VCE Specialist Mathematics 2023 Exam 1

a. [1 mark]. $(x + 2)(x - 1) = x^2 + x - 2$, so $x^2 + x - 6 = (x + 2)(x - 1) - 4$ and

b. [3 marks]. Asymptotes $x = 1$ and $y = x + 2$. Intercepts: $x^2 + x - 6 = (x + 3)(x - 2)$ gives $(-3, 0)$ and $(2, 0)$, and $f(0) = 6$ gives $(0, 6)$. Since $f'(x) = 1 + \tfrac{4}{(x - 1)^2} > 0$ there are no turning points. For $x < 1$ the graph is above the oblique asymptote; for $x > 1$ it is below.

From the report. Both asymptotes had to be correct and labelled; the oblique asymptote was sometimes missing. Some graphs moved away from their asymptotes instead of approaching them.

[3 marks]. Let $\theta = \arg(b - i)$. Since $b > 0$, $b - i$ is in the fourth quadrant, so $-\tfrac{\pi}{2} < \theta < 0$ and $-\tfrac{3\pi}{2} < 3\theta < 0$. We need $\arg(z) = 3\theta = -\tfrac{\pi}{2}$ (the only value in that range), so $\theta = -\tfrac{\pi}{6}$:

Check: $\sqrt3 - i = 2\,\text{cis}\left(-\tfrac{\pi}{6}\right)$, so $z = 8\,\text{cis}\left(-\tfrac{\pi}{2}\right) = -8i$.

Alternatively, expand: $(b - i)^3 = (b^3 - 3b) + (1 - 3b^2)i$. For argument $-\tfrac{\pi}{2}$ the real part must be 0 and the imaginary part negative, so $b^2 = 3$ and $b = \sqrt3$.

From the report. The report showed both the polar approach and expanding the cube; graphical approaches were also seen.

a. [2 marks]. Velocity is given in terms of $x$, so use $a = v\dfrac{dv}{dx}$:

At $x = 2$: $v = \tfrac83$ and $\tfrac{dv}{dx} = -\tfrac79$, so $a = \tfrac83 \times \left(-\tfrac79\right) = -\dfrac{56}{27}$ m s$^{-2}$.

b. [1 mark]. $v = \dfrac{3 + 2/x}{2 - 1/x} \to \dfrac32$ m/s.

From the report. Many students found $\tfrac{dv}{dx}$ at $x = 2$ and stopped: acceleration is $v\tfrac{dv}{dx}$ (or $\tfrac{d}{dx}\left(\tfrac12v^2\right)$), not $\tfrac{dv}{dx}$. Part b was answered correctly by only half the students.

[3 marks]. Product rule, with the chain rule on $\arcsin(y^2)$:

At $\left(6, \tfrac{1}{\sqrt2}\right)$: $y^2 = \tfrac12$, $\arcsin\tfrac12 = \tfrac{\pi}{6}$ and $\sqrt{1 - y^4} = \tfrac{\sqrt3}{2}$:

So $a = 6$ and $b = 144$.

From the report. The product and chain rules were often applied poorly; students who differentiated correctly usually completed the question.

[3 marks]. Integrate by parts, differentiating $\log_e x$ and antidifferentiating $x^2$:

From the report. Most did well. Some chose the parts the wrong way round, did not evaluate the definite integral consistently, or left $x$ in the final answer.

a. [2 marks]. Mean $20 + 8 + 12 = 40$ minutes. Variances add: $36 + 3 + 25 = 64$, so the standard deviation is $8$ minutes.

b. [2 marks]. The average wait $\bar X_w$ has mean 8 and standard deviation $\tfrac{\sqrt3}{\sqrt{12}} = \tfrac12$. So

giving $a = -\tfrac12$, $b = 1$ (by symmetry, $a = -1$, $b = \tfrac12$ is equivalent).

From the report. In part a many added the standard deviations ($6 + \sqrt3 + 5$) instead of the variances. Part b was not answered well; the usual error was using the wrong standard deviation for the sample mean.

[4 marks]. $\dot x = \tfrac t2$ and $\dot y = \sqrt3$, so

Let $u = \tfrac{t^2}{4} + 3$, so $du = \tfrac t2\,dt$ and $t\,dt = 2\,du$; the terminals become $u = 3$ and $u = 4$:

So $a = 64$, $b = 3$, $c = 3$, $d = 24$.

From the report. An explicit substitution was more reliable than trying to spot the antiderivative. A few students converted to Cartesian form and integrated successfully.

[4 marks]. Base case, $n = 1$. $f'(x) = e^{2x} + 2xe^{2x} = (2x + 1)e^{2x} = \left(2^1x + 1 \cdot 2^0\right)e^{2x}$. True.

Inductive step. Assume that for some $k \in Z^+$, $f^{(k)}(x) = \left(2^kx + k2^{k-1}\right)e^{2x}$. Differentiating with the product rule:

This is the statement for $n = k + 1$. Since it is true for $n = 1$, and truth for $n = k$ implies truth for $n = k + 1$, by mathematical induction it is true for all $n \in Z^+$.

From the report. Most set up the base case and assumption. The inductive step required differentiating $f^{(k)}$; many did not differentiate, or differentiated incorrectly, and some misapplied index laws when simplifying the powers of 2.

a. [1 mark]

$D(0, 2, 0)$.

b. [1 mark]

$\overrightarrow{AB} = (-1 - 1, -2 - 3, 4 + 2) = (-2, -5, 6)$ and $\overrightarrow{AD} = (0 - 1, 2 - 3, 0 + 2) = (-1, -1, 2)$.

c. [2 marks]

A normal is

Using $D$: $4x + 2y + 3z = 4$. (Check $A$: $4 + 6 - 6 = 4$.)

d. [1 mark]. $C$ lies on the plane: $4a - 2 + 15 = 4$, so $a = -\tfrac94$.

e. [1 mark]. Area $= \left|\overrightarrow{AB}\times\overrightarrow{AD}\right| = \sqrt{16 + 4 + 9} = \sqrt{29}$.

From the report. Parts a and b were done very well. In part c some slipped in the cross product or in substituting a point. In part e some gave the triangle's area (half the value), and a common error was multiplying the two side lengths.

a. [1 mark]. $\sin^2t = \tfrac{1 - \cos 2t}{2}$, so $5 - 6\sin^2t = 5 - 3 + 3\cos 2t = 2 + 3\cos(2t)$: $\alpha = 2$, $\beta = 3$.

b. [2 marks]. $x = 2 + 3\cos 2t$ and, since $6\sin t\cos t = 3\sin 2t$, $y = 1 + 3\sin 2t$. Then

c. [1 mark]. The particle moves around a circle of radius 3 with angle $2t$, so the arc length from $t = 0$ to $t = a$ is $3 \times 2a = 6a$. Then $6a = \tfrac{3\pi}{4}$ gives $a = \tfrac{\pi}{8}$.

d. [2 marks]. $\dot{\mathbf r} = -6\sin 2t\,\mathbf i + 6\cos 2t\,\mathbf j$. Then

so $\tan 2t = \tfrac12$ and

From the report. Part b needed a double angle formula and then a Pythagorean identity. Part c (36% correct) could be done with arc length $r\theta$ or by integrating the speed. In part d many set up $\mathbf r\cdot\dot{\mathbf r} = 0$ but only 7% reached the full general solution.