# STEP Maths I, II, III 2001 Solutions Watch

Announcements

(Updated as far as #40) SimonM - 28.03.2009

1: Solution by Glutamic Acid

2: Solution by Aurel-Aqua

3: Solution by Aurel-Aqua

4: Solution by SimonM

5: Solution by SimonM

6: Solution by Unbounded

7: Solution by brianeverit

8: Solution by SimonM

9: Solution by sonofdot

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by Mark13

14: Solution by SimonM

1: Solution by Glutamic Acid

2: Solution by ForGreatJustice

3: Solution by SimonM

4: Solution by Dadeyemi

5: Solution by Glutamic Acid

6: Solution by sonofdot

7: Solution by Glutamic Acid

8: Solution by Glutamic Acid

9: Solution by Glutamic Acid

10: Solution to tommm

11: Solution by Glutamic Acid

12: Solution by brianeverit

13: Solution by Aurel-Aqua

14: Solution by Aurel-Aqua

1: Solution by Aurel-Aqua

2: Solution by SimonM

3: Solution by tommm

4: Solution by tommm

5: Solution by SimonM

6: Solution by Dadeyemi

7: Solution by tommm

8: Solution by SimonM

9: Solution by tommm

10: Solution by brianeverit

11: Solution by tommm

12: Solution by brianeverit

13: Solution by Aurel-Aqua

14: Solution by SimonM

1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007

**STEP I:**1: Solution by Glutamic Acid

2: Solution by Aurel-Aqua

3: Solution by Aurel-Aqua

4: Solution by SimonM

5: Solution by SimonM

6: Solution by Unbounded

7: Solution by brianeverit

8: Solution by SimonM

9: Solution by sonofdot

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by Mark13

14: Solution by SimonM

**STEP II:**1: Solution by Glutamic Acid

2: Solution by ForGreatJustice

3: Solution by SimonM

4: Solution by Dadeyemi

5: Solution by Glutamic Acid

6: Solution by sonofdot

7: Solution by Glutamic Acid

8: Solution by Glutamic Acid

9: Solution by Glutamic Acid

10: Solution to tommm

11: Solution by Glutamic Acid

12: Solution by brianeverit

13: Solution by Aurel-Aqua

14: Solution by Aurel-Aqua

**STEP III:**1: Solution by Aurel-Aqua

2: Solution by SimonM

3: Solution by tommm

4: Solution by tommm

5: Solution by SimonM

6: Solution by Dadeyemi

7: Solution by tommm

8: Solution by SimonM

9: Solution by tommm

10: Solution by brianeverit

11: Solution by tommm

12: Solution by brianeverit

13: Solution by Aurel-Aqua

14: Solution by SimonM

**Solutions written by TSR members:**1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007

3

reply

Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions

0

reply

Report

#3

**I Question 2**

**Part (i) , .**

First it might help to find the points of equality: multiplying by x and rearranging, we get , . The roots are -1, 1 and 2. Now we must sketch the graph and notice that the inequality is satisfied for (the zero due to the asymptote) and .

**Part (ii) , .**

Because of the condition, squaring both sides still leaves an iff, giving: . Now by the condition, we can deduce that x > -1 because of the square root being always positive. Squaring again, , giving . From this, we can deduce that and , but our new condition, only gives us as the final solution.

3

reply

Report

#5

__STEP II Question 2__

Sketch not included, but pretty simple (y=0 for 0<x<N, and y=1 for N<x<2N)

i) for can be split into 3 regions:

for

for and

for

So by splitting the summation into these three ranges we get

as required

ii)

Splitting into 3 as before:

Using the formula for the sum of a geometric progression

we get

Which could probably be made simpler, so now we can see as then

2

reply

Report

#6

**I Question 2**

, . Well, we sketch the graph to show the two stationary points and crossing the x-axis at , , . We can conclude the two stationary points exist, so there must be a real x-value for them to be found.

.

Since the two stationary points are distinct and real, by , we get .

Considering , to have three roots, there must be a real solution to the quadratic, giving . This implies that the two other roots are and , and . Letting , , in any order (since they can be changed around due to triple-symmetry ). Knowing that and . Substituting into our previous result, we obtain: , as required. We can show that it is not necessary by letting , , .

