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# STEP Maths I, II, III 1996 Solutions

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1. (Original post by DFranklin)
Step QI, P5.

iii) Again, use the quadratic formula:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\

\implies z = 1 \pm \sqrt{1 - (1+i)(2\sqrt{3}-2)} = 1 \pm \sqrt{3-2\sqrt{3}+2(1-\sqrt{3})i)
.

Using (ii),
.
Just wondering, what happened to your 2a in the quadratic formula? I had a (1+i) in the bottom of the fraction, which would explain why they asked "express your answer in its simplest form", i.e multiplying through by the complex conjugate (1-i)/(1-i), I checked my solution with a ex Cambridge Maths guy and we both seemed to get the same bit. so your answer / (1+i)
2. Quite possible; I seem to recall losing the will to live with that question about half the way through...
3. What do you think the grade boundary for a 1 would be for this paper?
4. Shall we add the amendment to that then? Funny you mention that, This was a question which I got the knack of quite quickly (as opposed to many others which took me literally years)
5. (Original post by insparato)
Typing up STEP III question 2 Now

- eqn 1
- eqn 2
- eqn 3

eqn (2 - 1)

-eqn4

sub eqn 4 into 3

-eqn 5

sub eqn 4 into 2
-eqn6

subtracting 6 from 5

Sub into eqn 5

Sub these back into eqn 1 just to see they are indeed satisfy the system.

Therefore

Not complete, the cases A = 0, A = 1 have to be investigated as the standard solution blows up when A = 0 and A = 1. Ive done alot of maths today so i wont be doing this tonight but i might have a go tomorrow or something... Anyone who feels up to finishing it can do so freely .
Such a nice alternative solution with matricies i would type it out but i do ot know how to latex a matrix o.o sorry guys...
If anyone requests the solution i can always do a copy and scan in just message me
6. Here is an alternative approach to STEP 1996 I Q10.
Attached Images

7. (Original post by themaths)
Such a nice alternative solution with matricies i would type it out but i do ot know how to latex a matrix o.o sorry guys...
If anyone requests the solution i can always do a copy and scan in just message me

8. (Original post by DFranklin)
Step QI, P5.

(i) . Suppose . Then
would give us a rational expression for , contradiction.

So .

As r is rational, . (i.e. our solutions are .

(ii) . So
. Again, we set up the quadratic and get:
. Use the quadratic formula to get:

Now .
So .

As in (i), we need to find the square root of this, so look for rational with .

We find and end up with a quadratic . Solve to find and since we want rational deduce and so the square root is .

Plugging this back in, we have

.

Since p is real, we take the latter root, dividing by 2 and using (i) we find . Use to get final solutions .

(iii) Again, use the quadratic formula:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\

\implies z = 1 \pm \sqrt{1 - (1+i)(2\sqrt{3}-2)} = 1 \pm \sqrt{3-2\sqrt{3}+2(1-\sqrt{3})i)
.

Using (ii),
.
Sorry, ignore this post .
9. (Original post by DFranklin)
Step II, Q3:

F0 = 0, F1 =1, F2 = F0+F1 = 1, F3 = F1 + F2 = 2, F4 = F3 + F2 = 3, F5 = 5, F6 = 8, F7 = 13.

F0.F2-F1^2 = -1, F1.F3-F2^2 = 1, F2.F4-F3^2 = -1.

Claim . Proof by induction on n.
By above, true for n<=3. Assume true for n = k. Then

So true for n=k+1 and so true for all n by induction.

Want to show . Assume true for 1 < k <=m. Then
.

So true for k=m+1.

Explictly when k=1 we have . So true for k=1, so true for all k.
This question has been annoying me this afternoon, in the end I settled with what you have here, but I am not wholly satisfied.
For the second part, you have assumed that takes the same form as isn't that defeating the whole point of the induction?
We assume n=k, but in the proof it seems n=k-1 is assumed too, am I talking a load of rubbish?
10. The induction hypothesis was used in going from the bit in the little green circle to the bit in the green ellipse.

