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# STEP,I,II,III 1998 Mark Schemes

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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1. Re: STEP,I,II,III 1998 Mark Schemes
Good job guys.
2. Re: STEP,I,II,III 1998 Mark Schemes
I'll do Paper II, Q2, next.
3. Re: STEP,I,II,III 1998 Mark Schemes
Actually paper II question 2 and 3 is covered in Siklos' Advanced Problems in Core Mathematics booklet. (Question 10 is found in the booklet as well.) But I am sure everyone welcomes a look at a question from different perspective.

Since no one is doing question 6 ,7 paper II, I will be posting my solution.
4. Re: STEP,I,II,III 1998 Mark Schemes
STEP III Question 8

(i)Assume m is a unit vector. (Otherwise the claim simply doesn't hold, one could, in theory, have m so large (assuming b.m > 0, otherwise just reverse the direction of m) that a^2 > b.b - (b.m)^2 no matter the direction of m or the position of b)

Then, to find points on both the line and the sphere, we want to solve the below equation for lambda:

(since m is a unit vector, m.m = 1)

as required.

For there to be exactly one intersection, we have to have

The point of intersection, i.e. the solution to

is then

and so

We then have

as required.

(ii) For the sphere to be tangential, the minimum distance to the plane (i.e. the perpendicular distance) must be equal to the sphere's radius, (because the radius through a point p on the sphere's surface is always perpendicular to tangential plane through p (this is essentially what we just showed in (i) ). So the condition is

(iii) The points p on the second sphere are given by

Let now p be a point of intersection of the two sphere. Then p satisfies the above equation, and also the equation

Then:

Assume now the second sphere is such that as given in the problem. Then:

Let now

be a unit vector lying in the radius of the second sphere that's passing through p. Since the length of is a

Now the equation of the radius through p is given by

.

Sufficient condition that this radius intersects the first sphere in exactly one point is given by

equivalent to

and

which holds, and hence the radius of the second sphere passing through p intersects the first sphere in exactly one point, p. Then it also follows that

or,

,

i.e. the two radii through p are perpendicular, for any point of intersection p, QED.
Last edited by ukgea; 02-03-2007 at 13:51.
5. Re: STEP,I,II,III 1998 Mark Schemes
Question 6, STEP II 1998

For Curve 1,

For Curve 2,

To sketch the curve 1, note that y is always positive so the entire curve lies in the first quadrant for Also, from dy/dx above, you will notice that at (0,2) and (2pi,2) the curve is stationary. And at (pi,0), the gradient is 0/0 which is an indeterminate form, so the curve is not differentiable here (hence not continuous at this point). Substitute in a few more points if necessary to aid your sketching. Then put together all these information and proceed similarly for Curve 2. (refer to attachment for sketch of the 2 curves)

Equation of normal:

Substitute and solve for x where Hence normal to curve 1 intersects curve 2 at where the gradient of the tangent is

Hence, this normal is a tangent to the curve 2.
Attached Thumbnails

Last edited by khaixiang; 02-03-2007 at 15:26.
6. Re: STEP,I,II,III 1998 Mark Schemes
Question 7, STEP II 1998

a)

Since

,

It follows that f(x)>0 for 0<x<pi/2

b)

From (a),

It follows similarly that g(x)>0 for 0<x<pi/2

c)

From (b),

So h(x)>0 for 0<x<\pi/4

Also, rearrange h(x) to prove that

(d)
It can be seen that

Simplifying, we get , which from part (c) we know is true for 0<x<pi/4. So F'(x)<0
Last edited by khaixiang; 02-03-2007 at 16:48.
7. Re: STEP,I,II,III 1998 Mark Schemes
STEP III Question 6

(i) Tis' obvious with a diagram that the each of the four vertices has to be at a distance where a is the sidelength of the cube.

To see this slightly more rigorously, note that the distance between two different vertices of a cube is either , or . Now take two of the four vertices in question. Since they can't be adjacent, the distance between them has to be either or . But if it's , the two vertices are diametrically opposite each other. Then all of the other six vertices of the cube are adjacent to either of the first two vertices, which can't be, since we have four vertices, no two of which are adjacent. Hence the distance between any two of the four vertices is .

Now, the only configuration with four vertices all equidistant from each other is a regular tetrahedron.

For the second part of (i), a diagram would really, really be useful, but basically:

Consider a cube of side length . Then the tetrahedron formed by four of the vertices has unit side length. The tetrahedron can be obtained by "chopping off" four triangular pyramids from the cube, one for each of the vertices of the cube which is not a vertex in the tetraheron. The base of these pyramids (actually depending on what you call base, it can be different, but I will take one of the sides of the pyramid that is also part of a side of the cube) is a right isoceles triangle. The two equal sides of the base have length a, and hence the area of the base is . The height of the pyramid is a, and so the total volume of each pyramid is . The volume of the tetrahedron is then .

