Question 3 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 3 - 2011 - Paper 1

Using L'Hôpital's Rule, apply it directly to find the limit. We can differentiate the numerator and the denominator: [ \lim_{x \to 0} \frac{x}{\text{sin} \left( \frac{\pi}{2} x \right)} = \lim_{x \to 0} \frac{1}{\frac{\pi}{2} \text{cos}\left( \frac{\pi}{2} x \right)} = \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi}. ]

For the graph, note that this function is an even function, which implies that it will be symmetric about the y-axis. The graph will shoot up to infinity as $x$ approaches multiples of 2 in the interval $(0, 4)$ since $\text{sin}\left( \frac{\pi}{2} x \right) = 0$ at those points. Mark the points where the function becomes undefined as vertical asymptotes.

To find the volume of the solid, consider the area bounded by the curves $y = \text{cos} \ x$ and $y = -\text{cos} \ x$. The height of the cross-section at a given y is obtained from the intersection points of these curves. Then integrate the area of the triangular cross-sections from $0$ to $\frac{\pi}{2}$ to get the total volume: [ V = \int_{0}^{\frac{\pi}{2}} \text{Area} , dy = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \cdot b(h) , dy] where $b(h)$ is the length of the base based on the y-coordinates of the bounding curves.

For the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$, the eccentricity $\epsilon$ can be calculated using the formula: [\epsilon = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}. ]

The foci of the hyperbola can be found using the formula for the distance from the center $(0,0)$ to the foci: [c = a \epsilon = 4 \cdot \frac{5}{4} = 5.] The coordinates of the foci are thus $(\pm 5, 0)$.

The equations of the asymptotes for the hyperbola are given by: $y = \pm \frac{b}{a} x$, thus yielding: [y = \pm \frac{3}{4} x.]

To sketch the hyperbola, indicate the transverse axis along the x-axis. Mark the center at the origin, foci at $(\pm 5, 0)$, and draw the asymptotes to guide the curve's path. Plot points to illustrate the characteristics of the hyperbola opening left and right.

As the eccentricity $\epsilon$ approaches infinity, the hyperbola becomes more elongated along its transverse axis. The arms of the hyperbola move further apart, suggesting a flattening effect while maintaining their relationship to the asymptotes.