STEP maths I, II, III 1990 solutions Watch

Dystopia
Badges: 8
Rep:
?
#61
Report 11 years ago
#61
It's bizarre how after four days of no posting there are two independent posts within the space of two minutes.
0
reply
nota bene
Badges: 13
Rep:
?
#62
Report 11 years ago
#62
Haha, yeah nice timing!

I'm thinking of having a look at a few more questions later, seems to be a few that are doable.
0
reply
Dystopia
Badges: 8
Rep:
?
#63
Report 11 years ago
#63
Yup.

I think I've managed to do STEP II, Q16 (the amusing probability one), but since I haven't done any statistics whatsoever since my S1 exam in January, I won't risk writing up my solution unless someone else has got the same answer.

I got what the question asked me to show for the first part, and 11/63 for the second part.
0
reply
DFranklin
Badges: 18
Rep:
?
#64
Report 11 years ago
#64
(Original post by Dystopia)
Yup.

I think I've managed to do STEP II, Q16 (the amusing probability one), but since I haven't done any statistics whatsoever since my S1 exam in January, I won't risk writing up my solution unless someone else has got the same answer.

I got what the question asked me to show for the first part, and 11/63 for the second part.
Same here. This looks tedious to write out; it would at least have been courteous to make the colours start with different letters!

Edit: Q15 is a much nicer question (also rather easy!).
0
reply
insparato
Badges: 15
Rep:
?
#65
Report 11 years ago
#65
Bump. Thought people might want to try get a few more of these out .
0
reply
generalebriety
Badges: 14
Rep:
?
#66
Report 11 years ago
#66
Bump again.

2007 thread's had a bit of work done to it, but this one's had practically nothing; fresh meat! Rabite, you still up for updating the first post? :p:
0
reply
Dystopia
Badges: 8
Rep:
?
#67
Report 11 years ago
#67
(Original post by generalebriety)
Bump again.

2007 thread's had a bit of work done to it, but this one's had practically nothing; fresh meat! Rabite, you still up for updating the first post? :p:
I still can't be bothered to write up that stats question, but I was going through a 1991 paper the other day and realised that there was a rather nice alternate solution, so I think I'll go and write it up.
0
reply
generalebriety
Badges: 14
Rep:
?
#68
Report 11 years ago
#68
Well, I've done I/7. Might all be absolute rubbish, I'm a bit tired, and I can't be bothered latexing, but I shall scan.

Edit: oh, forgot the initial conditions. Sigh, tomorrowwwww.
Attached files
0
reply
Rabite
Badges: 8
Rep:
?
#69
Report 11 years ago
#69
Ooh I liked that question. I came out with something monstrous the first time though.

(Original post by The Billy)
Rabite, you still up for updating the first post?
Sure ~
I'll try to clean stuff up later today.
Oh, by the way...was it this thread that was littered with 1987 and 1989 discussion too, though? What do we do with those?
0
reply
Square
Badges: 12
Rep:
?
#70
Report 11 years ago
#70
EDIT: DAMMIT, just deleted my entire latex posting by clicking back by accident.

Sigh.
0
reply
Square
Badges: 12
Rep:
?
#71
Report 11 years ago
#71
I/5
i)
\displaystyle \text{I}=\int^3_1 \frac{1}{6x^2+19x+15}dx

Consider the quadratic on the denominator

6x^2+19x+15=0

Solve using the quadratic Formula:

\displaystyle x=\frac{-19\pm\sqrt{19^2-360}}{12}

\displaystyle\implies x=\frac{-19\pm1}{12}



\implies x=\frac{-18}{12},\frac{-20}{12}

Factorise using solutions from Quadratic Formula

\displaystyle6x^2+19x+15 = (3x+\frac{18}{4})(2x+\frac{20}{6  })

\displaystyle \text{I}=\int^3_1 \frac{1}{6x^2+19x+15}dx =\int^3_1 \frac{1}{(3x+\frac{18}{4})(2x+\f  rac{20}{6})}dx

Partial Fractions:

\displaystyle\frac{1}{(3x+\frac{  18}{4})(2x+\frac{20}{6})}=\frac{  \text{A}}{3x+\frac{18}{4}}+\frac  {B}{2x+\frac{20}{6}}

