HSC Maths Extension 1 2025

(a) [2 marks]

Let $y = 1 - \dfrac{1}{x-2}$. Make $x$ the subject:

$$y - 1 = -\frac{1}{x-2} \implies x - 2 = -\frac{1}{y-1} \implies x = 2 - \frac{1}{y-1}.$$

Swapping $x$ and $y$ for the inverse,

$$f^{-1}(x) = 2 - \frac{1}{x-1}.$$

(b) [3 marks]

Apply the double angle formula and factor (do not divide by $\sin\theta$):

$$2\sin\theta\cos\theta - \sin\theta = 0 \implies \sin\theta(2\cos\theta - 1) = 0.$$

Then $\sin\theta = 0$ gives $\theta = 0, \pi$; and $\cos\theta = \tfrac12$ gives $\theta = \tfrac{\pi}{3}$. So

$$\theta = 0,\ \tfrac{\pi}{3},\ \pi.$$

(c) [2 marks]

Use the product-to-sum identity $\sin A \cos B = \tfrac12[\sin(A+B) + \sin(A-B)]$:

$$\int \sin 3x \cos x\,dx = \frac12 \int \big(\sin 4x + \sin 2x\big)\,dx = -\frac{\cos 4x}{8} - \frac{\cos 2x}{4} + C.$$

(d) [2 marks]

Start from $y = \cos^{-1}(2x)$, which has domain $[-\tfrac12, \tfrac12]$ and range $[0, \pi]$. The factor $-\tfrac13$ reflects in the $x$-axis and compresses vertically, so the range becomes $[-\tfrac{\pi}{3}, 0]$. The curve runs from $(-\tfrac12, -\tfrac{\pi}{3})$ down to $(\tfrac12, 0)$, with $y$-intercept $-\tfrac{\pi}{6}$ at $x = 0$.

(e) [1 mark]

Parallel vectors are scalar multiples: $\begin{pmatrix} 2 \\ 6 \end{pmatrix} = 2\begin{pmatrix} 1 \\ 3 \end{pmatrix}$, so $\begin{pmatrix} 1 \\ m \end{pmatrix}$ is parallel when $m = 3$.

(f) [2 marks]

Combine over the common denominator $\alpha\beta\gamma$:

$$\frac{1}{\alpha\beta} + \frac{1}{\alpha\gamma} + \frac{1}{\beta\gamma} = \frac{\gamma + \beta + \alpha}{\alpha\beta\gamma} = \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}.$$

From the polynomial $2x^3 + 6x^2 + x - 1$: $\alpha+\beta+\gamma = -\tfrac{6}{2} = -3$ and $\alpha\beta\gamma = -\tfrac{-1}{2} = \tfrac12$. So the value is $\dfrac{-3}{1/2} = -6$.

(g) [3 marks]

Use $\cos^2(3x) = \tfrac12(1 + \cos 6x)$:

$$\int_{\pi/6}^{\pi/3} \cos^2(3x)\,dx = \frac12 \int_{\pi/6}^{\pi/3} \big(1 + \cos 6x\big)\,dx = \frac12 \left[x + \frac{\sin 6x}{6}\right]_{\pi/6}^{\pi/3}.$$

Since $\sin 2\pi = \sin\pi = 0$,

$$= \frac12 \left[\left(\frac{\pi}{3} + 0\right) - \left(\frac{\pi}{6} + 0\right)\right] = \frac12 \cdot \frac{\pi}{6} = \frac{\pi}{12}.$$

Marker's note. In (a) distinguish $f^{-1}(x)$ from $[f(x)]^{-1}$ and watch the negative signs when rearranging. In (b) do not divide through by $\sin\theta$ or you lose solutions; factor instead. In (c) recognise the product-to-sum form rather than expanding $\sin 3x$ as $\sin(2x+x)$. In (d) get the base inverse-cosine shape right and mark the $y$-intercept. In (e) the question asks for a value of $m$, not for perpendicular vectors. In (f) use the lowest common denominator and apply the sum and product of roots correctly, with care on signs. In (g) take the $\tfrac12$ out cleanly with brackets and use the correct double-angle identity.

