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# STEP Maths II, III 2009 Solutions

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1. (Updated as far as #57)

STEP II:
1: Solution by Zacken
2: Solution by IrrationalRoot
3: Soluions by Zacken
4: Solution by Zacken
5: Solution by Zacken
6: Solution by Zacken
7: Solution by B_9710
8: Solution by Farhan.Hanif93
9: Solution by A-cute-Angle
10: Solution by Zacken
11: Solution by The-Spartan
12: Solution by Farhan.Hanif3
13: Solution by Farhan.Hanif3

STEP III:
1: Solution by IrrationalRoot
2: Solution by Farhan.Hanif93
3: Solution by Zacken
4: Solution by EricPiphany
5: Solution by IrrationalRoot
6: Solution by Farhan.Hanif93
7: Solution by Farhan.Hanif93
8: Solution by IrrationalRoot
9: Solution by Farhan.Hanif93
10: Solution by Farhan.Hanif93
11: Solution by Farhan.Hanif93
12: Solution by Zacken
13: Solution by Farhan.Hanif3

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 - ...
2. STEP II, Question 1

First part

We note that since is unchanged if or as well as then the equation of symmetries are the axes (i.e: and ) and the lines .

Since is unchanged under then the equation of symmetry is

Second part
A quick sketch:
Spoiler:
Show
proves useful.

We can immediately identify as , as , as and as .

Third part
A quick check of the gradients reveals that , and likewise and = 1. Hence forms a rectangle.

Furthermore, the lengths are:

(since ).

Also:

.

The area is then .

Remember that lie on both curves, hence: and . So that:

.

Hence, the area is:

and verifying this in the given case and yields as required.
3. STEP II, Question 4

First part
Write for some fourth degree polynomial . Then .

Second part
Using the product rule, we have:

So is clearly divisible by .

Third part
Write for some fourth degree polynomial .

Then in a similar fashion as (ii). Also note that

But since is a th degree polynomial, then is an 8th degree polynomial.

However, we have that and .

So it must be the case that

From this, we can expand: , so that, if we integrate:

Now, using gives us: and

Adding both these equations yields and hence so that:

4. STEP II, Q5:

(i)
as expected.

Also, note that .

We can then state that and likewise

This means that our integral is nothing more than:

Where we take the positive root of the square root since it is positive in the interval that we are integrating over.

(ii)
Since there is a sign change in the integrand at we'll need to split the integral into two regions and flip the sign of the first region. That is:

.

Integrating , we get:

(iii)

Note that .

Hence .

Since the entirety of our integrand is always positive for all valid , and using the difference of two squares on the denominator:

We finally end up with:

5. STEP III, Question 1

p and q
Let denote the gradient of any line .

We have and .

Therefore the equation of is

,

so

,

as required.

To get , we note that we can simply replace with and with so that .

st and s+t
If and lie on the circle , then and .

Therefore a quadratic equation satisfied by both and is , that is, .

Considering the sum and product of the roots gives and .

p+q=0
We now also have and . Hence, and .

Therefore , so it suffices to show that

and so it suffices to show that

which is true, so we conclude that , as required.
6. (Original post by Zacken)
(Updated as far as #5)

STEP II:
1: Solution by Zacken
2:
3:
4: Solution by Zacken
5: Solution by Zacken
6:
7:
8:
9:
10:
11:
12:
13:

STEP III:
1: Solution by IrrationalRoot
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:

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 - ...
I'll take ii q11 for now. Will probably focus on the applied seeing as no-one else seems to like it :-P

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7. STEP III, Question 5

as required.

as required.

as required.

We require

.

It therefore suffices to find , and such that the cubic for ,,.

Therefore we can take , and .
8. Can I reserve STEP III Q4?
9. (Original post by EricPiphany)
Can I reserve STEP III Q4?
I was gonna do that one but I'm pretty new to Latex and so I'd be terribly slow with all those integrals and stuff so go ahead haha.

I'd like to reserve II Q2 (i.e. I'm doing it now) .

EDIT: Meant to say Q2 sorry.
10. STEP III Q4

(i).
Laplace transform of is
.

(ii).
Laplace transform of is
, making the substitution ,
.

(iii).
Laplace transform of is
, using integration by parts.

(iv).
Laplace transform of is

.
So, , so .

Now, is the derivative of , so by part (iii), Laplace transform of is .
And by part (ii), Laplace transform of is .
Thus, by part (i), Laplace transform of is .

What a mess, feel free to edit and repost.
11. (Original post by StrangeBanana)
Doing STEP III, Q5
Done a while ago!!
12. (Original post by IrrationalRoot)
Done a while ago!!
hmm

I'm tired

EDIT:
I'll come back tomorrow and see if any are still available
13. STEP II, Question 2

(i)
.

.

Therefore .

Therefore coordinates of stationary points are .

(ii)
(ii) , so for small ,

( is small)

as required.

(iii)
(iii) The graph should look sinusoidal, with maxima at and minima at .

These extrema should become closer together as increases. Note that for , is strictly increasing and as , .

(iv)
(iv) The maxima occur at for Therefore the required area is approximately

as required.
14. (Original post by Zacken)
STEP II, Question 4

So it must be the case that
Just a small typo
15. (Original post by IrrationalRoot)
...
Would have been simpler to note that is a bounded function, so that is maximum when is 1, giving rise to and giving rise to .
16. (Original post by EricPiphany)
Just a small typo
I'm being incredibly thick, but I can't see it?
17. (Original post by Zacken)
I'm being incredibly thick, but I can't see it?
coz I've tried to correct it
18. (Original post by EricPiphany)
coz I've tried to correct it
Ahhh, yeah lol, dumb mistake that one. Thanks!
19. STEP II, Question 3:

.

Which simplifies down to .

(i)
Set to get .

Hence, using the fact that:

as required.

(ii)

(iii)

Set , then:

But, we know that:

Hence: .

Giving us:

as required.

20. STEP II Q8

Solution
Straightforward diagram with , and non-collinear; lying on the line between and ; and lying on the line "beyond" . Note:

Furthermore, using and the definitions of and , observe that any point on the line has position vector:

By taking for any (and verifying by direct substitution into ), the coefficients of and collapse to and respectively. That is, always lies on the line , as required.

Finally, note that so that the (opposite) sides and are parallel and of equal length. It follows that the quadrilateral is a parallelogram.

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