HSC Maths Advanced 2023

[2 marks]. The first term is $a = 3$ and the common difference is $d = 7 - 3 = 4$. Using $T_n = a + (n-1)d$:

Marker's note. Use the arithmetic-sequence formula from the reference sheet, $T_n = a + (n-1)d$, and substitute carefully. If listing all fifteen terms instead, make sure every term is correct.

(a) [1 mark]. $E(X) = \sum x\,P(x)$:

(b) [2 marks]. First find $E(X^2) = \sum x^2 P(x)$:

Then $\text{Var}(X) = E(X^2) - [E(X)]^2 = 4.8 - 2^2 = 0.8$, so

Marker's note. In (a) a show question needs the full sum written out. In (b) use $\text{Var}(X) = E(X^2) - \mu^2$, then take the square root for the standard deviation; do not stop at the variance.

[2 marks]. Integrate with respect to $t$:

When $t = 0$, $P = 4000$, so $4000 = 1500e^{0} + C = 1500 + C$, giving $C = 2500$. Hence

Marker's note. Divide by the coefficient of $t$ when integrating the exponential, then substitute $t = 0$ and use $e^0 = 1$ to find $C$ by subtraction (not division).

[3 marks]. Differentiate using the chain rule:

At $x = 0$, $y' = 6(1)^2 = 6$, so the gradient of the tangent is 6. Using $y - y_1 = m(x - x_1)$ at $(0, 1)$:

Marker's note. The tangent needs a numerical gradient, not an expression in $x$, so substitute $x = 0$ before writing the line. Use the chain rule rather than expanding the cube.

(a) [2 marks]. For 10 periods at 6%, the table factor is $13.181$. The future value equals the contribution times this factor, so

(b) [3 marks]. Quarterly compounding gives rate $\tfrac{6\%}{4} = 1.5\%$ and $n = 10 \times 4 = 40$ periods. The table factor for 40 periods at 1.5% is $54.268$, so

Marker's note. In (a) divide the target by the factor (do not multiply). In (b) convert both the rate and the number of periods to quarterly values before reading the correct factor, then multiply by the deposit.

[4 marks]. The perimeter is the two sides $BC$ and $AD$, the side $DC$, the two straight segments of $AB$ outside the arc, and the arc $PQ$ in place of the chord $PQ$.

The arc length is

The chord $PQ$ (by the cosine rule) is

The straight part of $AB$ is $8.0 - 3.4404 = 4.5596$ m. So

Marker's note. The arc is $\tfrac{110}{360}$ of a circle, not a semicircle, and its radius is $2.1$ m. Find both the arc length and the chord, take the square root in the cosine rule, and add only the parts that form the perimeter (no area formulas are needed).

[2 marks]. This is of the form $\int f'(x)[f(x)]^n\,dx$ with $f(x) = x^2 + 1$, since $f'(x) = 2x$. Insert the factor of 2:

Marker's note. Recognise that the integrand is a derivative times a power, adjust by the constant $\tfrac12$, raise the power by one and divide. Use the reference sheet for the integral form.

(a) [3 marks]. Compute the means:

Plot the mean point $(5, 184)$ and the $y$-intercept $(0, 236)$ on the grid.

(b) [2 marks]. The line passes through $(0, 236)$ and $(5, 184)$, so its gradient is

With $y$-intercept 236, the equation is

(c) [1 mark]. At $x = 23$ the line gives $y = 236 - 10.4(23) = -3.2$ MW, a negative gas usage, which is not physically possible. This is also extrapolation well outside the data range $0$ to $10$ degrees C.

Marker's note. Read the initial value 236 as the $y$-intercept. Find the gradient from rise over run between the two known points, write an equation (not an expression), and link the part (c) answer to the context: a negative predicted usage is impossible.

(a) [2 marks]. The line $f(x) = x - 1$ crosses the $x$-axis at $(1, 0)$. The parabola $g(x) = (1 - x)(3 + x)$ is concave down (the coefficient of $x^2$ is negative) with $x$-intercepts at $x = 1$ and $x = -3$.

(b) [2 marks]. The curves meet where

Factorising, $(x + 4)(x - 1) = 0$, so $x = -4$ or $x = 1$. The line is below the parabola between these roots, so the solution is

Marker's note. Hence means use the sketch: the line is below the parabola between the intersection points. Expand the product, solve the quadratic, and express the answer as a combined inequality.

[3 marks]. Adjust the domain for the shifted angle: if $0\degree \le \theta \le 360\degree$, then $-60\degree \le \theta - 60\degree \le 300\degree$.

