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STEP Maths I, II, III 1989 solutions

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II/7:

t=xα⇒dtdy=αxα−1dxdy⇒dydx=αxα−1dydtt = x^{\alpha} \Rightarrow \dfrac{\text{d}t}{\text{d}y} = \alpha x^{\alpha - 1}\dfrac{\text{d}x}{\text{d}y} \Rightarrow \dfrac{\text{d}y}{\text{d}x} = \alpha x^{\alpha - 1} \dfrac{\text{d}y}{\text{d}t}

⇒d2ydx2=α(α−1)xα−2dydt+αxα−1ddx(dydt)\Rightarrow \dfrac{\text{d}^2y}{\text{d}x^2} = \alpha (\alpha - 1)x^{\alpha - 2} \dfrac{\text{d}y}{\text{d}t} + \alpha x^{\alpha - 1} \dfrac{\text{d}}{\text{d}x} \left( \dfrac{\text{d}y}{\text{d}t} \right)

Note that ddx(dydt)=ddt(dydt)dtdx=d2ydt2dtdx\dfrac{\text{d}}{\text{d}x} \left( \dfrac{\text{d}y}{\text{d}t} \right) = \dfrac{\text{d}}{\text{d}t} \left( \dfrac{\text{d}y}{\text{d}t} \right) \dfrac{\text{d}t}{\text{d}x} = \dfrac{\text{d}^2y}{\text{d}t^2} \dfrac{\text{d}t}{\text{d}x}.

⇒d2ydx2=α(α−1)xα−2dydt+α2x2α−2d2ydt2\Rightarrow \dfrac{\text{d}^2y}{\text{d}x^2} = \alpha (\alpha - 1)x^{\alpha - 2} \dfrac{\text{d}y}{\text{d}t} + \alpha^2 x^{2 \alpha - 2} \dfrac{\text{d}^2y}{\text{d}t^2}.

Substituting this horrendous crap leads to: α2x2αd2ydt2+(α(α−1)−bα)xαdydt+x2b+2y=0\alpha^2 x^{2 \alpha} \dfrac{\text{d}^2y}{\text{d}t^2} + (\alpha(\alpha - 1) -b \alpha)x^{\alpha} \dfrac{\text{d}y}{\text{d}t} + x^{2b + 2}y = 0.

Now choosing α=b+1\alpha = b + 1 and substituting will smash up dy/dt one time bro', leaving

(b+1)2t2d2ydt2+t2y=0⇒(b+1)2d2ydt2+y=0⇒y=Acos⁡(tb+1)+Bsin⁡(tb+1)(b+1)^2 t^2 \dfrac{\text{d}^2y}{\text{d}t^2} + t^2y = 0 \Rightarrow (b+1)^2 \dfrac{\text{d}^2y}{\text{d}t^2} + y = 0 \Rightarrow y = A \cos \left( \dfrac{t}{b+1} \right) + B \sin \left( \dfrac{t}{b + 1} \right)

=Acos⁡(xb+1b+1)+Bsin⁡(xb+1b+1)= A \cos \left( \dfrac{x^{b+1}}{b + 1} \right) + B \sin \left( \dfrac{x^{b+1}}{b + 1} \right).

If A = 1, then as x -> 0 then y -> and dy/dx -> 0, but as B can be anything, this is not a unique solution.
II/4:

f(x)=(x−a)(x−b)(x−c)(x−d)⇒ln⁡f(x)=ln⁡(x−a)+ln⁡(x−b)−ln⁡(x−c)−ln⁡(x−d)\text{f}(x) = \dfrac{(x-a)(x-b)}{(x-c)(x-d)} \Rightarrow \ln \text{f}(x) = \ln(x - a) + \ln(x - b) - \ln (x-c) - \ln(x - d)
⇒f’(x)f(x)=1x−a+1x−b−1x−c−1x−d\Rightarrow \dfrac{\text{f'}(x)}{\text{f}(x)} = \dfrac{1}{x - a} + \dfrac{1}{x - b} - \dfrac{1}{x - c} - \dfrac{1}{x - d}
⇒xf’(x)f(x)=(x−ax)−1+(x−bx)−1−(x−cx)−1−(x−dx)−1\Rightarrow \dfrac{x \text{f'}(x)}{\text{f}(x)} = \left( \dfrac{x-a}{x} \right)^{-1} + \left( \dfrac{x - b}{x} \right)^{-1} - \left( \dfrac{x - c}{x} \right)^{-1} - \left( \dfrac{x - d}{x} \right)^{-1}
=(1−ax)−1+(1−bx)−1−(1−cx)−1−(1−dx)−1= \left( 1 - \dfrac{a}{x} \right)^{-1} + \left( 1 - \dfrac{b}{x} \right)^{-1} - \left( 1 - \dfrac{c}{x} \right)^{-1} - \left( 1 - \dfrac{d}{x} \right)^{-1}

