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# STEP I, II, III 2002 Solutions

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1. Re: STEP I, II, III 2002 Solutions
STEP III, Question 1

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The area is

As , so the area tends to infinity as well

Volume of the solid of revolution is

As , the volume tends to 2\pi
Last edited by SimonM; 03-04-2009 at 14:35.
2. Re: STEP I, II, III 2002 Solutions
STEP III, Question 1

a)

Spoiler:
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ln(a) tends to infinty as a tends to infinty;

b)

Spoiler:
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Let and it follows that the integral is transformed to;

And IBP twice yields;

As a tends to infinty the volume tends to
Last edited by Oh I Really Don't Care; 05-04-2009 at 02:16.
3. Re: STEP I, II, III 2002 Solutions
is strictly increasing, but doesn't tend to infinity
4. Re: STEP I, II, III 2002 Solutions
STEP II 2002 Question 3

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If k=1, then and , therefore it is true for k=1.

If k=2, then and , therefore it is true for k=2.

If k=3, then and , therefore it is true for k=3.

Assume, for some k

Therefore by induction, for all positive integers k.

Now consider and with r<s. Assume that and have a common factor p

Since all Fermat numbers are odd, and cannot share a common factor greater than one, hence by contradiction, no two Fermat numbers have a common factor greater than 1.

All Fermat numbers are either prime or made up of prime factors. Since no Fermat numbers share a common factor greater than one, no Fermat numbers can have the same prime factor. Since there are infinitely many Fermat numbers, there must therefore be infinitely many prime numbers.
Last edited by sonofdot; 03-04-2009 at 20:50.
5. Re: STEP I, II, III 2002 Solutions
Question 1, STEP I, 2002
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We need to find the points of intersection between the two ellipses:

- ellipse 1
- ellipse 2

making y^2 the subject in the equation for ellipse 1:

and substitution into the equation for ellipse 2:

Subbing back into the equation for y^2:

Therefore the coordinates of intersection are (2,1) and (2,-1)

From a sketch, we can see that this means that the centre of a circle passing through those two points must lie on the x-axis.

Letting the coordinates of the centre of the circle be (a,0):

where r is the radius of the circle.

Substitute the point (2,1) in:

So the equation of the circle now becomes:

Last edited by Unbounded; 12-04-2009 at 10:09.
6. Re: STEP I, II, III 2002 Solutions
Some attached

Spoiler:
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(the end of the question 2 is the reason why I should not do maths late at night :/
Attached Thumbnails

Last edited by Dadeyemi; 05-06-2009 at 19:42.
7. Re: STEP I, II, III 2002 Solutions
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.
Attached Thumbnails

Last edited by Dadeyemi; 15-05-2009 at 23:01.
8. Re: STEP I, II, III 2002 Solutions
Question 6, STEP I, 2002
First Part
Let the height of the equilateral triangle base be x.

From pythag:

Therefore the area of this equilateral triangle base is given by:

Drawing a line down from the apex of the pyramid to the triangle base, letting h be the height, from pythagoras, we can see that:

The volume of the cone, V is given by,

Second Part
The area of the new base is to be found:

The height of the isoceles base, y, is given by:

And so the area of this isoceles base is

Now

Let the new height of the pyramid be k

Third Part
Considering the right angled triangle formed now with the height of this pyramid and the height of the equilateral triangle. Let denote the angle between the equilateral triangle and the horizontal.

Last edited by Unbounded; 13-04-2009 at 05:21.
9. Re: STEP I, II, III 2002 Solutions
STEP II 2002 Question 1

Part One
Part Two

Part Three

Last edited by sonofdot; 12-06-2012 at 16:03. Reason: Fixed mistake in part ii
10. Re: STEP I, II, III 2002 Solutions
STEP I 2002 Question 7

and with

(Note that in the interval , )

(i)
Let

(Note again that for sin(x) and cos(x) are never negative, so psin(x) + qcos(x) is never equal to 0)

Let

(ii)
Let and let

Last edited by sonofdot; 03-04-2009 at 20:54.
11. Re: STEP I, II, III 2002 Solutions
II/8:

For x < 0, dy/dx = -y ==> 1/y dy/dx = -1 ==> ln y = -x +c ==> y = Ae^(-x); as y = a when x = -1, a = Ae so A = a/e; therefore for x < 0.

For > 0, dy/dx = y ==> 1/y dy/dx = 1 ==> ln y = x + c ==> y = Be^x; as y = b when x = 1, b = Be so B = b/e therefore for x > 0.

