STEP Maths I, II, III 1988 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
TSR launches Learn Together! - Our new subscription to help improve your learning 16-05-2013
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. maltodextrin's Avatar
    61 14-04-2009  19:25
    • Respected Member
    • Location: Horsham
    • Posts: 236
    Re: STEP Maths I, II, III 1988 solutions
    (Original post by kabbers)
    III/1:

    Sketch y = \frac{x^2 e^{-x}}{x+1}

    I'm not going to bother posting most of the differentiating legwork, there arent too many tricks, its just an algebra bash.

    Differentiating (quotient rule is your friend), we get

    \frac{dy}{dx} = e^{-x}\frac{x(2-x^2)}{(x+1)^2}

    So we have turning points at x = 0, and x = +\sqrt{2}, -\sqrt{2}


    Differentiating again, we get

    \frac{d^2 y}{dx^2} = e^{-x}\frac{(2-3x^2)(x+1)^2 - 2(x+1)(2x-x^3) - (2x - x^3)(x+1)^2}{(x+1)^4}

    Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.


    Noticing the denominator of y, we find that there is an asymptote at x = -1.

    And because of the exponential properties of e^-x, y tends to zero as x \to +\infty

    y tends to -\infty as x \to -\infty since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.


    So the graph will look http://www.thestudentroom.co.uk/show...8&postcount=14



    Prove \displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

    First note that \frac{x^2}{x+1} = \frac{x^2+x}{x+1} - \frac{x}{x+1} = x - \frac{x}{x+1} = x - (\frac{x+1}{x+1} - \frac{1}{x+1}) = x - 1 + \frac{1}{x+1}

    Hence we may split the integral into \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

    Consider \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx

    = \displaystyle\int^{\infty}_0 xe^{-x} dx - \int^{\infty}_0 e^{-x} dx

    = \displaystyle([-xe^{-x}]^{\infty}_0 + \int^{\infty}_0 e^{-x} dx - \int^{\infty}_0 e^{-x} dx

    = 0


    Now consider \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

    I posit the inequality \frac{1}{x+1}e^{-x} < e^{-x} for x > 0

    \frac{1}{x+1} < 1
    1 < (x+1)
    0 < x

    So our inequality holds.

    So:

    \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < \int^{\infty}_0 e^{-x} dx

    \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < [-e^{-x}]^{\infty}_0

    \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

    Thus \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

    And therefore, \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1


    Now notice that the graph of y = \frac{x^2 e^{-x}}{1+x} is greater than 0 for x > 0, and hence so is the infinite integral.

    So, \displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

    please point out any mistakes
    I'm probably wrong but to show that the integral lies between 0 and 1 would it not be sufficient to say that

    (x^2)(e^-x)/(x + 1) < (x^2)(e^-x)/x = x(e^-x) for 0 < x < infinity

    then integrate x(e^-x), which equals 1. Seems a lot simpler?
  2. DFranklin's Avatar
    62 14-04-2009  20:11
    • TSR Royalty
    • Location: London
    • Posts: 18,041
    Re: STEP Maths I, II, III 1988 solutions
    Yeah, that's fine.

    While we're nitpicking - taking the 2nd derivative to deduce the nature of the turning points is a lot more work than simply sketching x(2-x^2) to argue whether the derivative is increasing or decreasing.
  3. brianeverit's Avatar
    63 03-05-2009  17:04
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    1988 Paper 2 number 4
    Attached Files
  4. File Type: pdf 1988PAPERII fma.4.pdf (40.0 KB, 112 views)
  5. brianeverit's Avatar
    64 03-05-2009  17:05
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    1988 Paper 2 number 8
    Attached Files
  6. File Type: pdf 1988PAPERII fma.8.pdf (37.9 KB, 63 views)
  7. brianeverit's Avatar
    65 03-05-2009  17:06
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    1988 Paper 2 numbers 10,11 & 12
    Attached Files
  8. File Type: pdf 1988PAPERII fma.10.pdf (45.8 KB, 49 views)
  9. File Type: pdf 1988PAPERII fma.11.pdf (32.2 KB, 42 views)
  10. File Type: pdf 1988PAPERII fma.12.pdf (30.5 KB, 43 views)
  11. brianeverit's Avatar
    66 03-05-2009  17:08
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    1988 Paper 2 numbers 13 & 14
    Attached Files
  12. File Type: pdf 1988PAPERII fma.13.pdf (33.6 KB, 42 views)
  13. File Type: pdf 1988PAPERII fma.14.pdf (43.2 KB, 32 views)
  14. brianeverit's Avatar
    67 03-05-2009  17:09
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    1988 Paper 2 numbers 15 & 16
    Attached Files
  15. File Type: pdf 1988PAPERII fma.15.pdf (32.6 KB, 34 views)
  16. File Type: pdf 1988PAPERII fma.16.pdf (36.5 KB, 34 views)
  17. SimonM's Avatar
    68 15-06-2009  18:35
    • PS Helper
    • TSR Idol
    • Posts: 9,190
    Re: STEP Maths I, II, III 1988 solutions
    Question 8, STEP III

