VCE Specialist Mathematics 2021 Exam 1

a. [1 mark]. Newton's second law, $\mathbf F = m\mathbf a$:

b. [2 marks]. The acceleration is constant, so antidifferentiate with respect to $t$:

$\mathbf v(0) = -3\mathbf j$ gives $\mathbf c = -3\mathbf j$, so

(This is $\mathbf v = \mathbf u + \mathbf a t$ for constant acceleration.)

c. [1 mark]. $\mathbf v(2) = \mathbf i + \left(\tfrac{12}{5} - 3\right)\mathbf j = \mathbf i - \tfrac35\mathbf j$, so the momentum is

From the report. Part a was answered well (80% correct), though some gave the magnitude of the acceleration instead of the vector. In part b (66% full marks) most made progress by integrating or with a constant acceleration formula; a common slip was wrongly combining the $\mathbf j$ terms into $-\tfrac95t\,\mathbf j$. In part c (58% correct) there were arithmetic errors, and some students gave the momentum's magnitude instead of the vector.

[3 marks]. Split the integrand into two fractions:

In the first, the numerator is the derivative of the denominator; the second is the derivative of $\arctan(x)$:

From the report. Well answered (63% full marks). Most students saw that the integrand splits into the sum of two rational functions. A substitution was unnecessary, and students who tried one sometimes introduced errors.

a. [1 mark]. Let $\mu$ be the mean lifetime of Shiny globes in weeks. The complaints are that globes last less than claimed, so

b. i. [2 marks]. Under $H_0$, $\bar X$ is normal with mean 200 and standard deviation $\tfrac{10}{\sqrt{36}} = \tfrac53$. So

By symmetry, $\Pr(Z < -3) = \dfrac{1 - 0.9973}{2} = 0.00135$, so $p = 0.001$ correct to three decimal places.

ii. [1 mark]. $p = 0.001 < 0.01$, so reject $H_0$. The company should be told there is evidence, at the 1% level of significance, that the mean lifetime of Shiny globes is less than 200 weeks: the complaints are supported.

c. [1 mark]. Here $\tfrac{\sigma}{\sqrt n} = \tfrac{10}{\sqrt{25}} = 2$, so the interval is

From the report. Part a was well answered (72%); errors were an alternative hypothesis with the wrong inequality ($\mu > 200$ or $\mu \ne 200$) and non-standard notation. In b.i (41% full marks, 47% scored zero) students had to use the given $\Pr(-3 < Z < 3) = 0.9973$ to find $\Pr(Z < -3) = 0.00135$. In b.ii (49% correct) some drew the wrong conclusion, sometimes because they confused the $p$ value (0.001) with the significance level (0.01). In part c (56% correct) many used $\mu = 200$ instead of the sample mean 250, and there were also arithmetic errors and uses of $z = 2$ instead of the given 1.96.

a. [3 marks]. Use $V = \pi\displaystyle\int y^2\,dx$ with the double angle identity $\sin^2(x) = \tfrac12\big(1 - \cos(2x)\big)$:

b. [1 mark]. The graph of $y = \sin(kx)$ is the graph of $y = \sin(x)$ dilated by a factor of $\tfrac1k$ from the $y$-axis. Its first two non-negative $x$-intercepts are $0$ and $\tfrac{\pi}{k}$, and the solid is the part a solid scaled by $\tfrac1k$ along the $x$-axis, with the same circular cross-sections. So its volume is

Check by integration: $\pi\displaystyle\int_0^{\pi/k}\sin^2(kx)\,dx = \pi \times \frac{\pi}{2k} = \frac{\pi^2}{2k} = \frac1k V_s$.

From the report. In part a (58% full marks) most wrote a correct integral for the volume, but some could not go further and incorrect attempts at integration were common; the double angle formula was the most effective method. Part b was poorly answered (30% correct): very few saw that the part b graph and solid are those of part a dilated by a factor of $\tfrac1k$, and many of those who did still did not write the answer in terms of $V_s$ as instructed.

