# STEP 2003 Solutions Thread

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1: Solution by Horizontal 8

2: Solution by Horizontal 8

3: Solution by Horizontal 8

4: Solution by Horizontal 8

5: Solution by Unbounded

6: Solution by Unbounded

7: Solution by Unbounded

8: Solution by Unbounded

9: Solution by Unbounded

10: Solution by Unbounded

11: Solution by Unbounded

12: Solution by Unbounded

13: Solution by AnonyMatt and Jkn

14: Solution by nota bene

1: Solution by tommm

2: Solution by tommm

3: Solution by Horizontal 8

4: Solution by tommm

5: Solution by Dadeyemi

6: Solution by Horizontal 8

7: Solution by SimonM

8: Solution by tommm

9: Solution by Elongar

10: Solution by TwoTwoOne

11: Solution by Rocious

12: Solution by SimonM

13: Solution by SimonM

14: Solution by TwoTwoOne

1: Solution by Adje

2: Solution by SimonM

3: Solution by Adje

4: Solution by tommm

5: Solution by SCE1912

6: Solution by SimonM

7: Solution by Daniel Freedman

8: Solution by Adje

9: Solution by brianeverit

10: Solution by tommm

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by toasted-lion

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 Horizontal 8

2: Solution by Horizontal 8

3: Solution by Horizontal 8

4: Solution by Horizontal 8

5: Solution by Unbounded

6: Solution by Unbounded

7: Solution by Unbounded

8: Solution by Unbounded

9: Solution by Unbounded

10: Solution by Unbounded

11: Solution by Unbounded

12: Solution by Unbounded

13: Solution by AnonyMatt and Jkn

14: Solution by nota bene

**STEP II:**1: Solution by tommm

2: Solution by tommm

3: Solution by Horizontal 8

4: Solution by tommm

5: Solution by Dadeyemi

6: Solution by Horizontal 8

7: Solution by SimonM

8: Solution by tommm

9: Solution by Elongar

10: Solution by TwoTwoOne

11: Solution by Rocious

12: Solution by SimonM

13: Solution by SimonM

14: Solution by TwoTwoOne

**STEP III:**1: Solution by Adje

2: Solution by SimonM

3: Solution by Adje

4: Solution by tommm

5: Solution by SCE1912

6: Solution by SimonM

7: Solution by Daniel Freedman

8: Solution by Adje

9: Solution by brianeverit

10: Solution by tommm

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by toasted-lion

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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**STEP III, Question 14**

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Given that's what I've written down on my whiteboard, I'll assume that was a typo

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#7

STEP I question 4

(Not 100% confident, since I always make mistakes with intervals)

(Not 100% confident, since I always make mistakes with intervals)

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#10

I would've never through of that

But does my solution check out?

PS: I think you missed post #5

But does my solution check out?

PS: I think you missed post #5

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#11

STEP II Q3

First Part:

Second part:

Last part:

First Part:

Spoiler:

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Second part:

Spoiler:

Suppose statement is true for n=k.

By the inductive hypothesis is irrational and it follow from our result in the first part of the question that the cube root of this must be irrational therefore it must be true for n=k+1.

The case where n=1 is true (from the question) therefore the result is true for all positive integral value of n.

Show

Suppose statement is true for n=k.

By the inductive hypothesis is irrational and it follow from our result in the first part of the question that the cube root of this must be irrational therefore it must be true for n=k+1.

The case where n=1 is true (from the question) therefore the result is true for all positive integral value of n.

Last part:

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#17

__Question 8, STEP I, 2003__
Spoiler:

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Let the B = yV. the rate at which B increases with respect to time is:

From the question, we see that:

where C is some constant

where

When A completely turns into B, y = 1.

and we have a contradiction and can conclude that A never completely turns into B.

I have a very strong feeling I've missed something, in the question

From the question, we see that:

where C is some constant

where

When A completely turns into B, y = 1.

and we have a contradiction and can conclude that A never completely turns into B.

I have a very strong feeling I've missed something, in the question

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#18

**STEP III 2003, Q4**

Spoiler:

, therefore

Using , we obtain the equation of the tangent:

This meets the curve again when . Substituting this in and rearranging to find a quadratic in , we obtain:

By the factor theorem, we find that is a root. We can factorise and divide by (because ) to obtain

.

Using the quadratic formula, we obtain

One of these gives , which we can disregard.

Therefore, as required.

The double angle formula for tangent is:

Therefore,

So, if , then

Therefore, if , then

We can also apply the formula similarly to the cotangent function, so

Using the identities and , we get

as required.

Another value of which satisfies the required properties is .

Show

, therefore

Using , we obtain the equation of the tangent:

This meets the curve again when . Substituting this in and rearranging to find a quadratic in , we obtain:

By the factor theorem, we find that is a root. We can factorise and divide by (because ) to obtain

.

Using the quadratic formula, we obtain

One of these gives , which we can disregard.

Therefore, as required.

The double angle formula for tangent is:

Therefore,

So, if , then

Therefore, if , then

We can also apply the formula similarly to the cotangent function, so

Using the identities and , we get

as required.

Another value of which satisfies the required properties is .

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#20

**STEP III 2003 Q10**

Spoiler:

n.b. I don't know how to LaTex the dots denoting differentiation w.r.t to time, so I'm just using x for displacement, v for speed and a for acceleration.

Integrating gives

and using the initial conditions gives

Separating variables gives

Using the initial conditions gives

Therefore, upon rearranging a bit, we get

As , the argument of tangent and hence as required.

In this second case, our first integration combined with the initial conditions gives

Now, for simplicity, let .

Separating variables we obtain

which becomes

Rearranging a lot, we get the rather nasty answer:

As , the fraction

Therefore .

Show

n.b. I don't know how to LaTex the dots denoting differentiation w.r.t to time, so I'm just using x for displacement, v for speed and a for acceleration.

Integrating gives

and using the initial conditions gives

Separating variables gives

Using the initial conditions gives

Therefore, upon rearranging a bit, we get

As , the argument of tangent and hence as required.

In this second case, our first integration combined with the initial conditions gives

Now, for simplicity, let .

Separating variables we obtain

which becomes

Rearranging a lot, we get the rather nasty answer:

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.

`x = \sqrt{d^2 - \frac{2U}{k}} \displaystyle\frac{1 + \frac{d - \sqrt{d^2 - \frac{2U}{k}}}}{d + \sqrt{d^2 - \frac{2U}{k}}} e^{\sqrt{d^2 - \frac{2U}{k}}kt}{1 - \frac{d - \sqrt{d^2 - \frac{2U}{k}}}{d + \sqrt{d^2 - \frac{2U}{k}}} e^{\sqrt{d^2 - \frac{2U}{k}}kt}`

As , the fraction

Therefore .

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