If |x| is much greater than |a|, |b|, |c| and |d| we can use the first two terms of the binomial expansion with sufficient* accuracy.

xf’(x)f(x)≈1+ax+1+bx−1−cx−1−dx\dfrac{x \text{f'}(x)}{\text{f}(x)} \approx 1 + \dfrac{a}{x} + 1 + \dfrac{b}{x} - 1 - \dfrac{c}{x} - 1 - \dfrac{d}{x}

⇒x2f’(x)f(x)≈a+b−c−d\Rightarrow \dfrac{x^2 \text{f'}(x)}{\text{f}(x)} \approx a + b - c - d.

As |x| gets very large, f(x) tends to 1 (can be seen by writing f(x) as (1−a/x)(1−b/x)(1−c/x)(1−d/x)\dfrac{(1 - a/x)(1 - b/x)}{(1 - c/x)(1 - d/x)}, and x^2 is always positive, so x^2/f(x) is positive. So f'(x) (the gradient of f(x)) has the same sign as a + b - c - d.

Let z=(x−a)(x−b)(x−c)(x−d)⇒z(x−c)(x−d)=(x−a)(x−b)z = \dfrac{(x -a)(x -b)}{(x-c)(x-d)} \Rightarrow z(x-c)(x-d) = (x-a)(x-b). Now, the value of z much be such that there is a root of this equation (i.e. a value of x). As it's a quadratic in x, the discriminant of this quadratic, the expression shown, will be greater than or equal to zero.
1989 Paper 1 numbers 10,12-15
1989PAPER1.no.10.pdf30.9KB
1989PAPER1.no.12.pdf38.8KB
1989PAPER1.no.13.pdf45.8KB
1989PAPER1.no.14.pdf32.0KB
1989PAPER1.no.15.pdf24.4KB
1989 Paper 1 number 16
1989PAPER1.no.16.pdf22.5KB
1989 Paper 2 nos. 4,7,8,9,10
1989PAPER2.no.4.pdf41.8KB
1989PAPER2.no.7.pdf37.4KB
1989PAPER2.no.8.pdf35.4KB
1989PAPER2.no.9.pdf35.1KB
1989PAPER2.no.10.pdf39.1KB
1989 Paper 2 nos. 12 & 13
1989PAPER2.no.12.pdf39.4KB
1989PAPER2.no.13.pdf55.0KB
1989 Paper 2 nos. 15 & 16
1989PAPER2.no.15.pdf48.9KB
1989PAPER2.no.16.pdf33.6KB
1989 Paper 3 nos. 3-7
1989PAPER III no.3.pdf39.3KB
1989PAPER III no.4.pdf46.9KB
1989PAPER III no.5.pdf32.4KB
1989PAPER III no.6.pdf38.7KB
1989PAPER III no.7.pdf41.4KB
1989 Paper 3 nos. 13-16
Now can ANYONE do numbers 11 and 12?
1989PAPER III no.13.pdf32.6KB
1989PAPER III no.14.pdf40.5KB
1989PAPER III no.15.pdf40.5KB
1989PAPER III no.16.pdf32.8KB
1989 Paper III
Sorry, there is an error in my solutoion for number 13 so here is an amended version.
1989PAPER III fmb.13.pdf32.6KB
brianeverit
1989 Paper 3 nos. 13-16
Now can ANYONE do numbers 11 and 12?


I have quite a different answer to the end of 15.

I get E(Z) = 0 as z(c^2 - z^2)^n is an odd function.

