# STEP 2006 Solutions Thread

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You can let someone else take control of the OP whenever.

1: Solution by Unbounded

2: Solution by nuodai

3: Solution by nuodai

4: Solution by Unbounded

5: Solution by darkness9999

6: Solution by SimonM

7: Solution by SimonM

8: Solution by Glutamic Acid

9: Solution by Unbounded

10: Solution by SimonM

11: Solution by brianeverit

12: Solution by Farhan.Hanif93

13: Solution by Aurel-Aqua

14: Solution by Aurel-Aqua

1: Solution by DeanK22

2: Solution by DeanK22 and etothepiiplusone

3: Solution by SimonM

4: Solution by sonofdot

5: Solution by Daniel Freedman

6: Solution by Elongar

7: Solution by SimonM

8: Solution by Daniel Freedman

9: Solution by brianeverit

10: Solution by Farhan.Hanif93

11: Solution by Farhan.Hanif93

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by Farhan.Hanif93

1: Solution by Daniel Freedman

2: Solution by Daniel Freedman

3: Solution by Daniel Freedman

4: Solution by Dadeyemi

5: Solution by Elongar

6: Solution by Anonymous

7: Solution by tommm

8: Solution by DeanK22

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

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 Unbounded

2: Solution by nuodai

3: Solution by nuodai

4: Solution by Unbounded

5: Solution by darkness9999

6: Solution by SimonM

7: Solution by SimonM

8: Solution by Glutamic Acid

9: Solution by Unbounded

10: Solution by SimonM

11: Solution by brianeverit

12: Solution by Farhan.Hanif93

13: Solution by Aurel-Aqua

14: Solution by Aurel-Aqua

**STEP II:**1: Solution by DeanK22

2: Solution by DeanK22 and etothepiiplusone

3: Solution by SimonM

4: Solution by sonofdot

5: Solution by Daniel Freedman

6: Solution by Elongar

7: Solution by SimonM

8: Solution by Daniel Freedman

9: Solution by brianeverit

10: Solution by Farhan.Hanif93

11: Solution by Farhan.Hanif93

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by Farhan.Hanif93

**STEP III:**1: Solution by Daniel Freedman

2: Solution by Daniel Freedman

3: Solution by Daniel Freedman

4: Solution by Dadeyemi

5: Solution by Elongar

6: Solution by Anonymous

7: Solution by tommm

8: Solution by DeanK22

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

**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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#3

__STEP III 2006, Question 3__
Spoiler:

i) tan is an odd function ie tan(-x) = - tan(x) and therefore can have no even powers of x in its series expansion

Prove

as required.

ii)

as required.

iii)

which is true by Pythagoras'.

EDIT: There was no point in factorising this way. I did the question whilst typing it up, and it seemed an obvious thing to do.

Equating coefficients gives:

as required.

Show

i) tan is an odd function ie tan(-x) = - tan(x) and therefore can have no even powers of x in its series expansion

Prove

as required.

ii)

as required.

iii)

which is true by Pythagoras'.

EDIT: There was no point in factorising this way. I did the question whilst typing it up, and it seemed an obvious thing to do.

Equating coefficients gives:

as required.

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#4

__STEP III 2006, Question 1__
Spoiler:

Therefore the curve has vertical asymptotes with equations and .

The curve has roots .

As , so the curve has an oblique asymptote with equation .

As . This is the equation of the tangent to the curve at the origin (can be verified by differentiating)

[SKETCH ATTACHED]

i)

[SKETCH ATTACHED]

The graph shows that the two curves have three points of intersection, which means the above equation has three real roots

ii)

[SKETCH ATTACHED]

The graph shows the two curves only have one point of intersection (using the fact that the line has the same gradient as the tangent to the curve at the origin). This means the above equation has one real root.

iii)

[SKETCH ATTACHED]

The graph shows the curves have six points of intersection, which means the above equation has six real roots (the curves don't meet again because there are only two solutions of

Show

Therefore the curve has vertical asymptotes with equations and .

