Question 8 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 8 - 2001 - Paper 1

Starting from the previously established result:

$2ab \leq c^2 + a^2 + b^2$

We can manipulate this to show:

1. Apply it cyclically for $ab$, $bc$, and $ca$ to gather terms into one inequality:
2. $2(ab + bc + ca) \leq (a^2 + b^2 + c^2)$.
3. Add $a^2 + b^2 + c^2$ to both sides.
4. This results in: $3(ab+bc+ca) \leq (a + b + c)^2$

Thus, the deduction is complete.

For a triangle with sides a, b, and c, we apply the triangle inequality:

1. The lengths of the sides mean that the difference between any two sides must be less than or equal to the length of the third side.
2. Therefore, by the triangle inequality, we get: $|b - c| \leq a$, which squares to give $(b-c)^2 \leq a^2$.

Using the result of part (ii):

1. We can sum the squares of the sides: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)$
2. Substituting ($(b-c)^2 \leq a^2$) gives a bound on each term.
3. Bring together the terms dynamically: $(b+c)^2 = b^2 + c^2 + 2bc$
4. Combine and manipulate these to yield the required result of: $(a+b+c)^2 \leq 4(ab+bc+ca)$

To show that: $\int_{0}^{1} x^\alpha e^x dx < \frac{3}{\alpha + 1}$

1. The function $e^x$ is monotonically increasing.
2. Therefore, $e^x \leq e$ under limits from 0 to 1, hence: $\int_{0}^{1} x^\alpha e^x dx < e \int_{0}^{1} x^\alpha dx$
3. This evaluates: $\int_{0}^{1} x^\alpha dx = \frac{1}{\alpha + 1}$. Thus:
4. Combining this yields: $\int_{0}^{1} x^\alpha e^x dx < e \cdot \frac{1}{\alpha + 1} < \frac{3}{\alpha + 1}$
5. This is due to the assumption that $e < 3$.

Induction Basis:

1. For $n=0$, compute: $\int_{0}^{1} e^x dx = e - 1$ This is a rational number.

Induction Step: 2. Assume true for n=k, such that: $\int_{0}^{1} x^k e^x dx = a_k + b_k$ 3. For $n = k + 1$, use integration by parts: $\int_{0}^{1} x^{k+1} e^x dx = \left[ x^{k+1} e^x \right]_{0}^{1} - \int_{0}^{1} (k+1)x^k e^x dx$ 4. Substitute the hypothesis into it hence proving step by induction. 5. Therefore, integers $a_n$ and $b_n$ exist for all n.

Consider the following:

1. Rewrite $|a + br|$ using $r = \frac{p}{q}$, hence: $|a + b \cdot \frac{p}{q}| = |\frac{aq + bp}{q}| = \frac{|aq + bp|}{q}$
2. As $aq + bp$ can take multiple values depending on integer a and b,
3. Either $|a + br| = 0$ holds when $a = -br$, or
4. Otherwise we apply the triangle inequality: $|a + br| \geq \frac{|a|}{q}$

To prove $e$ is irrational, consider:

1. Assume $e = \frac{p}{q}$ for integers p and q.
2. The series expansion is: $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ...$
3. Multiply through by $q!$: $q!e = q! + \frac{q!}{1!} + \frac{q!}{2!} + ...$
4. The left-hand side remains an integer. The right side contains terms reducing to: $\sum_{n=0}^{q-1} \frac{q!}{n!}\text{ which is an integer, plus a non-integer part.}$
5. This contradiction arises thus proving e cannot be expressed as a fraction, hence irrational.