    Spoiler:
    Show

    \displaystyle \frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt} = \frac{2a}{2at} = \frac{1}{t}

    Therefore the equation of the gradient and normal respectively are:

    \displaystyle \frac{y-2at}{x-at^2} = \frac{1}{t} and \displaystyle \frac{y-2at}{x-at^2} = -t

    For the triangle formed by the tangents, the vertices will be the intersection of the tangents. By symmetry we need only compute this for one vertex:

    \displaystyle \begin{cases} y = \frac{1}{t_1} x+ at_1 \\ y = \frac{1}{t_2} x + at_2 \end{cases} \Rightarrow x \left ( \frac{1}{t_1} - \frac{1}{t_2} \right ) = a(t_2 - t_1) \Rightarrow x = at_1t_2, y = a(t_1+t_2)

    Therefore the area of the triangle of tangents is:

    \displaystyle A_1 = \frac{1}{2} \begin{vmatrix} at_1t_2 & a(t_1+t_2) & 1 \\ at_2t_3 & a(t_2+t_3) & 1 \\ at_3t_1 & a(t_3+t_1) & 1 \end{vmatrix}

    For the triangle formed by normals, the vertices will be the intersection of the normals. Again by symmetry we need only consider one pair:

    \displaystyle \begin{cases} y = -t_1 x + at_1^3 + 2at_1 \\ y = -t_2 x + at_2^3 + 2at_2 \end{cases} \Rightarrow x(t_1 - t_2) = a(t_1^3 - t_2^3)+2a(t_1-t_2) \\ \Rightarrow x = a(t_1^2+t_1t_2 + t_2^2 + 2), y = -at_1t_2(t_1+t_2)

    This gives us an area

    \displaystyle A_2 = \begin{vmatrix} a(t_1^2+t_1t_2 + t_2^2 + 2) & -at_1t_2(t_1+t_2) & 1 \\ a(t_2^2+t_2t_3 + t_3^2 + 2) & -at_2t_3(t_2+t_3) & 1 \\ a(t_3^2+t_3t_1 + t_1^2 + 2) & -at_3t_1(t_3+t_1) & 1 \end{vmatrix}

    \displaystyle A_1 = \frac{1}{2} \begin{vmatrix} at_1t_2 & a(t_1+t_2) & 1 \\ at_2t_3 & a(t_2+t_3) & 1 \\ at_3t_1 & a(t_3+t_1) & 1 \end{vmatrix} = \frac{1}{2} \begin{vmatrix} at_1t_2 & a(t_1+t_2) & 1 \\ at_2(t_3-t_1) & a(t_3-t_1) & 0 \\ at_1(t_3-t_2) & a(t_3-t_2) & 0 \end{vmatrix} \\ = \frac{a^2}{2}|(t_3-t_1)(t_3-t_2)(t_2-t_1)|