Since $\sin$ takes the value $-\tfrac{\sqrt 3}{2}$ (related angle $60\degree$), the solutions for $\theta - 60\degree$ in the third and fourth quadrants, within the adjusted domain, are

Adding $60\degree$:

Marker's note. Adjust the domain to match $\theta - 60\degree$, then find the related angle and the correct quadrants. Give every solution in degrees, and remember $0\degree$ and $360\degree$ are both valid endpoints.

[3 marks]. With first term $a$ and ratio $r$:

Dividing (2) by (1):

If $r = \tfrac14$: $a\left(\tfrac14\right)^3 = 48$, so $a = 48 \times 64 = 3072$.

If $r = -\tfrac14$: $a\left(-\tfrac14\right)^3 = 48$, so $a = -3072$.

Marker's note. Set up two equations from the term formula, then divide to eliminate $a$. An even power gives two values of $r$, so state both ratios and their matching first terms.

[3 marks]. Work with the right-angled triangle $AEM$, where $AE = 8$ is vertical and $AM$ lies in the base.

In the base, $AM$ has perpendicular components $AD = 7$ and $DM = \tfrac12 \times 6 = 3$ (since $M$ is the midpoint of $CD$ and $CD = EF = 6$), so

In right-angled triangle $AEM$, with the right angle at $A$:

Marker's note. View the prism in three dimensions and pick the useful right-angled triangle. Find $AM$ first with Pythagoras, keep the exact surd, then use the tangent ratio. Do not assume any angles from the not-to-scale diagram.

[4 marks]. Standardise $x = 11.93$:

From the table $P(Z < 1.33) = 0.9082$, so

The expected number is

Marker's note. Calculate the $z$-score, then use the table for the probability below the value and subtract from 1 for the upper tail. Finish by multiplying by 400 to answer what the question actually asks for.

(a) [1 mark]. The garden has dimensions $(x - 2)$ by $(y - 1)$, because the 1 m path removes 1 m on each of two sides in the $x$-direction and 1 m on one side in the $y$-direction. So

(b) [4 marks]. The area of the path is the outer rectangle minus the garden, which simplifies to $2y + x - 2$. Substituting for $y$:

Differentiate:

Setting $A' = 0$ gives $(x - 2)^2 = 100$, so $x - 2 = \pm 10$ and $x = 12$ or $x = -8$. Since $x$ is a distance, take $x = 12$.

Testing the first derivative either side ($A' < 0$ at $x = 11$ and $A' > 0$ at $x = 13$) confirms a minimum turning point, so the path area is least when $x = 12$.

Marker's note. In (a) keep the brackets $(x - 2)$ and $(y - 1)$ and rearrange, rather than solving for a number. In (b) build the path-area expression, differentiate carefully with negative indices, reject the negative root, and justify the minimum with a first or second derivative test.

(a) [1 mark]. After one month, $A_1 = 10\,000(1.004) - M$. After two months the balance grows by another factor of $1.004$, then $M$ is withdrawn:

(b) [3 marks]. Extending the pattern, the withdrawals form a geometric series:

The bracket sums to $\dfrac{1.004^n - 1}{0.004}$, so

Since $\dfrac{1}{0.004} = 250$, expanding and grouping gives

(c) [2 marks]. To make at least 100 withdrawals, require $A_{100} > 0$:

With $(1.004)^{100} = 1.49063\ldots$:

The largest amount Jia could withdraw is $121.52.

Marker's note. In (a) and (b) build the series term by term and apply the geometric-series sum, using $\tfrac{1}{0.004} = 250$. In (c) substitute $n = 100$ with $A_{100} > 0$, then rearrange carefully to bound $M$.

(a) [2 marks]. Integrate with respect to $t$:

At $t = 0$, $x = 11.2$, so $11.2 = 1.2\cos 0 + k = 1.2 + k$, giving $k = 10$. Hence

(b) [2 marks]. The closest point occurs once per period. The period is

In 10 seconds there are $10 \div 1.6 = 6.25$ periods, so the swing reaches the closest point 6 times (the complete cycles).

Marker's note. In (a) divide by the coefficient of $t$ when integrating, watch the sign of the cosine, and use the initial condition for $k$. In (b) find the period and count the complete cycles within 10 seconds.

(a) [3 marks]. The vertex of the absolute-value graph is at the highest point $(6, 7)$, so the horizontal shift is $b = 6$ and the vertical shift is $c = 7$. Substitute a known point, $(3, -5)$:

So $a = -4$, $b = 6$, $c = 7$.

