads to connect all 4 villages. (The roads must connect at a village, and that only leaves 2 ends to connect 3 villages).
Even more obvious for 5 and 6 roads.

P(more than 3 roads down) = 15 p^4(1-p)^2 + 6p^5(1-p) + p^6 (binomial)
P(A cut off by having exactly 3 roads down) = p^3 (1-p)^3. So P(any village cut off by exactly 3 roads down) = 4p^3(1-p)^3

So P(villages are cut off) = p^3[4(1-p)^3 + 15p(1-p)^2+6p^5(1-p)+p^6]
So P(villages are connected) = 1 - p^3[4(1-p)^3 + 15p(1-p)^2+6p^5(1-p)+p^6]

When p = 1/2, this equals \dfrac{64 - 4 - 15 - 6 - 1}{64} = \dfrac{38}{64} = \dfrac{19}{32}

Wow - only 2 probablity questions out of 16. Hard times for probability specialists...
haven't you been just a little bit careless. Where you have taken out the factor of p^3 you have still left p^5 and p^6 inside the brackets.
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DFranklin
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(Original post by brianeverit)
haven't you been just a little bit careless. Where you have taken out the factor of p^3 you have still left p^5 and p^6 inside the brackets.
Looks that way, yeah.
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brianeverit
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(Original post by ben-smith)
Q12
A, B and C form a triangle. Let the angle at A be theta and the angle a C be phi. Let us further denote D to be the point on CB that also lies on the line perpendicular to CB and goes through A and T is the tension in the string.
The rod is in equilibrium so the forces vertically and the moments about A are balanced:
Mg=Tcos\phi

T(AD)=(L/2)sin\thetaMg
To find AD, notice that the are of triangle is 1/2(\alpha L (AD)) and also  \frac{1}{2}L^2\alpha sin(\pi-(\theta+\phi)).
Equating the two, we get:
AD=Lsin(\theta+\phi)
Substituting in from our first equations:
TLsin(\theta+\phi)=(L/2)sin\thetaMg \Rightarrow sin(\theta+\phi)=\frac{sin\theta  cos\phi}{2}

\Rightarrow sin\theta cos\phi+cos\theta sin\phi=\frac{sin\theta cos\phi}{2}

\Rightarrow cos\theta sin\phi=-\frac{sin\theta cos\phi}{2}
By the sine rule:
\frac{sin\phi}{L}=\frac{sin\thet  a}{\alpha L}\Rightarrow sin\phi=\frac{sin\theta}{\alpha}
So:
\frac{sin\theta cos\theta }{\alpha}=-\frac{sin\theta \sqrt{\alpha^2-sin^2\theta}}{2\alpha} \Rightarrow 2cos\theta=-\sqrt{\alpha^2-sin^2\theta}

\Rightarrow 4cos^2\theta=\alpha^2-sin^2\theta \Rightarrow cos^2\theta=\frac{\alpha^2-1}{3}
The function  cos^2\theta is greater that or equal to 0 and less than or equal to 1 but, in this case, we don't want the equality case as that would mean A, B and C would be collinear so:
1<\alpha<2
To find the tension:
 T=\frac{Mg}{cos\phi}

sin\phi=\frac{sin\theta}{\alpha} \Rightarrow cos\phi = 1/ \alpha(\sqrt{\alpha^2-\frac{4-\alpha^2}{3}})=\frac{2}{\alpha}(  \frac{\alpha^2-1}{3})^{1/2}
So:
T=\frac{Mg}{cos\phi)}=\frac{Mg \alpha}{2} (\frac{3}{\alpha^2-1})^{1/2}
As required.
The question asked for the inclination of the rod to the vertical. Where have you found that? I cannot see it anywhere.
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brianeverit
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Thought we should have a readable version of the solution for paper III question 10