2

reply

Report

#7

__STEP I - Q13__Given that plate is broken, the probability that it is broken by the mathematician is 0.25.

Therefore, the number of plates, , broken by the mathematician can be represented by the following binomial distribution:

~

We need to find when n=5.

0

reply

Report

#8

I shall get going.

Meikleriggs only has solutions since 2004 now

(Original post by

Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions

**SimonM**)Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions

0

reply

Report

#9

II/11:

Highest point of trajectory when v = 0, so 0 = v - gt and t = v/g. The horizontal distance travelled will be uv/g at this time, and the vertical will be .

P will hit the ground when its vertical displacement is , so one must solve the quadratic equation giving (using the quadratic formula). The horizontal distance travelled in this time will be .

Adding the distances: , as required.

If R > D then

, as required.

Highest point of trajectory when v = 0, so 0 = v - gt and t = v/g. The horizontal distance travelled will be uv/g at this time, and the vertical will be .

P will hit the ground when its vertical displacement is , so one must solve the quadratic equation giving (using the quadratic formula). The horizontal distance travelled in this time will be .

Adding the distances: , as required.

If R > D then

, as required.

0

reply

Report

#10

**III Question 1**

Case when :

.

Differentiating the given relation, we get:

.

Since it is true for n = 0, and it is also true for n + 1, by mathematical induction we conclude that it must also be true for n = 1, n = 2, and so forth for all positive values of n.

Substituting into the relation, we get . Thus , all even powers are zero. .

Maclaurin series with terms of even powers equal to zero is , so .

3

reply

Report

#11

(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).

On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).

I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.

2

reply

**Question 4, STEP I 2001**

Spoiler:

Show

Consider by De Moivre's Theorem.

Therefore

Dividing top and bottom by yields:

Since ,

Therefore

Therefore

The range of is . Considering that as the domain of , there are three places where . is periodic with period of

Therefore n =0,1,2

So

The range of is . This gives cos a smaller range between , so there will be two solutions, on negative and one positive, that is

Therefore

Dividing top and bottom by yields:

Since ,

Therefore

Therefore

The range of is . Considering that as the domain of , there are three places where . is periodic with period of

Therefore n =0,1,2

So

The range of is . This gives cos a smaller range between , so there will be two solutions, on negative and one positive, that is

3

reply

Report

#13

II/5:

dy/dx = 0 when x = 0, so there's a stationary point. , which equals 1 at x = 0, so it's a minimum point. Consider , letting u = x^2 gives du = 2x dx so it becomes

. From inspection, . No constant as the curve passes through the origin. At x = 1, y = and ; negative, so a maximum.

The other stationary point is as it's an even function (which is to be expected as the derivative is odd).

For C_2: . At x = 0, this equals 1; so minimum. At x = 1, this equals , negative, so maximum.

The derivatives of C_1 and C_2 are equal at x = 0 and x = 1. If the derivative of C_2 is greater than C_1 for 0 < x < 1 then it follows that . This relies on showing that for 0 < x < 1, which can be shown with a very simple sketch. As e^(x^2) and e^(x^3) < 1 for 0 < x < 1, when taking logarithms the inequality is reversed.

dy/dx = 0 when x = 0, so there's a stationary point. , which equals 1 at x = 0, so it's a minimum point. Consider , letting u = x^2 gives du = 2x dx so it becomes

. From inspection, . No constant as the curve passes through the origin. At x = 1, y = and ; negative, so a maximum.

The other stationary point is as it's an even function (which is to be expected as the derivative is odd).

For C_2: . At x = 0, this equals 1; so minimum. At x = 1, this equals , negative, so maximum.

The derivatives of C_1 and C_2 are equal at x = 0 and x = 1. If the derivative of C_2 is greater than C_1 for 0 < x < 1 then it follows that . This relies on showing that for 0 < x < 1, which can be shown with a very simple sketch. As e^(x^2) and e^(x^3) < 1 for 0 < x < 1, when taking logarithms the inequality is reversed.

2

reply

Report

#14

**STEP III 2001, QUESTION 3**

Consider the equation

where b and c are real numbers.

