The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2007 - Paper 1

To sketch $f(|x|)$, we reflect the portion of the graph for $x extgreater 0$ across the y-axis to create the graph for $x extless 0$. This will produce a graph that is symmetric about the y-axis.

The graph of $f(x) - x$ can be obtained by taking the graph of $f(x)$ and shifting it downward by $x$. The new intercepts can be found by solving $f(x) = x$.

Given that $\alpha, \beta, \gamma$ are zeros of the polynomial $P(x) = x^3 - 5x + 3$, we can find the new polynomial by scaling the roots. A polynomial with roots $r_1, r_2, r_3$ is given by:

$Q(x) = k(x - r_1)(x - r_2)(x - r_3)$

Thus, substituting the new roots $2\alpha$, $2\beta$, and $2\gamma$:

$Q(x) = k\left(x - 2\alpha\right)\left(x - 2\beta\right)\left(x - 2\gamma\right).$

Expanding this will yield the required polynomial with integer coefficients.

The volume $V$ can be calculated using the formula for cylindrical shells:

$V = 2\pi \int_{a}^{b} y \cdot r \, dx,$

where $y$ is the height and $r$ is the radius of the shell. Here, the bounds $a = 1$ and $b = e$ should be established based on intersection with the graph of $y = \log \frac{x}{x}$.

By resolving the forces in the vertical direction, we have:

$N + mg \cos \theta - F = 0,$

and in the horizontal direction, we find:

$F_r - mr\omega^2 = 0.$

Thus, substituting these values into the equations gives the required relationship.

Setting the equation $N = mg \cos \theta - mr\omega^2 \sin \theta > 0$ leads to:

$mg \cos \theta > mr\omega^2 \sin \theta.$

From this, isolating $\omega$ yields $\omega < \sqrt{\frac{g \cos \theta}{r \sin \theta}}.$

This expression provides the conditions for which $N$ remains positive.