HSC Maths Advanced 2022

(a) [2 marks]. The cumulative frequency adds each frequency in turn. After stock shortage ($98$) and delivery fee ($62$):

The cumulative percentage is the cumulative frequency over the total. For the damaged-item row the cumulative frequency is $192$:

(b) [1 mark]. Reading across the Pareto chart, the cumulative percentage first reaches $80\%$ at the delivery-fee column, so the manager will address stock shortage and delivery fee.

Marker's note. Build the cumulative frequency one row at a time, then turn it into a percentage by dividing by the total of $200$. In (b) read the cumulative line of the Pareto chart, not the individual bars, and stop at the columns that together make up $80\%$.

(a) [2 marks]. Inverse variation means $M = \dfrac{k}{T}$. Substitute $T = 15$, $M = 12$:

(b) [2 marks]. Complete the table from the equation:

Plot the three points and join them with a single smooth curve (a hyperbola) that falls steeply and then levels off as $T$ increases.

Marker's note. Write the reciprocal relationship $M = k/T$ first, then use the given pair to find $k = 180$. In (b) read the grid scale carefully and join the plotted points with one smooth curve, not two ruled straight segments.

[2 marks]. Two applications need three function values at $x = 0, 1, 2$ with strip width $h = 1$:

Apply the trapezoidal rule $\dfrac{h}{2}\big[(y_0 + y_2) + 2y_1\big]$:

Marker's note. Two applications require three function values; substitute $x = 0, 1, 2$ and show the substitution into the formula. The trapezoidal rule estimates the integral directly, so no antiderivative is needed.

[2 marks]. The amplitude is the value of $k$, read from the graph as $k = 4$. The period of $\sin(ax)$ is $\dfrac{2\pi}{a}$, and the graph shows a period of $6\pi$:

So $k = 4$ and $a = \tfrac13$.

Marker's note. The vertical dilation equals the amplitude (a distance, so positive), and the horizontal scale comes from the period formula $\dfrac{2\pi}{a}$. Set that equal to $6\pi$ to solve for $a$.

(a) [1 mark]. Choosing an ordinary die has probability $\tfrac23$ (then $P(5) = \tfrac16$); choosing the special die has probability $\tfrac13$ (then $P(5) = \tfrac36$):

(b) [1 mark]. By the conditional probability formula:

Marker's note. In (a) split into the two ways a $5$ can appear and add the products of the probabilities. In (b) use the value from (a) as the denominator and the special-die path as the numerator.

[3 marks]. Between $x = -1$ and $x = 3$ the line is above the parabola, so

Evaluate at each limit:

The enclosed area is $\dfrac{32}{3} = 10\tfrac23$ square units.

Marker's note. Subtract the curves in the right order (line minus parabola) before integrating, then substitute the limits using brackets, taking care with the negative lower limit.

(a) [2 marks]. The rows form an arithmetic series with $a = 3$, $d = 3$. The total of $12$ rows is

(b) [3 marks]. Let $n$ be the number of rows. Set $S_n = 828$:

Factorising, $(n - 23)(n + 24) = 0$, so $n = 23$ (rejecting the negative root). There are 23 rows.

Marker's note. Write a few terms to confirm the series is arithmetic, then use the reference-sheet sum formula with $a = 3$, $d = 3$. In (b) form and solve the quadratic, accept the positive integer root, and remember that "total" means the sum, not a single term.

(a) [2 marks]. By the chain rule with inner function $x^2 + 1$:

(b) [1 mark]. From part (a), $\displaystyle\int 8x(x^2 + 1)^3\,dx = (x^2 + 1)^4 + C$, so dividing by $8$:

Marker's note. Apply the chain rule in (a). In (b) the word "hence" signals you should reverse part (a): the integrand is $\tfrac18$ of the derivative found, so the antiderivative is $\tfrac18(x^2+1)^4 + C$.

[3 marks]. Applying the transformations in order gives

Complete the square on the given quadratic:

Comparing $k(x - m)^2 - 5$ with $3(x - 2)^2 - 5$ gives $k = 3$ and $m = 2$.

Marker's note. Write $g(x)$ in the transformed form $k(x-m)^2 - 5$ first, then either expand and equate coefficients or complete the square of the non-monic quadratic. The vertex at $x = 2$ identifies $m$.

(a) [1 mark]. At $t = 0$, $N(0) = 200e^0 = 200$ bacteria.

