HSC Maths Extension 1 2023

(a) [2 marks]

From $x = 1 + 3t$, $t = \dfrac{x - 1}{3}$. Substitute into $y = 4t$:

$$y = 4 \cdot \frac{x - 1}{3} = \frac{4}{3}x - \frac{4}{3}.$$

(b) [2 marks]

CONDOBOLIN has 10 letters with 3 Os and 2 Ns repeated:

$$\frac{10!}{3!\,2!} = \frac{3\,628\,800}{12} = 302\,400.$$

(c) [3 marks]

By the factor theorem $P(-1) = 0$:

$$-1 + a - b - 12 = 0 \implies a - b = 13. \quad (1)$$

By the remainder theorem $P(2) = -18$:

$$8 + 4a + 2b - 12 = -18 \implies 4a + 2b = -14 \implies 2a + b = -7. \quad (2)$$

Adding $(1)$ and $(2)$: $3a = 6$, so $a = 2$, and then $b = a - 13 = -11$.

(d) [2 marks]

Rewrite to match the inverse-sine form $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\tfrac{x}{a}$:

$$\int \frac{1}{\sqrt{4 - 9x^2}}\,dx = \frac13 \int \frac{3}{\sqrt{2^2 - (3x)^2}}\,dx = \frac13 \sin^{-1}\!\left(\frac{3x}{2}\right) + C.$$

(e) [3 marks]

Use the auxiliary-angle form. Write $\cos\theta + \sin\theta = \sqrt{2}\,\sin\!\left(\theta + \tfrac{\pi}{4}\right)$, so

$$\sqrt{2}\,\sin\!\left(\theta + \tfrac{\pi}{4}\right) = 1 \implies \sin\!\left(\theta + \tfrac{\pi}{4}\right) = \frac{1}{\sqrt{2}}.$$

For $0 \le \theta \le 2\pi$, the argument $\theta + \tfrac{\pi}{4}$ ranges over $[\tfrac{\pi}{4}, 2\pi + \tfrac{\pi}{4}]$, giving $\theta + \tfrac{\pi}{4} = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{9\pi}{4}$. Hence

$$\theta = 0, \ \frac{\pi}{2}, \ 2\pi.$$

(f)(i) [2 marks]

Let $p = 0.3$, so $1 - p = 0.7$ and $n = 900$. The sample proportion $\hat{p}$ is approximately normal with

$$\text{Var}(\hat{p}) = \frac{p(1-p)}{n} = \frac{0.3 \times 0.7}{900} = \frac{0.21}{900} = \frac{7}{30\,000}.$$

(f)(ii) [2 marks]

The standard deviation is $\sqrt{\tfrac{7}{30\,000}} \approx 0.01528$. The $z$-score is

$$z = \frac{0.31 - 0.3}{0.01528} \approx 0.65.$$

From the table $P(Z \le 0.65) = 0.7422$, so $P(\hat{p} \le 0.31) \approx 0.74$.

Marker's note. In (a) change the subject for $t$ and eliminate the parameter cleanly. In (b) identify the repeated letters (the Os and Ns) and divide by their factorials. In (c) apply both the factor and remainder theorems to form two equations, then solve them simultaneously. In (d) rewrite the integrand to match the Reference Sheet inverse-sine form before integrating. In (e) name the correct auxiliary or $t$ formula and solve over the full domain. In (f) connect the given information to $\tfrac{p(1-p)}{n}$, then compute the $z$-score and read the table correctly.

(a) [3 marks]. With $u = x - 3$, $x = u + 3$, $du = dx$, and the limits change to $u = 0$ (at $x = 3$) and $u = 1$ (at $x = 4$):

Integrating:

(b) [3 marks]. Let $P(n)$ be the given statement.

Base case $n = 1$: LHS $= 1 \times 2 = 2$; RHS $= 2 + (1 - 1)2^2 = 2$. So $P(1)$ holds.

Inductive step. Assume $P(k)$: $(1 \times 2) + \cdots + (k \times 2^k) = 2 + (k - 1)2^{k+1}$. Then

Combine the last two terms:

This is $P(k+1)$, so by induction $P(n)$ holds for all integers $n \ge 1$.

