STEP Maths I, II, III 1988 solutions

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

STEP III - Question 2

$au_{n+2} + bu_{n+1} + cu_{n} = 0$

If $u_{n} = A\alpha^{n} + B\beta^{n}$, where $\alpha$ and $\beta$ satisfy $ax^{2} + bx +c = 0 (*)$, then $au_{n+2} + bu_{n+1} + cu_{n} = aA\alpha^{n+2} + aB\beta^{n+2} + bA\alpha^{n+1} + bB\beta^{n+1} + cA\alpha^{n} + cB\beta^{n} = A\alpha( a\alpha^{2} + b\alpha + c) + B\beta^{n}(a\beta^{2} + b\beta + c) = 0$. (as alpha and beta satisfy (*).)

So the difference equation is satisfied by $u_{n} = A\alpha^{n} + B\beta^{n}$, for all A,B.

We also need:

$u_{0} = A+B$ and $u_{1} = A\alpha + B\beta$, and so:

$B = \frac{u_{1} - \alpha u_{0}}{ \beta - \alpha}$ and $A = \frac{u_{1} - \beta u_{0}}{\alpha - \beta}$.

~~~~~~~~~

Define $u'_{n}$ to be a possible solution, so that $u'_{n} \neq u_{n}$. And define also: $w_{n} = u_{n} - u'_{n}$.

Suppose that $w_{k-1} = w_{k}=0$. We shall show that $w_{k+1} =0$

We have: and .

Now: $au_{k+1} + bu_{k} + cu_{k} = 0$ and $au'_{k+1} + bu'_{k} + cu'_{k} = 0.$

$\Rightarrow aw_{k+1} + bw_{k} + cw_{k} = 0$

$\Rightarrow w_{k+1} = 0$, as $w_{k} = w_{k-1} = 0$.

Also, for the determined values of A and B, the fist two terms are $u_{0} and u_{1}$, and so there is no alternative $u'_{0}$ and $u'_{1}$. Hence by induction, $u_{n}$ are the only solutions.

~~~~~~~~

Put $v_{n} = (n+1)t_{n}$.

Then you get: $(n+1)(n+2)(n+3)[ 8t_{n+2} - 2t_{n+1} - t_{n}] = 0$

$\Rightarrow 8t_{n+2} - 2t_{n+1} - t_{n} = 0$.

Let the roots of this difference equation be p and q. So from the first part, p and q must satisfy: $8x^{2} -2x -1 = 0$, whose roots are: $\frac{1}{2} and \frac{-1}{4}$.

So let p be 0.5, and so q = -0.25. Then $t_{n} = C(\frac{1}{2})^{n} + D(\frac{-1}{4})^{n}$. Where C = $\frac{t_{1} - q t_{0}}{ p -q}$ and D = $\frac{t_{1} - p t_{0}}{ q-p}$.

So $t_{n} = \frac{4}{3}(\frac{1}{2})^{n} -\frac{4}{3}(\frac{-1}{4})^{n}$.

Hence: $v_{n} = (n+1)[\frac{4}{3}(\frac{1}{2})^{n} -\frac{4}{3}(\frac{-1}{4})^{n}]$.

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.

STEP III Q12
Let $\mathbf{u_1}/u_1, \mathbf{v_1}/v_1, \mathbf{v_2}/v_2$ be the velocities/speeds of ball 1 and 2 with u being before the collision and v after.
Momentum is conserved in a closed system so:
$\mathbf{u_1}=\mathbf{v_1}+\mathbf{v_2}$
The coefficient of restitution is < 1 so kinetic energy is not conserved amongst the two balls, i.e:
$\mathbf{u_1}.\mathbf{u_1}> \mathbf{v_1}.\mathbf{v_1}+ \mathbf{v_2}.\mathbf{v_2}$
Combining the two:
$(\mathbf{v_1}+\mathbf{v_2})^2 > \mathbf{v_1}^2+\mathbf{v_2}^2 \Rightarrow 2\mathbf{v_1}.\mathbf{v_2}>0[br]\Rightarrow 2v_2v_1cos\phi>0 \Rightarrow \phi < \frac{\pi}{2}$
Where $\phi$ is the angle between the two velocities as desired.
For the next part, let $\theta$ be the angle between $\mathbf{u_1}\\and\\\mathbf{v_2}$.Separating momentum into two components and and considering restitution in the direction of $\mathbf{v_2}$ we obtain 3 equations, sufficient to eliminate the three speeds and obtain an equation in terms of our two angles, thus confirming the intuitive idea that the initial velocity of the first ball is irrelevant:
$u_1cos\theta=v_2+v_1sin\phi[br]u_1sin\theta=v_1sin\phi[br]eu_1cos\theta=v_2-v_1cos\theta$
By adding the first and last and substituting, we obtain:


The angle of deflection is $\phi-\theta$ which we shall call k. Consider
Notice that the maximum of Tank will occur for the same value of theta as the maximum of k so, differentiating both sides w.r.t. theta we get:

$D[tan(k)]=\frac{(1+(2/(1-e))Tan^2\theta)((1+e)sec^2\theta)-((1+e)tan\theta)(2/(1-e)2tan\theta sec^\theta)}{(1+(2/(1-e))Tan^2\theta)^2}$
Which, if we equate it to zero, gives us:
$(1+\frac{2}{1-e}Tan^2\theta)(1+e)sec^2\theta=(1+e)tan\theta \frac{2}{1-e}2tan\theta sec^2\theta[br]\Rightarrow 1+\frac{2}{1-e}Tan^2\theta=\frac{4}{1-e}Tan^2\theta[br]\Rightarrow tan^2\theta=\frac{1-e}{2}$
So:
$Tan(max[k])=max[Tan(k)]=\sqrt{\frac{1-e}{2}}\frac{(1+e)/(1-e)}{2}=\frac{sin(k_{max})}{\sqrt{1-sin^2(k_{max})}}[br][br]\Rightarrow \frac{(1-sin^2(k_{max}))}{2}=\frac{4sin^2(k_{max})(1-e)}{(1+e)^2}[br]\Rightarrow (\frac{4(1-e)}{(1+e)^2}+\frac{1}{2})sin^2(k_{max})=\frac{1}{2}[br]\Rightarrow sin^2(k_{max})=(\frac{2(1+e)^2}{8(1-e)+(1+e)^2})\frac{1}{2}= \frac{(1+e)^2}{(3-e)^2} [br]\Rightarrow k_{max}=\sin^{-1}(\frac{1+e}{3-e})$
as required.

Comment: I found this Q particularly difficult and it has taken me many attempts to get it right. Comments are welcome.

I am afraid I disagree with this result...see attached file...

I think the derivative of what braineverit get is right? f'(b)=pi-2/a*√a²+b² rather than having the 2/a*a√2 term for we should differentiating w.r.t. b
but for the final result, pi²/4-1>1 so we should choose a???

I think you are absolutely right...cannot recall what on earth I was thinking about!!

🤨😀

only trouble is that the given answer makes b greater then a!!!!

Is that an official answer of just someone did it?
(P.S. I think this question is a good idea.Begin daydreamingope this year STEP questions will have this kind of questions haha.stop daydreaming))

Ihave corrected my version, pointed out the contradion of the answer and offered an alternative solution....see if you like it..!

Why aren't group theory questions in STEP any more :'(

1988 Paper 2 number 8

Here are 10, 11,12,13,14, and 16 on Paper I