    \displaystyle A_2 = \begin{vmatrix} a(t_1^2+t_1t_2 + t_2^2 + 2) & -at_1t_2(t_1+t_2) & 1 \\ a(t_2^2+t_2t_3 + t_3^2 + 2) & -at_2t_3(t_2+t_3) & 1 \\ a(t_3^2+t_3t_1 + t_1^2 + 2) & -at_3t_1(t_3+t_1) & 1 \end{vmatrix} \\ =\begin{vmatrix} a(t_1^2+t_1t_2 + t_2^2 + 2) & -at_1t_2(t_1+t_2) & 1 \\ a(t_2(t_3-t_1) + t_3^2-t_1^2 ) & -at_2(t_1-t_3)(t_1+t_2+t_3) & 0 \\ a(t_3^2-t_2^2+t_1(t_3-t_2)) & -at_1(t_2-t_3)(t_1+t_2+t_3) & 0 \end{vmatrix} \\ =\\ \begin{vmatrix} a(t_1^2+t_1t_2 + t_2^2 + 2) & -at_1t_2(t_1+t_2) & 1 \\ a(t_3-t_1)(t_1+t_2+t_3) & -at_2(t_1-t_3)(t_1+t_2+t_3) & 0 \\ a(t_3-t_2)(t_1+t_2+t_3) & -at_1(t_2-t_3)(t_1+t_2+t_3) & 0 \end{vmatrix} \\ = \frac{a^2}{2} |(t_3-t_1)(t_3-t_2)(t_2-t_1)(t_1+t_2+t_3)^2| = (t_1+t_2+t_3)^2 A_1

    As required.

