VCE Specialist Mathematics 2022 Exam 1

a. [1 mark]. Complete the square. Half the coefficient of $z$ is $3i$, and $(3i)^2 = -9$:

So $a = 3$ and $b = -16$.

b. [2 marks]. From part a, $p(z) = 0$ when $(z + 3i)^2 = 16$, so $z + 3i = \pm 4$ and

The quadratic formula gives the same result: $z = \dfrac{-6i \pm \sqrt{(6i)^2 + 100}}{2} = \dfrac{-6i \pm \sqrt{64}}{2} = -3i \pm 4$.

From the report. Part a was answered well (79% correct): most students saw that completing the square was needed, though there were some arithmetic slips. In part b (70% full marks) students could use part a, as above, or the quadratic formula.

[3 marks]. The equation is separable:

At $x = 2$, $y = 0$: $\arcsin(0) = 0 = -2 + c$, so $c = 2$. Taking the sine of both sides,

(The step from $\arcsin$ to $\sin$ holds while $2 - \tfrac{x^2}{2}$ stays between $-\tfrac{\pi}{2}$ and $\tfrac{\pi}{2}$, which includes the initial point $x = 2$.)

From the report. Well answered (60% full marks): most students recognised the separable differential equation and tried to solve it. A few worked correctly through to an inverse cosine version of the solution, which gives the equivalent form $y = 2\cos\left(\tfrac{x^2}{2} + \tfrac{\pi}{2} - 2\right)$.

a. [2 marks]. Let $T = X_1 + X_2 + X_3 + X_4$ be the total time for four cups, where the $X_i$ are independent, each normal with mean 10 and variance $1.5^2 = 2.25$. Then $T$ is normal with

(This is a sum of four separate cups, not $4X$ for one cup, which would have variance $16 \times 2.25 = 36$.) So

About 95% of a normal distribution lies within 2 standard deviations of the mean, so $\Pr(Z < -2) \approx 0.025$ and

From the report. Only 31% scored full marks and 59% scored zero. Many students got the standard deviation of the total wrong, so could not reduce the problem to $\Pr(Z > -2)$; others reached $\Pr(Z > -2)$ but could not evaluate it. Students who got it right often sketched the normal curve and knew the approximate probabilities within 1, 2 and 3 standard deviations of the mean.

[4 marks]. Use partial fractions with a linear and an irreducible quadratic factor:

Putting $x = 0$ gives $4A = 12$, so $A = 3$. Comparing coefficients of $x^2$: $A + B = 3$, so $B = 0$; of $x$: $C = 4$. So

A shortcut avoids partial fractions: split the numerator as $\left(3x^2 + 12\right) + 4x = 3\left(x^2 + 4\right) + 4x$, so the integrand is $\dfrac{3}{x} + \dfrac{4}{x^2 + 4}$ straight away.

From the report. 36% scored full marks. The shortcut above was spotted by only a small number of students. Many mixed some initial algebra with one or more rounds of partial fractions, which was inefficient and often much longer than needed. Some omitted the absolute value in the logarithm or the constant of integration, and a few integrated $\tfrac{4}{x^2 + 4}$ incorrectly.

a. [2 marks]. From $\tan(\theta) = \tfrac13$, draw a right triangle with opposite side 1 and adjacent side 3; the hypotenuse is $\sqrt{10}$, so $\sin(\theta) = \dfrac{1}{\sqrt{10}}$.

The plane is smooth, so the only force with a component along the plane is the weight, whose component down the plane is $mg\sin(\theta)$. By Newton's second law the acceleration down the plane is

The acceleration is constant and $u = 0$, so $v = u + at$ gives

b. [1 mark]. Constant velocity means zero acceleration, so the net force along the plane is zero: $mg\sin(\theta) - R = 0$ and

From the report. Part a was answered well (55% full marks), with several equivalent forms of the answer accepted; successful students often marked the forces on the diagram before resolving them. In part b (48% correct) equivalent answers were also accepted.

a. [2 marks]. $\mathbf a \cdot \mathbf b = 2 - 6 + 12 = 8$, $|\mathbf a| = \sqrt{4 + 9 + 36} = 7$ and $|\mathbf b| = \sqrt{1 + 4 + 4} = 3$, so

The dot product is positive, so this angle is already acute.

b. i. [1 mark]. The semicircle has centre $(a, 0)$ and radius $a$, so $Q$ is the point $(2a, 0)$. Then

where $y = \sqrt{a^2 - (x - a)^2}$.

ii. [3 marks]. Using $y^2 = a^2 - (x - a)^2$,

The dot product is zero, so $\overrightarrow{OP}$ is perpendicular to $\overrightarrow{QP}$ for every point $P$ on the semicircle other than $O$ and $Q$ (where one of the vectors is zero). This is the vector proof that the angle in a semicircle is a right angle.