(c)(i) [2 marks]

Each treadmill is independently in use with probability $0.65$, so the count in use is binomial with $n = 5$:

$$\binom{5}{3}(0.65)^3(0.35)^2.$$

(c)(ii) [1 mark]

No rowing machine in use has probability $(0.6)^4$, independent of the treadmills:

$$\binom{5}{3}(0.65)^3(0.35)^2 \times (0.6)^4.$$

(d) [2 marks]

Combine the first two terms using the given identity:

$$\binom{2022}{80} + \binom{2022}{81} = \binom{2023}{81}.$$

Also $\binom{2023}{1943} = \binom{2023}{2023 - 1943} = \binom{2023}{80}$. Then

$$\binom{2023}{81} + \binom{2023}{80} = \binom{2024}{81}.$$

So $p = 2024$ and $q = 81$ is one possible solution.

(e) [4 marks]

Rotating about the $y$-axis, take the total as the hyperbola's solid plus a cylinder. From $y = \dfrac{60}{x + 5}$, $x = \dfrac{60}{y} - 5$, and the curve runs from $y = 4$ (at $x = 10$) to $y = 12$ (at $x = 0$). The hyperbola volume is

$$V_1 = \pi \int_4^{12} x^2\,dy = \pi \int_4^{12} \left(\frac{60}{y} - 5\right)^2 dy = \pi \int_4^{12} \left(\frac{3600}{y^2} - \frac{600}{y} + 25\right)dy.$$

Integrating:

$$V_1 = \pi \left[-\frac{3600}{y} - 600\ln y + 25y\right]_4^{12} = \pi\big(800 - 600\ln 12 + 600\ln 4\big) = \pi(800 - 600\ln 3).$$

The cylinder below $y = 4$ has radius $10$ and height $4$: $V_2 = \pi \times 10^2 \times 4 = 400\pi$. Hence

$$V = V_1 + V_2 = \pi(1200 - 600\ln 3)\text{ units}^3.$$

Marker's note. In (a) keep brackets around $(u + 5)$, change the limits to $u$ values, and integrate the fractional powers carefully. In (b) test $n = k + 1$ on both sides, use the index laws, and write a clear conclusion. In (c) apply binomial probability and, in (ii), include the independent rowing-machine factor. In (d) use the symmetry $\binom{n}{r} = \binom{n}{n-r}$ together with Pascal's rule. In (e) split the solid into the hyperbola revolution plus a cylinder, write $x$ in terms of $y$, and apply the logarithm laws.

(a)(i) [2 marks]

Differentiate $V$ with respect to $h$: $\dfrac{dV}{dh} = \pi(2Rh - h^2) = \pi h(2R - h)$. By the chain rule $\dfrac{dV}{dt} = \dfrac{dV}{dh}\cdot\dfrac{dh}{dt}$, so

$$k(2R - h) = \pi h(2R - h)\frac{dh}{dt}.$$

Since $0 \le h \le R$, $2R - h \ne 0$; dividing gives $\dfrac{dh}{dt} = \dfrac{k}{\pi h}$.

(a)(ii) [2 marks]

Separate the variables, with $h = 0$ at $t = 0$ and $h = R$ when the tank is full at $t = T$:

$$\int_0^R \pi h\,dh = \int_0^T k\,dt \implies \left[\frac{\pi h^2}{2}\right]_0^R = kT \implies \frac{\pi R^2}{2} = kT.$$

Hence $T = \dfrac{\pi R^2}{2k}$.

(a)(iii) [3 marks]

Once inflow stops, $\dfrac{dV}{dt} = -kh$ (only the drainage term remains). Then

$$-kh = \frac{dV}{dh}\cdot\frac{dh}{dt} = \pi h(2R - h)\frac{dh}{dt} \implies \frac{dh}{dt} = \frac{-k}{\pi(2R - h)}.$$

Separate, with $h = R$ at $t = 0$ and $h = 0$ at $t = T_E$:

$$\int_R^0 \pi(2R - h)\,dh = \int_0^{T_E} -k\,dt \implies \left[\pi\left(2Rh - \frac{h^2}{2}\right)\right]_R^0 = -kT_E.$$

The left side is $-\pi\left(2R^2 - \tfrac{R^2}{2}\right) = -\tfrac{3\pi R^2}{2}$, so $\tfrac{3\pi R^2}{2} = kT_E$, giving $T_E = \dfrac{3\pi R^2}{2k} = 3T$. The tank takes 3 times as long to empty.

