STEP Maths I, II, III 1996 Solutions

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

Announcements Posted on
Enter our travel-writing competition for the chance to win a Nikon 1 J3 camera 21-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
Search Thread
Advanced Search
Sign in to Reply
  1. disco1000's Avatar
    121 16-06-2010  11:16
    • Full Member
    • Posts: 92
    Re: STEP Maths I, II, III 1996 Solutions
    (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),
    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)
  2. DFranklin's Avatar
    122 16-06-2010  11:20
    • TSR Royalty
    • Location: London
    • Posts: 18,046
    Re: STEP Maths I, II, III 1996 Solutions
    Quite possible; I seem to recall losing the will to live with that question about half the way through...
  3. cmhcgs815's Avatar
    123 17-06-2010  12:10
    • Respected Member
    • Posts: 183
    Re: STEP Maths I, II, III 1996 Solutions
    What do you think the grade boundary for a 1 would be for this paper?
  4. disco1000's Avatar
    124 20-06-2010  21:45
    • Full Member
    • Posts: 92
    Re: STEP Maths I, II, III 1996 Solutions
    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. themaths's Avatar
    125 21-06-2011  16:53
    • New Member
    • Posts: 15
    Re: STEP Maths I, II, III 1996 Solutions
    (Original post by insparato)
    Typing up STEP III question 2 Now

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

    eqn (2 - 1)

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

    sub eqn 4 into 3

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

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

    subtracting 6 from 5

    y + ay - y = 2-b

    y = \frac{2-b}{a}

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

    x + \frac{2-b}{a} = b

    ax + 2-b = ab

    ax = ab + b - 2

    x = \frac{ab+b-2}{a}

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

    \frac{ab+b-2}{a} + \frac{2-b}{a} + \frac{2a-ab}{a} = \frac{2a}{a} = 2

    Therefore

    x = \frac{ab+b-2}{a}

    y = \frac{2-b}{a}

    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 .
    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
    Post rating:
    1
  6. waxwing's Avatar
    126 26-07-2011  14:49
    • Junior Member
    • Location: Shenzhen
    • Posts: 65
    Re: STEP Maths I, II, III 1996 Solutions
    Here is an alternative approach to STEP 1996 I Q10.
    Attached Thumbnails
    Click image for larger version.  Name: STEP1996IQ10_1.gif  Views: 29  Size: 36.3 KB  ID: 108869   Click image for larger version.  Name: STEP1996IQ10_2.gif  Views: 26  Size: 27.7 KB  ID: 108870  
  7. brianeverit's Avatar
    127 31-05-2012  12:55
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1996 Solutions
    (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
    \text{If }a=0 \text{ there are no solutions unless }b=2
    \text{ in which case we have }x+y=x+z=2 \text{ so}x=\lambda, y=\lambda, z=2-\lambda
    \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
    \text{i.e. }x=2(b-1),y=\lambda,z=4-2b-\lambda
    Last edited by brianeverit; 31-05-2012 at 12:59.
  8. brittanna's Avatar
    128 06-08-2012  15:42
    • Peer Of The TSR Realm
    • Posts: 1,502
    Re: STEP Maths I, II, III 1996 Solutions
    (Original post by DFranklin)
    Step QI, P5.

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

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

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

    (ii) (p+qi)^2 = p^2-q^2+2pqi = 3-2\sqrt{3} + 2(1-\sqrt{3})i. So
    p^2-q^2 = 3-2\sqrt{3}, pq = 1-\sqrt{3}. Again, we set up the quadratic and get:
    p^4-(3-2\sqrt{3})p^2-(1-\sqrt{3})^2 = 0. Use the quadratic formula to get:
    2p^2 = (3-2\sqrt{3}) \pm \sqrt{(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2}
    Now (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 (3-2\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 \alpha, \beta with (\alpha+\beta\sqrt{3})^2 = 37 - 20\sqrt{3}.

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

    Plugging this back in, we have

    2p^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 = \pm (1-\sqrt{3}). Use pq = (1-\sqrt{3}) to get final solutions \pm (1-\sqrt{3}+i).

