STEP Maths I, II, III 1988 solutions

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?

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1988 STEP III number 3.
I'm afraid I disagree with the solutions published so far. Here is my solution.

$|z-i|=2 \Rightarrow |x+(y-1)i|=2 \text{ where }z=x+iy$
$\Rightarrow x^2+(y-1)^2=4 \text{ so parametric equations are }x=2\cos\theta, y=2\sin\theta+1$

$\Rightarrow |2i|=2|w-1| \text{ i.e. }|w-1|=1 \text{ so it is a circle, centre }w=1 \text{ radius }1$
$(ii) z \text{ real } \Rightarrow \text {Im}\left( \dfrac{(u+1)i-v}{(u-1)i-v} \right)= \text{ Im}\left( \dfrac{((u+1)i-v)((u-1)i-v)}{(u-1)^2+v^2} \right)= -\dfrac{2uv}{(u-1)^2+v^2}=0$
$\text{hence, }2uv=0 \text{ i.e. It is the two axes}$
$(iii) z\text{ imaginary }\Rightarrow \text{Re} \left( \dfrac{w+1}{w-=1} \right)i=0 \text{ i.e. } \dfrac {v^2-(u^2-1)}{(u-1)^2+v^2}=0 \Rightarrow v^2-u^2+1=0 \text{ i.e. A hyperbola}$

I think that the three parts to this question are entirely separate and do not include the condition of being on the original circle...that is where we disagree. As regards the other question (4) can you see where we differ on the diffentiation...that is where the theorem comes in....

I/7.

Therefore
$f(1)(x+1/2) + f(-1)(x - 1/2) - 2f(0)x = (a+b+c)(x + 1/2) + (a-b+c)(x - 1/2) - 2cx$
$= ax + bx + cx + 1/2a + 1/2b + 1/2c + ax - bx + cx - 1/2a + 1/2b - 1/2c - 2cx = 2ax + b$

If |f(x)| <= 1, then -1 <= f(x) <= 1.
If |f'(x)| <= 4, then -4 <= f(x) <= 4.

f'(x) = f(1)(x + 1/2) + f(-1)(x - 1/2) - 2f(0)x.

The largest values of f'(x) will be when x = -1 or 1 to maximize the values of (x + 1/2) and -(x-1/2). The largest value of f(1)(x + 1/2) can be 3/2 when f(1) = 1. The largest value of f(-1)(x - 1/2) can be 1/2 when f(-1) = 1. The largest value of -2f(0)x can be 2 when f(0) = -1. 3/2 + 1/2 + 2 = 4, so |f'(x)| <= is satisfied. When x = -1, the largest value can be (-1)(-1/2) + (-1)(-3/2) - (2)(1)(-1) = 4.

The smallest values of f'(x) will be when x = -1 or 1 likewise. When x = 1, be (-1)(3/2) + (-1)(1/2) - 2(1) with f(0) = 1, f(1) = -1 and f(-1) = -1. This gives -4. When x = -1, it will be (1)(-1/2) + (1)(-3/2) - 2(1) = -4, with f(1) = 1, f(-1) = 1 and f(0) = -1, giving -4. Therefore |f'(x) <= 4 is still satisfied.

Letting f(0) = -1, this gives c = -1. Letting f(1) = 1, this gives a + b - 1 = 1 hence a + b = 2. Letting f(-1) = 1, this gives a - b = 2. Hence 2a = 4, a = 2 and b = 0.
f(x) = 2x^2 - 1.

Sorry for any errors.

(Updated as far as #68) SimonM - 15.06.2009
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(Updated as far as #68) SimonM - 15.06.2009
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Question 8, STEP III
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Here is my answer for Question 11, which curiously doesn't seem to have been done yet (?) for Paper 3 (i.e. 1988 Paper FM B Q11).

Just for novelty I've posted my answer as a youtube video, although warning it is 40+ minutes long and the tone will be a bit too didactic (i.e. boring) for most of the kind of people viewing a thread like this (It's primarily intended for students at my school).



Edit: For anyone just double checking their answer, I got (i) $\mu = \frac{\sqrt{3}}{3}$ and (ii) $\mu = \frac{3\sqrt{3}}{5}$

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/showpost.php?p=11883448&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

$x = y = z = \frac{2}{3} a$ is also a valid solution

relatively straightforward q, pointing out any mistakes will be appreciated as always

Ok first note that at the end of day n, the firm will owe everything it owed the previous day, the interest on this plus everything it has borrowed that day.

This may be expressed as $\displaystyle S_{n+1} = (1+k)S_n + c$

By evaluating the first few terms of this sequence, it becomes easy to 'guess' and then prove a total value for Sn:

$\displaystyle S_0 = 0$

$\displaystyle S_1 = c$

$\displaystyle S_2 = (1+k)c + c = c((1+k) + 1)$

$\displaystyle S_3 = (1+k)((1+k)c + c) + c = c((1+k)^2 + (1+k) + 1)$


ok my guess is that $\displaystyle S_n = c \sum^{n-1}_{r=0} (1+k)^r = c \frac{(1+k)^n - 1}{(1+k) - 1} = c \frac{(1+k)^n - 1}{k}$ (by geometric series)

this fits the questions, but we have to prove it (induction!)