(i) Show that the roots of the equation are real and positive if and only if and , and sketch the region of the b-c plane in which these conditions hold.

Spoiler:

let the roots be

then by the quadratic formula:

Clearly , so both roots are positive iff

therefore

(1)

If , then , which cannot be true if the roots are real. Therefore (first condition), and thus we can square both sides of equation (1):

or (second condition)

Finally, (third condition), otherwise the roots would clearly be complex.

The three conditions obtained are equivalent to those stipulated in the question.

Pathetic sketch, where the curve drawn is the positive half of and the shaded area underneath (inclusive of the curve itself) is where the conditions hold:

Show

let the roots be

then by the quadratic formula:

Clearly , so both roots are positive iff

therefore

(1)

If , then , which cannot be true if the roots are real. Therefore (first condition), and thus we can square both sides of equation (1):

or (second condition)

Finally, (third condition), otherwise the roots would clearly be complex.

The three conditions obtained are equivalent to those stipulated in the question.

Pathetic sketch, where the curve drawn is the positive half of and the shaded area underneath (inclusive of the curve itself) is where the conditions hold:

(ii) Sketch the region of the b-c plane in which the roots are real and less than 1 in magnitude.

Spoiler:

By assumption, we can take , therefore these conditions become

, as these both imply the other inequalities involved. These lead to:

The right hand sides of these must clearly be positive, therefore . On this assumption we can square both inequalities, which (after some manipulation) become:

Plotting these on a graph, together with , we get something looking vaguely like:

Show

By assumption, we can take , therefore these conditions become

, as these both imply the other inequalities involved. These lead to:

The right hand sides of these must clearly be positive, therefore . On this assumption we can square both inequalities, which (after some manipulation) become:

Plotting these on a graph, together with , we get something looking vaguely like:

(Apologies for any dodgy copypasting, notepad is a ***** to work with sometimes.)

0

reply

Report

#15

(Original post by

It seems to me you are assuming independance between the plates - that is, knowing that a particular plate was broken by the mathematician doesn't affect the probability that another plate was broken by the mathematician. It's not at all obvious to me such an assumption is justified.

(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).

On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).

I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.

**DFranklin**)It seems to me you are assuming independance between the plates - that is, knowing that a particular plate was broken by the mathematician doesn't affect the probability that another plate was broken by the mathematician. It's not at all obvious to me such an assumption is justified.

(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).

On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).

I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.

0

reply

**STEP I 2001, Question 5**

Which happens to be true. (For more info, look into differentiation under the integral sign)

2

reply

Report

#17

(Original post by

First it might help to find the points of equality: multiplying by x and rearranging, we get , . The roots are -1, 1 and 2. Now we must sketch the graph and notice that the inequality is satisfied for (the zero due to the asymptote) and .

Because of the condition, squaring both sides still leaves an iff, giving: . Now by the condition, we can deduce that x > -1 because of the square root being always positive. Squaring again, , giving . From this, we can deduce that and , but our new condition, only gives us as the final solution.

**Aurel-Aqua**)**I Question 2****Part (i) , .**First it might help to find the points of equality: multiplying by x and rearranging, we get , . The roots are -1, 1 and 2. Now we must sketch the graph and notice that the inequality is satisfied for (the zero due to the asymptote) and .

**Part (ii) , .**Because of the condition, squaring both sides still leaves an iff, giving: . Now by the condition, we can deduce that x > -1 because of the square root being always positive. Squaring again, , giving . From this, we can deduce that and , but our new condition, only gives us as the final solution.

2

reply

Report

#18

(Original post by

Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.

**Glutamic Acid**)Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.

0

reply

**STEP I, Question 14**

Spoiler:

Show

Spoiler:

Show

The probability the best amongst the first n candidates is best or second best is the probability it is the best is there + probability that the second best is there - probability that both are there. This is

Let the candidates be where ability at maths is measured by one's initial.

Possibilities for interviews are

So the probability that the best guy is there is

The probability that one of the best two are there is

Let the candidates be where ability at maths is measured by one's initial.

Possibilities for interviews are

So the probability that the best guy is there is

The probability that one of the best two are there is

0

reply

X

### Quick Reply

Back

to top

to top