(ii) An octahedron can be considered to be two square pyramids glued together base to base. The height of these pyramids is the distance from the square's centre to the top, which by the Pythagorean theorem is (if the sides are of unit length)

The square has unit side length, and thus area 1, and so the volume of the octahedron is

.

(iii)Assume the sides of the cube have length a. For symmetry reasons, the eight faces formed by connected the six centres will all be equilateral triangles of the same size. Hence it's an octahedron. It's side length is the distance between the centres of two adjacent faces, which is equal to

The volume of this octahedron is then times that of the octahedron with unit length, i.e. the volume is

The volume of the tetrahedron was found previously to be , which is twice the octahedron's volume, as required.
Last edited by ukgea; 02-03-2007 at 19:10.
8. Re: STEP,I,II,III 1998 Mark Schemes
Step II 1998, Q2:

Thus

It's probably easiest to multiply this by 5/4 now to get

Comment: Nice concept, but incredibly tedious arithmetic. I used the fact that I knew 1/7 was a cyclic decimal quite a bit, another approach would have been just a long division by 7 at the end; that would have been very tedious to typeset hwoever.

I don't know if you'd say it's good or bad question design, but it is very easy to miss getting 6 good decimal places - I ended up redoing both parts once. For the first part, the true value of is 1.41421356, so if you're unlucky with choices of when to truncate, you actually only need a large error in the eighth decimal place to end up with 1.414213 instead of 1.414214.

For the second part, it's rather tempting to stop 1 term earlier; the last term is only 1.067 x 10^{-6} and it's easy to think it won't matter.
Last edited by DFranklin; 02-03-2007 at 19:45.
9. Re: STEP,I,II,III 1998 Mark Schemes
STEP III Q7.
The sketch is manageable. Since f(3) = 3 > 0, we need only try m=1,2 - and both fail.

If b'(x) = 0, then
3x^2 (e^(x/T) - 1) = x^3 (1/T) e^(x/T)

Since x>0, then
e^(x/T) (x/T - 3) + 3 = 0
=> f(x/T) = 0

So it has one turning point, namely the root of f(x/T) = 0. So the graph of y = b(x) for x>=0 looks like it increasing to the turning point, then decreases and has an asymptote x=0.

Now we do the integral. (Prepare for some dirty maths.)

(!!!)

So n=4.

For the last bit, we consider a triangle of base 3 and height b(m) = b(3). Then:
K T^4 = 2 * 1/2 * 3 * 27/(e^(3/T) - 1)
= 81/(e^(3/T) - 1)

Therefore,
K = 81/(T^4 (e^(3/T) - 1))
----------------------------

Of course, all of this could be wrong!

I think we have all the pure maths STEP III questions now.
Last edited by dvs; 02-03-2007 at 20:15.
10. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by dvs)
I think we have all the pure maths STEP III questions now.
And all the pur STEP II as well. I'm gonna start a thread for 1997
11. Re: STEP,I,II,III 1998 Mark Schemes
STEP II Q 10 is covered by Siklos on page 125 of his booklet.
12. Re: STEP,I,II,III 1998 Mark Schemes
So is there actually any interest in the Applied/Stats? It would be quite nice to do the papers completely, but I'm not sure I want to put myself down for the lot...
13. Re: STEP,I,II,III 1998 Mark Schemes
Hello!

Here's my attempt at Q12 from STEP II (one of the stats ones). Please scream at me if I've done/said something stupid. My answers seem sensible, but...

Attached as a Word document as I have no idea how I would go about drawing a tree diagram on TSR and have not the foggiest about LATEX.

love danniella

(I have removed the attachment - or at least attempted to - as I'd miscopied from one line to the next. Shoot me now. At least it wasn't in an exam. and many thanks to datr for spotting it. Corrected attachment is a couple of posts further on. )
Last edited by danniella; 02-03-2007 at 21:32.
14. Re: STEP,I,II,III 1998 Mark Schemes
For the first part of the question doesn't that fraction come out as 1/11 not 10/11?
15. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by datr)
For the first part of the question doesn't that fraction come out as 1/11 not 10/11?
Which fraction? The final answer to a)?
Um. I did go back and check the lot on a calculator and I still get 10/11.
*confuzzled*

love danniella
16. Re: STEP,I,II,III 1998 Mark Schemes
Hello!

Corrected version below. One day I will learn not to make stupid mistakes like writing 9/1000 in one line and 9/100 below it. Thanks!
Attached Files
17. STEP II 1998 Q12.doc (32.5 KB, 259 views)
18. Re: STEP,I,II,III 1998 Mark Schemes
Step III, 1998, Q12. I confess, my curiousity was piqued by the "WRONG" scrawled next to the formula you are supposed to derive.