(a) [3 marks]

The sector area is $\tfrac12 r^2 \theta = 10$, so $\theta = \dfrac{20}{r^2}$ and $\dfrac{d\theta}{dr} = -\dfrac{40}{r^3}$. By the chain rule,

$$\frac{dr}{dt} = \frac{dr}{d\theta} \cdot \frac{d\theta}{dt} = -\frac{r^3}{40} \times 2 = -\frac{r^3}{20}.$$

When $r = 4$: $\dfrac{dr}{dt} = -\dfrac{64}{20} = -3.2$ cm s$^{-1}$. The radius is decreasing at $3.2$ cm s$^{-1}$.

(b) [2 marks]

Rotating about the $y$-axis with $x = \tfrac1y$, the volume is

$$V = \pi \int_1^a x^2\,dy = \pi \int_1^a \frac{1}{y^2}\,dy = \pi \left[-\frac{1}{y}\right]_1^a = \pi\left(1 - \frac{1}{a}\right).$$

(c) [3 marks]

Let $P(n)$ be the statement $1 \times (1!) + 2 \times (2!) + \cdots + n \times (n!) = (n+1)! - 1$.

Base case $n = 1$: LHS $= 1 \times (1!) = 1$ and RHS $= 2! - 1 = 1$, so $P(1)$ holds.

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

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

(d) [3 marks]

The right side is a difference of squares, $(2-y)(2+y) = 4 - y^2$. Separate the variables:

$$\int \frac{dy}{4 - y^2} = \int dx \implies \sin^{-1}\!\frac{y}{2} = x + C.$$

At $x = 0$, $y = 1$: $\sin^{-1}\tfrac12 = C$, so $C = \tfrac{\pi}{6}$. Then $x = \sin^{-1}\tfrac{y}{2} - \tfrac{\pi}{6}$, that is

$$y = 2\sin\!\left(x + \frac{\pi}{6}\right).$$

(e)(i) [1 mark]

$$\sqrt{3}\sin x - \cos x = 2\left(\frac{\sqrt{3}}{2}\sin x - \frac12\cos x\right) = 2\sin\!\left(x - \frac{\pi}{6}\right).$$

(e)(ii) [2 marks]

Rewrite $\sqrt{3}\sin x = \cos x + 1$ as $\sqrt{3}\sin x - \cos x = 1$, then use part (i):

$$2\sin\!\left(x - \frac{\pi}{6}\right) = 1 \implies \sin\!\left(x - \frac{\pi}{6}\right) = \frac12.$$

For $0 \le x \le 2\pi$ the argument $x - \tfrac{\pi}{6}$ ranges over $[-\tfrac{\pi}{6}, \tfrac{11\pi}{6}]$, so $x - \tfrac{\pi}{6} = \tfrac{\pi}{6}$ or $\tfrac{5\pi}{6}$, giving

$$x = \frac{\pi}{3} \quad\text{or}\quad x = \pi.$$

Marker's note. In (a) write an appropriate chain rule, treat $\theta$ and $r$ as variables (not constants), use the sector-area formula from the Reference Sheet, and read the negative sign as a decreasing rate. In (b) keep the factor $\pi$ and rotate about the $y$-axis, integrating with respect to $y$. In (c) test the base case correctly, set out the $n=k$ and $n=k+1$ steps clearly, and conclude LHS $=$ RHS. In (d) recognise the difference of squares and use the Reference Sheet to integrate to $\sin^{-1}$, then solve for the constant carefully. In (e) write the exact given form in radians and add angles in radians correctly.

(a) [3 marks]

Separate the variables:

$$\int y\,dy = \int 5\,dx \implies \frac{y^2}{2} = 5x + C.$$

At $x = 0$, $y = -4$: $\tfrac{16}{2} = C$, so $C = 8$ and $y^2 = 10x + 16$. Since $y = -4 < 0$ at the initial point, take the negative root:

$$y = -\sqrt{10x + 16}.$$

(b) [2 marks]

Around a round table there are $(8-1)! = 7!$ unrestricted arrangements. Treating the two who refuse to sit together as a single block gives $6! \times 2$ arrangements where they are adjacent. So the number where they are not adjacent is

$$7! - 6! \times 2 = 5040 - 1440 = 3600.$$

(c) [4 marks]