If a = b, there is no jump.

Note that e^x - 1 is positive if x > 0, and negative if x < 0.

for x > 0.
Given that y = e^e when x = 1, e^e = Ae^(e - 1) so A = e;

for x < 0.

At x = 0, y = using the solution for x > 0. So for continuity, this must the same result when substituted into x < 0, so

(i)

(ii) .
12. Re: STEP I, II, III 2002 Solutions
II/7: (Scary scary vectors.)
??????????

Let the lines have direction vector which I'll denote by r, and without loss of generality let |r| = 1 so a^2 + b^2 + c^2 = 1;

.

b = 1 - a; c = 1 - a so
The solutions are a =1, b = 0, c = 0 and a = 1/3, b = 2/3, c = 2/3.

m3 has direction ; m4 direction .

Considering gives .

(i) A has position vector ; B .
Let P = , and Q =

AQ . BP = 0 so

. Showing that the latter is > 0; 1 - sqrt(6)/3 > 0 therefore 1 > sqrt(6)/3 therefore 1 > 6/9, which is true.

(ii)

So

The second equation gives . Substituting into the third gives . Substituting these into the first:
; multiplying out gives , so no non-zero solutions.
13. Re: STEP I, II, III 2002 Solutions
Question 2, STEP I, 2002
First Part

At stationary points, f'(x) = 0

which is the point that satisfies the conditions.
Second Part

Now the first and third brackets are always positive. The middle bracket alternates signs. When n is even, the middle bracket is negative, which implies f''(x) < 0 which implies it is a maximum. When n is odd, the middle bracket is positive, which implies it is a minimum. Q.E.D.
Third Part
Last edited by Unbounded; 05-04-2009 at 02:18.
14. Re: STEP I, II, III 2002 Solutions
I'm not checking if anyone posted the solutions, cause I completed these not long ago.

2002 II Question 1

as required.

and

, as required.

Using , , we obtain:

Last edited by Aurel-Aqua; 10-04-2009 at 12:08.
15. Re: STEP I, II, III 2002 Solutions
2002 II Question 11

Total weight (to check our calculus)

Total moment
.

We know that total moment , so , as required.

Draw out a diagram: A touching the ground, with R_A going up, F_A going right. B touching the wall, with F_B going up, R_B going left. Somewhere ebtween A and B, we have the centre of mass. We denote to be the distance of c.o.m. from A. Angle going from the horizontal to the line AB.

Equating the horizontal forces: . Noting that , we get . Noting that , we get
Equating the vertical forces: .
Equating the moments:

as required.

Not sure if I'm right, but I'll give it a go:
, and
Last edited by Aurel-Aqua; 10-04-2009 at 12:22.
16. Re: STEP I, II, III 2002 Solutions
2002 II Question 12

probability such that more than l coins of the L coins will produce less than m heads

Estimating q: , where z is the probability of a coin producing less than m heads.
We can say that , which gives:
for small z.
We then substitute to get as required.

z is the probability that less than m heads are made after M throws.

.

as required.

K = 7, k = 2, L = 500, l = 4, M = 100, m = 48, p = 0.6. , which is a large z, so we are not allowed to approximate as needed.
17. Re: STEP I, II, III 2002 Solutions
2002 II Question 13

,

since since it's a cumulative distribution function.

We know

We know

Using the condition found before, we deduce that

.

We know , so .

We also know that

So
, as required.
18. Re: STEP I, II, III 2002 Solutions
Question 10, STEP I, 2002
First Part
By Newton's law of restitution, for the first collision:

Assume for some n=k it is true, ie:

Looking at the (k+1)th collision:

And so by induction, we have shown it to be true:

Second Part
[With a diagram] we can see that:

[Also with a diagram] we can see that:

substituting for into , we get:

Third Part

u = v,

Similarly,

By we have:

Last edited by Unbounded; 11-04-2009 at 01:14.
19. Re: STEP I, II, III 2002 Solutions
Question 11, STEP I, 2002
First Part
Let the initial speed of be u and let and leave the collision at speeds and respectively.

Looking at momentum in the direction of initially:

Looking at momentum perpendicular to the initial direction of :

And considering the energy of the system:

Squaring we get:

And substituting in

Making the subject in we get:

Substituting this result into we get:

Applying product-sum formulae:

By double angle formulae:

I fear I've taken some unnecessarily long route
Second Part
If the angle between the particles after the collision is a right angle:

And also and

Applying this to the equation

we get

Last edited by Unbounded; 04-05-2009 at 12:16.

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