(a) If  n is an odd number then it may be weritten as 4k\pm1 for some integer  k
Hence  n^2=16k^2 \pm 8k +1 and so leaves a remainder of 1 when divided by 8.
If  n is an even number then we may write it as  4k \mathrm {\ or\ } 4k \pm 2
So n^2=16k^2 \mathrm{\ or\ } 16k^2 \pm16k+4 and so leaves a remainder of 0 or 4 when divided by 8.
A number of the form  8n+7  obviously leaves a remainder of 7 when divided by 8.
The sum of any 3 square numbers must leave a remainder of  a+b+c where  a,b,c take values from the set  (0,1,4) and so cannot be equal to 7.
I.e. No number of the form  8n+7 can be the sum of 3 squares.
(b)  n=1 \Rightarrow 2^{3n+1}+3(5^{2n+1})=2^4+3\times  5^3=16+375=391=17 \times 23
So true for n=1
Assume that  2^{3k+1}+3(5^{2k+1})=17X for some integer X
\mathrm{\ replacing\ } k \mathrm{\ with\ }k+1 \mathrm{\ we\ have\ }
2^{3k+4}+3(5^{2k+3})=25(2^{3k+1}  +3(5{2k+1})- 25 \times 2^{3k+1}+2^{3k+4}
= 25 \times 17X-2^{3k+1}(25-2^3)=17(25X-2^{3k+1})
25X-2^{3k+1} is an integer hence true for n=k implies true for  n=k+1 hence true for all integers by induction.
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brianeverit
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(Original post by DFranklin)
STEP III, Q8.

(i) I'm going to be lazy and use capitals to denote vectors. Let the vertices of the tetrahedron be A, B, C, D. Sum of squares of edges = |A-B|^2+|A-C|^2+|A-D|^2+|B-C|^2+|B-D|^2+|C-D|^2.
Sum of squares of midpoints = [(A+B-C-D)^2 + (A+C-B-D)^2 + (A+D-B-C)^2] / 4.
So 4 x Sum of squares of midpoints = (A+B-C-D)^2 + (A+C-B-D)^2 + (A+D-B-C)^2 = ((A-D)+(B-C))^2 + ((A-B)+(C-D))^2 +((A-C)+(D-B))^2
= (A-D)^2+(B-C)^2 + 2(A-D).(B-C)+(A-B)^2+(C-D)^2+2(A-B).(C-D)+(A-C)^2+(D-B)^2 + (A-C).(D-B).
So, sufficient to show (A-D).(B-C) + (A-B).(C-D) + (A-C).(D-B) = 0. Some tedious expansion gives us A.B-A.C+C.D-B.D+A.C-A.D+B.D-B.C+A.D-A.B+B.C-C.D and everything cancels.

(ii) Take a = 3I + 2J + 6K, b = xI + yJ + K.
Then we have  (3x+2y+6) \leq \sqrt{3^2+4^2+6^2} \sqrt{x^2+y^2+1} = 7\sqrt{x^2+y^2+1}.
a.b = cos t |a| |b, where cos t is the angle between the vectors.

So a.b = |a| |b if and only if cos t = 1.
So we require 3I+2J+6K and xI+yJ+1 to be in the same direction. So x = 3/6, y=2/6, or x = 1/2, y=1/3.
(a) is easier if you take the origin to be one of the vertices.
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brianeverit
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An alternative solution for III/4
Taking the equations of the two tangents to be  py=x+ap^2,\ qy=x+aq^2
They intersect the directrix where y=\frac{ap^2-a}{p} \mathrm{\ and\ }y=\frac{aq^2-a}{q} respectively.
So  \frac{ap^2-a}{p}-\frac{aq^2-a}{q}=l \Rightarrow  ap^2q-aq+apq^2-ap=pql
 \Rightarrow a(p-q)+apq(p-q)=pql \Rightarrow a(p-q)(1+pq)=pql
The tangents intersect where  q(x+ap^2)=p(x+aq^2) \Rightarrow (p-q)x=apq(p-q) \Rightarrow x=apq,\ y=a(p+q)
 pq=\frac{x}{a},\ p+q=\frac{y}{a},\ p-q=\sqrt((p+q)^2-4pq)=\sqrt((\frac{y}{a})^2-\frac{4x}a})=\frac{1}{a}\sqrt(y^  2-4ax)
Subbing in a(p-q)(1+pq)=pql we thus have  \sqrt(y^2-4ax)(1+\frac{x}{a})=\frac{x}{a}l
Squaring both sides then gives the required result.
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DFranklin
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#88
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(Original post by brianeverit)
(a) is easier if you take the origin to be one of the vertices.
I prefer having the symmetry (i.e. not having a favoured vertex), but it's probably a case where the stylistic preference does result in slightly more writing.
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brianeverit
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(Original post by ben-smith)
STEP III Q11
 \frac{dx}{dt}=k/\alpha z ,\frac{dz}{dt}=k \Rightarrow z=kt+h

