VCE Math Methods 2024 Exam 1

a. [1 mark]. Product rule, with the chain rule on $\cos(3x)$:

b. [2 marks]. Derivative of $\log_e(u)$ is $\tfrac{u'}{u}$:

From the report. In part a many students left untidy negative signs, dropped brackets, or changed the argument of the sine term. In part b some omitted the numerator $3x^2 - 3$ when applying the chain rule, or left out brackets around the quadratic term, which caused slips when substituting $x = 3$.

[3 marks]. No solution means the lines are parallel (same gradient) but distinct (different intercepts). The coefficient determinant must be zero:

Now test each value.

• $k = -2$: the equations are $-6x - 2y = 2$ and $-6x - 2y = 2$. They are the same line, so there are infinitely many solutions.
• $k = \tfrac43$: the equations become $y = 2x - \tfrac83$ and $y = 2x - 1$. Same gradient, different $y$-intercepts: no solution.

So $k = \dfrac43$.

From the report. Students using the determinant often found both $k = \tfrac43$ and $k = -2$ but did not justify which one gives no solution. Some chose $k = -2$ and rejected $\tfrac43$, confusing "infinite solutions" with "no solution". Many used the quadratic formula when the quadratic factorised by inspection.

a. [3 marks]. This is a truncus $y = \tfrac{1}{x^2}$ translated 3 left and 2 down.

• Asymptotes: $x = -3$ and $y = -2$.
• $y$-intercept: $g(0) = \tfrac19 - 2 = -\tfrac{17}{9}$, so $\left(0, -\tfrac{17}{9}\right)$.
• $x$-intercepts: $(x+3)^2 = \tfrac12$, so $x = -3 \pm \tfrac{1}{\sqrt2} = -3 \pm \tfrac{\sqrt2}{2}$, giving $\left(-3 - \tfrac{\sqrt2}{2}, 0\right)$ and $\left(-3 + \tfrac{\sqrt2}{2}, 0\right)$.

The graph is symmetric about $x = -3$ and both branches lie above $y = -2$.

b. [2 marks]. On $[-2, 0]$ the graph is below the $x$-axis (the nearest $x$-intercept, $-3 + \tfrac{\sqrt2}{2} \approx -2.29$, is to the left of $-2$). Integrate and take the positive value:

The area is $\dfrac{10}{3}$ square units.

From the report. Part a was generally well drawn; errors were unlabelled asymptotes, wrong intercept coordinates, a graph that was not symmetric, or a hyperbola instead of a truncus. In part b some used an $x$-intercept as a terminal instead of $-2$ and $0$. Many got $-\tfrac{10}{3}$ and simply wrote $\tfrac{10}{3}$ without a reason; state that area is positive or use an absolute value.

a. [1 mark]. $\text{sd}(X) = \sqrt{np(1-p)} = \sqrt{4 \times 0.9 \times 0.1} = \sqrt{0.36} = 0.6 = \dfrac35$.

b. [2 marks]. $\Pr(X < 2) = \Pr(X = 0) + \Pr(X = 1)$:

From the report. In part a some gave the variance instead of the standard deviation or slipped multiplying decimals. In part b some students added or lost a zero when converting $(0.1)^4$ and $(0.1)^3$, or gave only one of the two terms.

a. [1 mark]. $h(4) = \dfrac{3000}{5} = 600$ people.

b. [2 marks]. Half the rate means $h_1'(t) = \tfrac12h'(t)$, so $h_1(t) = \tfrac12h(t) + c$. Then $h_1(0) = 1500 + c = 3000$ gives $c = 1500$:

Transformations, in this order:

1. A dilation by a factor of $\tfrac12$ from the horizontal ($t$) axis.
2. A translation of 1500 units in the positive vertical direction (up).

(Equivalently: translate up 3000, then dilate by $\tfrac12$ from the $t$-axis.)

c. i. [2 marks]. $\hat p = 0.6$ and

The interval is $\hat p \pm 2 \times \tfrac{\sqrt6}{50}$:

ii. [1 mark]. $\sqrt{\dfrac{0.24}{n}} = \dfrac{\sqrt2}{50}$ gives $\dfrac{0.24}{n} = \dfrac{2}{2500}$, so $n = 300$.