From the report. Part a was answered very well (76% full marks), mostly by the method above, with some arithmetic errors. Part b.i (61%) was answered well apart from occasional sign errors. In b.ii (47% full marks) students had to use the dot product; some made algebraic errors or drew the wrong conclusion.

[3 marks]. Differentiate implicitly with respect to $x$, using the product rule and then the chain rule on $\cos(x + y)$. Keep $\cos(x + y)$ as it is rather than expanding it:

At the point, $x + y = \tfrac{8\pi}{24} = \tfrac{\pi}{3}$, so $\cos(x + y) = \tfrac12$ and $\sin(x + y) = \tfrac{\sqrt3}{2}$:

So

with $a = 8$ and $b = 3$. (The point is on the curve: $\tfrac{\pi}{24}\cos\left(\tfrac{\pi}{3}\right) = \tfrac{\pi}{48}$.)

From the report. 34% scored full marks and 33% scored zero. Many students expanded $\cos(x + y)$ with the compound angle formula before differentiating; few of them reached the right answer, as the number of terms became unmanageable.

[4 marks]. Acceleration is given in terms of $x$, so write it as $\dfrac{d}{dx}\left(\dfrac12v^2\right)$:

At $x = 0$, $v = -2$: $\tfrac12 \times 4 = c$, so $c = 2$. Then

The body is at rest when $v = 0$, that is at $x = \pm1$. It starts from rest at $x = 1$, moves in the negative direction through the origin (where $v = -2 < 0$) and comes to rest at $x = -1$, so $v \le 0$ throughout. Take the negative square root:

From the report. 37% scored full marks. Many students used a suitable form of acceleration, either $\tfrac{d}{dx}\left(\tfrac12v^2\right)$ or $v\tfrac{dv}{dx}$. A common error was the wrong sign in the final answer: the negative root is needed because $v = -2$ at $x = 0$.

[4 marks]. Substitute $u = \sin(2x)$, so $\dfrac{du}{dx} = 2\cos(2x)$ and $\cos(2x)\,dx = \tfrac12\,du$:

At $x = \tfrac{\pi}{8}$, $\sin(2x) = \sin\left(\tfrac{\pi}{4}\right) = \tfrac{1}{\sqrt2}$, so $\sin^2(2x) = \tfrac12$ and

Hence

From the report. 34% scored full marks and 32% scored zero; other correct forms were accepted. Some students tried to rewrite the integrand with trigonometric identities before integrating, usually without success. Among those who chose a suitable substitution, a frequent slip was antidifferentiating $u^{-3}$ as $-2u^{-2}$ or $-2u^{-4}$ instead of $-\tfrac12u^{-2}$.

a. [3 marks]. $\sec(4x) = \dfrac{1}{\cos(4x)}$, whose period is $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$, so the interval covers exactly one period.

• Asymptotes where $\cos(4x) = 0$: $4x = \pm\tfrac{\pi}{2}$, so $x = -\tfrac{\pi}{8}$ and $x = \tfrac{\pi}{8}$.
• Turning point where $\cos(4x) = 1$: the local minimum $(0, 1)$.
• Endpoints: $\cos(\pm\pi) = -1$, so $\left(-\tfrac{\pi}{4}, -1\right)$ and $\left(\tfrac{\pi}{4}, -1\right)$, both included.

Between the asymptotes the graph is a U shape above $y = 1$; outside them it lies on or below $y = -1$, falling from each endpoint towards $-\infty$ as it approaches the nearer asymptote. The graph is symmetric about the $y$-axis and has no horizontal asymptote.

b. [3 marks]. $V = \pi\displaystyle\int y^2\,dx$, and $\sec^2(4x)$ has antiderivative $\tfrac14\tan(4x)$:

Here $\tan\left(-\tfrac{\pi}{6}\right) = -\tfrac{1}{\sqrt3}$. For $\tan\left(\tfrac{\pi}{12}\right)$ use the difference formula:

(Alternatively, with $t = \tan\left(\tfrac{\pi}{12}\right)$ the double angle formula gives $\tfrac{2t}{1 - t^2} = \tfrac{1}{\sqrt3}$, so $t^2 + 2\sqrt3\,t - 1 = 0$ and $t = -\sqrt3 \pm 2$; take the positive root, $2 - \sqrt3$.) So

So $a = 3$, $b = 3$ and $c = 6$ (other forms such as $\tfrac{\left(6 - \sqrt{12}\right)\pi}{12}$ are equivalent).

From the report. In part a (13% full marks) many students found the vertical asymptotes, the turning point and the endpoints correctly, but some implied a horizontal asymptote, and some graphs had the wrong shape or were not symmetric about the $y$-axis. In part b (25% full marks) most wrote a correct volume integral and many recognised $\tfrac14\tan(4x)$ as an antiderivative of $\sec^2(4x)$. The sticking point was finding $\tan\left(\tfrac{\pi}{12}\right)$: those using the double angle formula sometimes could not solve the resulting quadratic or kept the wrong root, and those using the difference formula often made arithmetic errors.