(a) A model for the population, $P$, of elephants in Serengeti National Park is $$P = \frac{21000}{7 + 3e^{-\frac{t}{3}}}$$ where $t$ is the time in years from today - HSC - SSCE Mathematics Extension 2 - Question 5 - 2008 - Paper 1

To find the population today, substitute $t = 0$ into the population model:

$P(0) = \frac{21000}{7 + 3e^{0}} = \frac{21000}{10} = 2100.$ Thus, the population today is 2100.

As $t \to \infty$, the exponential term approaches zero. Therefore, the model predicts the eventual population:

$P = \frac{21000}{7 + 3 \cdot 0} = \frac{21000}{7} = 3000.$ Hence, the eventual population will be 3000.

To find the growth rate, evaluate the derivative at $t = 0$. Taking the derivative we established:

$\frac{dP}{dt} = \frac{1}{3000} \left( 1 - \frac{P}{3000} \right) P.$ Substituting $P(0) = 2100$ gives:

$\frac{dP}{dt} = \frac{1}{3000} \left( 1 - \frac{2100}{3000} \right) \cdot 2100 = \frac{1}{3000} \cdot \frac{900}{3000} \cdot 2100 = \frac{63000}{9000000} = 0.007.$ Thus, the annual percentage growth rate today is approximately 0.7%.

To show that $p(x)$ has a double zero at $x = 1$, calculate $p(1)$:

$p(1) = 1^{n+1} - (n + 1) \cdot 1 + n = 1 - (n + 1) + n = 0.$

Next, find the derivative $p'(x)$ and evaluate at $x = 1$:

$p'(x) = (n + 1)x^n - (n + 1).$
Evaluating at $x = 1$ gives: $p'(1) = (n + 1)(1)^n - (n + 1) = 0.$
Hence, $p(x)$ has a double zero at $x = 1$.

For $x \geq 0$, consider $p(0)$:

$p(0) = 0^{n+1} - (n + 1) \cdot 0 + n = n \geq 0.$
For $x=1$, we have already shown $p(1) = 0$. The polynomial $p(x)$ is continuous and follows a general trend of being non-negative for $x \geq 0$ due to its structure.

Now consider $p(3)$:

$p(3) = x^4 - 4x + 3.$
To factor, find the roots or use synthetic division. The factorization is: $p(3) = (x - 1)^2 (x - 3).$

To find $x_1$ and $x_2$, solve the equation of the circle:

$\left( x - a^{2} \right)^{2} + h^{2} = b^{2}$ or rearranging gives:

$x - a^{2} = \sqrt{b^{2} - h^{2}}.$
Thus, $x_1 = a^{2} - \sqrt{b^{2} - h^{2}}, \, x_{2}= a^{2} + \sqrt{b^{2} - h^{2}}.$

The area $A$ of the annulus is given by the difference of the inner and outer circle areas:

$A = \pi \left( x_2^2 - x_1^2 \right) = \pi \left( (a^{2} + \sqrt{b^{2} - h^{2}})^2 - (a^{2} - \sqrt{b^{2} - h^{2}})^2 \right).$ Expanding gives:

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The volume $V$ of the torus is given by the area of the cross-section multiplied by the circumference:

$V = A \cdot 2\pi a = 2\pi a \cdot \pi \left( 4a^{2}\sqrt{b^{2} - h^{2}} \right) = 8\pi^{2} a^{3}\sqrt{b^{2} - h^{2}}.$