[3 marks]. Write $e^xe^{2y} = e^{x + 2y}$ and differentiate implicitly with respect to $x$. The right side is a constant, so its derivative is 0:

Substitute the point straight away. At $(2, 1)$, $e^{x + 2y} = e^4$ and $e^{4y^2} = e^4$, so

(The point is on the curve: $e^2e^2 + e^4 = 2e^4$.)

From the report. Answered well (53% full marks). Common errors were not differentiating the constant $2e^4$ to zero and substituting $x = 2$, $y = 1$ incorrectly. When $\tfrac{dy}{dx}$ is not needed in terms of $x$ and $y$, it can be easier to substitute the values of $x$ and $y$ straight after differentiating.

[4 marks]. First find when the vectors are linearly dependent. Since $\mathbf a$ and $\mathbf b$ are not parallel, the set is dependent exactly when $\mathbf c = \alpha\mathbf a + \beta\mathbf b$ for some real $\alpha$ and $\beta$:

• $\mathbf i$: $-\alpha + 2\beta = 3$
• $\mathbf j$: $6\alpha - 8\beta = 2$
• $\mathbf k$: $-3\alpha + 5\beta = \left|1 - p^2\right|$

From the first equation $\alpha = 2\beta - 3$; substituting into the second gives $12\beta - 18 - 8\beta = 2$, so $\beta = 5$ and $\alpha = 7$. The third equation then needs

The first gives $p^2 = -3$ (no real solution) and the second gives $p^2 = 5$, so the vectors are dependent only when $p = \pm\sqrt5$. They are linearly independent for

Equivalently, $\mathbf a \cdot (\mathbf b \times \mathbf c) = 16 - 4\left|1 - p^2\right|$, which is zero exactly when $\left|1 - p^2\right| = 4$.

From the report. Most students realised they had to set up and solve a system of linear equations to find when the vectors are dependent (average 2.5 out of 4; 26% full marks). Many found $p = \pm\sqrt5$ for dependence but did not go on to conclude $p \in R \setminus \left\{-\sqrt5, \sqrt5\right\}$ (or equivalent) for independence.

a. [3 marks]. Separate the variables:

At $t = 0$, $x = 1$: $0 = -1 + c$, so $c = 1$. Since $x = 1 > 0$ initially, $\log_e(x) = 1 - \cos(t)$ and

b. [2 marks]. $e^{1 - \cos(t)}$ is largest when $\cos(t)$ is smallest, that is when $\cos(t) = -1$. The maximum displacement is $e^{1 - (-1)} = e^2$ cm, and it occurs when

that is at $t = \pi, 3\pi, 5\pi, \ldots$ seconds.

From the report. Part a was well answered (66% full marks); most separated the variables correctly, though there were sign errors when integrating the trigonometric function. Part b was poorly answered (8% full marks, 58% scored zero): the maximum could be found by inspection, but most students did not attempt, or could not give, the correct times at which it occurs.

a. [1 mark]. Complete the square: $(z + 1)^2 + 1 = 0$, so $(z + 1)^2 = -1 = i^2$ and

b. [3 marks]. The equation involves $\bar z$, so it is not a polynomial equation in $z$: the quadratic formula does not apply, and the solutions to part a do not carry over. Let $z = x + iy$ with $x, y \in R$, so $\bar z = x - iy$:

Equate real and imaginary parts:

• Real: $x^2 - y^2 + 2x + 2 = 0$
• Imaginary: $2xy - 2y = 0$, so $2y(x - 1) = 0$ and $y = 0$ or $x = 1$.

If $y = 0$, then $x^2 + 2x + 2 = 0$, which has no real solutions (its discriminant is $4 - 8 < 0$). If $x = 1$, then $1 - y^2 + 2 + 2 = 0$, so $y^2 = 5$ and $y = \pm\sqrt5$. Hence

Check: $\left(1 + \sqrt5\,i\right)^2 = -4 + 2\sqrt5\,i$ and $2\bar z = 2 - 2\sqrt5\,i$, and $-4 + 2\sqrt5\,i + 2 - 2\sqrt5\,i + 2 = 0$.

