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STEP Maths I, II, III 1996 Solutions

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DFranklin
Step QI, P5.

iii) Again, use the quadratic formula:
Unparseable latex formula:

2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\[br]\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),
z=1±(1−3+i)=3−i,2−3+i z = 1 \pm (1-\sqrt{3}+i) = \sqrt{3} - i, 2-\sqrt{3}+i.


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)
Quite possible; I seem to recall losing the will to live with that question about half the way through...
What do you think the grade boundary for a 1 would be for this paper?
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)
Reply 124
Avatar for themaths
themaths
Original post by insparato
Typing up STEP III question 2 Now

x+y+az=2 x + y + az = 2 - eqn 1
x+ay+z=2 x + ay + z = 2 - eqn 2
2x+y+z=2b 2x + y + z = 2b - eqn 3

eqn (2 - 1)

ay−y+z−az=0 ay-y + z-az = 0
y(a−1)=z(a−1) y(a-1) = z(a-1)
y=z y = z -eqn4

sub eqn 4 into 3

2x+2y=2b 2x+2y = 2b
x+y=b x + y = b -eqn 5

sub eqn 4 into 2
x+(a+1)y=2 x + (a+1)y = 2 -eqn6

subtracting 6 from 5

y+ay−y=2−b y + ay - y = 2-b

y=2−ba y = \frac{2-b}{a}

Sub y=2−ba y = \frac{2-b}{a} into eqn 5

x+2−ba=b x + \frac{2-b}{a} = b

ax+2−b=ab ax + 2-b = ab

ax=ab+b−2 ax = ab + b - 2

x=ab+b−2a x = \frac{ab+b-2}{a}

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

ab+b−2a+2−ba+2a−aba=2aa=2 \frac{ab+b-2}{a} + \frac{2-b}{a} + \frac{2a-ab}{a} = \frac{2a}{a} = 2

Therefore

x=ab+b−2a x = \frac{ab+b-2}{a}

y=2−ba y = \frac{2-b}{a}

z=2−ba z = \frac{2-b}{a}

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 :biggrin:.


Such a nice alternative solution with matricies :biggrin: 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 :smile: just message me :smile:
Reply 125
Avatar for waxwing
waxwing
5
Here is an alternative approach to STEP 1996 I Q10.
STEP1996IQ10_1.gif37.2KB
STEP1996IQ10_2.gif28.4KB
Original post by themaths
Such a nice alternative solution with matricies :biggrin: 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 :smile: just message me :smile:


If a=0 there are no solutions unless b=2\text{If }a=0 \text{ there are no solutions unless }b=2
 in which case we have x+y=x+z=2 sox=λ,y=λ,z=2−λ \text{ in which case we have }x+y=x+z=2 \text{ so}x=\lambda, y=\lambda, z=2-\lambda
 If a=1 then x+y+z=2 and 2x+y+z=2b so x=2(b−1),y+z=4−2b\text{ If }a=1\text{ then }x+y+z=2 \text{ and }2x+y+z=2b \text{ so }x=2(b-1), y+z=4-2b
i.e. x=2(b−1),y=λ,z=4−2b−λ\text{i.e. }x=2(b-1),y=\lambda,z=4-2b-\lambda
(edited 11 years ago)
Original post by DFranklin
Step QI, P5.

(i) (r+s3)=4−23  ⟹  r2+3s2+2rs3=4−23(r+s\sqrt{3})=4-2\sqrt{3} \implies r^2+3s^2+2rs\sqrt{3} = 4-2\sqrt{3}. Suppose rs≠−1rs \neq -1. Then
r2+3s2−4=−2(1+rs)3r^2+3s^2-4 = -2(1+rs)\sqrt{3} would give us a rational expression for 3\sqrt{3}, contradiction.

So
Unparseable latex formula:

rs=1, \text{ so } s = -1/r \text { and so } r^2+3/r^2 = 4 \implies r^4-4r^2+3 = 0\\[br]\implies r^2 = 1, 3

.

As r is rational, r2=1, so r=±1,s=−rr^2 = 1, \text{ so } r = \pm 1, s = -r. (i.e. our solutions are ±(1−3)\pm(1-\sqrt{3}).

(ii) (p+qi)2=p2−q2+2pqi=3−23+2(1−3)i(p+qi)^2 = p^2-q^2+2pqi = 3-2\sqrt{3} + 2(1-\sqrt{3})i. So
p2−q2=3−23,pq=1−3p^2-q^2 = 3-2\sqrt{3}, pq = 1-\sqrt{3}. Again, we set up the quadratic and get:
p4−(3−23)p2−(1−3)2=0p^4-(3-2\sqrt{3})p^2-(1-\sqrt{3})^2 = 0. Use the quadratic formula to get:
2p2=(3−23)±(3−23)2+4(1−3))22p^2 = (3-2\sqrt{3}) \pm \sqrt{(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2}
Now (3−23)2=32+4⋅3−123=21−123,(1−3)2=1+3−23=4−23 (3-2\sqrt{3})^2 = 3^2+4\cdot 3 - 12\sqrt{3} = 21 - 12\sqrt{3}, \quad (1-\sqrt{3})^2 = 1 + 3 - 2 \sqrt{3} = 4-2\sqrt{3}.
So (3−23)2+4(1−3))2=37−203(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2 = 37 - 20 \sqrt{3}.

