Hey there Sign in to join this conversationNew here? Join for free

STEP I, II, III 2002 Solutions

Announcements Posted on
    • 6 followers
    Offline

    ReputationRep:
    (Original post by adrienne_om)
    Above from --- http://www.thestudentroom.co.uk/show...1#post19333071

    Maybe a correction here -- I'm not sure -- but I got a positive sign in the RHS of the second equation ---

     ab^2 + (a-1)b +(a-1)=0

     \iff b^2 + \dfrac{a-1}{a}b = \dfrac{1 - a}{a}

     \iff b^2 + \dfrac{a-1}{a}b + \left(\dfrac{a - 1}{2a})^2\right = \dfrac{1 - a}{a} + \left(\dfrac{a - 1}{2a}\right)^2

     \iff \left( b + \dfrac{a-1}{2a} \right)^2 = \dfrac{1-a}{a} + \dfrac{a^2 - 2a +1}{4a^2}
    \text{I agree with you. So we finally have }

    b=\dfrac{1}{2a}\left[1-a\pm \sqrt{(a-1)^2-4a(a-1)} \text{ or }k=\dfrac{b}{1-a}=\dfrac{1}{2a} \pm\dfrac{\sqrt{1+2a-3a^2}}{2a(1-a)}
    • 3 followers
    Offline

    ReputationRep:
    Seems Paper II Q1 has a bad solution here:

    http://www.thestudentroom.co.uk/show...9#post18023769

    No need to actually look at the question paper, Part ii) to this solution has some funky algebra!

    The last few lines should be

    Part Two
    \begin{array}{rl}

\therefore \displaystyle\int_{\frac{\sqrt3}  {2}}^1 \frac{1}{1-\sqrt{1-x^2}} \, dx

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \frac{2\cos 2\theta}{1-\sqrt{1-\sin^2 \theta}} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \frac{2\cos 2\theta}{1-\cos 2\theta} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \frac{2\cos^2 - 2\sin^2 \theta}{2\sin^2 \theta} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \mathrm{cosec}^2 \theta - 2 \, d\theta \\ \br \\

& \displaystyle = \sqrt3 - 1 + \frac{\pi}{3} - \frac{\pi}{2} \\ \br \\

& \displaystyle = \boxed{\sqrt3 - 1 - \frac{\pi}{6}} \end{array}


    The jump from the third line to the fourth still isn't all that clear but at least now it seems correct

    EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct.
    • 13 followers
    Offline

    ReputationRep:
    (Original post by Xero Xenith)
    I think the result only holds for complex numbers modulus 1.
    Not even for all numbers of modulus 1. e.g. z = i, n = 2.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Xero Xenith)
    Seems Paper II Q1 has a bad solution here:

    http://www.thestudentroom.co.uk/show...9#post18023769

    No need to actually look at the question paper, Part ii) to this solution has some funky algebra!

    The last few lines should be

    Part Two
    \begin{array}{rl}

\therefore \displaystyle\int_{\frac{\sqrt3}  {2}}^1 \frac{1}{1-\sqrt{1-x^2}} \, dx

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \frac{2\cos 2\theta}{1-\sqrt{1-\sin^2 \theta}} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \frac{2\cos 2\theta}{1-\cos 2\theta} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \frac{2\cos^2 - 2\sin^2 \theta}{2\sin^2 \theta} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \mathrm{cosec}^2 \theta - 2 \, d\theta \\ \br \\

& \displaystyle = \sqrt3 - 1 + \frac{\pi}{3} - \frac{\pi}{2} \\ \br \\

& \displaystyle = \boxed{\sqrt3 - 1 - \frac{\pi}{6}} \end{array}


    The jump from the third line to the fourth still isn't all that clear but at least now it seems correct

    EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct.
    Good spot, fixed
    • 19 followers
    Offline

    ReputationRep:
    (Original post by Xero Xenith)
    Seems Paper II Q1 has a bad solution here:

    http://www.thestudentroom.co.uk/show...9#post18023769

    No need to actually look at the question paper, Part ii) to this solution has some funky algebra!

