Use the Question 11 Writing Booklet (a) Solve the quadratic equation $$z^2 - 3z + 4 = 0,$$ where $z$ is a complex number - HSC - SSCE Mathematics Extension 2 - Question 11 - 2023 - Paper 1

To find the angle between two vectors, we apply the formula: $\cos(\theta) = \frac{q \cdot b}{|q||b|},$ where $q \cdot b$ is the dot product of the vectors $q$ and $b$, and $|q|$, $|b|$ are their magnitudes. First, calculate the dot product: $q \cdot b = (1)(-1) + (2)(4) + (-3)(2) = -1 + 8 - 6 = 1.$ Next, we compute the magnitudes: $|q| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14},$ $|b| = \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}.$ Now substitute into the cosine formula: $\cos(\theta) = \frac{1}{\sqrt{14}\sqrt{21}}.$ Hence, calculating $heta$ gives: $\theta = \cos^{-1}\left(\frac{1}{\sqrt{14} \sqrt{21}}\right) \approx 87^{\circ}.$

The vector equation of the line can be expressed in the form: $\mathbf{r} = \mathbf{a} + t \mathbf{b},$ where $\mathbf{a}$ is a position vector on the line, and $\mathbf{b}$ is a direction vector. Using points $A$ and $B$, we have: $\mathbf{a} = \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 0 - (-3) \\ 2 - 1 \\ 3 - 5 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}.$ Thus, the vector equation of the line is: $\mathbf{r} = \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix} + t \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}, \; t \in \mathbb{R}.$

To show that $CDFE$ is a parallelogram, we need to demonstrate that pairs of opposite sides are equal in length and parallel. Given that quadrilaterals $ABCD$ and $ABEF$ are both parallelograms, we know: $AB = DC \quad \text{and} \quad AB = FE.$ Since $CD$ is equal to $AB$ and $EF$ is equal to $AB$, we have: $DC = FE.$ This shows that one pair of opposite sides of quadrilateral $CDFE$ is equal in length and parallel, thus confirming that $CDFE$ is a parallelogram.

The differential equation for simple harmonic motion is: $\dot{x} = -9(x - 4).$ Rearranging gives: $\dot{x} + 9x = 36.$ The standard form represents a simple harmonic oscillator, and solving it gives:

• The coefficient of $x$ is related to angular frequency $\omega = 3$.
• Since angular frequency $= \frac{2\pi}{T}$, the period $T$ can be found: $T = \frac{2\pi}{n} = \frac{2\pi}{3}.$
• The central point of motion occurs at $x = 4$. Thus, we find:
• Period: $T = \frac{2\pi}{3}$.
• Central point of motion: $x = 4.$

To evaluate the integral, we first perform partial fraction decomposition: $\frac{5x - 3}{(x + 1)(x - 3)} = \frac{A}{x + 1} + \frac{B}{x - 3}.$ Multiplying through by the denominator gives: $5x - 3 = A(x - 3) + B(x + 1).$ Setting up the equations yields:

• $A + B = 5$
• $-3A + B = -3$. Solving these equations leads to:
• $A = 2$ and $B = 3.$ Thus, we have: $\frac{5x - 3}{(x + 1)(x - 3)} = \frac{2}{x + 1} + \frac{3}{x - 3}.$ Now we can integrate: $\int_0^{1} \left(\frac{2}{x + 1} + \frac{3}{x - 3}\right)dx = \left[2\ln|x + 1| + 3\ln|x - 3|\right]_0^{1}.$ Evaluating this from 0 to 1 gives the final result.