    If they are congruent, the area of the triangle will be zero: t_1+t_2+t_3 = 0
    Last edited by SimonM; 15-06-2009 at 18:39.
  18. r2enigma's Avatar
    69 10-04-2010  07:07
    • Full Member
    • Posts: 121
    Re: STEP Maths I, II, III 1988 solutions
    Can somebody please post the solution for STEP III Q3?
  19. hannah.j's Avatar
    70 07-02-2011  18:12
    • Junior Member
    • Posts: 35
    Re: STEP Maths I, II, III 1988 solutions
    (Original post by SimonM)
    Done
    Am I missing something or is the solution, well, missing?
  20. Mensorah's Avatar
    71 12-05-2011  09:51
    • Full Member
    • Posts: 142
    Re: STEP Maths I, II, III 1988 solutions
    Where can i get STEP 2 solution 2
  21. ben-smith's Avatar
    72 19-05-2011  23:54
    • Overlord in Training
    • Location: Hilbert Space
    • Posts: 2,366
    Re: STEP Maths I, II, III 1988 solutions
    (Original post by SimonM)
    ....
    STEP III Q10
    Consider the positions of the dogs as time goes on. They will form a square of decreasing side length that rotates about the centre. If we take the position of one of the dogs to be (x,y) from the centre, then we see by symmetry that the position of one the next dogs along is (-y,x). We can also see that the tangent at any moment goes through the position of the next dog along, so:
    \frac{dy}{dx}=\frac{y-x}{x-(-y)}=\frac{\frac{y}{x}-1}{1+\frac{y}{x}}
    let:
    u=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u \therefore \frac{du}{dx}x+u=\frac{u-1}{1+u} \Rightarrow \frac{du}{dx}x=\frac{-u^2-1}{1+u} \Rightarrow \frac{dx}{du} \frac{1}{x}=\frac{1+u}{-u^2-1} \Rightarrow \frac{1}{x}dx=\frac{1+u}{-u^2-1}du
    let \theta be the angle between the x axis and the straight line to the point (x,y). Since u=\frac{y}{x} and, by trigonometry, \frac{y}{x}=Tan\theta \Rightarrow u=Tan\theta. So, substituting into the RHS and integrating we get:
    lnx+c_1=-ln(sec\theta)-\theta+c_2.
    By trigonometry, again, x=rcos\theta where r is the distance from the origin to the point (x,y) \Rightarrow ln(rcos\theta)+ln(sec\theta)=-\theta +c \Rightarrow ln(r)=-\theta +c \Rightarrow r=e^{-\theta}e^c
    e^c is constant so r=e^{-\theta}\lambda for some constant \lambda as required.
    To generalise to n-dogs held at the vertices of an n-gon it must be noted that the next dog along will have the coordinates of the first dog bur rotated \frac{2\pi}{n} about the centre of the field i.e:
    \begin{pmatrix} cos(\frac{2\pi}{n}) & -sin(\frac{2\pi}{n})\\sin(\frac{2 \pi}{n}) & cos(\frac{2\pi}{n}) \end{pmatrix}\begin{pmatrix}x\\y \end{pmatrix}=\begin{pmatrix}xco s\frac{2\pi}{n}-ysin\frac{2\pi}{n}\\xsin\frac{2 \pi}{n}+ycos\frac{2\pi}{n} \end{pmatrix}
    So, like the previous part,:
    \frac{dy}{dx}=\frac{y-xsin\frac{2 \pi}{n}-ycos\frac{2 \pi}{n}}{x-xcos\frac{2 \pi}{n}+ysin\frac{2 \pi}{n}}=\frac{yx^{-1}-sin\frac{2 \pi}{n}-yx^{-1}cos\frac{2 \pi}{n}}{1-cos\frac{2 \pi}{n}+yx^{-1}sin\frac{2 \pi}{n}}
    let:
    u=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u \therefore \frac{du}{dx}x=\frac{-sin\frac{2 \pi}{n}-u^2sin\frac{2 \pi}{n}}{1-cos\frac{2 \pi}{n}+usin\frac{2 \pi}{n}} \Rightarrow \frac{dx}{du}\frac{1}{x}=-cosec(\frac{2 \pi}{n})\frac{1-cos\frac{2 \pi}{n}+usin\frac{2 \pi}{n}}{u^2+1}
    let:
    u=Tan\theta \Rightarrow \frac{du}{d\theta}=sec^2\theta \Rightarrow \frac{dx}{du}\frac{1}{x}=-cosec(\frac{2 \pi}{n})\frac{1-cos\frac{2 \pi}{n}+tan\theta sin\frac{2 \pi}{n}}{sec^2\theta} \Rightarrow \frac{1}{x}dx=(-cosec(\frac{2 \pi}{n})+cot\frac{2 \pi}{n}-Tan\theta)d\theta
    Integrating both sides we get:
    lnx+c=(-cosec(\frac{2 \pi}{n})+cot\frac{2 \pi}{n})\theta-ln\sec\theta +c
    And now, just like in the previous part:
    ln(rcos\theta)+ln(sec\theta)+c=(-cosec(\frac{2 \pi}{n})+cot\frac{2 \pi}{n})\theta+c \Rightarrow r=e^{-\theta(cosec(\frac{2 \pi}{n})-cot\frac{2 \pi}{n})}e^c= e^{-\theta (\frac {1-cos\frac{2 \pi}{n}}{sin\frac{2 \pi}{n}})}\lambda=e^{-\theta Tan(\frac{\pi}{n})}\lambda for some constant \lambda. Letting n=4 we can check and see that it gives the same answer as the first part.
    EDIT: Just realised that the final result simplified to the much nicer form that it is currently in.
    Last edited by ben-smith; 01-09-2011 at 12:37.
  22. Dzwx777's Avatar
    73 20-05-2011  06:25
    • New Member
    • Posts: 6
    Re: STEP Maths I, II, III 1988 solutions
    Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!
  23. ben-smith's Avatar
    74 20-05-2011  08:01
    • Overlord in Training
    • Location: Hilbert Space
    • Posts: 2,366
    Re: STEP Maths I, II, III 1988 solutions
    (Original post by Dzwx777)
    Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!
    http://www.mathshelper.co.uk/oxb.htm
  24. mikelbird's Avatar
    75 05-06-2011  17:38
    • Respected Member
    • Location: Gloucestershire
    • Posts: 191
    Re: STEP Maths I, II, III 1988 solutions
    (Original post by insparato)
    III/3

    |z - i| = 2 is the circle on the complex plane with radius 2 and centre (0,i)

    so x^2 + (y - 1)^2 = 2

    and therefore in paramatrised form

    x = \sqrt2 cos\theta

    y = 1 + \sqrt2 sin \theta

    I've omitted sketches however I have described them in sufficient detail to make an easy sketch if one chooses.