    The last few lines should be

    Part Two
    \begin{array}{rl}

\therefore \displaystyle\int_{\frac{\sqrt3}  {2}}^1 \frac{1}{1-\sqrt{1-x^2}} \, dx

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \frac{2\cos 2\theta}{1-\sqrt{1-\sin^2 \theta}} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \frac{2\cos 2\theta}{1-\cos 2\theta} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \frac{2\cos^2 - 2\sin^2 \theta}{2\sin^2 \theta} \, d\theta \\ \br \\

& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi}  {4}} \mathrm{cosec}^2 \theta - 2 \, d\theta \\ \br \\

& \displaystyle = \sqrt3 - 1 + \frac{\pi}{3} - \frac{\pi}{2} \\ \br \\

& \displaystyle = \boxed{\sqrt3 - 1 - \frac{\pi}{6}} \end{array}


    The jump from the third line to the fourth still isn't all that clear but at least now it seems correct

    EDIT: deleted my post below because I was wrong and the solution to Q2 seems correct.
    Probably being a bit stupid here but how did you go from 1-sqrt(1-cos^2 x) to 1-cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally. Ofcourse doesn't affect final answer, but could lead to a little confusion
    • 3 followers
    Offline

    ReputationRep:
    (Original post by Extricated)
    Probably being a bit stupid here but how did you go from 1-sqrt(1-cos^2 x) to 1-cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally.
    Good spot yourself, pretty sure you're right!


    (Original post by sonofdot)
    Good spot, fixed
    Seems like there's still just a minor thing on the third line (the one with the three dots) - square root sign should be 1- \cos^2 2 \theta not just 1- \cos^2 \theta.
    • 18 followers
    Online

    ReputationRep:
    (Original post by Extricated)
    Probably being a bit stupid here but how did you go from 1-sqrt(1-cos^2 x) to 1-cos2x?(denominator line 1, to denominator line2), I think the sin^2(2x) has been replaced by sin^2x in line 1 accidentally. Ofcourse doesn't affect final answer, but could lead to a little confusion
    nah you are being stupid here :lol:
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Glutamic Acid)
    II/7: (Scary scary vectors.)
    ??????????

    Let the lines have direction vector \left( \begin{array}{c} a \\ b \\ c \end{array} \right) which I'll denote by r, and without loss of generality let |r| = 1 so a^2 + b^2 + c^2 = 1;

    \Rightarrow \left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right) . \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \sqrt{2} \cos \dfrac{\pi}{4} \Rightarrow a + b = 1

    \Rightarrow \left( \begin{array}{c} a \\ b \\ c \end{array} \right) . \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right) = \sqrt{2} \cos \dfrac{\pi}{4} \Rightarrow a + c = 1.

    b = 1 - a; c = 1 - a so a^2 + (1 - a)^2 + (1 - a)^2 = 1 \Rightarrow (3a - 1)(a - 1) = 0
    The solutions are a =1, b = 0, c = 0 and a = 1/3, b = 2/3, c = 2/3.

    m3 has direction \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right); m4 direction \left( \begin{array}{c} 1/3 \\ 2/3 \\ 2/3 \end{array} \right).

    Considering \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) . \left( \begin{array}{c} 1/3 \\ 2/3 \\ 2/3 \end{array} \right) gives \cos \theta = 1/3.

    (i) A has position vector \left( \begin{array}{c} \lambda \\ \lambda \\ 0 \end{array} \right); B \left( \begin{array}{c} \lambda \\ 0 \\ \lambda \end{array} \right).
    Let P = \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right), and Q = \left( \begin{array}{c} 1/3 \\ 2/3 \\ 2/3 \end{array} \right)

    AQ . BP = 0 so \left( \begin{array}{c} \lambda - 1/3 \\ \lambda - 2/3 \\ -2/3 \end{array} \right) . \left( \begin{array}{c} \lambda - 1 \\ 0 \\ \lambda \end{array} \right) = 0

    (\lambda - 1)(\lambda - 2/3) - 2/3 \times \lambda = 0 \text{giving} \lambda = 1 \pm \dfrac{\sqrt{6}}{3}. Showing that the latter is > 0; 1 - sqrt(6)/3 > 0 therefore 1 > sqrt(6)/3 therefore 1 > 6/9, which is true.