(b) [2 marks]. The line $y = mx$ passes through the origin. The line through $(0, 0)$ and the vertex $(6, 7)$ has gradient $\tfrac{7}{6}$; for the line to cut the graph twice, $m$ must be less than $\tfrac{7}{6}$. The right arm of the graph has gradient $-4$, so $m$ must be greater than $-4$ to meet the graph twice. Hence

Marker's note. In (a) read the translations from the vertex, then substitute a point to find the dilation $a$ (which is negative here). In (b) reason with the two boundary gradients from the origin rather than solving simultaneous equations.

[4 marks]. The tangent at $T$ has gradient 1, so at $R$ the gradient is also 1:

So $x = 3$ or $x = -1$; since $x = -1$ is point $T$, the point $R$ has $x = 3$.

To find $y$, integrate the gradient function:

Using $T(-1, 6)$: $6 = -1 - 3 + 8 + k$, so $k = 2$. Then at $x = 3$:

So $R = (3, -22)$.

Marker's note. Parallel tangents share a gradient, so set the gradient function equal to 1 and solve, discarding the root at $T$. Integrate (remembering $+k$) and use the known point to find $k$ before substituting the $x$-coordinate of $R$.

(a) [2 marks]. The mode is where $f(x)$ is greatest. Expand and differentiate:

$f'(x) = 0$ at $x = 0$ and $x = \tfrac23$. Since $f(0) = 0$, discard it, so the mode is $x = \tfrac23$.

(b) [2 marks]. Integrate from 0 to $x$:

with $F(x) = 0$ for $x < 0$ and $F(x) = 1$ for $x > 1$.

(c) [2 marks]. Substitute the mode $x = \tfrac23$ into $F$:

Since $F\!\left(\tfrac23\right) = 0.59 > 0.5$, more than half the probability lies below the mode, so the median (where $F = 0.5$) is less than the mode. Hence the mode is greater than the median.

Marker's note. In (a) expand before differentiating and reject $x = 0$. In (b) integrate the density with limits 0 to $x$. In (c) compare $F(\text{mode})$ with 0.5: a value above 0.5 means the mode lies past the median.

(a) [3 marks]. By the product rule:

Since $e^{-x} \ne 0$, set $\cos x = \sin x$, which on $0 \le x \le 2\pi$ gives $x = \tfrac{\pi}{4}$ or $x = \tfrac{5\pi}{4}$. The $y$-values are

So the stationary points are about $\left(\tfrac{\pi}{4}, 0.322\right)$ and $\left(\tfrac{5\pi}{4}, -0.014\right)$.

(b) [2 marks]. The intercepts occur where $\sin x = 0$, that is $x = 0, \pi, 2\pi$. Sketch a smooth curve starting at the origin, rising to the maximum near $\tfrac{\pi}{4}$, falling through $(\pi, 0)$ to a small minimum near $\tfrac{5\pi}{4}$, then rising back to $(2\pi, 0)$.

Marker's note. Use the product rule and note $e^{-x} = 0$ has no solutions, so solve $\cos x = \sin x$ over the whole radian domain and give exact or accurate $y$-values. In (b) draw one smooth curve through the intercepts at $0, \pi, 2\pi$ and label the turning points.

(a) [1 mark]. The events are not independent, because $P(F\mid S) = \tfrac18 \ne P(F) = \tfrac{3}{10}$. Knowing Kim is available on Saturday changes the probability he is available on Friday.

(b) [2 marks]. From $P(S\mid F) = \dfrac{P(S \cap F)}{P(F)}$:

Since $P(S \cap F) = P(F \cap S)$, use $P(F\mid S) = \dfrac{P(F \cap S)}{P(S)}$:

(c) [2 marks]. Each student is available on Saturday with probability $\tfrac45$. The complement of all four being available is at least one not available:

Marker's note. In (a) compare $P(F\mid S)$ with $P(F)$. In (b) find the intersection from one conditional, then use that $P(S \cap F) = P(F \cap S)$ in the other. In (c) use the complement of all four available, raising $\tfrac45$ to the fourth power.

(a) [3 marks]. From $x = 0$ to $x = \ln 2$ the curve $y = e^{-2x}$ is above $y = e^{-x} - \tfrac14$, so

Integrating:

At $x = \ln 2$, $e^{-x} = \tfrac12$ and $e^{-2x} = \tfrac14$:

(b) [3 marks]. Set $e^{-2x} = e^{-x} + k$. Let $u = e^{-x}$ (so $u > 0$):

For two real solutions the discriminant must be positive: $1 + 4k > 0$, so $k > -\tfrac14$. For both roots to be positive (so each gives a real $x$), the product of roots $-k$ must be positive, so $k < 0$. Hence