\displaystyle x=-\frac{10}{6} \implies B=-2

\displaystyle x=-\frac{6}{4} \implies A=3

\displaystyle\text{I}=\int^3_1 \frac{1}{(3x+\frac{18}{4})(2x+\f  rac{20}{6})}dx= \int^3_1 \frac{3}{(3x+\frac{18}{4})}-\frac{2}{(2x+\frac{20}{6})}dx

\displaystyle\text{I}=[\text{ln}|3x+\frac{9}{2}|-\text{ln}|2x+\frac{10}{3}|]^3_1

\displaystyle=[\text{ln}|\frac{3x+\frac{9}{2}}{  2x+\frac{10}{3}}|]^3_1

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\displaystyle=\text{ln}|\frac{9+ \frac{9}{2}}{6+ \frac{10}{3}}-\text{ln}|{\frac{3+\frac{9}{2}}{ 2+\frac{10}{3}}|


\displaystyle=\text{ln}\frac{81}  {56}-\text{ln}\frac{45}{32}

\displaystyle=\text{ln}\frac{36}  {35}

ii) I'm writing up now, not so sure about this bit though.
0
reply
Rabite
Badges: 8
Rep:
?
#72
Report 11 years ago
#72
*pat pat*
That's why I always hit ctrl+A and ctrl+C every ten seconds.
Every few minutes would suffice I suppose, but I'm neurotic.
0
reply
ukgea
Badges: 6
Rep:
?
#73
Report 11 years ago
#73
Hey! II/5 is exactly the same as 2007 III/6, except the 2007 one gives one extra middle step as a hint.

*Insert rant about STEP getting easier*

So er, as there's a solution in

http://www.thestudentroom.co.uk/show...67&postcount=6

maybe you could just link that, Rabite?

Though there's a slight difference in notation, (Seventeen years down the road, L, M has become R, S and A_i' has become B_i. Isn't it fascinating how fast maths is progressing?) so perhaps a link and an explanation is in order?
0
reply
generalebriety
Badges: 14
Rep:
?
#74
Report 11 years ago
#74
(Original post by Rabite)
Ooh I liked that question. I came out with something monstrous the first time though.


Sure ~
I'll try to clean stuff up later today.
Oh, by the way...was it this thread that was littered with 1987 and 1989 discussion too, though? What do we do with those?
No, I think that was my 1991 thread. Um, I'm ignoring it all to be honest. :p: If anyone wants to start 1987 and 1989 threads (but please god not yet!) and nick stuff from that thread, then meh, why not. :p:
0
reply
ukgea
Badges: 6
Rep:
?
#75
Report 11 years ago
#75
I/1

Let \varphi = \angle CDY. Then

\theta + \alpha + \varphi = \pi (1)

Consider the lengths of the altitudes dropped

from Y to CD,
from Y to AB and
from X to AD.

They are

r\sin\varphi,
r\sin\alpha and
r\sin\theta respectively.

Thus the areas of the triangles DCY, DYX, DXA are \frac{1}{2}r^2\sin\varphi, \frac{1}{2}r^2\sin\alpha and \frac{1}{2}r^2\sin\theta respectively, and the area of the quadrilateral is

\displaystyle K =  \frac{r^2}{2}\left(\sin \theta + \sin \alpha + \sin \varphi\right) (2)

When \alpha is fixed, and we wish to maximise this area:

\displaystyle K =  \frac{r^2}{2}\left(\sin \theta + \sin \alpha + \sin (\pi - \alpha - \theta)\right)

\displayste \frac{dK}{d\theta} = \frac{r^2}{2}\left(\cos \theta - \cos (\pi - \alpha - \theta)\right)

Setting the derivative to zero, we get

\cos \theta - \cos (\pi - \alpha - \theta) = 0

Now, both \theta and \pi - \alpha - \theta clearly belong in the interval (0, \pi) (the latter because it is the angle marked \varphi), and thus

\theta = \pi - \alpha - \theta

\displaystyle \theta = \frac{1}{2}(\pi - \alpha)

which is the maximum. This can be seen by working out the second derivative,

\displaystyle \frac{d^2K}{d\theta^2} = \frac{r^2}{2}\left(- \sin \theta - \sin (\pi - \alpha - \theta)\right)

which is clearly negative at  \theta = \frac{1}{2}(\pi - \alpha).