For Var(Z):

Note that ∫−cc(c2−z2)n dz=(n!)2(2c)2n+1(2n+1)!\displaystyle \int_{-c}^c (c^2 - z^2)^n \, \text{d}z = \frac{(n!)^2 (2c)^{2n+1}}{(2n+1)!}. Now, let z=csin⁡θz = c \sin \theta and we get ∫−π/2π/2c2n+1cos⁡2n+1θ dθ=(n!)2(2c)2n+1(2n+1)!\displaystyle \int_{-\pi/2}^{\pi/2} c^{2n+1} \cos^{2n+1} \theta \, \text{d} \theta = \frac{(n!)^2 (2c)^{2n+1}}{(2n+1)!}

For Var(Z), we want to evaluate (2n+1)!(n!)2(2c)2n+1∫z2(c2−z2)n dz\displaystyle \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int z^2(c^2 - z^2)^n \, \text{d}z. Letting z=csin⁡θz = c \sin \theta as before transforms this to

c2(2n+1)!(n!)2(2c)2n+1∫−π/2π/2c2n+1cos⁡2n+1θ dθ−(2n+1)!(n!)2(2c)2n+1∫−π/2π/2c2n+3cos⁡2n+3θ dθ\displaystyle c^2 \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-\pi/2}^{\pi/2} c^{2n+1} \cos^{2n+1} \theta \, \text{d} \theta - \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-\pi/2}^{\pi/2} c^{2n+3} \cos^{2n+3} \theta \, \text{d}\theta. Using what we know for the first integral, and replacing n by n + 1 for the second, we get

c2−(2n+1)!(n!)2(2c)2n+1⋅((n+1)!)2(2c)2n+3(2n+3)!=c22n+3c^2 - \dfrac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \cdot \dfrac{((n+1)!)^2 (2c)^{2n+3}}{(2n+3)!} = \dfrac{c^2}{2n+3}.
Quite a nice answer, makes me wonder if there's an easier way.
Having looked at number Paper III 15 again I agree with your solution.
Dystopia
STEP 1, Q3

x=xa,  y=(1−x)b\textbf{x} = x\textbf{a}, \; \textbf{y} = (1-x)\textbf{b}

c=13(2a+b),d=13(a+2b)\textbf{c} = \frac{1}{3}(2\textbf{a} + \textbf{b}), \textbf{d} = \frac{1}{3}(\textbf{a} + 2\textbf{b})

(By the ratio theorem.)

CY→=(23−x)b−23a\overrightarrow{CY} = (\frac{2}{3} - x)\textbf{b} - \frac{2}{3}\textbf{a}
DX→=(x−13)a−23b\overrightarrow{DX} = (x - \frac{1}{3})\textbf{a} - \frac{2}{3}\textbf{b}

These vectors are perpendicular, so the scalar product is zero.

(29+x−x2)a.b−29=0(\frac{2}{9} + x - x^{2})\textbf{a}.\textbf{b} - \frac{2}{9} = 0

(Note that the scalar product is commutative and distributive.)

a.b=−29x2−9x−2\textbf{a}.\textbf{b} = \frac{-2}{9x^{2} - 9x - 2}

Let f(x)=−29x2−9x−2,  0≤x≤1f(x) = \frac{-2}{9x^{2} - 9x - 2}, \; 0 \leq x \leq 1

f′(x)=2(18x−9)(9x2−9x−2)2f'(x) = \frac{2(18x - 9)}{(9x^{2}-9x-2)^{2}}

Minimum and maximum values occur at the endpoint of a range or at a turning point. There is a turning point when x = 1/2.

f(0)=1,  f(12)=817,  f(1)=1f(0) = 1, \; f(\frac{1}{2}) = \frac{8}{17}, \; f(1) = 1

So 817≤a.b≤1\frac{8}{17} \leq \textbf{a}.\textbf{b} \leq 1

Note that a.b=∣a∣∣b∣cos⁡θ=cos⁡θ\textbf{a}.\textbf{b} = |a||b|\cos\theta = \cos\theta

Where θ=∠AOB,  0<θ<π\theta = \angle AOB, \; 0 < \theta < \pi

The maximum value is θ\theta occurs when cos⁡θ=817\cos\theta = \frac{8}{17}

Suppose that λCY→=CE→,  μDX→=DE→\lambda \overrightarrow{CY} = \overrightarrow{CE},\; \mu \overrightarrow{DX} = \overrightarrow{DE}

Then λ(16b−23a)=13(b−a)+μ(16a−23b)\lambda (\frac{1}{6} \textbf{b} - \frac{2}{3} \textbf{a}) = \frac{1}{3} ( \textbf{b} - \textbf{a} ) + \mu ( \frac{1}{6} \textbf{a} - \frac{2}{3} \textbf{b} )