From the report. Part b was poorly done (2% full marks): most found the dilation but named the wrong axis or direction. Use the study design wording, "from the $t$-axis" or "parallel to the vertical axis". In c.i the errors were in simplifying surds with decimals, and some left out the interval brackets.

[4 marks]. The equation needs $x > 4$. Use log laws:

Expand: $x^3 - 8x^2 + 16x - 9 = 0$. Test $x = 1$: $1 - 8 + 16 - 9 = 0$, so $(x - 1)$ is a factor:

Check the domain $x > 4$: $x = 1$ fails, $\tfrac{7 - \sqrt{13}}{2} \approx 1.7$ fails, and $\tfrac{7 + \sqrt{13}}{2} \approx 5.3$ works.

From the report. Most students knew the log laws but not always the right combination. Some identified $(x + 1)$ or $x = -1$ as the factor. Many found all three roots and did not reject the two that fall outside $x > 4$.

a. [3 marks]. The $x$-values are $0, \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \pi$:

b. i. [1 mark]. $f'(x) = \sin(x) + x\cos(x)$.

ii. [1 mark]. $f'\!\left(\tfrac{\pi}{2}\right) = 1 + 0 = 1$ and $f'\!\left(\tfrac{2\pi}{3}\right) = \tfrac{\sqrt3}{2} - \tfrac{\pi}{3}$. On this interval $f''(x) = 2\cos(x) - x\sin(x) < 0$, so $f'$ is decreasing and its range is

iii. [1 mark]. $f'$ is continuous, $f'\!\left(\tfrac{\pi}{2}\right) = 1 > 0$, and $f'\!\left(\tfrac{2\pi}{3}\right) = \tfrac{\sqrt3}{2} - \tfrac{\pi}{3} < 0$ (because $\tfrac{\pi}{3} > 1 > \tfrac{\sqrt3}{2}$). So $f'(x) = 0$ somewhere in the interval: $f$ has a stationary point there.

c. [3 marks]. $f$ is even, so $f'$ is odd. Endpoints: $f'(-\pi) = 0 + (-\pi)(-1) = \pi$ and $f'(\pi) = 0 + \pi(-1) = -\pi$. The graph of $f'$ crosses the $x$-axis where $f$ is stationary: at $x = 0$ and at $x \approx \pm 2.03$ (the peaks of $f$).

From the report. In part a some left out the factor 2 on the middle terms, tried to integrate instead, or did not evaluate $\sin\!\left(\tfrac{\pi}{3}\right)$ and $\sin\!\left(\tfrac{2\pi}{3}\right)$ correctly and combine the terms. Only 20% got b.ii (the interval was often reversed or badly bracketed) and 12% got b.iii: "verify" needs the two specific values and the observation that $\tfrac{\sqrt3}{2} - \tfrac{\pi}{3} < 0$, not a general comment. In part c many placed the endpoints wrongly or on the $x$-axis; the $x$-intercepts of $f'$ must line up with the turning points of $f$.

a. [1 mark]. $g(0) = \sqrt[3]{-k} + m = m - \sqrt[3]{k}$ (the cube root is an odd function), so $P = \left(0,\ m - \sqrt[3]{k}\right)$.

b. [2 marks]. $g'(x) = \tfrac13(x - k)^{-\frac23}$, so

The gradient is positive: $(-k)^2 = k^2$.

c. [1 mark]. $g(0) = 0$ gives $m - \sqrt[3]{k} = 0$, so $k = m^3$.

d. [3 marks]. Equal gradients: $\tfrac13(x - k)^{-\frac23} = \tfrac13(-k)^{-\frac23}$, so $(x - k)^2 = k^2$ and $x - k = \pm k$. That gives $x = 0$ (the point $P$) or $x = 2k$.

So $Q = \left(2m^3, 2m\right)$. This also follows from symmetry: the graph is symmetric about its point of inflection $(k, m) = (m^3, m)$, and $Q$ is the image of $P = (0, 0)$.

From the report. In part a the $y$-value was often right but not written as a coordinate pair. In part b some integrated instead of differentiating, or pulled out a negative sign to get $-\tfrac{1}{3k^{2/3}}$. Part d was the hardest on the paper (average 0.5 out of 3); the few who used the symmetry of the graph were usually successful.