Basis case: $\displaystyle S_0 = c \frac{(1+k)^0 - 1}{k} = 0$, therefore the basis case works.

Inductive step: $\displaystyle S_{n+1} = (1+k)S_n + c = (1+k)c \frac{(1+k)^n - 1}{k} + c = c((1+k)\frac{(1+k)^n - 1}{k} + 1)$

$\displaystyle = c(\frac{(1+k)^{n+1} - (k+1)}{k} + 1) = c(\frac{(1+k)^{n+1} - (k+1) + k}{k}) = c\frac{(1+k)^{n+1} - 1}{k})$

This is in the same form as the above, so it works.


for part 2, firstly we can assert that $\displaystyle S_T = c\frac{(1+k)^{T} - 1}{k})$ (this is the initial amount owed)

the firm must pay interest on what it owes then pay back what it has earned.

$\displaystyle S_{T+m+1} = (1+k)S_{T+m} - d$

Again, let's try a few values:

$\displaystyle S_{T+1} = (1+k)S_T - d$

$\displaystyle S_{T+2} = (1+k)((1+k)S_T - d) - d = (1+k)^2 S_T - d(1+k) - d$

$\displaystyle S_{T+3} = (1+k)((1+k)((1+k)S_T - d) - d) - d = (1+k)^3 S_T - d(1+k)^2 - d(1+k) - d$

hypothesis: $\displaystyle S_{T+m} = (1+k)^m S_T - d\sum^{m-1}_{r=0}(1+k)^r$

$\displaystyle = (1+k)^m S_T - d\frac{(1+k)^m - 1}{k} = (1+k)^m(S_T - \frac{d}{k}) + \frac{d}{k}$ (again by geometric)


we need to prove this, again by induction.

basis case: when m = 0, $\displaystyle S_T = (1+k)^0 (S_T - \frac{d}{k}) + \frac{d}{k} = S_T$ basis works

inductive step: $\displaystyle S_{T+m+1} = (1+k)S_{T+m} - d = (1+k)((1+k)^m(S_T - \frac{d}{k}) + \frac{d}{k}) - d$

$\displaystyle = (1+k)^{m+1}(S_T - \frac{d}{k}) + (1+k)\frac{d}{k} - d$

$\displaystyle = (1+k)^{m+1}(S_T - \frac{d}{k}) + \frac{d}{k} + d - d$

$\displaystyle = (1+k)^{m+1}(S_T - \frac{d}{k}) + \frac{d}{k}$

Hence the sum works. However the question did not ask for us to express S_T, so we must substitute in (see our above def. for S_T)


$\displaystyle S_{T+m} = (1+k)^m(c\frac{(1+k)^{T} - 1}{k} - \frac{d}{k}) + \frac{d}{k}$

$\displaystyle S_{T+m} = (1+k)^m(c\frac{(1+k)^{T} - 1 - d}{k}) + \frac{d}{k}$

$\displaystyle S_{T+m} = \frac{c}{k} (k+1)^{T+m} - \frac{c+d}{k}(k+1)^m + \frac{d}{k}$



for part 3, note that if the sequence is decreasing, the firm will pay off its debt. so the conditions are:

$\displaystyle S_{T+m+1} < S_{T+m}$

$\displaystyle \frac{c}{k} (k+1)^{T+m+1} - \frac{c+d}{k}(k+1)^{m+1} + \frac{d}{k} < \frac{c}{k} (k+1)^{T+m} - \frac{c+d}{k}(k+1)^m + \frac{d}{k}$

Letting x = k+1


$\displaystyle \frac{c}{k} x^{T+m+1} - \frac{c+d}{k}x^{m+1} + \frac{d}{k} < \frac{c}{k} x^{T+m} - \frac{c+d}{k}x^m + \frac{d}{k}$

$\displaystyle \frac{c}{k} x^{T+m+1} - \frac{c+d}{k}x^{m+1} < \frac{c}{k} x^{T+m} - \frac{c+d}{k}x^m$

taking everything on to one side


$\displaystyle x^{T+m}(\frac{c}{k}x - \frac{c}{k}) - x^m(\frac{c+d}{k}x-\frac{c+d}{k}) < 0$

$\displaystyle (x-1)(\frac{c}{k}x^{T+m} - \frac{c+d}{k}x^m) < 0$

$\displaystyle x^m(x-1)(\frac{c}{k}x^{T} - \frac{c+d}{k}) < 0$

k+1 > 0, so both x^m and (x-1) are greater than zero. Therefore, for the inequality to be satisfied,

$\displaystyle \frac{c}{k}x^{T} - \frac{c+d}{k} < 0$

$\displaystyle cx^{T} - (c+d) < 0$

$\displaystyle x^{T} - 1 < \frac{d}{c}$

$\displaystyle (1+k)^{T} - 1 < \frac{d}{c}$ as required.

I don't think it is... $\frac{2}{3} a + \frac{2}{3} a + \frac{2}{3} a \neq a$

I get the same as you....see file...