(b) [1 mark]. At $t = 24$:

(c) [2 marks]. Differentiate: $\dfrac{dN}{dt} = 200\times 0.013\,e^{0.013t} = 2.6e^{0.013t}$. At $t = 24$:

Marker's note. Initial means $t = 0$ and $e^0 = 1$. In (c) the instantaneous rate is the derivative, so differentiate the exponential keeping the coefficient $0.013$, then substitute $t = 24$.

(a) [2 marks]. Monthly rate $= \dfrac{0.012}{12} = 0.001$, over $N = 10\times 12 = 120$ periods:

(b) [2 marks]. Option 2 compounds quarterly: $r = \dfrac{0.024}{4} = 0.006$ over $N = 10\times 4 = 40$ periods. From the table the factor is $45.05630$, so

The difference is $45\,097.17 - 45\,056.30 = \$40.87$.

Marker's note. In (a) adjust the rate and number of periods to monthly compounding and substitute clearly into the compound interest formula. In (b) read the correct cell ($r = 0.006$, $N = 40$), multiply by $\$1000$, and subtract to find the difference.

[4 marks]. Differentiate and find stationary points:

Evaluate $y$ at the stationary points and the endpoints:

• $x = -1$: $y = -1 - 6 + 8 = 1$
• $x = 0$: $y = 8$
• $x = 4$: $y = 64 - 96 + 8 = -24$
• $x = 7$: $y = 343 - 294 + 8 = 57$

The global maximum value is $57$ (at $x = 7$) and the global minimum value is $-24$ (at $x = 4$).

Marker's note. Global extremes occur at stationary points or endpoints, so compare the $y$-values of $x = 0, 4$ and the two endpoints. Show the substitutions as numerical expressions, and do not confuse global with local extremes.

(a) [2 marks]. The cosine ranges from $-1$ to $1$, so

(b) [1 mark]. The period is

(c) [3 marks]. Set $d = 1$ and solve:

So $\dfrac{4\pi}{25}t = \dfrac{\pi}{3}$ or $\dfrac{5\pi}{3}$, giving

The depth is at least $1$ metre between these times:

Marker's note. Use the amplitude to find the tide extremes and $\dfrac{2\pi}{a}$ for the period. In (c) write the equation for $d = 1$, find both relevant solutions in radians, and subtract carefully to get the length of time.

[4 marks]. Four distinct, contextual observations:

• The relationship is positive and roughly linear, but the correlation of $0.4564$ is only moderate, so age of character explains some, not all, of the variation in actor age.
• The gradient $1.85$ means that for each extra year of character age, the actor's age tends to rise by almost two years on average.
• Actors playing teenagers need not be teenagers: character ages span about $14$ to $17$, while actor ages span about $14$ to $30$, so older people often play teenage roles.
• The youngest actors tend to play characters close to their own age, while the wider spread of older actors shows casting is flexible above the youngest ages.

Marker's note. Use correct bivariate language (strength, direction, type) and refer to the correlation coefficient, the gradient and the context. A four-mark question expects about four accurate, unique statements rather than a single statistical measure.

[3 marks]. Differentiate twice: $f'(x) = 2\cos 2x$ and $f''(x) = -4\sin 2x$. Work in the range $0 < 2x < 2\pi$.

From $f'(x) = -\sqrt3$:

From $f''(x) = 2$:

The value satisfying both conditions is $x = \dfrac{7\pi}{12}$.

Marker's note. Differentiate $\sin 2x$ twice using the reference sheet, change the domain to $0 < 2x < 2\pi$, and find both solutions of each equation. The word AND means you take the common value, here $\dfrac{7\pi}{12}$. Answer in radians.

[3 marks]. Since $60\%$ have a life span below $860$ hours and $50\%$ lie below the mean of $840$, about $10\%$ lie between $840$ and $860$. By symmetry $10\%$ also lie between $820$ and $840$.

The score $920$ is one standard deviation above the mean ($840 + 80$). About $68\%$ lie within one standard deviation ($760$ to $920$), so $34\%$ lie between $840$ and $920$.

Adding the piece below the mean:

Marker's note. Use the given $60\%$ to find the $10\%$ between $820$ and $840$, and the empirical $68\%$ (halved to $34\%$) for the slice from $840$ to $920$. Sketch the normal curve, and distinguish percentage from number of batteries.

(a) [2 marks]. Differentiate $f'(x) = e^{-2x} - 2xe^{-2x}$, using the product rule on the second term:

(b) [2 marks]. Stationary points where $f'(x) = 0$:

Then $f(\tfrac12) = \tfrac12 e^{-1} = \dfrac{1}{2e}$. Testing with the second derivative, $f''(\tfrac12) = 4(-\tfrac12)e^{-1} = -2e^{-1} < 0$, so $\left(\tfrac12, \dfrac{1}{2e}\right)$ is a local maximum.