    (i)

    w = \frac{z + i}{z - i}

    wz - wi = z + i

    wz - z = i + wi

    z(w - 1) = i(w + 1)

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

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

    = \frac{1 + i}{w - 1}

    |z - i| = \left|\frac{1+i}{w-1}\right|

    2 = \frac{|1 + i|}{|w - 1|}

    |w - 1| = \frac{\sqrt2}{2}

    Okay so its a circle of radius \frac{\sqrt2}{2} and the centre is (1,0) [/latex]

    (ii)

    If z = x + iy and is real then y must = 0

    so

    w = u + iv = \frac{x + i(y + 1)}{x + i(y - 1)} = \frac{x + i}{x - i}

    (u + iv)(x - i) = x + i

    ux - ui + vxi + v = x + i

    ux + v + i(-u + vx - 1) = x

    -u + vx - 1 must = 0 (this is important bit on)

    so

    ux + v = x

    x(1 - u) = v

    x = \frac{v}{1 - u}

    Now from before

    -u + vx - 1 = 0

    vx = u + 1

    x = \frac{u+1}{v}

    so

    \frac{v}{1 - u} = \frac{u+1}{v}

    v^2 = (u + 1)(1 - u)

    u^2 + v^2 = 1

    So in the W plane, when z is real, you get a circle centre(0,0) with radius 1

    (iii)

    if z = x + iy and z is imaginary then x = 0

    u + vi = \frac{x + i(y + 1)}{x + i(y - 1)}

    u + vi = \frac{i(y + 1)}{i(y - 1)}

    (u + vi)(iy - i) = yi + i

    uyi - ui - vy + v = yi + i

    yi - uyi = -ui - vy + v - i

    y(i-ui) = i(-1 -u) - vy + v

    -vy + v = 0

    y = \frac{i(-1-u)}{i(1 - u)}

    y = \frac{-1 -u}{1 - u}

    1 = \frac{-1 - u}{1 - u}

    -1 - u = 1 -u

    0 = 2

    Wahey! wtf ? I'm sure i got something different when jotting it down, something that made sense! hmm have i gone wrong somewhere? I think I'm doing the right method.
    erm....I have something different....
    Attached Files
  25. File Type: pdf Step1988Paper3Question3.pdf (69.9 KB, 63 views)
  26. brianeverit's Avatar
    76 16-06-2011  12:15
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    1988 STEP III number 3.
    I'm afraid I disagree with the solutions published so far. Here is my solution.

    |z-i|=2 \Rightarrow |x+(y-1)i|=2 \text{ where }z=x+iy
    \Rightarrow x^2+(y-1)^2=4 \text{ so parametric equations are }x=2\cos\theta, y=2\sin\theta+1
    (i) w=\dfrac{z+i}{z-i} \Rightarrow z=\left( \dfrac{w+1}{w-1} \right)i \text{ and }|z-i|=2 \Rightarrow \left|\dfrac{w+1}{w-1}i-i\right|=2 \Rightarrow left| \dfrac{2i}{w-1} \right|=2
    \Rightarrow |2i|=2|w-1| \text{ i.e. }|w-1|=1 \text{ so it is a circle, centre }w=1 \text{ radius }1
    (ii) z \text{ real } \Rightarrow \text {Im}\left( \dfrac{(u+1)i-v}{(u-1)i-v} \right)= \text{ Im}\left( \dfrac{((u+1)i-v)((u-1)i-v)}{(u-1)^2+v^2} \right)= -\dfrac{2uv}{(u-1)^2+v^2}=0
    \text{hence, }2uv=0 \text{ i.e. It is the two axes}
    (iii) z\text{ imaginary }\Rightarrow \text{Re} \left( \dfrac{w+1}{w-=1} \right)i=0 \text{ i.e. } \dfrac {v^2-(u^2-1)}{(u-1)^2+v^2}=0 \Rightarrow v^2-u^2+1=0 \text{ i.e. A hyperbola}
    Attached Thumbnails
    Click image for larger version.  Name: 1988.III.3 Picture.jpg  Views: 22  Size: 17.0 KB  ID: 105975  
    Last edited by brianeverit; 16-06-2011 at 15:54.
  27. brianeverit's Avatar
    77 18-06-2011  20:46
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    1988 STEP I question 1 Alternative Solution