\Rightarrow  \frac{dx}{dt}=\frac{k}{\alpha(kt  +h)} 

\Rightarrow x=k/ \alpha(\frac{1}{k}ln(kt+h)+C). 

t=0, x=0 \Rightarrow C=-\frac{lnh}{k}

\therefore \alpha x=ln(\frac{kt}{h}+1) \Rightarrow t=\frac{h}{k}(e^{\alpha x}-1)
as required.
For the second plough:
z=k(time it takes for 2nd plough to go y metres-time taken for first plough to reach y)
=k(t-(e^{\alpha y}-1)\frac{h}{k}
is there a closing bracket missing here.
And since
\frac{dy}{dt}=k/\alpha z \Rightarrow 1/ \alpha \frac{dt}{dz}=z/k= t-(e^{\alpha y}-1)\frac{h}{k}
As required.
Differentiating this with respect to time we get:
Differentiating what with respect to t?
1/ \alpha \frac{dt}{dz}

=1/ \alpha (\frac{h\alpha}{k}e^{\alpha y}+ \alpha e^{\alpha y}(T-\frac{\alpha hy}{k})-\frac{h\alpha}{k}e^{\alpha y})=e^{\alpha y}(T-\frac{\alpha hy}{k})=(e^{\alpha y}-1)\frac{h}{k}+e^{\alpha y}(T-\frac{\alpha hy}{k})-e^{\alpha y}(T-\frac{\alpha hy}{k})=t-(e^{\alpha y}-1)\frac{h}{k}
I don't understand how this last step follows.
Which means it is a solution as it satisfies the equation.
They will collide when their times are equal and x=y. Notice how the time for second plough is the same as the one for the 1st +e^{\alpha y}(T-\frac{\alpha hy}{k})
This should disappear when they collide so:
T-\frac{\alpha hx}{k}=0 \Rightarrow x=\frac{Tk}{\alpha h}
As required.
Please see comments in red.
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brianeverit
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(Original post by ben-smith)
Q12
A, B and C form a triangle. Let the angle at A be theta and the angle a C be phi. Let us further denote D to be the point on CB that also lies on the line perpendicular to CB and goes through A and T is the tension in the string.
The rod is in equilibrium so the forces vertically and the moments about A are balanced:
Mg=Tcos\phi

T(AD)=(L/2)sin\thetaMg
To find AD, notice that the are of triangle is 1/2(\alpha L (AD)) and also  \frac{1}{2}L^2\alpha sin(\pi-(\theta+\phi)).
Equating the two, we get:
AD=Lsin(\theta+\phi)
Substituting in from our first equations:
TLsin(\theta+\phi)=(L/2)sin\thetaMg \Rightarrow sin(\theta+\phi)=\frac{sin\theta  cos\phi}{2}

\Rightarrow sin\theta cos\phi+cos\theta sin\phi=\frac{sin\theta cos\phi}{2}

\Rightarrow cos\theta sin\phi=-\frac{sin\theta cos\phi}{2}
By the sine rule:
\frac{sin\phi}{L}=\frac{sin\thet  a}{\alpha L}\Rightarrow sin\phi=\frac{sin\theta}{\alpha}
So:
\frac{sin\theta cos\theta }{\alpha}=-\frac{sin\theta \sqrt{\alpha^2-sin^2\theta}}{2\alpha} \Rightarrow 2cos\theta=-\sqrt{\alpha^2-sin^2\theta}

