VCE Math Methods 2021 Exam 1

a. [1 mark]. Chain rule on the exponent:

b. [2 marks]. Product rule, with the chain rule on $\sqrt{2x + 1} = (2x + 1)^{\frac12}$:

So

From the report. Part a was well answered. In part b the common error was not differentiating the $2x$ inside $\sqrt{2x + 1}$ (losing the factor 2 from the chain rule), and some students slipped when evaluating the square roots.

[2 marks]. Antidifferentiate, keeping the constant:

Use $f(1) = 2$: $\tfrac14 + \tfrac12 + c = 2$, so $c = \tfrac54$.

From the report. Well answered, but students who left out the constant of integration could not finish: here $c$ is what picks out the one correct antiderivative. Write fractions clearly; "1/4x4" can be misread.

a. [1 mark]

The amplitude is 2, so the range is $[-2, 2]$.

b. [1 mark]

Period $= \dfrac{2\pi}{2} = \pi$.

c. [3 marks]

$\sin(2x) = \dfrac{\sqrt3}{2}$. The reference angle is $\tfrac{\pi}{3}$, and sine is positive in the first and second quadrants, so

Divide by 2:

Equivalent general forms are also correct, as long as $k$ is stated to be an integer.

From the report. In part a, common errors were $(-2, 2)$ (wrong brackets) or just "2". In part c most students knew to look for more than one solution, but some gave only the solutions in one period, and the general solution was often done poorly. Common errors were ending with $+2k\pi$ (the period here is $\pi$) and writing $k \in R$, $k \in R^+$ or $k \in Z^+$ (or not saying what $k$ is) instead of $k \in Z$.

a. [3 marks]. This is the hyperbola $y = \tfrac1x$ dilated by 2 from the $x$-axis, reflected in the $x$-axis, then translated 2 right and 1 up.

• Asymptotes: $x = 2$ and $y = 1$.
• $y$-intercept: $y = 1 - \tfrac{2}{-2} = 2$, so $(0, 2)$.
• $x$-intercept: $1 = \tfrac{2}{x - 2}$ gives $x = 4$, so $(4, 0)$.

Because of the reflection, the left branch sits above $y = 1$ and the right branch below it.

b. [1 mark]. From the graph, $y \ge 3$ only on the left branch, between the point where $y = 3$ and the asymptote. Solve $1 - \tfrac{2}{x - 2} = 3$: $\tfrac{2}{x - 2} = -2$, so $x = 1$. The graph is above $y = 3$ from $x = 1$ up to (but not including) $x = 2$:

From the report. Part a was mostly well drawn; errors were unlabelled or wrongly labelled asymptotes, and intercepts such as an $x$-intercept labelled $(4, 0)$ but drawn at $(3, 0)$. Part b was not done well (32% correct): most students tried algebra instead of reading the graph and found only the lower bound, or wrote the endpoints in the wrong order or with the wrong brackets.

a. [2 marks]. Solve $g(x) = 0$:

The other intercept is $(0, 0)$.

b. [2 marks]. Dilation by $\tfrac12$ from the vertical axis replaces $x$ with $2x$; translating 2 right then replaces $x$ with $x - 2$:

The intercepts can be found by transforming the intercepts of $f$, $(\pm 2, 0)$: halve the $x$-coordinate, then add 2. That gives $(1, 0)$ and $(3, 0)$. (Check: $4(x - 2)^2 = 4$ gives $x = 1$ or $3$.)

From the report. In part a some students solved $f(x) = g(x)$ instead of $g(x) = 0$, and some made the working harder with the quadratic formula. Part b was poorly done (74% scored 0): many left out the brackets and wrote $h(x) = (2x - 2)^2 - 4$, and few realised the intercepts could be found by simply transforming points. Many did not find the intercepts at all.

a. [1 mark]. Custard: $\tfrac12$. Glazed with custard: $\tfrac1{10}$. So not glazed with custard is

A two-way table (Karnaugh map) makes this quick:

b. [2 marks]. $\tfrac{3}{10}$ of 20 is 6 glazed doughnuts. Box A has $g$ of them, so Box B has $6 - g$. With $G$ = "glazed":

c. [3 marks]. The population is very large, so the number under 25 in the sample is $X \sim \text{Bi}\left(5, \tfrac12\right)$ and $\hat P = \tfrac{X}{5}$. Then $\hat P \ge 0.8$ means $X \ge 4$:

From the report. In part a a common wrong answer came from assuming independence: $\tfrac12 \times \tfrac{7}{10} = \tfrac{7}{20}$. Part b was not well answered (11% full marks): many did not use $g$ as instructed, few found $6 - g$, and some used $6 - g$ as a probability instead of $\tfrac{6 - g}{10}$. Students who drew a tree diagram or a table did better. In part c most identified $n = 5$ and $p = \tfrac12$; common errors were $n = 4$ or $p = 0.8$.

a. [1 mark]. The total area must be 1:

b. [2 marks].

