VCE Math Methods 2022 Exam 1

a. [1 mark]. Product rule, with $\tfrac{d}{dx}e^{2x} = 2e^{2x}$:

b. [2 marks]. Write $f(x) = e^{-x}\cos(x)$ and use the product rule (the quotient rule works too):

From the report. In part a a number of students stopped at $3x \cdot 2e^{2x} + 3e^{2x}$ without simplifying, some did not use the product rule, and many factorised and made an error. Factorising was not required, and a correct answer followed by an incorrect "simplification" cannot get full marks. In part b some did not simplify, or cancelled only one of the $e^x$ terms.

a. [1 mark]. Using $\int \tfrac{1}{ax + b}\,dx = \tfrac1a\log_e(ax + b)$, and $2x - 3 > 0$ on the domain:

(Any constant can be added; one antiderivative is enough.)

b. [3 marks]. Expand the integrand, then use linearity of the definite integral:

From the report. In part a it was fine to call the antiderivative $G(x)$, $\int g(x)\,dx$ or give no name, but not $g'(x)$. Common wrong answers were $3\log_e(2x - 3)$ and $6\log_e(2x - 3)$, and some used the domain values as terminals of a definite integral. Part b was not answered well by many: some wrote the integral of a product as a product of two integrals, and others substituted $\tfrac13$ for $f(x)$ and then integrated constants. Brackets were often missing.

[3 marks]. Infinitely many solutions means the two equations describe the same line: equal gradients and equal $y$-intercepts.

Step 1: equal gradients. The determinant of the coefficients must be zero:

Step 2: test each value.

• $k = -3$: the equations are $-3x - 5y = 1$ and $3x + 5y = -1$. The second is the first multiplied by $-1$, so they are the same line: infinitely many solutions.
• $k = -5$: the equations are $-5x - 5y = -1$ (so $x + y = \tfrac15$) and $3x + 3y = -1$ (so $x + y = -\tfrac13$). Parallel but distinct: no solution.

So $k = -3$.

(Equivalently, equate the gradients $\tfrac{k}{5} = -\tfrac{3}{k + 8}$ and the $y$-intercepts $-\tfrac{k + 4}{5} = -\tfrac{1}{k + 8}$; the second gives $(k + 3)(k + 9) = 0$, and $k = -3$ is the only value satisfying both.)

From the report. Students equated gradients and intercepts, used a determinant, formed ratios, or tried to solve simultaneously, with varying success. Those who knew the two lines had to be identical generally succeeded. Students using the determinant often reached $k = -5$ and $k = -3$ but did not justify which one was valid. Setting the two equations equal to each other led to too many variables to solve for $k$.

a. [2 marks]. $X \sim \text{Bi}\left(4, \tfrac12\right)$, so $\Pr(X = x) = \binom4x\tfrac{1}{16}$:

Check: the probabilities sum to $\tfrac{16}{16} = 1$.

b. [1 mark]. The draws are independent, so the first card being blue does not affect the next three. Exactly two red in three draws: RRB, RBR or BRR, each with probability $\left(\tfrac12\right)^3$:

c. [2 marks]. Same three arrangements, now each with probability $\left(\tfrac23\right)^2\left(\tfrac13\right)$:

From the report. Part a was generally done well, although a significant number did not recognise that the probabilities must sum to 1. In part b, tree diagrams worked well, and treating it as a conditional probability was acceptable. Part c was not answered well: many who set it up as a conditional probability did not divide by $\tfrac13$, or divided by $\tfrac12$ instead. Some found only $\Pr(RRB)$ and gave $\tfrac{4}{27}$.

a. [2 marks]. Write $100 = 10^2$ and equate indices:

b. [3 marks]. The argument of a logarithm must be positive:

This upright parabola is positive outside its roots $-1$ and $3$, so $x < -1$ or $x > 3$:

From the report. Part a was generally done well, though some wrote 100 incorrectly as a power of 10. In part b many factorised correctly but could not turn that into the right intervals; those who sketched the parabola usually could. A common error was writing an intersection, $(-\infty, -1) \cap (3, \infty)$, instead of a union.

a. [2 marks]. Reflecting in the horizontal axis gives $g(x) = -f(x) = 1 - 2\sin(2x)$ on $[0, 2\pi]$. Every point $(x, y)$ becomes $(x, -y)$:

• Endpoints: $(0, 1)$ and $(2\pi, 1)$.
• Minimums of $-1$ at $x = \tfrac{\pi}{4}$ and $\tfrac{5\pi}{4}$ (where $f$ had its maximums of 1).
• Maximums of 3 at $x = \tfrac{3\pi}{4}$ and $\tfrac{7\pi}{4}$ (where $f$ had its minimums of $-3$).
• The same four $x$-intercepts as $f$ (found in part b).