\Rightarrow 4cos^2\theta=\alpha^2-sin^2\theta \Rightarrow cos^2\theta=\frac{\alpha^2-1}{3}
The function  cos^2\theta is greater that or equal to 0 and less than or equal to 1 but, in this case, we don't want the equality case as that would mean A, B and C would be collinear so:
1<\alpha<2
To find the tension:
 T=\frac{Mg}{cos\phi}

sin\phi=\frac{sin\theta}{\alpha} \Rightarrow cos\phi = 1/ \alpha(\sqrt{\alpha^2-\frac{4-\alpha^2}{3}})=\frac{2}{\alpha}(  \frac{\alpha^2-1}{3})^{1/2}
So:
T=\frac{Mg}{cos\phi)}=\frac{Mg \alpha}{2} (\frac{3}{\alpha^2-1})^{1/2}
As required.
The angle between the rod and the vertical can be found very much more easily by elementary trigonometry as follows.
Since there are only 3 forces acting on the rod they must be concurrent and since rod is uniform it follows that D and B are the mid-points of AC and CE respectively. (see diagram)
Hence  \sin \phi=\frac{AE}{\alpha L} \mathrm{\ and\ } \sin \theta=\frac{AE}{L} \Rightarrow \sin \theta=\alpha \sin \phi
 \cos \phi=\frac{CE}{\alpha L} \mathrm{\ and\z } \cos \theta=\frac{BE}{L} \Rightarrow \cos \theta=\frac{\alpha}{2}\cos \phi
Hence,  \frac{1}{\alpha^2}\sin^2 \theta+\frac{4}{\alpha^2} \cos^2 \theta=\sin^2 \phi+\cos^2 \phi=1
\Rightarrow \sin^2 \theta+4\cos^2 \theta=1 \Rightarrow 1-\cos^2 \theta+4\cos^2 \theta=1 \Rightarrow 3\cos^2 \theta=\alpha^2-1
 \Rightarrow \theta=\cos^{-1} \sqrt \Left(\frac{\alpha^2-1}{3}\Right)
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brianeverit
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(Original post by DFranklin)
STEP III, Q16:

(i) If X_i is the number of extra children needed for the ith male child (that is, the number of children since the i-1th male child), then C = X_1 + ... + X_r, and so E[C] = E[X_1]+...+E[X_r]. The X_i are iid geometric distributions with p = 1/2, so E[X_i] = 2 and C = 2r.

(ii) Suppose the king ignores the rules and just has 2r-1 children. If he has r or more boys, clearly C < 2r, and conversely, if he has < r boys, C>=2r.
So P(C < 2r) = P(there are at least r boys from 2r-1 children). Since boys and girls are equally likely, P(there are at least r girls from 2r-1 children) = P(there are at least r boys from 2r-1 children).
But these two events are mutually exclusive, and 1 of them must occur. So P(there are at least r girls from 2r-1 children) = 1 - P(there are at least r boys from 2r-1 children).
So P(there are at least r boys from 2r-1 children) = 1 - P(there are at least r boys from 2r-1 children), and so P(there are at least r boys from 2r-1 children) = 1/2. That is, P(C < 2r) = 1/2.
Shouldn't that be E[C]=2r at the end of part (i)?
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DFranklin
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(Original post by brianeverit)
Shouldn't that be E[C]=2r at the end of part (i)?
Yesl.
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brianeverit
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Specimen Paper II
7. I=\int_0^{\lnK}=\Sigma_{r=1}^K[(r-1)(\lnr-\ln(r-1)]
=(\ln2-\ln1)+2(\ln3-\ln2)+3(\ln4-\ln3)+...+(K-1)( \ln K-\ln(K-1)
=(K-1) \ln K-(\ln2+\ln3+\ln4+...+\ln(K-1))
=K \ln K-(\ln2+\ln3+\ln4+...+\lnK)=K \ln K-\ln(K!)
=N \ln K-\ln{(N!)} if N=|K|
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physicsmaths
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(Original post by Farhan.Hanif93)
STEP I - Q11
Let the cross section of the cylinder have centre O and radius a. 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:

