For every integer $m \\geq 0$ let \ \\ I_m = \int_0^1 m x^{m-1} (x^2 - 1)^5 \, dx - HSC - SSCE Mathematics Extension 2 - Question 8 - 2011 - Paper 1

To prove the relationship between $I_m$ and $I_{m-2}$, we start with the definition:

$I_m = \int_0^1 m x^{m-1} (x^2 - 1)^5 \, dx.$\n Using integration by parts or properties of definite integrals, we can manipulate this integral.
We convert the term $(x^2 - 1)^5$ using the binomial theorem. Then we'll find that:

1. Substitute for $x$, leading to two parts: one for $x^m$ and one for $x^{m-2}$.
2. Evaluate separately for each component.
3. Collect and simplify to show $I_m = \frac{m - 1}{m + 1} I_{m-2}$.

(b)

(i) What is the probability that each ball is selected exactly once?

The total number of selections is 7, and we want exactly one of each ball to be selected once. There are $7!$ ways to arrange the selection and the total selections is $7^7$. Therefore, the probability is:

$P = \frac{7!}{7^7}.$

(ii) What is the probability that at least one ball is not selected?

Using the complement principle: The probability that all balls are selected at least once can be computed using the inclusion-exclusion principle, leading finally to:

$P = 1 - P(\text{all selected}).$

(iii) What is the probability that exactly one of the balls is not selected?

To find this probability, consider choosing one ball not to be selected and then arranging the remaining 6 balls over 7 selections:

$P = \binom{7}{1} \frac{6^7}{7^7} = 7 \left( \frac{6}{7} \right)^7.$

(c)

(i) Show that |β|n≤M( |β|n−1 + |β|n−2 + ... + |β| + 1).

Using the triangle inequality, we express:

$P(z) = (z - \beta) (z^{n-1} + a_{n-2} z^{n-2} + \ldots + a_0).$

Combining results yields the required inequality.

(ii) Hence, show that for any root β of P(z), |β| < 1 + M.

By applying the previously shown inequality and isolating |β|, we find:

$|\beta| < 1 + M.$

(d)

Using part (c), show that S(x) = ∑k=0nc_k(x+1/x)k has no real solutions.

This requires proving by contradiction. Assuming it has a real solution for $S(x)$ would imply that the maximum value relates, leading to contradictions with the properties of polynomials and defined bounds on c_k. Thus, showing $S(x) = 0$ has no real solutions.