(b)(i) [2 marks]

Since $\tan\theta = 2$ with $\theta$ in the first quadrant, the right triangle has opposite $2$, adjacent $1$ and hypotenuse $\sqrt{5}$, so $\cos\theta = \dfrac{1}{\sqrt{5}}$. The particles collide, so their $x$-coordinates are equal: $vt\cos\theta = ut$, giving $u = v\cos\theta = \dfrac{v}{\sqrt{5}}$. Hence $v = \sqrt{5}\,u$.

(b)(ii) [1 mark]

At collision the $y$-coordinates are equal:

$$vt\sin\theta - \tfrac12 gt^2 = H - \tfrac12 gt^2 \implies vt\sin\theta = H.$$

With $\sin\theta = \dfrac{2}{\sqrt{5}}$ and $v = \sqrt{5}\,u$, $vt\sin\theta = \sqrt{5}\,u \cdot t \cdot \dfrac{2}{\sqrt{5}} = 2ut$, so $2uT = H$ and $T = \dfrac{H}{2u}$.

(b)(iii) [3 marks]

The velocity vectors are $\underline{v}_A = \begin{pmatrix} v\cos\theta \\ v\sin\theta - gt \end{pmatrix}$ and $\underline{v}_B = \begin{pmatrix} u \\ -gt \end{pmatrix}$. Perpendicular at $t = T$ means $\underline{v}_A \cdot \underline{v}_B = 0$:

$$uv\cos\theta - gT(v\sin\theta - gT) = 0.$$

Using $v\cos\theta = u$, $v\sin\theta = 2u$:

$$u^2 - gT(2u - gT) = 0 \implies u^2 - 2ugT + g^2T^2 = 0 \implies (u - gT)^2 = 0,$$

so $gT = u$. With $T = \dfrac{H}{2u}$, $\dfrac{gH}{2u} = u$, giving $H = \dfrac{2u^2}{g}$.

(b)(iv) [2 marks]

The vertex of A's path is where the vertical velocity is zero: $v\sin\theta - gt = 0$, so $t = \dfrac{v\sin\theta}{g} = \dfrac{2u}{g}$. The height there is

$$y = vt\sin\theta - \tfrac12 gt^2 = 2u\cdot\frac{2u}{g} - \frac12 g\left(\frac{2u}{g}\right)^2 = \frac{4u^2}{g} - \frac{2u^2}{g} = \frac{2u^2}{g} = H.$$

Marker's note. In (a)(i) find $\tfrac{dV}{dh}$, link the rates by the chain rule, and cancel $2R - h$ with justification. In (a)(ii) and (iii) separate the variables matching the integration variable, evaluate the constants, and show every step. In (b)(i) find $\cos\theta$ in exact form and equate horizontal displacements. In (b)(iii) differentiate to get the velocities, remember $H$ is constant so it drops out, and simplify the dot product to a perfect square. In (b)(iv) set the vertical velocity to zero and simplify the height carefully in terms of $H$.

(a)(i) [1 mark]

For $x > 0$, $f'(x) = 2 + \dfrac{1}{x} > 0$, so $f$ is strictly increasing and therefore one-to-one. A one-to-one function has an inverse that is itself a function.

(a)(ii) [2 marks]

Since $f(1) = 2(1) + \ln 1 = 2$, we have $g(2) = 1$. Using $g'(x) = \dfrac{1}{f'(g(x))}$,

$$g'(2) = \frac{1}{f'(g(2))} = \frac{1}{f'(1)} = \frac{1}{2 + 1} = \frac{1}{3}.$$

(b)(i) [1 mark]

A point of intersection satisfies $y = \dfrac{1}{x}$ and $(x - c)^2 + y^2 = c^2$. Substituting,

$$(x - c)^2 + \frac{1}{x^2} = c^2 \implies x^2 - 2cx + c^2 + \frac{1}{x^2} = c^2.$$

Multiply by $x^2$ and simplify:

$$x^4 - 2cx^3 + 1 = 0,$$

so the $x$-coordinates are zeros of $P(x) = x^4 - 2cx^3 + 1$.