    \text{h}(x)=\dfrac{\ln x}{x} \Rightarrow \text{h}'(x)=\dfrac{1- \ln x}{x^2}=0 \text{ when } \ln x=1\Rightarrow x=\text{e}
    \text{for small }h, \text{h}'(1-h)&gt;0 \text{ while h}'(1+h)&lt;0 \text { so a maximum point at }(\text{e,e}^{-1}
    x=1 \Rightarrow \text{h}(x)=0
    \text {h}(x) \rightarrow 0 \text{ as }x \rightarrow \infty
    \text {h}(x) \rightarrow -\infty \text{ as }x \rightarrow 0

    n^m=m^n \Rightarrow m \ln n=n \ln m \Rightarrow \dfrac{\ln n}{n}=\dfrac{\ln m}{m}
    \text{Clearly, }m \text{ and }n\text{ must lie on either side of the turning point}
    \text{ so the only possible values for the smaller of }m \text{ and }n \text{ are } 1 \text{ and }2
    \text{ there is obviously no other integer with }1^m=m^1 \text{ but } 2^4=4^2
    \text{ so only pair of integers is }2 \text { and }4
    Attached Thumbnails
    Click image for larger version.  Name: 1988.I.1.jpg  Views: 10  Size: 11.5 KB  ID: 106338  
  28. brianeverit's Avatar
    78 19-06-2011  11:29
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    1988 STEP I question 2 Alternative solution

    \text{g}(x)= \text{ f}(x)+\dfrac{1}{\text{f}(x)} \Rightarrow \text{ g}'(x)=\text{ f}'(x) -\dfrac{\text{f}'(x)}{(\text{f}(x ))^2}
    \text{and g}''(x)=\text{ f}''(x)- \dfrac{(\text{f}(x))^2\text{f}'' (x)-\text{f}'(x).2\text{f}(x)\text{f }'(x)}{(\text{f}(x))^4}
    \text{i.e. g}''(x)=\text{ f}''(x)- \dfrac{\text{f}''(x)}{(\text{f}( x))^2}+ \dfrac{2(\text{f}'(x))^2}{(\text {f}(x))^3}
    \text{f}(x)=4+ \cos2x+2 \sin x \Rightarrow \text{f}'(x)=2\cosx-2 \sin2x \text{ and f}''(x)=-2 \sin x-4\cos2x
    \text{so g}'(x)=2\cos x-2\sin x-\dfrac{2\cos x-2\sin 2x}{(4+\cos 2x+2\sin x)^2}
    \text{g}'(x)=0 \text{ when }}2\cos x-2\sin 2x=0 \text{ or }(4+\cos 2x+2\sin x)^2=1
    4+\cos 2x+2\sin 2x&gt;1 \text{ for all }x \text{ so solutions are given by }2 \cos x-2 \sin 2x=0
    \Rightarrow \cos x-2\sin x\cos x=0 \Rightarrow \cos x=0 \text{ or }\sin x =\dfrac{1}{2} \text{ so }x=\dfrac{\pi}{6},\dfrac{\pi}{2} ,\dfrac{5\pi}{6} \text{ or }\dfrac{3\pi}{2}
    \text{g}(x) \text{ is a continuous function since }4+\cos 2x+2\sin x \text{ cannot be zero }
    \text{ so we need only consider one of the turning points }
    x=\dfrac{\pi}{2} \Rightarrow \text{ f}(x)=5,\text{ f}'(x)=0\text{ and f}''(x)=2 \Rightarrow \text{ g}''(x)&gt;0
    \text{so a minimum at }x=\dfrac{\pi}{2}
    \text{ g}\left(\dfrac{\pi}{6}\right)= \dfrac{11}{2}+ \dfrac{2}{11}= \dfrac{125}{22}, \text{ g}\left(\dfrac{\pi}{2}\right)=5+ \dfrac{1}{5}= \dfrac{26}{5}, \text{ g} \left( \dfrac{5\pi}{6} \right)= \dfrac{11}{2}+ \dfrac{2}{11}= \dfrac{125}{22} \text{ and g} \left( \dfrac{3\pi}{2} \rihjt)=1+1=2
    \text{so stationary points are maxima at } \left( \dfrac{\pi}{6}, \dfrac{125}{22} \right) \text{ and }\left( \dfrac{5\pi}{6}, \dfrac{125}{22} \right) \text{ and minima at }\left( \dfrac{\pi}{2}, \dfrac{26}{5} \right) \text{ and } \left( \dfrac{3\piu}{2}, 2\right)
  29. brianeverit's Avatar
    79 19-06-2011  12:02
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    1988 STEP I question 7 Alternative solution