    (ii) AQ = \left( \begin{array}{c} 1/3 \\ 2/3 \\ 2/3 \end{array} \right) + \alpha \left( \begin{array}{c} \lambda - 1/3 \\ \lambda - 2/3 \\ -2/3 \end{array} \right)

    BP = \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) + \beta \left( \begin{array}{c} \lambda - 1 \\ 0 \\ \lambda \end{array} \right)

    So 1/3 + \alpha(\lambda - 1/3) = 1 + \beta(\lambda - 1)
    2/3 + \alpha(\lambda - 2/3) = 0
    2/3 - \alpha 2/3 = \lambda \beta

    The second equation gives \alpha = \dfrac{2}{2 - 3 \lambda}. Substituting into the third gives \beta = \dfrac{2}{2 - 3 \lambda}. Substituting these into the first:
    1/3 + \dfrac{2}{2 - 3 \lambda}(\lambda - 1/3) = 1 + \dfrac{2}{3 \lambda - 2}(\lambda - 1); multiplying out gives 2 \lambda = 0 \Rightarrow \lambda = 0, so no non-zero solutions.
    sorry im really confused that why does the modulus of r equal to 1?
    • 7 followers
    Offline

    ReputationRep:
    (Original post by yukki0822)
    sorry im really confused that why does the modulus of r equal to 1?
    We're only interested in the direction.
    • 16 followers
    Offline

    ReputationRep:
    For II, Q10, I believe there's an error in the final part - the -\dfrac{29}{2} should have an x attached, which makes the final answer 81/40. I also get this answer through an entirely different approach to the question:

    Spoiler:
    Show


     13t + (T-t)(\dfrac{14T+2t}{T}) = \dfrac{339}{8}

     13Tt + 14T^2+2Tt-14Tt-2t^2 = \dfrac{339}{8}T

     14T^2 + Tt - 2t^2 = \dfrac{339}{8}T

    Differentiate implicitly:

      28T\dfrac{dT}{dt}+T+t\dfrac{dT}{  dt}-4t = \dfrac{339}{8}\dfrac{dT}{dt}

    When T is stationary,  \dfrac{dT}{dt} = 0 \Rightarrow T - 4t = 0 , so  t = \dfrac{1}{4}T

    Plug this back into the original equation:

     14T^2+\dfrac{1}{4}T^2-\dfrac{1}{8}T^2 = \dfrac{339}{8}T

     113T^2 - 339T = 0

    This gives maxima/minima at T = 0 and T = 3. T = 0 is obviously impossible, so T must equal 3, as required.

    For the second part, use a trapezium to calculate the final speed of the second competitor, and use this to work out the acceleration:

     \frac{3}{2}(16+v) = \frac{339}{8} \Rightarrow V = \frac{49}{4}

    And so the acceleration is then -1.25 kph.

     T = 3 \Rightarrow t = 0.75 , which means that the speed in the second part is 14.5. Next, find the time when they are moving at equal speed, as this is when maximum displacement occurs:

     16-1.25t = 14.5 \Rightarrow t = 1.2

    This is greater than 0.75 so is fine. Then, calculate the distances for each:

     x_1 = 16(1.2)-\frac{5}{8}(1.2)^2  = \frac{183}{10}

     x_2 = 0.75(13)+0.45(14.5) = \frac{651}{40}

    Subtracting one from the other yields the final answer  \frac{81}{40}

    • 0 followers
    Offline

    ReputationRep:
    (Original post by SimonM)
    STEP I, Question 4

    Spoiler:
    Show


    \displaystyle y' = - \frac{2x}{(1+x^2)^2}

    Therefore the equation of the tangent is \displaystyle y = - \frac{2a}{(1+a^2)^2} x + \frac{1}{1+a^2} +\frac{2a^2}{(1+a^2)^2}

