# STEP Maths I, II, III 2001 Solutions

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Thread starter 12 years ago
#1
(Updated as far as #40) SimonM - 28.03.2009

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
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Thread starter 12 years ago
#2
Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions
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12 years ago
#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.
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12 years ago
#4
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12 years ago
#5
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12 years ago
#6
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12 years ago
#7
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12 years ago
#8
I shall get going. (Original post by SimonM)
Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions
Meikleriggs only has solutions since 2004 now 0
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12 years ago
#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.
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12 years ago
#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 .
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12 years ago
#11
(Original post by Mark13)
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:
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.
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Thread starter 12 years ago
#12
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12 years ago
#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.
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12 years ago
#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:
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:
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.)
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12 years ago
#15
(Original post by 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.
Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.
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Thread starter 12 years ago
#16
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12 years ago
#17
(Original post by 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.
You're genius you know, I've got 98% in my C2 - but I was clueless in this question - you're really a maths genius 2
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12 years ago
#18
(Original post by Glutamic Acid)
Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.
Yeah, I'd actually expect Mark13's method to give the right answer, I just don't think it's justified. I was a little surprised to get a different answer doing it by hand, and I probably have made a mistake - that's why I said I'd need to check.
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12 years ago
#19
Can someone do 2001 III Q8?
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Thread starter 12 years ago
#20
STEP I, Question 14

Spoiler:
Show
The probability that the best candidate is amongst the first n candidates, is the same as the probability that any candidate is amongst the first n candidates. This probability is Spoiler:
Show
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