\Re (\mathrm{up} \mathrm{the} \mathrm{ramp}): \dfrac{P}{\sqrt 2} - \dfrac{\mu P}{\sqrt 2} - \dfrac{Mg}{\sqrt 2} - \nu R =0 (1)

\Re (\mathrm{perpendicular} \mathrm{to} \mathrm{ramp}): R - \dfrac{Mg}{\sqrt 2} - \dfrac{P}{\sqrt 2} - \dfrac{\mu P}{\sqrt 2} = 0 (2)

Taking moments about O: a\nu R = a\mu P (3)

Subbing (3) into (1): P(1-\mu -\sqrt 2 \mu) - Mg=0 \implies \boxed{P=\dfrac{Mg}{1-\mu(1+\sqrt 2)}}, as required.

Note that P&gt;0 \implies 1-\mu (1+\sqrt 2) &gt; 0 \implies \boxed{\mu &lt; \dfrac{1}{\sqrt 2 + 1} = \sqrt 2 -1}, 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): R = \dfrac{Mg(2- \sqrt 2 \mu)}{\sqrt 2(1-\mu (1+\sqrt 2))}

We require \nu R \geq \mu P

\implies \dfrac{\nu (2-\sqrt 2\mu)}{\sqrt 2(1-\mu (1+\sqrt 2))} \geq \dfrac{\mu}{1-\mu (1+\sqrt 2)}

\implies  \nu (\sqrt 2 - \mu)  \geq  \mu

By the earlier result, \sqrt 2 - \mu &gt; 1 so it's safe to divide through by it to obtain the required result \boxed{\nu \geq \dfrac{\mu}{\sqrt 2 - \mu}}.
in part a i keep on getting mg/(1-root2 mu-2mu) inthe first part the friction between the cylinder and the pushing object, which way is it acting i thought it is acting directly dowward perpendicular to the resultant at p is acting. Can a diagram be drawn please. It will make my day i keep on amking a little mistake somewhere and i think its to do with my frictional force at the left side of the circle!!!!!!!
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A Slice of Pi
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STEP I - Q5 (I'm aware there is already a solution, but mine may be slightly different in some respects)
Spoiler:
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(a) The first 9 positive integers contain 1 digit each. The next 90 contain 2 digits each. The next 900 contain 3 digits each and so on. The sum of these digits forms a rather aptly named arithmetico-geometric series (that is, the sum has characteristics of both an arithmetic and geometric series). We require "n" such that
(9 \times 1) + (90 \times 2) + (3 \times n) = 1000. A quick rearrangement shows that n = 270\frac{1}{3}. This tells us that we must stop writing down the digits part-way into the number 370 (after the 3), so we will have written down 369 complete numbers. Between 1 and 100, there are 20 number 7s. The same is true for 100 to 200 and 200-300. From 300-370 there are 8 number 7s. However, we stop counting just before the "7" in 370, so there are in fact only 7 that we count. The total number of 7s is then 20 + 20 + 20 + 7 = 67.

(b) 365! = 365 \times 364 \times 363 \times \ldots \times 3 \times 2 \times 1.
We now need to think of the different ways we can make up factors of 10. One way this is achieved is by multiplying 5 with different even numbers. There are obviously less multiples of 5 than multiples of 2 in this factorial expansion, so we will just consider the multiples of 5 for the moment. There are 73 multiples of 5 between 1 and 365. 14 of these are also multiples of 25, and 2 of these are also multiples of 125. Therefore, the number of zeros will be 73 + 14 + 2 = 89.
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MTripos
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Specimen Paper I - Q1 (first STEP question ever!)

