Questions 1 (ii) — (vii) from here.

With all of these questions the first thing we want to do is plug in the value; i.e. if we are evaluating we should first try . If we find that is undefined (e.g. division by zero), we will usually have to ‘factor out the bad stuff’. The reason this works is because of the following theorem:

### Theorem

*Suppose that except perhaps at . Then*

.

Take for example

.

Now if , then — which is undefined. However we can cancel an above and below… well what we really do is as follows:

.

However this is only true in the case where ; i.e. we have that for all — and use the above theorem to say that

.

## (ii)

.

First off plugging in gives — undefined. What we want to do is factorise above and below and hopefully there will be a term which we can cancel above and below (for ), and use the above theorem. First off the denominator is simply a difference of two squares:

.

I can think of three ways to factorise the numerator.

#### Formula

The numerator has the form which I know has factors via a formula. I never learnt this formula ‘off’ but instead have a quick derivation (using the Binomial Theorem I suppose):

.

Hence I have

.

#### Factor Theorem

The factor theorem states that:

*A complex number is a root of a polynomial if and only if is a factor of .*

i.e. if , then for some polynomial (and vice versa).

When I plugged in into I got zero — hence is a factor of . This means I can perform polynomial long division to get

;

that is

as before.

#### Polynomial Division

You can just straight away do polynomial long division to get

,

for .

Taking one of the first two factorisations leads to:

,

for .

Using this or the will allow us then to plug in to get

.

## (iii)

Plugging in yields division by zero — undefined.

,

for . Hence:

.

## (iv)

Plugging in gives . Done and dusted!

## (v)

Plugging in gives — done.

## (vi)

Plugging in gives division by zero. These type need a manipulation called ‘multiplying by the conjugate’ (which is just multiplying by — which changes nothing of course):

.

The numerator here is a difference of two squares: so we have:

.

Hopefully the top will factorise with an term that we could cancel out:

.

Thence we have

,

for . Therefore

.

## (vii)

This one is not so straightforward and beyond exam standard. I can’t see a simple solution I will simply present *a* solution.

At , the expression in undefined. Let and consider

,

.

By inspection is a root of so hence is a factor. Polynomial long division yields:

.

Therefore, after a cancellation of the terms, and division by a :

,

Now plugging in gives .

## Question 2 Summer 2010.

### (a)

We want to find positive integers (whole numbers) and such that:

.

There are two main methods of doing this.

#### Closed Interval Method.

*Suppose that a function is continuous. Then attains its absolute maximum and minimum on and further can only be attained at *criticial points* — i.e. the endpoints , points where the derivative is zero (), or points where the function is not differentiable ( does not exist).*

In this case we want to look at . We know that the numerator and denominator are continuous as polynomials and their quotient is continuous for — which is certainly the case for . Hence we can use the closed interval method to find the absolute maximum/ minimum of .

First we find the critical points. As the endpoints, we have are critical. Next we find points where . By the quotient rule:

.

Now a fraction is equal to zero if and only if the numerator is zero and as (!), there are no points where .

Finally, by the quotient rule, is differentiable for all which is satisfied for all so the function is differentiable on .

Hence the only critical points are and so we test the function at these points:

.

.

Now these are the absolute maximum and minimum of on so we can safely conclude that

on as required.

#### Using Inequalities

A quick manipulation using the properties of the absolute value function:

.

This problem requires the following fact:* *

*For , if and then we may divide the smaller by the larger and the larger by the smaller to preserve the inequality, i.e.*

*.*

So to find an upper bound (), I want to find an upper bound for and a lower bound for .

By the triangle inequality, and using properties of the absolute value function:

.

Clearly this is maximised for large so we have (i.e. is the largest we are concerned with in ):

.

By the reverse triangle inequality:

,

for similar reasons.

Putting these two bounds together, using the above fact:

.

Now we do the opposite. First a lower bound for :

.

A upper bound for :

.

So we have:

.

So, by this method, all and all:

.

(Note that the closed interval method gives exactly sharp bounds (well until we round to the smaller/ larger integer)).

### (b)

Now let and . We want to show that for all , there exists a such that if , then .

How we start this is by looking at the term.

.

We would hope at this stage to see a term and a term. Sure enough if we factorise the numerator we see so we write:

,

using the properties of the absolute value function:

.

This is exactly where we want to be now because we can use the upper bound above ( — be it or depending on which method we used) to say that

.

Hence now if we make small so that

,

,

i.e. , then we are done.

Now we write:

*Let . Choose ( *could be or or whatever upper bound you found in part (a)*). Now let . Then*

.

That is,

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