    Since (0,1) is a solution

    \displaystyle 1 = \frac{1}{1+a^2} +\frac{2a^2}{(1+a^2)^2} \Rightarrow 1+2a^2+a^4 = 1+a^2 + 2a^2 \Rightarrow a^4 -a^2 = 0 \Rightarrow a \in \{-1,0,1\} \Rightarrow a = 1

    This gives us the equation, \displaystyle y = -\frac{1}{2} x+1

    We have \displaystyle 1 = \left (1 - \frac{1}{2} x \right) (1+x^2) \Rightarrow  0 = x(x-1)^2 so the only intersections are 0 and 1.

    We have

    \displaystyle \int_0^1 y \, dx = \frac{\pi}{4} (given)

    The area under the tangent is a trapezium with area

    \displaystyle \frac{1}{2} \left ( 1+ \frac{1}{2} \right ) \cdot 1 = \frac{3}{4}

    Since the graph is always above it, we get \frac{3}{4} < \frac{\pi}{4} which is what we want to show.

    Considering the half closest to the y axis, and revolving it around that axis we get:

    \displaystyle \pi \int_{1/2}^1 x^2 \, dy = \pi \left [ \ln y - y\right ]_{1/2}^1 = \pi \left ( \ln 2 - \frac{1}{2} \right )
    and

    \displaystyle \pi \int_{1/2}^1 (2(y-1))^2\, dy = 4\pi \left [ \frac{y^3}{3} \right ]_{-1/2}^0 = \pi \frac{1}{6}

    Therefore \ln 2 > \frac{1}{2} + \frac{1}{6} = \frac{2}{3}
    Does this graph have an oblique asymptote?
    • 2 followers
    Offline

    ReputationRep:
    (Original post by Glutamic Acid)
    II/8:

    y = e^{3 + x - e^x} \text{for} x < 0
    I think this should be:

    y = e^{2e -1 + x - e^x} ​ instead?
    • 0 followers
    Offline

    ReputationRep:
    Regarding STEP III / 2002 / Q8, i had thought that Dadeyemi gave an excellent answer, but a student of mine was seeking clarification for the first part regarding the periodicities of the argument and the tangent. I looked at the problem more closely just now, and agree that it needs to be tightened up a little. Here is my little "patch":-

    Let  \alpha = arg(z - w) . It has already been shown that  \tan \alpha = \tan \frac{x+y-\pi}{2}
    Hence  \alpha = \frac{x+y-\pi}{2} + n \pi (*)

    Since  x, y \in (-\pi, \pi] , we have that
     x+y-\pi \in (-3\pi, \pi] and so
    Spoiler:
    Show
     \frac{x+y-\pi}{2} + n \pi \in D_n := (-\frac{3\pi}{2}+n\pi, \frac{\pi}{2}+n\pi] .



    Spoiler:
    Show
    When  \tan \alpha > 0 ,  \alpha \in (-\pi,-\frac{\pi}{2}) \cup (0,\frac{\pi}{2}) \subseteq D_0 = (-\frac{3\pi}{2}, \frac{\pi}{2}] , hence (*) holds with n = 0.

    Spoiler:
    Show
    When  \tan \alpha < 0 ,  \alpha \in (-\frac{\pi}{2},0) \cup (\frac{\pi}{2},\pi) \subseteq D_1 = (-\frac{\pi}{2}, \frac{3\pi}{2}] , hence (*) holds with n = 1.

    We check the remaining cases where  \tan \alpha = 0 as follows (draw your own diagrams):-
    Spoiler:
    Show
    When  x = -\frac{3\pi}{4}, y = \frac{-\pi}{4} ,  \alpha = 0 \in D_0 , hence (*) holds with n = 0.