11. (Original post by BabyMaths)
The induction hypothesis was used in going from the bit in the little green circle to the bit in the green ellipse.

Sorry i didn't specify i was talking about the last part, but I have been sorted out on the step prep thread
12. (Original post by DFranklin)
Step QI, P5.

(i) . Suppose . Then
would give us a rational expression for , contradiction.

So .

As r is rational, . (i.e. our solutions are .

(ii) . So
. Again, we set up the quadratic and get:
. Use the quadratic formula to get:

Now .
So .

As in (i), we need to find the square root of this, so look for rational with .

We find and end up with a quadratic . Solve to find and since we want rational deduce and so the square root is .

Plugging this back in, we have

.

Since p is real, we take the latter root, dividing by 2 and using (i) we find . Use to get final solutions .

(iii) Again, use the quadratic formula:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\

\implies z = 1 \pm \sqrt{1 - (1+i)(2\sqrt{3}-2)} = 1 \pm \sqrt{3-2\sqrt{3}+2(1-\sqrt{3})i)
.

Using (ii),
.
Hi, I think you forgot to divide by 2(1+i) to obtain the final answer in part III, and then you might want to simplify it
13. For III Q2.
You can show that those are the answers are correct with the exception of a=1,0. Here's how:
The three equation can be rewritten as three matrices as shown below:

We can now solve it because of the result
Ainverse (A)R = Ainverse B
which goes to :
R = Ainverse B
now this means that the equations have no solution when A is non-invertible.
A is invertible when
so when
so when or
14. (Original post by DFranklin)
Quite possible; I seem to recall losing the will to live with that question about half the way through...
I think I've got a nicer solution to this problem which I've posted below:

Q5 STEP I:

Part i)

Spoiler:
Show

So the only combination which results in rational r and s is:

Part ii)

Spoiler:
Show

Part iii)

Spoiler:
Show

or

15. (Original post by DFranklin)
Solved it, yes, written it up, no - nice job. I didn't see: did you manage the last bit? I thought it surprisingly tricky; needing to find the value of tan(3pi/8) was a tad annoying...
You missed a trick if you needed the value of tan(3pi/8).
Notice that you're letting a=cos(3pi/4), use t substitutions with t=tan(3pi/8) and it all cancels down very quickly to 2arctan(1)
16. (Original post by DFranklin)
Step II, Q3:

F0 = 0, F1 =1, F2 = F0+F1 = 1, F3 = F1 + F2 = 2, F4 = F3 + F2 = 3, F5 = 5, F6 = 8, F7 = 13.

F0.F2-F1^2 = -1, F1.F3-F2^2 = 1, F2.F4-F3^2 = -1.

Claim . Proof by induction on n.
By above, true for n<=3. Assume true for n = k. Then

So true for n=k+1 and so true for all n by induction.

Want to show . Assume true for 1 < k <=m. Then
.

So true for k=m+1.

Explictly when k=1 we have . So true for k=1, so true for all k.
In your final proof by induction the proof for k=m+1 assumes the result for k=m and k=m-1. Doesn't that mean that it can not be used to prove the result for k=2, since that would rely on the result for k=0 ,for which the result is undefined? Therefore you have to explicitly show the result to hold for k=1 and k=2 to complete the proof?
17. (Original post by Nick_)
..
I explicitly show the result for n<=3 in the first few lines of the proof.
18. (Original post by DFranklin)
I explicitly show the result for n<=3 in the first few lines of the proof.
I meant the second proof by induction
19. Fair point.
20. (Original post by Rainingshame)
For III Q2.
You can show that those are the answers are correct with the exception of a=1,0. Here's how:
The three equation can be rewritten as three matrices as shown below:

We can now solve it because of the result
Ainverse (A)R = Ainverse B
which goes to :
R = Ainverse B
now this means that the equations have no solution when A is invertible.
A is invertible when
so when
so when or
You mean non-invertible.

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