Note that in this case, the maximum area is given by

\displaystyle K_{max} = \frac{r^2}{2}\left(2\sin \frac{\pi - \alpha}{2} + \sin \alpha\right) (3)

If we wish to find the maximum when both \alpha and \theta vary, we can proceed by finding the value of \alpha for which the K_{max} is at a maximum. Thus,

\displaystyle \frac{dK_{max}}{d\alpha} = \frac{r^2}{2}\left(-\cos \frac{\pi - \alpha}{2} + \cos \alpha}\right)

Setting the derivative to zero:

\displaystyle -\cos \frac{\pi - \alpha}{2} + \cos \alpha = 0

Now, both \frac{1}{2}(\pi - \alpha) and \alpha are in (0, \pi), and thus we can write

\displaystyle - \frac{\pi - \alpha}{2} + \alpha = 0

from which follows

\displaystyle \alpha = \frac{\pi}{3} (4)

Again, to make that that this is a maximum, let us take the second derivative:

\displaystyle \frac{d^2 K_{max}}{d\alpha^2} = \frac{r^2}{2}\left( -\frac{1}{2} \sin \frac{\pi - \alpha}{2} - \sin \alpha\right)

which again can be seen to be negative, thus ensuring that (4) indeed gives us the maximum K_max.

Now, insertion into (3) gives us the maximum area as

\displaystyle K = 3\frac{r^2}{2}\sin \frac{\pi}{3}, i.e.

\displaystyle K = \frac{3\sqrt{3}r^2}{4}

Just for clarification, note that this happens when

\displaystyle \theta = \alpha = \frac{\pi}{3}.



Now you could stop reading here, but for the extra educational benefit or something, I couldn't resist pointing out that this question follows very easily from Jensen's inequality, introduced to us avid STEP-solvers in 2007 II/7:

Let f(x) be a concave function on some interval and let x_1, x_2, \ldots, \x_n lie in that interval. Jensen's inequality states that

\displaystyle \frac{1}{n}\sum_{k=1}^n f(x_k) \leq f\left(\frac{1}{n}\sum_{k=1}^n x_k\right)

and that equality holds if and only if x_1 = x_ 2 = \cdots = x_n
Now, in 2007 II/7, it was proven that \sin x was a concave function on (0, \pi).

Setting, in Jensen's inequality, n = 3 , x_1 = \theta , x_2 = \alpha, x_3 = \varphi , and using (2) and (1) we get

\displaystyle K = \frac{3r^2}{2} \cdot \frac{1}{3}\left(\sin \alpha + \sin \theta + \sin \varphi\right) \leq \frac{3r^2}{2}\sin \frac{\pi}{3}

with equality iff \theta = \alpha = \varphi.

This immediately gives us the maximum.

(Note that what we done here was really use the result in (i) of 2007 II/7, give or take a factor r^2 / 2 ).
0
reply
insparato
Badges: 15
Rep:
?
#76
Report 11 years ago
#76
(Original post by Speleo)
1987 III/6

\frac{dx}{dt} + 2x - 5y = 0
\frac{dy}{dt} + x - 2y = 2\cos t

at t=0; x=0, \frac{dy}{dt}=0

Can anyone get me started?
Laplace Transforms should sort this out, however i don't know whether in 1987 it was on the A level course.
0
reply
ukgea
Badges: 6
Rep:
?
#77
Report 11 years ago
#77
(Original post by Speleo)
1987 III/6

\frac{dx}{dt} + 2x - 5y = 0
\frac{dy}{dt} + x - 2y = 2\cos t

at t=0; x=0, \frac{dy}{dt}=0

Can anyone get me started?
I might just be incredibly stupid, but can't you just eliminate a variable (thus turning it into a second-order ODE) more or less right away (it's not might not be that right away, but easy to find once you have the idea)? I get

x'' + x = 10 \cos t

which seems very nice indeed.

I think the method by which you eliminate a variable is the exact same as the method you use to solve the linear equation system after you have Laplace-transformed (in the sense that a Laplace transform takes every step of my solution into a corresponding step in solving a linear equation system).
0
reply
Dystopia
Badges: 8
Rep:
?
#78
Report 11 years ago
#78
STEP III, Q6.

Spoiler:
Show

\left(\begin{array}{c c}X\\Y \end{array}\right) = \frac{2}{5}\left(\begin{array}{c c}9 & -2 \\-2 & 6 \end{array}\right) \left(\begin{array}{c c}x\\y \end{array}\right)

So applying the transformations to the vectors (1, 2) and (2, -1), we get (2, 4) and (8, -4), both of which are scalar multiples of the original vectors.