    Spoiler:
    Show
    When  x = \frac{\pi}{4}, y = \frac{3\pi}{4} ,  \alpha = \pi \in D_1 , hence (*) holds with n = 1.


    all the best for your STEP exams tomorrow!
    • 6 followers
    Offline

    ReputationRep:
    (Original post by Generic Name)
    Does this graph have an oblique asymptote?
    No. The only asymptote is the x axis.
    • 1 follower
    Offline

    ReputationRep:
    (Original post by DFranklin)
    I think this is your problem:
    Spoiler:
    Show
    When \theta = 0,\, y = \sqrt{a^2} + \sqrt{b^2} = |a| + |b|, not a+b.
    this was for question 3 step 1- I am having some trouble with it because i do not not know how to interpret the power  ^{1/2}
    do you have to assume that the  ^{1/2} means take the positive root beacause you want it to be a function?

    It is also confusing for me because after every step the possibilities are many for example the step nota bene has just made.

    \theta = 0,\, y = \sqrt{a^2} + \sqrt{b^2} = |a| + |b|

    It could also be:
    \theta = 0,\, y = \sqrt{a^2} + \sqrt{b^2} = \pm a \pm b

    Summary if this question meant take the power of half to mean take the positive square root- i would have no problems
    but i am not sure if this is so.

    Could you please clarify?
    • 20 followers
    Offline

    ReputationRep:
    (Original post by nahomyemane778)
    this was for question 3 step 1- I am having some trouble with it because i do not not know how to interpret the power  ^{1/2}
    do you have to assume that the  ^{1/2} means take the positive root beacause you want it to be a function?
    Yes - you've answered your own question
    • 1 follower
    Offline

    ReputationRep:
    (Original post by davros)
    Yes - you've answered your own question
    thanks again
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Dadeyemi)
    Some more;

    Did these quite a while ago I'm afraid some may be partial solutions.
    You missed (0,1) and (0,-1) for question 4 I believe.
    • 0 followers
    Offline

    ReputationRep:
    Re: 2002/III/Q8

    Dadeyemi gave an excellent solution and SingaporeCantab had a point on the periodicity. Notice that the periodicity of complex numbers is 2n(pi) because of the specific signs for the real part and the imaginary part. Need extra care in using the tan function where the periodicity is n(pi), e.g cis(pi/4) is not equal to cis(pi/4 + pi) though tan(pi/4) = tan(pi/4+pi). I think the patch is not entirely correct.

    My patch to the "patch": I think it is sufficient to say that (theta1 + theta2 - pi)/2 has a range (-3pi/2, pi/2). When
    (theta1 + theta2 - pi)/2 is in the range (-3pi/2, -pi], we need to add 2pi (i.e. n=1) to bring it back to the domain of (-pi,pi]. Otherwise n=0.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by brianeverit)
    \text{I agree with you. So we finally have }

    b=\dfrac{1}{2a}\left[1-a\pm \sqrt{(a-1)^2-4a(a-1)} \text{ or }k=\dfrac{b}{1-a}=\dfrac{1}{2a} \pm\dfrac{\sqrt{1+2a-3a^2}}{2a(1-a)}
    II Q5 So
    k= 1/2a x {1 +/- sqrt[(1+3a)/(1-a)]}
    = 1/2a x {1 +/- sqrt[1+ 4a/(1-a)]}
    since 4a/(1-a)>0 so the sqrt >1, and k>0, hence cannot take the negative sqrt.
    k = 1/2a x {1 + sqrt[1+ 4a/(1-a)]}

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?

    this is what you'll be called on TSR

  2. this can't be left blank
    this email is already registered. Forgotten your password?

    never shared and never spammed

  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide the button to the right to create your account

    Slide to join now Processing…

Updated: April 6, 2014
New on TSR

Naughtiest thing you did at school

Did you get away with it or were you punished?

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.