From the report. In part a (a "show that") most set up the integral correctly; occasionally the antiderivative was wrongly written as a logarithm. In part b some integrated $f(x)$ instead of $x f(x)$, and a common slip was not recognising that $\log_e(1) = 0$.

a. [3 marks]. Antidifferentiate term by term, using $\int (x + 6)^{\frac12}\,dx = \tfrac23(x + 6)^{\frac32}$:

Substitute $\left(3, \tfrac{29}{4}\right)$. Since $9^{\frac32} = 27$:

b. [2 marks]. Test the sign of the gradient on each side of $x = 3$, using values that give perfect squares under the root:

The gradient changes from positive to negative (and $x = 3$ is the only stationary point), so $\left(3, \tfrac{29}{4}\right)$ is a local maximum.

Alternatively, $\dfrac{d^2y}{dx^2} = \dfrac{1}{2\sqrt{x + 6}} - \dfrac12$, which is $\tfrac16 - \tfrac12 = -\tfrac13 < 0$ at $x = 3$.

From the report. In part a most got the right powers but often the wrong coefficients, and some lost the $(x + 6)$ and just integrated $\sqrt x$. Finding $c$ caused trouble, especially evaluating terms like $\sqrt{9^3}$. The rule, not just $c$, was the required answer. In part b most knew to check the slope on each side of $x = 3$, but not all showed the substitution clearly; the report describes many responses as "muddled scribble". Set out a sign table.

a. [2 marks]. A tangent is perpendicular to the radius, so triangle $OPA$ has a right angle at $P$, with $OP = 1$ and $OA = 2$. Let $\theta = \angle AOP$:

So $P = \left(\cos\tfrac{\pi}{3}, \sin\tfrac{\pi}{3}\right) = \left(\tfrac12, \tfrac{\sqrt3}{2}\right)$. The radius $OP$ has gradient $\tan\tfrac{\pi}{3} = \sqrt3$, so the tangent has gradient $-\tfrac{1}{\sqrt3}$. Through $A(2, 0)$:

(Check with $P$: $-\tfrac{1}{2\sqrt3} + \tfrac{2}{\sqrt3} = \tfrac{3}{2\sqrt3} = \tfrac{\sqrt3}{2}$.)

b. i. [1 mark]. $T$ maps $(x, y)$ to $(x, qy)$: a dilation by factor $q$ from the $x$-axis (a reflection too if $q < 0$). So

a line through $A(2, 0)$ with gradient $-\tfrac{q}{\sqrt3}$. When $|q| = 1$ the line is a tangent (the original line, or its reflection in the $x$-axis). A smaller $|q|$ gives a flatter line that cuts the circle; a larger $|q|$ gives a steeper line that misses it. (Formally, the distance from $O$ to the line is $\tfrac{2|q|}{\sqrt{q^2 + 3}} \le 1$ exactly when $q^2 \le 1$.) Since $q \ne 0$:

ii. [1 mark]. Two intersections needs $q \ne \pm 1$, and positive coordinates needs a negative gradient, so $0 < q < 1$. Starting from $A$ the line enters the circle in the first quadrant; the far intersection stays in the first quadrant only if the line's $y$-intercept $\tfrac{2q}{\sqrt3}$ lies above the circle, that is $\tfrac{2q}{\sqrt3} > 1$, so $q > \tfrac{\sqrt3}{2}$. (At $q = \tfrac{\sqrt3}{2}$ the line passes through $(0, 1)$, which has a zero coordinate.)

c. i. [2 marks]. Using area $= \tfrac12 ab\sin C$ with $OA = 2$ and $OP' = 1$ (a radius):

For the domain: at $q = 1$ the line is the tangent, $P' = P$ and $\theta = \tfrac{\pi}{3}$; as $q \to 0^+$ the line flattens and $P'$ approaches $(1, 0)$, so $\theta \to 0$. Hence

ii. [2 marks]. $\sin\theta$ is increasing on $\left(0, \tfrac{\pi}{3}\right]$, so the maximum is at the endpoint $\theta = \tfrac{\pi}{3}$:

Setting $g'(\theta) = 0$ would give $\theta = \tfrac{\pi}{2}$ and an area of 1, but that angle is outside the domain.

From the report. Many students found Question 9 challenging. In part a many did not show how they found the angle, the gradient or the point $P$; some differentiated the semicircle and tried to evaluate at $x = 2$, which is outside its domain. Label angles and sides on a diagram. Only 6% got b.i (many gave $[-1, 1]$, forgetting to exclude 0) and 4% got b.ii. In c.i many wrote $g(\theta) = \sin\theta$ but very few stated the domain, and some wrote $g(x) = \sin\theta$. Because of that, in c.ii many differentiated and gave the incorrect maximum of 1 instead of using the endpoint $\theta = \tfrac{\pi}{3}$.