    (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),
    z = 1 \pm (1-\sqrt{3}+i) = \sqrt{3} - i, 2-\sqrt{3}+i.
    Sorry, ignore this post .
    Last edited by brittanna; 06-08-2012 at 15:44.
  9. bogstandardname's Avatar
    129 30-12-2012  14:48
    • Respected Member
    • Posts: 218
    Re: STEP Maths I, II, III 1996 Solutions
    (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 F_{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
    F_{k+2}F_k - F_{k+1}^2 \\ = (F_{k+1}+F_k)F_k - F_{k+1}^2 \\ = F_k^2 +F_{k+1}F_k - F_{k+1}^2 = F_k^2 - F_{k+1}(F_{k+1} - F_k) \\ = [F_{k+1}F_{k-1}+(-1)^{k+1}] - F_{k+1}(F_{k+1} - F_k) \\ = (-1)^{k+1} + F_{k+1}F_{k-1} - F_{k+1}(F_{k-1})\\ = (-1)^{k+1}F_{k+1}^2

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

    Want to show F_{n+k} = F_kF_{n+1}+F_{k-1}F_n. Assume true for 1 < k <=m. Then
    F_{n+k+1} = F_{n+k}+F_{n+k-1}\\ = F_kF_{n+1}+F_{k-1}F_n + F_k-1F_{n+1}+F_{k-2}F_n\\ =(F_k+F_{k-1})F_{n+1}+(F_{k-1}+F_{k-2}F_n \\ =F_{k+1}F_{n+1}+F_k F_n.

    So true for k=m+1.

    Explictly when k=1 we have F_{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 F_{n+k-1} takes the same form as F_{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?
  10. BabyMaths's Avatar
    130 30-12-2012  15:17
    • Peer Of The TSR Realm
    • Posts: 1,567
    Re: STEP Maths I, II, III 1996 Solutions
    The induction hypothesis was used in going from the bit in the little green circle to the bit in the green ellipse.

    Click image for larger version.  Name: induction DF.png  Views: 31  Size: 4.7 KB  ID: 189127
  11. bogstandardname's Avatar
    131 31-12-2012  12:03
    • Respected Member
    • Posts: 218
    Re: STEP Maths I, II, III 1996 Solutions
    (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.

    Click image for larger version.  Name: induction DF.png  Views: 31  Size: 4.7 KB  ID: 189127
    Sorry i didn't specify i was talking about the last part, but I have been sorted out on the step prep thread
  12. mathsymathsy's Avatar
    132 04-02-2013  19:35
    • Respected Member
    • Posts: 176
    Re: STEP Maths I, II, III 1996 Solutions
    (Original post by DFranklin)
    Step QI, P5.

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

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

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

    (ii) (p+qi)^2 = p^2-q^2+2pqi = 3-2\sqrt{3} + 2(1-\sqrt{3})i. So
    p^2-q^2 = 3-2\sqrt{3}, pq = 1-\sqrt{3}. Again, we set up the quadratic and get:
    p^4-(3-2\sqrt{3})p^2-(1-\sqrt{3})^2 = 0. Use the quadratic formula to get:
    2p^2 = (3-2\sqrt{3}) \pm \sqrt{(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2}
    Now (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 (3-2\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 \alpha, \beta with (\alpha+\beta\sqrt{3})^2 = 37 - 20\sqrt{3}.

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

    Plugging this back in, we have

    2p^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 = \pm (1-\sqrt{3}). Use pq = (1-\sqrt{3}) to get final solutions \pm (1-\sqrt{3}+i).

    (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),
    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
  13. Rainingshame's Avatar
    133 05-03-2013  19:26
    • Exalted and Worshipped Member
    • Posts: 1,239
    Re: STEP Maths I, II, III 1996 Solutions
    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:
    \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 det A =0
    so when 2a-2a^2 =0
    so when a=0 or a=1
    Last edited by Rainingshame; 12-04-2013 at 16:24.
  14. metaltron's Avatar
    134 01-04-2013  00:29
    • Exalted and Worshipped Member
    • Posts: 908
    Re: STEP Maths I, II, III 1996 Solutions
    (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
    (r + s\sqrt3)^2 = 3 -2\sqrt3 + 1 = (\sqrt3 -1)^2

    r + s\sqrt3 = \pm(\sqrt3 -1)

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

    r = \pm1 , s= \mp1


    Part ii)