(b)(ii) [3 marks]

A single intersection point means $P(x)$ has a double root, where $P(x) = 0$ and $P'(x) = 0$. Now

$$P'(x) = 4x^3 - 6cx^2 = 2x^2(2x - 3c),$$

so $P'(x) = 0$ at $x = 0$ or $x = \dfrac{3c}{2}$. Since $P(0) = 1 \ne 0$, the double root is $x = \dfrac{3c}{2}$. Substituting into $P(x) = 0$:

$$\left(\frac{3c}{2}\right)^4 - 2c\left(\frac{3c}{2}\right)^3 + 1 = 0 \implies \frac{81c^4}{16} - \frac{54c^4}{8} + 1 = 0.$$

This gives $-\dfrac{27c^4}{16} + 1 = 0$, so $c^4 = \dfrac{16}{27}$ and, taking $c > 0$,

$$c = \sqrt[4]{\frac{16}{27}} = \frac{2}{\sqrt[4]{27}}.$$

(c)(i) [3 marks]

Let $\theta$ be the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$, so the area of triangle $OAB$ is $\tfrac12 |\overrightarrow{OA}||\overrightarrow{OB}|\sin\theta$. Let $\overrightarrow{OC} = \begin{pmatrix} a_2 \\ -a_1 \end{pmatrix}$, which by the given information is perpendicular to $\overrightarrow{OA}$ with the same magnitude. Then

$$\overrightarrow{OB} \cdot \overrightarrow{OC} = b_1 a_2 + b_2(-a_1) = a_2 b_1 - a_1 b_2.$$

The angle between $\overrightarrow{OB}$ and $\overrightarrow{OC}$ is $\tfrac{\pi}{2} \pm \theta$, so $\overrightarrow{OB} \cdot \overrightarrow{OC} = |\overrightarrow{OB}||\overrightarrow{OC}|\cos(\tfrac{\pi}{2} \pm \theta) = \mp|\overrightarrow{OA}||\overrightarrow{OB}|\sin\theta$. Taking magnitudes,

$$|a_2 b_1 - a_1 b_2| = |\overrightarrow{OA}||\overrightarrow{OB}|\sin\theta = 2 \times \text{Area}.$$

Hence the area of triangle $OAB$ is $\dfrac12 |a_1 b_2 - a_2 b_1|$.

(c)(ii) [4 marks]

From the geometry, $\overrightarrow{OP} = \begin{pmatrix} r + r\cos t \\ r\sin t \end{pmatrix}$ and $\overrightarrow{OQ} = \begin{pmatrix} -R + R\cos 2t \\ R\sin 2t \end{pmatrix}$. By part (i),

$$\text{Area} = \frac12\left|(r + r\cos t)(R\sin 2t) - (r\sin t)(-R + R\cos 2t)\right| = \frac{Rr}{2}\left|\sin 2t + \sin t\right|,$$

after expanding and using $\sin 2t\cos t - \sin t\cos 2t = \sin t$. Let $f(t) = \sin 2t + \sin t$. Then

$$f'(t) = 2\cos 2t + \cos t = 2(2\cos^2 t + \cos t - 1) = 2(2\cos t - 1)(\cos t + 1).$$

Since $\cos t + 1 \ge 0$, the sign of $f'(t)$ matches $2\cos t - 1$, which is positive for $\cos t > \tfrac12$, i.e. $-\tfrac{\pi}{3} < t < \tfrac{\pi}{3}$. The function $f$ is odd, so the area $\tfrac{Rr}{2}|f(t)|$ is maximised at the turning points

$$t = \frac{\pi}{3} \quad\text{and}\quad t = -\frac{\pi}{3}.$$

Marker's note. In (a)(i) explain clearly that $f' > 0$ makes $f$ one-to-one, using proper terms. In (a)(ii) relate $f(1) = 2$ to $g(2) = 1$ and apply the inverse-derivative rule. In (b)(i) show full working with correct signs before reaching the given polynomial. In (b)(ii) recognise that a single intersection means a double root, find it from $P'(x) = 0$, and reject $x = 0$. In (c)(i) label the perpendicular vector and the angle cases carefully, ending on exactly the expression asked for. In (c)(ii) build the area from part (i), differentiate, factor, and use the odd symmetry to locate both maximising values of $t$.