OK, the formula is clearly wrong, since we must have P=0 when p=1 and vice versa. Instead, suppose P to be the opposite: that is, P is the probability that there is some pair of villages that you cannot get between. We will find a formula for P which bears a striking resemblance to the expression on the exam paper.

Now the way I organised my thoughts was to draw little graphs for ABCD:

Code:
    A
/|\
/ D \
/ / \ \
B-------C
and mark out various scenarios.

It is easy to see that if only 2 roads are down, then all 4 villages ABCD must be connected. (e.g. suppose AC is down, there is no other road we can remove that leaves a village disconnected. By symmetry this holds regardless of which road we choose instead of AC).

Slightly trickier is the case with 3 roads down. Note that 3 roads have 6 ends, so at least one village will have 2 roads down. Wlog suppose these roads are AB and AC. Looking at the resultant graph, we can see the only road we can remove to leave the graph disconnected is AD. So the only way 3 roads down can leave a village disconnected is if they are the 3 roads connecting that village. There are 4 ways of choosing that village, so the probability for this scenario is

Now suppose there are 4 roads down. Then there are 2 roads open. Those 2 roads have 4 ends, so the only way every village has at least one road open is if the two roads don't share a common village. In other words, a scenario like A joined to B, C joined to D. But then you can't get from A to C.

Clearly if there are more than 4 roads down the villages are also disconnected and so if there are 4 or more roads down, the villages are disconnected. The probability of this is just

Thus

Note that this result is almost, but not quite the expression given in the exam paper. I believe it is correct, having checked against a computer simulation.
Last edited by DFranklin; 19-03-2007 at 13:32.
19. Re: STEP,I,II,III 1998 Mark Schemes
STEP III, 1998, Q14

Write M = Max{X1,...,Xn}. M < k => Xi < k for each Xi. Now the Xi are independent, and P(Xi < k) = k/N. Thus P(M < k) = (k/N)^n.

Thus M has p.d.f.

(We are told to treat the range as continuous, so this differentiation is justifiable, though it makes your hair curl a bit!).

Thus
.

Thus .

The other estimator is much easier to deal with: E(Xi) = N/2 for each n, so .

Moving on,

So
So

Looking at Z2, Var(Xi)^2 = N^2/12, so .

So Z1 has smaller variance for n > 1 (and equal for n = 1) and so is the preferred estimator.
20. Re: STEP,I,II,III 1998 Mark Schemes
STEP III, 1998 Q11.

Let v be the speed of the pendulum bob when it hangs at an angle . Then , while by conservation of energy . So

Now , so . So

The time taken for the pendulum to reach the bottom position is:

and this is a quarter of the period. So the period

as desired.

Set . Then , where

Substitute . Then

Also, , so that

Now we do the "" bit.

Recall , so

At this point, it seems we're almost back where we started, but the difference is that if is small, we can expand this out as a binomial series (in ). Thus

so . Finally, note , so

Comment: this is a fairly fiddly and tedious question, with not a lot of actual mechanics content. Without knowing what expression you're trying to get to (in order to expand using a binomial), it's very easy to give up and assume you're not making any progress. The actual result is a fairly classic one, however.
21. Re: STEP,I,II,III 1998 Mark Schemes
STEP III, 1998, Q10

This question has probably the worst choice of notation I've ever seen. Not only are variables distinguished by suffix, but also by dashes. Plus, later on, we'll introduce a variable V, and we've got lots of squared terms, so we have to distinguish between . That's just asking for stupid errors!

To (slightly!) reduce the insanity, I will write for the velocities before the collision, . The mass of the spheres is actually irrelevant, but I guess I'd better call it m and show it doesn't matter what m is.

Conservation of momentum gives
Conservation of K.E. gives

Square the conservation of momentum equation to get:
Subtract twice this twice the conservation of KE equation to get:

and so the relative speeds are preserved.

In the case where , the above gives us , while conservation of momentum gives us and so we must have . Expanding both sides, we find and so and so the particles move at right angles after the collision.

For the second part, we again have and , but this time the energy equation gives us . We have . Write W for the final speed of the 2nd sphere, so .

We know . Squaring both sides we have:

.

Use and the standard result to get:

.

Now we can use the energy equation to eliminate , since and so .

So we find
.

Consider this as a quadratic in W. Since it must have real roots, we must have since .

Comment: To repeat myself, this question has the most annoying notational choices I've ever seen. Aside from everything else, having the "stationary sphere" be "sphere 1" instead of "sphere 2" is totally unnatural IMHO. I am embarrassed to say I spent about an hour trying (and failing) to form a quadratic in instead of , when as you can see it was actually quite easy once you did what the question asked!

This also seemed to be one of those STEP III questions where the first bit seems to expect a high degree of detail, but to sustain that level throughout the proof would require a ridiculously long answer. So as it went on, I started cutting down more and more on the explanation.

I'd love to see the examiners' report for this question. I bet there were an obscene number of stupid mistakes caused solely by the notation.

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