So 16λ=13−23μ\frac{1}{6} \lambda = \frac{1}{3} - \frac{2}{3}\mu

−23λ=−13+16μ-\frac{2}{3}\lambda = -\frac{1}{3} + \frac{1}{6}\mu

λ=μ=25\lambda = \mu = \frac{2}{5}

e=c+25CY→=25a+25b\textbf{e} = \textbf{c} + \frac{2}{5}\overrightarrow{CY} = \frac{2}{5}\textbf{a} + \frac{2}{5}\textbf{b}

By the cosine rule, AB2=1817AB^{2} = \frac{18}{17}
Let F be the midpoint of AB. By Pythagoras, OF2=1−934=2534,  OF=534OF^{2} = 1 - \frac{9}{34} = \frac{25}{34}, \; OF = \frac{5}{\sqrt{34}}

OE=35×534=334OE = \frac{3}{5} \times \frac{5}{\sqrt{34}} = \frac{3}{\sqrt{34}}


OF=1/2(a+b) and OE=2/5 (a+b) so surely OE=4/5(a+b)=4/5OF not 3/5?
I made a sign error at one point (accidently multiplied by A sinh t instead of -A sinh t) so there's a bit of a mess on the last little bit (I'd have wrote it out again in a real exam) but it should be understandable - STEP II 1989 Q9 is attatched (It'd probably take a very long time to latex)
step21989q9part1.jpg276.2KB
step21989q9part2.jpg207.8KB
step21989q9part3.jpg179.9KB
Reply 74
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r2enigma
brianeverit
1989 Paper 3 nos. 13-16
Now can ANYONE do numbers 11 and 12?


I did 12 and did the first part of 11 but have no idea about the last part
Reply 75
Avatar for r2enigma
r2enigma
brianeverit
1989 Paper III
Sorry, there is an error in my solutoion for number 13 so here is an amended version.


holy **** how did you think of that?:eek3:
Haha 1989 ... I bet the STEP papers were monstrous back then! :eek:
Original post by squeezebox
...

'SimonM'
I wasn't sure if the OP was still active.

STEP I - Q1

solution

(edited 13 years ago)
Original post by brianeverit
1989 Paper 1 numbers 10,12-15


Hi, I´ve been trying to do Q10 and I just saw your solution which is different to my method.

Couldn´t you just calculate the loss in kinetic energy and then equate that to the work done by the force? i.e. :
ΔK.E=∫0dMω(v2+V2)vdx\Delta K.E = \int^d_0 \frac{M\omega (v^2 +V^2)}{v}dx
where x is the distance from the start of the field and v=x˙v=\dot{x}

I was having a bit of trouble with the integration so I was wondering if anyone could give me a hand:colondollar:

It might just all be wrong but any help would be appreciated.
Original post by ben-smith
Hi, I´ve been trying to do Q10 and I just saw your solution which is different to my method.

Couldn´t you just calculate the loss in kinetic energy and then equate that to the work done by the force? i.e. :
ΔK.E=∫0dMω(v2+V2)vdx\Delta K.E = \int^d_0 \frac{M\omega (v^2 +V^2)}{v}dx
where x is the distance from the start of the field and v=x˙v=\dot{x}

I was having a bit of trouble with the integration so I was wondering if anyone could give me a hand:colondollar:Well, you can't really integrate that without ending up doing what Brian did.

Let's write v(x) for v, to emphasise that v is a function of x.

Then you have

12v(0)2−12v(d)2=∫0dω(v(x)2+V2)v(x)dx\frac{1}{2}v(0)^2 - \frac{1}{2}v(d)^2 = \int_0^d \frac{\omega (v(x)^2 +V^2)}{v(x)}dx

So, here's the problem - at this point you have absolutely no idea what v(x) is, which is going to make finding the integral pretty tricky.

So, let's relabel some of the variables: I'm going to write 'x' instead of d, and 't' instead of 'x'.

12v(0)2−12v(x)2=∫0dω(v(t)2+V2)v(t)dt\frac{1}{2}v(0)^2 - \frac{1}{2}v(x)^2 = \int_0^d \frac{\omega (v(t)^2 +V^2)}{v(t)}dt

Now diff both sides w.r.t. x:

v(x)dvdx=ω(v(x)2+V2)v(x)v(x)\frac{dv}{dx} = \frac{\omega (v(x)^2 +V^2)}{v(x)}

Which is Brian's first line. So from here you can solve for v as a function of x, which will give you the desired result.

Obviously it's quicker to just use the v dv/dx form of acceleration to start with.
(edited 13 years ago)

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