    Spoiler:
    Show
    (p+qi)^2 = i^2 + 2(1-\sqrt3)i + 4 - 2\sqrt3 = i^2 + 2(1/\sqrt3)i + (\sqrt3 -1)^2 = (i+1-\sqrt3)^2

    p + qi = \pm(i+1-\sqrt3)

    p = \pm(1-\sqrt3) , q = \pm1


    Part iii)

    Spoiler:
    Show
    By the quadratic formula:

    z = \frac{2\pm\sqrt{4-4(i+1)(2\sqrt3-2)}}{2(1+i)} = \frac{1\pm\sqrt{1 - (1+i)(2\sqrt3 - 2)}}{1+i} = \frac{1\pm\sqrt{(3-2\sqrt3)+2i(1-\sqrt3)}}{1+i} = \frac{1\pm(i+1-\sqrt3)}{1+i}

    z = \frac{\sqrt3 - i}{1+i} = \frac{(\sqrt3 - 1) - (\sqrt3 + 1)i}{2}

    or

    z = \frac{2+i-\sqrt3}{1+i} = \frac{(3-\sqrt3) + (\sqrt3 - 1)i}{2}
  15. Nick_'s Avatar
    135 10-04-2013  12:20
    • Respected Member
    • Posts: 201
    Re: STEP Maths I, II, III 1996 Solutions
    (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. Nick_'s Avatar
    136 11-04-2013  13:12
    • Respected Member
    • Posts: 201
    Re: STEP Maths I, II, III 1996 Solutions
    (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 F_{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
    F_{k+2}F_k - F_{k+1}^2 \\ = (F_{k+1}+F_k)F_k - F_{k+1}^2 \\ = F_k^2 +F_{k+1}F_k - F_{k+1}^2 = F_k^2 - F_{k+1}(F_{k+1} - F_k) \\ = [F_{k+1}F_{k-1}+(-1)^{k+1}] - F_{k+1}(F_{k+1} - F_k) \\ = (-1)^{k+1} + F_{k+1}F_{k-1} - F_{k+1}(F_{k-1})\\ = (-1)^{k+1}F_{k+1}^2

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

    Want to show F_{n+k} = F_kF_{n+1}+F_{k-1}F_n. Assume true for 1 < k <=m. Then
    F_{n+k+1} = F_{n+k}+F_{n+k-1}\\ = F_kF_{n+1}+F_{k-1}F_n + F_k-1F_{n+1}+F_{k-2}F_n\\ =(F_k+F_{k-1})F_{n+1}+(F_{k-1}+F_{k-2}F_n \\ =F_{k+1}F_{n+1}+F_k F_n.

    So true for k=m+1.

    Explictly when k=1 we have F_{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?
  17. DFranklin's Avatar
    137 11-04-2013  13:39
    • TSR Royalty
    • Location: London
    • Posts: 18,046
    Re: STEP Maths I, II, III 1996 Solutions
    (Original post by Nick_)
    ..
    I explicitly show the result for n<=3 in the first few lines of the proof.
  18. Nick_'s Avatar
    138 11-04-2013  17:04
    • Respected Member
    • Posts: 201
    Re: STEP Maths I, II, III 1996 Solutions
    (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. DFranklin's Avatar
    139 11-04-2013  17:32
    • TSR Royalty
    • Location: London
    • Posts: 18,046
    Re: STEP Maths I, II, III 1996 Solutions
    Fair point.
  20. iMadRichard's Avatar
    140 12-04-2013  15:44
    • New Member
    • Posts: 16
    Re: STEP Maths I, II, III 1996 Solutions
    (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:
    \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 det A =0
    so when 2a-2a^2 =0
    so when a=0 or a=1
    You mean non-invertible.
    Post rating:
    1
Sign in to Reply
Search Thread
Advanced Search
Share this discussion:  
Useful resources

Articles:

Study Help rules and posting guidelinesStudy Help Member of the Month nominations

Quick Link:

Unanswered Maths Exams Threads

Groups associated with this forum:

View associated groups
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is moderated by:

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22

Registered Office: International House, Queens Road, Brighton, BN1 3XE

Shortcuts

Get Started

TSR Group

Using TSR

Info

Connect with TSR

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.