As in (i), we need to find the square root of this, so look for rational α,β\alpha, \beta with (α+β3)2=37−203(\alpha+\beta\sqrt{3})^2 = 37 - 20\sqrt{3}.

We find β=−10/α\beta = -10/\alpha and end up with a quadratic α4−37α2+300=0\alpha^4 - 37\alpha^2+300=0. Solve to find α2=12,25\alpha^2 = 12, 25 and since we want α\alpha rational deduce α=±5\alpha = \pm 5 and so the square root is ±(5−23)\pm(5-2\sqrt{3}).

Plugging this back in, we have

2p2=(3−23)±(5−23)=−2,8−432p^2 = (3-2\sqrt{3}) \pm (5-2\sqrt{3}) = -2, 8-4\sqrt{3}.

Since p is real, we take the latter root, dividing by 2 and using (i) we find p=±(1−3)p = \pm (1-\sqrt{3}). Use pq=(1−3)pq = (1-\sqrt{3}) to get final solutions ±(1−3+i)\pm (1-\sqrt{3}+i).

(iii) Again, use the quadratic formula:
Unparseable latex formula:

2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\[br]\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),
z=1±(1−3+i)=3−i,2−3+i z = 1 \pm (1-\sqrt{3}+i) = \sqrt{3} - i, 2-\sqrt{3}+i.


Sorry, ignore this post :smile:.
(edited 11 years ago)
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 Fn+1Fn−1−Fn2=(−1)nF_{n+1}F_{n-1}-F_n^2 = (-1)^n. Proof by induction on n.
By above, true for n<=3. Assume true for n = k. Then
Unparseable latex formula:

F_{k+2}F_k - F_{k+1}^2 \\[br]= (F_{k+1}+F_k)F_k - F_{k+1}^2 \\ [br]= F_k^2 +F_{k+1}F_k - F_{k+1}^2 = F_k^2 - F_{k+1}(F_{k+1} - F_k) \\[br]= [F_{k+1}F_{k-1}+(-1)^{k+1}] - F_{k+1}(F_{k+1} - F_k) \\[br]= (-1)^{k+1} + F_{k+1}F_{k-1} - F_{k+1}(F_{k-1})\\[br]= (-1)^{k+1}F_{k+1}^2



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

Want to show Fn+k=FkFn+1+Fk−1FnF_{n+k} = F_kF_{n+1}+F_{k-1}F_n. Assume true for 1 < k <=m. Then
Unparseable latex formula:

F_{n+k+1} = F_{n+k}+F_{n+k-1}\\[br]= F_kF_{n+1}+F_{k-1}F_n + F_k-1F_{n+1}+F_{k-2}F_n\\[br]=(F_k+F_{k-1})F_{n+1}+(F_{k-1}+F_{k-2}F_n \\[br]=F_{k+1}F_{n+1}+F_k F_n

.

So true for k=m+1.

Explictly when k=1 we have Fn+1=F1Fn+1+F0Fn=Fn+1F_{n+1} = F_1F_{n+1}+F_0F_n = F_{n+1}. 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 Fn+k−1F_{n+k-1} takes the same form as Fn+kF_{n+k} 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?
Reply 129
Avatar for BabyMaths
BabyMaths
The induction hypothesis was used in going from the bit in the little green circle to the bit in the green ellipse.

induction DF.png
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.

induction DF.png


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

(i) (r+s3)=4−23  ⟹  r2+3s2+2rs3=4−23(r+s\sqrt{3})=4-2\sqrt{3} \implies r^2+3s^2+2rs\sqrt{3} = 4-2\sqrt{3}. Suppose rs≠−1rs \neq -1. Then
r2+3s2−4=−2(1+rs)3r^2+3s^2-4 = -2(1+rs)\sqrt{3} would give us a rational expression for 3\sqrt{3}, contradiction.

So
Unparseable latex formula:

rs=1, \text{ so } s = -1/r \text { and so } r^2+3/r^2 = 4 \implies r^4-4r^2+3 = 0\\[br]\implies r^2 = 1, 3

.

As r is rational, r2=1, so r=±1,s=−rr^2 = 1, \text{ so } r = \pm 1, s = -r. (i.e. our solutions are ±(1−3)\pm(1-\sqrt{3}).