b. [3 marks]. $2\sin(2k) - 1 = 0$ gives $\sin(2k) = \tfrac12$. For $k \in [0, 2\pi]$, $2k \in [0, 4\pi]$, so look for four solutions (two per revolution, in quadrants 1 and 2):

c. i. [1 mark]

The translated graph is $y = 2\sin\big(2(x - a)\big) - 1 + b$. To match $g(x) = 1 - 2\sin(2x)$, the centre line must move from $y = -1$ to $y = 1$: $b = 2$.

ii. [1 mark]

We need $2\sin(2x - 2a) = -2\sin(2x)$. Since $\sin(\theta - \pi) = -\sin(\theta)$, take $2a = \pi$: $a = \tfrac{\pi}{2}$. (Half a period, $\tfrac{\pi}{2}$, is the smallest positive shift that flips the sine curve.)

iii. [1 mark]

The image must have domain $[0, 2\pi]$, the domain of $g$. Translating right by $\tfrac{\pi}{2}$ adds $\tfrac{\pi}{2}$ to every $x$-value, so

From the report. In part a some graphs ended at the wrong endpoint or had the wrong curvature, many were close to but not exactly on the correct $x$-intercepts, and many reflected $f$ in its centre line $y = -1$ instead of the horizontal axis. Use the grid. In part b most knew there were four solutions for $k$; the reference angle $\tfrac{\pi}{6}$ was common, but some used $\tfrac{\pi}{3}$ or $\tfrac{\pi}{4}$ or got the third and fourth angles wrong. Know the exact values for $0, \tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}, \tfrac{\pi}{2}$. In c.i and c.ii some confused the vertical and horizontal translations ($b = -2$, $b = \tfrac{\pi}{2}$, $a = \tfrac{\pi}{4}$). Part c.iii was not answered well (12%): students commonly translated the wrong way and gave $\left[\tfrac{\pi}{2}, \tfrac{5\pi}{2}\right]$.

a. i. [1 mark]. $20 \times 20 = 400$ cm$^2$.

ii. [1 mark]. The shaded area under $f$ must be half the tile, 200 cm$^2$. The sine term has period $\tfrac{2\pi}{\pi/10} = 20$, so over $[0, 20]$ it completes one full cycle and its integral is 0:

(So the curve is centred on the horizontal midline $y = 10$ of the tile.)

b. [3 marks].

The tile's area is 400 cm$^2$, and $200 = \tfrac12 \times 400$, so the region under $g$ is exactly half the front surface: a Type B tile meets Condition 1.

c. [2 marks]. Evaluate both rules at both edges of a tile (with $a = 10$):

• $f(0) = 4\sin(0) + 10 = 10$ and $f(20) = 4\sin(2\pi) + 10 = 10$.
• $g(0) = 10$ and $g(20) = -80 + 120 - 40 + 10 = 10$.

Every tile's dividing curve starts and ends at height 10, so the right edge of any tile meets the left edge of any other at the same point. The tiles can go in any order (AA, AB, BA, BB) and the pattern is continuous.

From the report. Part a.i was well done. In a.ii some found $a = 6$ by treating $\sin(0)$ or $\sin(2\pi)$ as 1, and some set up an integral for half the area; let the mark allocation guide how much work a 1-mark part needs. Part b is a "show that": it needed the link to half the tile's area, not just 200, and some wrongly tried to show the integral equals $\tfrac12$. In part c many checked only the right-hand end of each tile ($f(20)$ and $g(20)$) and not $f(0)$ and $g(0)$; others tried to match derivatives, which was not what was asked.

a. [1 mark].

b. [2 marks]. $A(k) = \int_0^k f(x)\,dx$, so by the fundamental theorem of calculus $f(k) = A'(k)$. Product rule:

c. [2 marks]. The average value of $f$ over $[0, k]$ is

$\sin(k)$ is largest when $k = \tfrac{\pi}{2}$, and $\tfrac{\pi}{2} \approx 1.57$ lies in $[0, 2]$. So $k = \dfrac{\pi}{2}$.

From the report. Part a was mostly well answered, but answers like $\tfrac{\pi}{3}\sin\tfrac{\pi}{3}$ had to be simplified. Part b depended on seeing that $f(k) = A'(k)$; students who did, and used the product rule, generally succeeded, while a common error was $A'(k) = k\cos(k)$. Part c was not answered well: many got $\sin(k)$ and then differentiated unnecessarily, or solved $\cos(k) = 0$ and wrote $k = 1$. Others found an average rate of change instead of an average value, integrated $A$ instead of $f$, or used terminals 0 and 2.