The matrix transforms the general point (x, y) to (X, Y) = (2/5)(9x-2y, 6y-2x).

5X = 18x - 4y
5Y = 12y - 4x

25X2 = 324x2 - 144xy + 16y2
25XY = 232xy - 72x2 - 48y2
25Y2 = 144y2 - 96xy + 16x2

Therefore the function x2 + y2 = 1 is transformed to:

8X^{2} + 12XY + 17Y^{2} = \frac{2000x^{2} + 2000y^{2}}{25} = 80

As required.

The area is equal to the original area multiplied by the determinant of the matrix transformation. The original area is \pi; the determinant is 8. Therefore the area is 8\pi.

To find the maximum value of X we find \frac{dX}{dY}

Differentiating implicitly, we get \frac{dX}{dY} = - \frac{6x + 17y}{8x + 6y}

There is a stationary point when Y = \frac{-6x}{17}

Treating the curves equation as a quadratic in Y, we get Y = \frac{-6x \pm \sqrt{1360 - 100X^{2}}}{17}

At the stationary point, 1360 - 100X^{2} = 0 \Rightarrow X^{2} = \frac{68}{5}

\Rightarrow X = 2\sqrt{\frac{17}{5}}

The point (0.6, 0.8) is transformed to the point (\frac{38}{25}, \frac{36}{25}).

At this point, \frac{dY}{dX} = -\frac{13}{21}

The equation of the tangent at this point is therefore 21Y + 13X = 50.
0
reply
Dystopia
Badges: 8
Rep:
?
#79
Report 11 years ago
#79
Attempt at STEP III, Q8.

Spoiler:
Show
Py'' + Qy' + Ry = \frac{d}{dx}(py' + qy) = py'' + (p' + q)y' + q'y

\Leftrightarrow P = p, \; Q = p' + q, \; R = q'
\Leftrightarrow Q' = p'' + q' = P'' + R
\Leftrightarrow P'' - Q' + R = 0

(x - x^{4})y'' + (1 - 7x^{3})y' - 9x^{2}y = (x^{3} + 3x^{2})e^{x}

P = x - x^{4}, \; Q = 1-7x^{3}, \; R = -9x^{2}
P'' - Q' + R = - 12x^{2} - (-21x^{2}) - 9x^{2} = 0

\Rightarrow Py'' + Qy' + Ry = \frac{d}{dx}((x-x^{4})y' - 3x^{3}y)

\Rightarrow (x-x^{4})y' - 3x^{3}y = \int{(x^{3} + 3x^{2})e^{x}} \; \mathrm{d}x = x^{3}e^{x} + c

\Rightarrow (1 - x^{3})y' - 3x^{2}y = x^{2}e^{x} + \frac{c}{x}

\Rightarrow y' + \frac{3x^{2}}{x^{3} - 1}y = \frac{x^{2}e^{x}}{1-x^{3}} + \frac{c}{x-x^{4}}

The integrating factor is x^{3}-1.

\Rightarrow (x^{3}-1)y' + 3x^{2}y = -x^{2}e^{x} - \frac{c}{x}

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\Rightarrow (x^{3}-1)y = -\int{x^{2}e^{x} + \frac{c}{x} \; \mathrm{d}x = -x^{2}e^{x} + 2xe^{x} - 2e^{x} - c\ln{x} + k


\Rightarrow y = e^{x}(\frac{x^{2}-2x +2}{1-x^{3}}) - \frac{c\ln{x}}{x^{3}-1} + \frac{k}{x^{3}-1}

Initial conditions: x=2, y=2e2, y'=0

From this I get y = e^{x}(\frac{x^{2} - 2x + 2}{1-x^{3}}) + \frac{56e^{2} \ln x}{x^{3}-1} + \frac{16e^{2}-56e^{2}\ln2}{x^{3}-1}

Which is utterly horrific, so presumably I've made an error.
1
reply
insparato
Badges: 15
Rep:
?
#80
Report 11 years ago
#80
12 + 2 + 2/3 = 44/3
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Arts University Bournemouth
    Art and Design Foundation Diploma Further education
    Sat, 25 May '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Wed, 29 May '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Thu, 30 May '19

How did your AQA GCSE Physics Paper 1 go?

Loved the paper - Feeling positive (115)
29.34%
The paper was reasonable (161)
41.07%
Not feeling great about that exam... (63)
16.07%
It was TERRIBLE (53)
13.52%

Watched Threads

View All