From the report. Part a was answered well (70%); completing the square or the quadratic formula both work. Part b was not (10% full marks, 50% scored zero). Successful students let $z = x + iy$ and equated real and imaginary parts, but algebraic errors were common in solving the resulting equations. Some assumed the solutions to part a also solved part b, and some confused the conjugate with the reciprocal. Starting from the polar form $z = r\,\text{cis}(\theta)$ is possible, but few took that approach and those who did rarely made significant progress.

a. i. [1 mark]. From $x = -1 + 4\cos(t)$ and $y = \tfrac{2}{\sqrt3}\sin(t)$, $\cos(t) = \tfrac{x + 1}{4}$ and $\sin(t) = \tfrac{\sqrt3y}{2}$. Since $\cos^2(t) + \sin^2(t) = 1$:

ii. [1 mark]. Multiply by 16: $(x + 1)^2 + 12y^2 = 16$, so

In the first quadrant $y \ge 0$, so take the positive square root:

b. i. [1 mark]. The particles collide if they are at the same point at the same time. Equate the $\mathbf i$ components:

$\mathbf s(t)$ is only defined where $\sec(t)$ is, so on $0 \le t \le c$ we have $t < \tfrac{\pi}{2}$ and $\cos(t) > 0$. So $\cos(t) = \tfrac{\sqrt3}{2}$ and $t = \tfrac{\pi}{6}$. Check the $\mathbf j$ components at $t = \tfrac{\pi}{6}$:

Both components agree at the same time, $t = \tfrac{\pi}{6}$, so the particles collide.

ii. [1 mark]. At $t = \tfrac{\pi}{6}$, $x = -1 + 4 \times \tfrac{\sqrt3}{2} = 2\sqrt3 - 1$ and $y = \tfrac{1}{\sqrt3}$. The point of collision is

c. i. [2 marks]. Let $Q = -x^2 - 2x + 15$ and note that $16 - (x + 1)^2 = Q$. By the chain rule,

By the product rule, with $\dfrac{d}{dx}\sqrt Q = \dfrac{-2x - 2}{2\sqrt Q} = -\dfrac{x + 1}{\sqrt Q}$,

Add the two, using $(x + 1)^2 = 16 - Q$:

ii. [2 marks]. By part c.i, $8\arcsin\left(\tfrac{x + 1}{4}\right) + \tfrac{(x + 1)\sqrt{-x^2 - 2x + 15}}{2}$ is an antiderivative of $\sqrt{-x^2 - 2x + 15}$, so

• At $x = 2\sqrt3 - 1$: $\tfrac{x + 1}{4} = \tfrac{\sqrt3}{2}$ and $-x^2 - 2x + 15 = 16 - (x + 1)^2 = 16 - 12 = 4$, giving $8 \times \tfrac{\pi}{3} + \tfrac{2\sqrt3 \times 2}{2} = \tfrac{8\pi}{3} + 2\sqrt3$.
• At $x = 1$: $\tfrac{x + 1}{4} = \tfrac12$ and $-x^2 - 2x + 15 = 12$, giving $8 \times \tfrac{\pi}{6} + \tfrac{2 \times 2\sqrt3}{2} = \tfrac{4\pi}{3} + 2\sqrt3$.

The $2\sqrt3$ terms cancel:

So $a = 2$ and $b = 9$.

From the report. Part a.i was very well answered (83%), using $\sin^2(\theta) + \cos^2(\theta) = 1$. In a.ii (41% correct) a common error was not justifying the choice of the positive square root ($y \ge 0$ in the first quadrant). Part b.i (62%) was answered well by showing that both components agree when $\cos(t) = \tfrac{\sqrt3}{2}$; in b.ii (52%) some gave the point as $\left(\tfrac{\pi}{6}, \tfrac{1}{\sqrt3}\right)$, confusing the parameter $t$ with the $x$-coordinate. Few students scored full marks on either part of c (15% and 22%). In c.i the product and chain rules had to be applied and the result simplified with every step shown, since it is a show-that question; many missed steps or made algebraic errors. In c.ii most realised part c.i should be used, but many could not finish the arithmetic.