VCE Math Methods 2023 Exam 1

a. [2 marks]. Write $y = (x^2 - x)e^{-x}$ and use the product rule:

b. [2 marks]. $f'(x) = \cos(x)e^{2x} + 2\sin(x)e^{2x} = e^{2x}\big(\cos(x) + 2\sin(x)\big)$. At $x = \tfrac{\pi}{4}$:

From the report. Part a asked for a simplified answer: many students simplified the quadratic but left the exponential terms messy, and missing brackets led to sign errors when collecting like terms. In part b some mixed up the derivatives, or could not handle the surd arithmetic.

[3 marks]. Rearrange into a quadratic in $e^x$ and factorise:

$e^x = -2$ has no solution because $e^x > 0$. So $e^x = 6$ and $x = \log_e(6)$.

From the report. Most set up the quadratic correctly, but some did not go on to factorise and solve, and some gave $\log_e(-2)$ as a second answer. Logarithms are only defined for positive numbers.

a. [3 marks]. A hyperbola reflected in the $x$-axis (because of the minus sign), dilated by 3, and translated 1 right and 2 up.

• Asymptotes: $x = 1$ and $y = 2$.
• $y$-intercept: $f(0) = 2 + 3 = 5$, so $(0, 5)$.
• $x$-intercept: $\tfrac{3}{x - 1} = 2 \implies x = \tfrac52$, so $\left(\tfrac52, 0\right)$.

b. [1 mark]. From the graph, $f(x) = 1$ when $\tfrac{3}{x-1} = 1$, that is $x = 4$. The graph is at or below $y = 1$ only on the right branch between the asymptote and $x = 4$:

From the report. Part a was well done; the common slips were labelling the $y$-intercept as $(5, 0)$ and mislabelling the asymptotes. Part b (38% correct) was often answered $x \le 4$: use your graph to see that the left branch is always above 2.

[2 marks]. Width $h = 1$, with $f(1) = 2$, $f(2) = \tfrac52$ and $f(3) = \tfrac{10}{3}$:

From the report. Integrating was not acceptable because the question specifies trapeziums. Common errors were $f(2) = \tfrac32$ instead of $\tfrac52$, a wrong value for $f(3)$, using $\tfrac13$ for the strip-width factor, and fraction arithmetic with different denominators. Adding two separate trapezium areas worked well for many students.

a. [1 mark].

b. [3 marks]. $\displaystyle\int_k^{\pi/2}\cos(x)\,dx = \Big[\sin(x)\Big]_k^{\pi/2} = 1 - \sin(k)$. Setting this equal to $\tfrac12$ gives $\sin(k) = \tfrac12$.

Solutions are $k = \tfrac{\pi}{6} + 2n\pi$ or $k = \tfrac{5\pi}{6} + 2n\pi$. Those in $(-3\pi, 2\pi)$ are

(The next ones down, $-\tfrac{19\pi}{6}$ and $-\tfrac{23\pi}{6}$, are less than $-3\pi$.)

From the report. In part a some antidifferentiated $\sin(x)$ to $\cos(x)$ or used $\cos\!\left(\tfrac{\pi}{3}\right) = \tfrac{\sqrt3}{2}$. In part b, an error in part a led to an impossible or non-standard value of $\sin(k)$ and no reference angle; remember $\sin(k)$ can only lie in $[-1, 1]$.

a. [1 mark]. The interval is centred on $\hat p$: $\hat p = \dfrac{0.04 + 0.16}{2} = 0.1$.

b. [2 marks]. The margin of error is $0.06$:

c. [1 mark]. The width is proportional to $\tfrac{1}{\sqrt n}$. Multiplying $n$ by 4 multiplies the width by $\tfrac{1}{\sqrt4} = \tfrac12$: the width is halved.

From the report. In part a some gave values of $\hat p$ outside $[0, 1]$. In part b arithmetic slips led to the common wrong answers $n = 10$ and $n = 1000$. Part c was poorly done (23% correct); a factor of $\tfrac14$ was a common wrong answer.

a. [1 mark]

$f(x) = (x - 1)^2 - 1$ is decreasing on $(-\infty, 1]$ with minimum $f(1) = -1$. Range $[-1, \infty)$.

b. [2 marks]

Reflect the graph of $f$ in the line $y = x$. The endpoint $(1, -1)$ becomes $(-1, 1)$, the origin stays at $(0, 0)$, and, for example, $(-1, 3)$ becomes $(3, -1)$. The graph of $f^{-1}$ starts at $(-1, 1)$ and decreases slowly to the right. (It is drawn in orange in the figure for part d.)

c. [2 marks]

Swap $x$ and $y$ in $y = (x - 1)^2 - 1$:

The range of $f^{-1}$ is the domain of $f$, $y \le 1$, so take the negative root:

d. [2 marks]. The line $y = -x$ passes through both endpoints $(1, -1)$ and $(-1, 1)$ and the origin. It encloses two regions, mirror images in $y = x$, so they have equal area. On $[0, 1]$ the line is above $f$:

From the report. In part a the interval was sometimes written backwards or with the wrong brackets. In part c the most common error was the positive branch $1 + \sqrt{x + 1}$; use the graph from part b to choose the branch, and name the inverse $f^{-1}$. Part d was poorly done (19% full marks): students needed to see there are two equal areas, and many lost the signs by not bracketing $x^2 - 2x$ when subtracting it from $-x$. Some found an area without the line $y = -x$.

a. [1 mark]. Total probability is 1:

b. [2 marks].

c. [3 marks]. This is $\Pr(T > 2 \mid T > 1) = \dfrac{\Pr(T > 2)}{\Pr(T > 1)}$. Using $\int_a^4\left(16t - t^3\right)dt = \left[8t^2 - \tfrac{t^4}{4}\right]_a^4$:

From the report. In part a (a "show that") some left $dt$ out of the integral or mixed variables. In part c (average 0.7 out of 3) many read "already queued for one minute" incorrectly, used the wrong terminals, or struggled with the fractions; cancel factors of 64 early.

a. [1 mark]. $f(0) = a - 0 = a = 12$, and $g(1) = 12 + b = 9$ gives $b = -3$.

b. [2 marks]. $g'(x) = 12 - 6x = 0$ at $x = 2$, and $g(2) = 24 - 12 = 12$. For $f$:

so $f'(2) = 0$, and $f(2) = 12 - 0 = 12$. Both curves have a turning point at $P(2, 12)$.

c. [3 marks]. The triangle has base $k$ and height $g(k) = 12k - 3k^2$:

$A'(k) = 0$ for $k = \tfrac83$ in $(0, 4)$, and $A'$ changes from positive to negative there, so it is a maximum:

From the report. In part b it was not enough to show the curves meet at $P$: both derivatives had to be zero there. Some differentiated $x(x - 2)^2$ incorrectly or only checked $g$. Part c was poorly done (13% full marks); most who found $k = \tfrac83$ then made arithmetic errors substituting it into the area.