(ii) (p+qi)2=p2−q2+2pqi=3−23+2(1−3)i(p+qi)^2 = p^2-q^2+2pqi = 3-2\sqrt{3} + 2(1-\sqrt{3})i. So
p2−q2=3−23,pq=1−3p^2-q^2 = 3-2\sqrt{3}, pq = 1-\sqrt{3}. Again, we set up the quadratic and get:
p4−(3−23)p2−(1−3)2=0p^4-(3-2\sqrt{3})p^2-(1-\sqrt{3})^2 = 0. Use the quadratic formula to get:
2p2=(3−23)±(3−23)2+4(1−3))22p^2 = (3-2\sqrt{3}) \pm \sqrt{(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2}
Now (3−23)2=32+4⋅3−123=21−123,(1−3)2=1+3−23=4−23 (3-2\sqrt{3})^2 = 3^2+4\cdot 3 - 12\sqrt{3} = 21 - 12\sqrt{3}, \quad (1-\sqrt{3})^2 = 1 + 3 - 2 \sqrt{3} = 4-2\sqrt{3}.
So (3−23)2+4(1−3))2=37−203(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2 = 37 - 20 \sqrt{3}.

As in (i), we need to find the square root of this, so look for rational α,β\alpha, \beta with (α+β3)2=37−203(\alpha+\beta\sqrt{3})^2 = 37 - 20\sqrt{3}.

We find β=−10/α\beta = -10/\alpha and end up with a quadratic α4−37α2+300=0\alpha^4 - 37\alpha^2+300=0. Solve to find α2=12,25\alpha^2 = 12, 25 and since we want α\alpha rational deduce α=±5\alpha = \pm 5 and so the square root is ±(5−23)\pm(5-2\sqrt{3}).

Plugging this back in, we have

2p2=(3−23)±(5−23)=−2,8−432p^2 = (3-2\sqrt{3}) \pm (5-2\sqrt{3}) = -2, 8-4\sqrt{3}.

Since p is real, we take the latter root, dividing by 2 and using (i) we find p=±(1−3)p = \pm (1-\sqrt{3}). Use pq=(1−3)pq = (1-\sqrt{3}) to get final solutions ±(1−3+i)\pm (1-\sqrt{3}+i).

(iii) Again, use the quadratic formula:
Unparseable latex formula:

2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\[br]\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),
z=1±(1−3+i)=3−i,2−3+i z = 1 \pm (1-\sqrt{3}+i) = \sqrt{3} - i, 2-\sqrt{3}+i.


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 :smile:
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:
[11a1a1211]⋅[xyz]=[222b] \begin{bmatrix} 1 & 1 & a \\ 1 & a & 1 \\ 2 & 1 & 1 \end{bmatrix}\cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 2b \end{bmatrix}
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 detA=0 det A =0
so when 2a−2a2=0 2a-2a^2 =0
so when a=0 a=0 or a=1 a=1
(edited 11 years ago)
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



Part ii)

Spoiler



Part iii)

Spoiler

Reply 134
Avatar for Nick_
Nick_
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)
Reply 135
Avatar for Nick_
Nick_
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 Fn+1Fn−1−Fn2=(−1)nF_{n+1}F_{n-1}-F_n^2 = (-1)^n. Proof by induction on n.
By above, true for n<=3. Assume true for n = k. Then
Unparseable latex formula:

F_{k+2}F_k - F_{k+1}^2 \\[br]= (F_{k+1}+F_k)F_k - F_{k+1}^2 \\ [br]= F_k^2 +F_{k+1}F_k - F_{k+1}^2 = F_k^2 - F_{k+1}(F_{k+1} - F_k) \\[br]= [F_{k+1}F_{k-1}+(-1)^{k+1}] - F_{k+1}(F_{k+1} - F_k) \\[br]= (-1)^{k+1} + F_{k+1}F_{k-1} - F_{k+1}(F_{k-1})\\[br]= (-1)^{k+1}F_{k+1}^2



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

Want to show Fn+k=FkFn+1+Fk−1FnF_{n+k} = F_kF_{n+1}+F_{k-1}F_n. Assume true for 1 < k <=m. Then
Unparseable latex formula:

F_{n+k+1} = F_{n+k}+F_{n+k-1}\\[br]= F_kF_{n+1}+F_{k-1}F_n + F_k-1F_{n+1}+F_{k-2}F_n\\[br]=(F_k+F_{k-1})F_{n+1}+(F_{k-1}+F_{k-2}F_n \\[br]=F_{k+1}F_{n+1}+F_k F_n

.

So true for k=m+1.

Explictly when k=1 we have Fn+1=F1Fn+1+F0Fn=Fn+1F_{n+1} = F_1F_{n+1}+F_0F_n = F_{n+1}. 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?
Nick_
..
I explicitly show the result for n<=3 in the first few lines of the proof.
Reply 137
Avatar for Nick_
Nick_
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
Fair point.
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:
[11a1a1211]⋅[xyz]=[222b] \begin{bmatrix} 1 & 1 & a \\ 1 & a & 1 \\ 2 & 1 & 1 \end{bmatrix}\cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ 2b \end{bmatrix}
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 detA=0 det A =0
so when 2a−2a2=0 2a-2a^2 =0
so when a=0 a=0 or a=1 a=1


You mean non-invertible. :smile:

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