The curve has roots .

As , so the curve has an oblique asymptote with equation .

As . This is the equation of the tangent to the curve at the origin (can be verified by differentiating)

[SKETCH ATTACHED]

i)

[SKETCH ATTACHED]

The graph shows that the two curves have three points of intersection, which means the above equation has three real roots

ii)

[SKETCH ATTACHED]

The graph shows the two curves only have one point of intersection (using the fact that the line has the same gradient as the tangent to the curve at the origin). This means the above equation has one real root.

iii)

[SKETCH ATTACHED]

The graph shows the curves have six points of intersection, which means the above equation has six real roots (the curves don't meet again because there are only two solutions of

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#5

**STEP III 8**

We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows;

We will now prove property 1

Spoiler:

Show

We now prove property 2 is met by this operation

We prove property three is met by the above operation.

We now prove the final property;

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

Spoiler:

Show

We proceed by induction.

Assume that for some k in

P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains

Assume that for some k in

**N**that delta(x^k) = kx^(k-1). Assume that the result holds for k + 1.P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains

**N**. So delta(f(x)) is equivalent to d/dx (f(x)) if f(x) is a polynomial in x.As required

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#6

**I/8**:

(i) Let the base be OAC, so area = 1/2ca. Height is b, so the area is b*1/3*(1/2ca) = abc/6.

(ii) is the angle between BC and AC.

Area of triangle = "". Note that , so

The area will clearly be equal to abc/6, only that the height of the new triangle is d.

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#7

(Original post by

We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows;

We will now prove property 1

We now prove property 2 is met by this operation

We prove property three is met by the above operation.

We now prove the final property;

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

As required

**DeanK22**)**STEP III 8**We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows;

We will now prove property 1

Spoiler:

Show

We now prove property 2 is met by this operation

We prove property three is met by the above operation.

We now prove the final property;

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

Spoiler:

Show

We proceed by induction.

Assume that for some k in

P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains

Assume that for some k in

**N**that delta(x^k) = kx^(k-1). Assume that the result holds for k + 1.P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains

**N**. So delta(f(x)) is equivalent to d/dx (f(x)) if f(x) is a polynomial in x.As required

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#9

**2006 I Question 13**

**Part (i)**

The number of diamonds in one kilogram has a Poisson distribution of . We find that per 100 grams, . For 100T grams, . is the distribution of scores of the die.

Drawing 1, we have .

Similarly,

(recognising a finite geometric sum).

**Part (i) - expectation**

**Part (ii)**

This time, is a geometric distribution with . Thus, .

This time, our probability is:

, as required.

**Part (ii) - expectation**

.

Consider: .

Thus, .

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#10

**2006 I Question 14**

Either I missed something crucial, or this question is incredibly easy. I'll write up the incredibly-easy interpretation of this question:

**Part (i)**

This is a geometric progression, where , where , , so .

Differentiating, we get: . Equating to zero, . We know that this is the maximum as r tends to infinity, P tends to zero.

**Part (ii)**

. Clearly maximum at n = 1.

Did I make a mistake here? It just looks too simple for a STEP question, I'm afraid.

Edit: see GHOSH-5's post here: http://www.thestudentroom.co.uk/show...3&postcount=41

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#11

(Original post by

I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that , using the four rules. Then let , and apply the operation to this function, making use of rules ii) and iii) to show that .

**Daniel Freedman**)I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that , using the four rules. Then let , and apply the operation to this function, making use of rules ii) and iii) to show that .

They've said "given that the operator has these properties; show that...".

Dean has said "A particular operator has those properties, and for , we have ...."

But he

*hasn't*shown that is the only possible operator with the properties.

This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a

**ton**of marks for this (even if the Examiner sympathised).

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#12

(Original post by

Actually, it's worse than that.

They've said "given that the operator has these properties; show that...".

Dean has said "A particular operator has those properties, and for , we have ...."

But he

This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a

**DFranklin**)Actually, it's worse than that.

They've said "given that the operator has these properties; show that...".