a) Completing the square we have (y+8x)^2-64x^2+4x^2+24x=0, and so (y+8x)^2=60x^2-24x. This implies 12x(5x-2) \geq 0 which leads to x \geq 2/5 or x \leq 0. Similarly, to find restrictions on y, we wish to write the equation in the form [f(x,y)]^2=g(y), meaning g(y) \geq 0. Writing the original equation in the form 4x^2+8x(2y+3)+y^2=0 we have, once again, by completing the square that (2x+(4y+6))^2-(4y+6)^2+y^2=0. Rearranging and expanding eventually yields (5y+6)(y+2) \geq 0 and therefore y \geq -6/5 or y \leq -2.

b) First, decompose into partial fractions. This is quite standard and yields \frac{9}{(2-x)^2(1+x)}=\frac{1}{1+x}+\frac{1  }{2-x}+\frac{3}{(2-x)^2}. \frac{1}{1+x} can be written as \frac{1}{1-(-x)}, which is an infinite geometric series with first term 1 and common ratio -x, valid for |x|&lt;1. Hence, the coefficient of x^n for this fraction is simply (-1)^n. Similarly, \frac{1}{2-x} can be written as \frac{1/2}{1-(x/2)}, which is an infinite geometric series with first term 1/2 and common ratio x/2, valid for |x|&lt;2. It is not hard to see that the coefficient of x^n for this fraction is \frac{1}{2^{n+1}}. Lastly, it could be shown that 3(2-x)^{-2}=(3/4)(1-(\frac{x}{2}))^{-2}. Using the expansion of (1+x)^n we obtain the coefficient for this fraction as (3/4)(-1/2)^n(\frac{(-2)(-3)...(-2-n+1)}{n!}) (the last bracket is, in a way, \displaystyle \binom{-2}{n}). Putting everything together, the coefficient of x^n of the original fraction is (-1)^n+\frac{1}{2^{n+1}}+(3/4)(-1/2)^n(\frac{(-2)(-3)...(-2-n+1)}{n!}. This could be simlified further as (-2)(-3)...(-2-n+1)=(n+1)!(-1)^n, and so (3/4)(-1/2)^n(\frac{(-2)(-3)...(-2-n+1)}{n!})=\frac{(3/4)(n+1)}{2^n}. The final answer is (-1)^n+\frac{1}{2^{n+1}}+\frac{(3/4)(n+1)}{2^n}.

This is valid for |x|&lt;2 (not 100% sure about this).
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solC
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(Original post by MTripos)
x.
A slightly different way of doing part a):

Treating the equation as a quadratic in y we have

 y^2 + 16xy + 4x(x+6) = 0

Since it is given that y is real, the discriminant is greater than or equal to zero. That is,
 (16x)^2 - 16x(x+6) \geq 0

\Rightarrow 16x(15x-6) \geq 0

 \Rightarrow x \leq 0 or x \geq \frac{2}{5} as required

Similarly, treating the equation as a quadratic in x gives

 4x^2 + 8(2y+3)x + y^2 = 0

It is given that x is also real, hence

64(2y+3)^2 - 16y^2 \geq 0

 \Rightarrow 5y^2 + 16y + 12 \geq 0

 \Rightarrow (5y+6)(y+2) \feq 0

 \Rightarrow y \leq -2 or  y \geq \frac{-6}{5}
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cxs
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A lot of questions of paper 2 are unfinished!
I'll try them.
papers are herehttps://pmt.physicsandmathstutor.com...urther%20A.pdf
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#99
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#99
Q1,2,3,7,10,11,14 are finished by braineverit, Q9 can be found on the STEP vector module
13,16 can be found on the Stephen SIklos book
Last edited by cxs; 6 days ago
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#100
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#100
Q4
(a)

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let θ be sin-1(tanh x), and β be tan-1(sinh x). then sinθ=tanhx, tanβ=sinhx
[cos(θ)]2=1-sin2(θ)=1-tanh2x=sech2x
cos2(β)=1/(1 tan2(β))=1/(1 sinh2x)=sechx
so |cos(θ)|=|cos(β)|
a quick check of sign, from the original equation, θ and β should both larger than0, so θ=β. i.e.

(b)
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The same method as above, not hard, this question is held for you who open this spoiler. (lol I just don't want to type it haha)
for the last part, just consider the sign. easy.
Last edited by cxs; 6 days ago
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