(c) [3 marks]. A point of inflection occurs where $f''(x) = 0$, that is $x = 1$, with $f(1) = e^{-2} = \dfrac{1}{e^2}$; concavity changes there. The curve passes through the origin $(0, 0)$, rises to the maximum at $\left(\tfrac12, \dfrac{1}{2e}\right)$, has an inflection at $\left(1, \dfrac{1}{e^2}\right)$, and approaches the positive $x$-axis as $x \to \infty$.

Marker's note. In (a) apply the product rule and handle the negative signs carefully before factorising. In (b) classify the point as a maximum (not just "stationary"). In (c) show the positive $x$-axis is an asymptote, distinguish the inflection from a horizontal inflection, and draw a smooth labelled curve.

(a) [2 marks]. The point $(1, 1)$ gives $\theta = \dfrac{\pi}{4}$, and the radius is $\sqrt2$. The shaded region is the sector minus the triangle:

(b) [2 marks]. Substitute $(0, 0)$:

Substitute $(1, 1)$ and use $a = b$:

So $a = b = 2$.

(c) [3 marks]. With $a = b = 2$ the hyperbola is $y = \dfrac{2}{2 - x} - 1$. The area under it from $x = 0$ to $x = 1$ is

Adding the circular region from part (a):

Marker's note. In (a) work in radians, find $\theta = \dfrac{\pi}{4}$, and subtract the triangle from the sector. In (b) do not use the given answer to prove itself; substitute both points. In (c) the primitive of $-1$ is $-x$, and "using part (a)" means add that area without recomputing it.

(a) [1 mark]. The rectangle areas are $2^0 + 2^{-1} + 2^{-2} + \cdots$, a geometric series with first term $a = 1$ and ratio $r = \tfrac12$:

(b) [2 marks]. Using the reference sheet, $\displaystyle\int 2^{-x}\,dx = \dfrac{2^{-x}}{-\ln 2}$:

(c) [2 marks]. The first four rectangles (from $x = 0$ to $x = 4$) overestimate the area under the decreasing curve, and their total is less than the full limiting sum $2$. So

Taking exponentials of both sides gives $e^{15} < 2^{32}$.

Marker's note. State $a$ and $r$ clearly in (a). In (b) integrate the exponential from the reference sheet and manage the negative index and limits. In (c) compare the rectangle sum and the integral, then rearrange with index and logarithm laws.

(a) [1 mark]. A cumulative distribution function reaches $1$ at the top of its range, so $F(e^3) = 1$:

(b) [2 marks]. Using $P(X > c) = 1 - P(X < c)$:

So $F(c) = \dfrac23$:

Marker's note. In (a) recognise that a CDF equals $1$ at the upper limit; solve for $k$ rather than substituting $k = 3$. In (b) use $P(X > c) = 1 - P(X < c)$, simplify to $P(X < c) = \tfrac23$, then solve the logarithmic equation.

(a) [2 marks]. The points $X$, $P$ and $Y$ are collinear, so the gradient from $X$ to $P$ equals the gradient from $P$ to $Y$:

Rearranging:

(b) [4 marks]. The triangle has area $A = \tfrac12\,x\,y = \dfrac{x^2}{x - 1}$. Differentiate using the quotient rule:

Setting $\dfrac{dA}{dx} = 0$ gives $x = 0$ or $x = 2$; reject $x = 0$ since $x > 1$. The gradient changes from negative to positive at $x = 2$, so it is a minimum. The minimum area is

Marker's note. In (a) equate two equivalent gradient (or ratio) expressions and manipulate to the required form without using the answer. In (b) write $A$ in terms of $x$, differentiate with the quotient rule, test the nature of the stationary point, then substitute back to find the area.

(a) [2 marks]. The bracket is a geometric series with ratio $1.0025$ over $180$ terms. Setting $A_{180} = 0$:

Making $M$ the subject:

(b) [3 marks]. With $P = 100\,032$, $r = 1.0035$ and $M = 1381.16$, set the amount owing to $0$:

Expanding and collecting the $(1.0035)^n$ terms:

So Jane makes 83 full monthly repayments.

(c) [2 marks]. After $83$ repayments the amount owing is

This accrues one more month of interest, so the final ($84$th) payment is

Marker's note. In (a) recognise the geometric sum, select the reference-sheet formula, make $M$ the subject, and confirm the given value (do not separately prove the supplied formula). In (b) substitute the new rate and principal, collect the exponential terms, and take logarithms to find $n = 83$ full payments. In (c) compute the residual owing after $83$ payments and add one month of interest.