5. (a) Find $\int \cos^2 3x \, dx$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2003 - Paper 1

$f(x) = x^2 - 4x + 5$ is a quadratic function that opens upwards. It has a minimum point, indicating that it is not one-to-one. In other words, there are multiple values of $x$ that yield the same $f(x)$ output. Therefore, it fails the horizontal line test, which is necessary for a function to have an inverse.

To sketch $g^{-1}(x)$, first identify the vertex of $g(x)$, which is at the point (2, 1). The graph is a parabola opening upwards starting from this point. For $x \leq 2$, it is decreasing. The inverse graph will be a reflection of this curve about the line $y = x$. Therefore, we reflect the coordinates; (1, 2) becomes (2, 1) in general terms, and retains its decreasing nature.

The domain of $g^{-1}(x)$ corresponds to the range of $g(x)$. Since $g(x)$ has a minimum value of 1 (at $x=2$) and approaches infinity, the domain of $g^{-1}(x)$ is $[1, \infty)$.

The function $g(x) = x^2 - 4x + 5$ can be rewritten in vertex form:

$g(x) = (x - 2)^2 + 1$

To find the inverse, swap $x$ and $y$:

$x = (y - 2)^2 + 1$

This simplifies to

$(y - 2)^2 = x - 1$

Taking the square root yields:

$y - 2 = \pm\sqrt{x - 1}$

Since we consider values for $x \leq 2$, we take the negative root:

$y = 2 - \sqrt{x - 1}$
Thus,

$g^{-1}(x) = 2 - \sqrt{x - 1}$.

We start from Newton's law:

$\frac{dT}{dt} = k(T - A)$

Substituting $T = A + Be^{kt}$, we differentiate:

$\frac{d}{dt}(A + Be^{kt}) = k(A + Be^{kt} - A)\Rightarrow kBe^{kt} = kBe^{kt}$

As both sides are equal, $T = A + Be^{kt}$ satisfies the equation.

Using the temperature data: After 6 minutes, $T(6) = 80°C$, substituting into $T = 20 + Be^{6k}$ gives:

$80 = 20 + Be^{6k}$

Thus, $B e^{6k} = 60$.

After 8 minutes, $T(8) = 50°C$:

$50 = 20 + Be^{8k}$

Thus, $Be^{8k} = 30$.
Dividing these two equations gives:

$\frac{60}{30} = \frac{Be^{6k}}{Be^{8k}} \Rightarrow 2 = e^{-2k}$
From here, taking the natural logarithm yields

$-2k = \log(2) \Rightarrow k = -\frac{\log(2)}{2}$

Using $T(6) = 80°C$ again, we have $B e^{6k} = 60$. We have $k = -\frac{\log(2)}{2}$ from the previous calculation.

Thus:

$B e^{6(-\frac{\log(2)}{2})} = 60$

This provides the equation for $B$:

$B = 60 \cdot e^{3\log(2)} = 60 \cdot 2^3 = 60 \cdot 8 = 480$
Thus, the value of $B$ is 480.