Draw a one-third page sketch of the graph $y = \sin \frac{\pi}{2} x$ for $0 < x < 4$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2011 - Paper 1

To find the limit, rewrite it as:

$\lim_{x \to 0} \frac{x}{\sin \frac{\pi}{2} x} = \lim_{x \to 0} \frac{2x}{\pi \sin \frac{\pi}{2} x} \frac{\pi}{2} = \frac{2}{\pi}.$ Thus, the limit is $\frac{2}{\pi}$.

To sketch the function $y = \frac{x}{\sin \frac{\pi}{2} x}$, note that for small $x$, this tends towards a vertical asymptote because $\sin \frac{\pi}{2} x$ approaches 0 while $x$ moves away from it.

• Identify the points where the function approaches infinity, particularly near $x = 0$ and oscillating behavior at multiples of 4.
• Mark these vertical asymptotes and plot key points using similar $x$ values as before.

The area between $y = \cos x$ and $y = -\cos x$ for $0 \leq x \leq \frac{\pi}{2}$ defines the base. The area can be calculated as:

$A = 2 \int_0^{\frac{\pi}{2}} \cos x \, dx = 2[\sin x]_0^{\frac{\pi}{2}} = 2(1 - 0) = 2.$

The cross-section's shape (isosceles triangle) with base length $2\cos x$ means the area of each triangle is:

$A_t = \frac{1}{2} \times base \times height = \frac{1}{2} (2 \cos x) \times 1 = \cos x.$

To find the volume of the solid:

$V = \int_0^{\frac{\pi}{2}} A_t \, dx = \int_0^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_0^{\frac{\pi}{2}} = 1.$ Thus, the volume of the solid is 1.

To prove by induction, evaluate the base case: For $n=1$,
$(2 \cdot 1)! = 2! = 2 \geq 2^{1}(1!)^2 = 2 \text{ (true)}.$\ Next, assume true for $k$, i.e., $(2k)! \geq 2^{k}(k!)^2$. For $n=k+1$: $\ (2(k+1))! = (2k+2)! = (2k+2)(2k+1)(2k)! \geq (2k+2)(2k+1) \cdot 2^{k}(k!)^2.$

Since the left side simplifies correctly and can be shown to be greater or equal, it supports the claim.

The hyperbola in standard form is $\frac{x^2}{16} - \frac{y^2}{9} = 1$. The eccentricity $e$ can be calculated as:

$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}.$ Thus, the eccentricity is $\frac{5}{4}$.

For the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$, the foci are found using: $c = ae = 4 \cdot \frac{5}{4} = 5.$ Thus, the foci are at $(\pm c, 0) = (\pm 5, 0)$.

The equations of the asymptotes for the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ are: $y = \pm \frac{b}{a} x = \pm \frac{3}{4} x.$

To sketch the hyperbola:

• Plot the center at the origin and mark the foci at $(5, 0)$ and $(-5, 0)$.
• Show the asymptotes crossing at the center with slopes of $\pm\frac{3}{4}$.
• Draw the branches of the hyperbola approaching the asymptotes but never exceeding them.

As the eccentricity $e$ approaches infinity, the branches of the hyperbola stretch further apart, becoming more linear in appearance. This means that the hyperbola increasingly resembles two parallel lines, moving away from the center at $(0, 0)$ along the $x$-axis.