2023 VCE Maths Methods Mini Test 7

Question 1 [2023 Exam 2 Section A Q11]

Two functions, \( f \) and \( g \), are continuous and differentiable for all \( x \in \mathbb{R} \). It is given that
\( f(-2) = -7, \quad g(-2) = 8, \quad f'(-2) = 3, \quad g'(-2) = 2 \)
The gradient of the graph \( y = f(x) \times g(x) \) at \( x = -2 \) is

• A. −10
• B. −6
• C. 0
• D. 6
• E. 10

Question 2 [2023 Exam 2 Section A Q12]

The probability mass function for the discrete random variable \( X \) is shown below:

X	−1	0	1	2
Pr(\(X = x\))	\(k^2\)	\(3k\)	\(k\)	\(-k^2 - 4k + 1\)

The maximum possible value for the mean of \( X \) is:

• A. 0
• B. \( \frac{1}{3} \)
• C. \( \frac{2}{3} \)
• D. 1
• E. 2

Question 3 [2023 Exam 2 Section A Q13]

The following algorithm applies Newton’s method using a For loop with 3 iterations...

\[ \begin{array}{l} \textbf{Inputs:} \quad f(x), \text{ a function of } x \\ \quad\quad\quad\quad df(x), \text{ the derivative of } f(x) \\ \quad\quad\quad\quad x_0, \text{ an initial estimate} \\ \\ \textbf{Define } \texttt{newton}(f(x), df(x), x_0) \\ \quad \textbf{For } i \text{ from } 1 \text{ to } 3 \\ \quad\quad \textbf{If } df(x_0) = 0 \textbf{ Then} \\ \quad\quad\quad \textbf{Return } \text{``Error: Division by zero''} \\ \quad\quad \textbf{Else} \\ \quad\quad\quad x_0 \leftarrow x_0 - f(x_0) \div df(x_0) \\ \quad \textbf{EndFor} \\ \quad \textbf{Return } x_0 \end{array} \]

The return value of the function newton(x³ + 3x − 3, 3x² + 3, 1) is closest to

• A. 0.83333
• B. 0.81785
• C. 0.81773
• D. 1
• E. 3