Dean has said "A particular operator has those properties, and for , we have ...."

But he

*hasn't*shown that is the only possible operator with the properties.This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a

**ton**of marks for this (even if the Examiner sympathised).
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#13

Well, that's effectively what the question is asking you to show. I would:

(1) Prove by induction that (using (i) and (iv)).

(2) Then use (iii) to prove

(3) Then use (ii) to prove .

[Which effectively proves uniqueness of for polynomials].

Then simply observe this is the same as the derivative in the case of polynomials.

(1) Prove by induction that (using (i) and (iv)).

(2) Then use (iii) to prove

(3) Then use (ii) to prove .

[Which effectively proves uniqueness of for polynomials].

Then simply observe this is the same as the derivative in the case of polynomials.

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#14

I have different answers to

**STEP I Q 14**Im not to sure who is wrong...**Part (i)**
Spoiler:

Show

**Part (ii)**
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#16

__STEP III 2006, Q5__We wish to show that form an equilateral iff

Multiplying by two and factorising, we wish to show that,

holds for all equilateral triangles.

Note that are the sides of the triangle. If they are shown on an Argand diagram, the angle between any two consecutive sides is . Note also that the lengths of the sides are all of equal length, so the magnitudes of the complex numbers that represent them will also be equal.

Using this, we reconstruct the equation:

or, more simply,

The bracket evaluates to 0, so we are done.

For the next part, we take the roots of the cubic to be , and expand, yielding Vieta's formula:

Algebra shows that:

and the result follows immediately.

Now we write,

The transformation has the effect of rotating our equilateral triangle by the angle , magnifying it by a factor of and translating it through the vector represented by (would I need to show this?). Under these transformations, the equilateral triangle formed by the roots of the initial cubic remains an equilateral triangle.

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#18

RE; Noudai

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.

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#19

(Original post by

RE; Noudai

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.

**DeanK22**)RE; Noudai

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.

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#20

**STEP III 2006 Q7**

Spoiler:

Using the quadratic formula we obtain

If we let , we obtain the differential equation they're asking about, so:

Using the condition that at , we see that the must be a +, hence

Integrating we obtain

and using y = 0 when x = 0 we find that c = 1.

Using the quadratic formula again to find dy/dx, we obtain (apologies for not writing it out in full):

We can integrate the left hand side using recognition: the top is the derivative of the bottom, so:

Now, if the sign were a minus, then the condition x = 0 when y = 0 would be impossible, as this would mean that the arbitrary constant were infinite. We can therefore deduce that the must be a +. Therefore:

Using x = 0 when y = 0, we find that c = ln2. Therefore

, which is the solution of the differential equation.

Writing the cosh in full exponentials and multiplying by 2 gives:

To find the asymptotes, we consider two cases:

1) when , we can disregard and because they are comparatively small, so . Rearranging gives the equation of the asymptote, .

1) when , we can disregard and because they are comparatively small, so . Rearranging gives the equation of the other asymptote, .

Show

**Part i):**

Using the quadratic formula we obtain

If we let , we obtain the differential equation they're asking about, so:

Using the condition that at , we see that the must be a +, hence

Integrating we obtain

and using y = 0 when x = 0 we find that c = 1.

**Part ii):**

Using the quadratic formula again to find dy/dx, we obtain (apologies for not writing it out in full):

We can integrate the left hand side using recognition: the top is the derivative of the bottom, so:

Now, if the sign were a minus, then the condition x = 0 when y = 0 would be impossible, as this would mean that the arbitrary constant were infinite. We can therefore deduce that the must be a +. Therefore:

Using x = 0 when y = 0, we find that c = ln2. Therefore

, which is the solution of the differential equation.

Writing the cosh in full exponentials and multiplying by 2 gives:

To find the asymptotes, we consider two cases:

1) when , we can disregard and because they are comparatively small, so . Rearranging gives the equation of the asymptote, .

1) when , we can disregard and because they are comparatively small, so . Rearranging gives the equation of the other asymptote, .

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