    \text{f}(x)=ax^2+bx+c \Rightarrow \text{ f}'(x)=2ax+b
    \text{f}(1)(x+ \frac{1}{2})+\text{ f}(-1)(x- \frac{1}{2})-2\text{ f}(0)x=(a+b+c)(x+ \frac{1}{2})+(a-b+c)(x- \frac{1}{2})-2cx
    =2ax+2cx+b-2cx=2ax+b=\text{ f}'(x)
    \text{Since f}'(x) \text{ is linear it must attain it's maximum value at either }x=1 \text{ or }x=-1
    \text{f}'(-1)=-\frac{1}{2}\text{f}(1)- \frac{3}{2}\text{f}(-1)-2\text{f}(0) \text{ and f}'(1)= \frac{3}{2}\text{f}(1)+ \frac{1}{2}\text{f}(-1)-2\text{f}(0)
    \text{f}(-1),\tewxt{ f}(0) \text{ and f}(1) \text{ are all between }\pm{1} \Rightarrow |\text{f}'(x)| \leq4
    \text{Consider }2x^2-1 \text{ then }-1 \leq \text{ f}(x0) \leq1 \text{ i.e. }\text{f}(x)| \leq1 \text{ for all }x \text { with }|x| \leq1
    \text{and f}'(x)=4x \text{ lies between }\pm4 \text{ so this function satisfies the requirements}
  30. brianeverit's Avatar
    80 19-06-2011  12:59
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP Maths I, II, III 1988 solutions
    (Original post by nota bene)
    I/9

    i) \displaystyle\int_1^e \frac{\log x}{x^2} dx Recall that \int \log x dx= x(\log (x)-1)+c, now we can use this to integrate by parts.
    I=\displaystyle\int \frac{\log x}{x^2}dx=\frac{1}{x^2}x(\log(x)-1)-\displaystyle\int-2\frac{x(\log(x)-1)}{x^3}dx
    -I=\frac{1}{x}(\log(x)-1)+2\frac{1}{x}=\frac{\log x +1}{x}
    Putting in the limits gives -I=\frac{2}{e}-1= So I=1-\frac{2}{e}

    ii) \displaystyle\int \frac{\cos(x)}{\sin(x)\sqrt{1+\s in(x)}}dx Let u^2=1+\sin(x)\Rightarrow 2u\frac{du}{dx}=\cos(x)\Leftrigh tarrow dx=\frac{2u}{\cos(x)}du
    So the integral becomes \displastyle\int \frac{2}{\sin x}du=\displaystyle\int\frac{2}{u ^2-1}du=\displaystyle\int\frac{1}{u-1}du-\displaystyle\int\frac{1}{u+1}du =\log|u-1|-\log|u+1|+c=\log|\frac{u-1}{u+1}|+c
    Going back to x we have \log|\frac{u-1}{u+1}|+c=\log|\frac{\sqrt{1+\s in x}-1}{\sqrt{1+\sin x}+1}|+c
    Alternative solution for part (b)

    \int \dfrac { \cos x}{ \sin x \sqrt{1+ \sin x}}dx
    \text{ so putting }\sin x= \tan^2 \theta \text{ so } \cos x \dfrac {dx}{d \theta}=2 \tan x\sec^2 x
    \text{integral becomes } \int \dfrac{2\tan \theta \sec^2 \theta d\theta}{\tan^2 \theta \sec\theta }=\int \dfrac{2\sec\theta}{\tan\theta}d \theta=\int 2\csc\theta d\theta=-\ln{|\csc\theta+\cot\theta|}+c
    \tan\theta=\sqrt{x} \Rightarrow \cot\theta=\dfrac{1}{\sqrt{x}},\ sin\theta= \dfrac{\sqrt{x}}{\sqrt{1+x}} \text{ so }\csc\theta=\dfrac{\sqrt{1+x}}{ \sqrt{x}}
    \text{Hence }\int \dfrac{ \cos x}{ \sin x \sqrt{1+\sin x}}dx=-\ln A \left| \dfrac{1+ \sqrt{1+x}}{\sqrt{x}} \right|=\ln A\left| \dfrac{\sqrt x}{1+\sqrt{1+x}} \right|
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.