Differentiate the position vector:

$$\underline{v}(t) = \underline{i} + \frac{2t}{9}\underline{j}, \qquad \underline{a}(t) = \frac{2}{9}\underline{j}.$$

The scalar product is $\underline{v} \cdot \underline{a} = \dfrac{2}{9} \cdot \dfrac{2t}{9} = \dfrac{4t}{81}$, while $|\underline{a}| = \dfrac29$ and $|\underline{v}| = \sqrt{1 + \dfrac{4t^2}{81}}$. With the angle $\tfrac{\pi}{4}$,

$$\underline{v}\cdot\underline{a} = |\underline{v}|\,|\underline{a}|\cos\frac{\pi}{4} \implies \frac{4t}{81} = \sqrt{1 + \frac{4t^2}{81}} \cdot \frac{2}{9} \cdot \frac{1}{\sqrt{2}}.$$

For $t > 0$, square both sides:

$$\frac{16t^2}{81^2} = \left(1 + \frac{4t^2}{81}\right) \cdot \frac{4}{81} \cdot \frac12 \implies \frac{8t^2}{81} = 1 + \frac{4t^2}{81} \implies \frac{4t^2}{81} = 1.$$

So $t^2 = \tfrac{81}{4}$ and, since $t > 0$, $t = \dfrac92$.

(d) [4 marks]

With $p$ the probability a counter is green, $E(X) = 100p = 20$ gives $p = \tfrac15$. The variance is

$$\sigma^2 = 100 \times \frac15 \times \frac45 = 16, \qquad \sigma = 4.$$

Using $Z = \dfrac{X - 20}{4} \sim N(0, 1)$, the condition $P(X \ge k) = 0.0668$ means $P(X < k) = 0.9332$. From the table $P(Z \le 1.5) = 0.9332$, so

$$\frac{k - 20}{4} = 1.5 \implies k = 26.$$

(e)(i) [1 mark]

Rearrange the given relation as $\dbinom{n-1}{r-1} = \dbinom{n}{r} - \dbinom{n-1}{r}$. Setting $n = m+1$ and $r = R+1$,

$$\binom{m}{R} = \binom{m+1}{R+1} - \binom{m}{R+1}.$$

(e)(ii) [2 marks]

Note $\dbinom{2000}{2000} = \dbinom{2001}{2001}$. Apply part (i) to each term (with $R = 2000$), so $\dbinom{m}{2000} = \dbinom{m+1}{2001} - \dbinom{m}{2001}$, and the sum telescopes:

$$\binom{2001}{2001} + \sum_{m=2001}^{2050} \binom{m}{2000} = \binom{2001}{2001} + \left[\binom{2051}{2001} - \binom{2001}{2001}\right] = \binom{2051}{2001}.$$

Hence the left side equals $\dbinom{2051}{2001}$, as required.

Marker's note. In (a) separate and integrate, then use the given point $(0, -4)$ to fix the constant and to choose the negative square root. In (b) start from $(n-1)!$ for a circle and subtract the block-together arrangements. In (c) derive $\underline{v}$ and $\underline{a}$ from $\underline{r}$, apply the scalar product in component form, then solve the resulting equation and reject the negative time. In (d) read the $z$-score table to get $1.5$ from $0.9332$, and substitute the right values into the $z$-score formula. In (e)(i) make the correct substitutions and justify each step; in (e)(ii) use part (i) to telescope and note the binomial-coefficient simplifications.

(a) [2 marks]

Pair the factors of $60$ into six pairs that each multiply to $60$:

$$\{1, 60\},\ \{2, 30\},\ \{3, 20\},\ \{4, 15\},\ \{5, 12\},\ \{6, 10\}.$$

These six pairs act as six pigeonholes. Choosing seven distinct factors forces, by the pigeonhole principle, at least two factors from the same pair, and their product is $60$. Hence the product of all seven factors is a multiple of $60$.

(b) [4 marks]

Let $OB = h$, so $OA = kh$. Measure time $t$ from when the first particle (from $A$) is launched; the second (from $B$) is launched at $t = T$. Horizontal and vertical motion give, for the particle from $A$,

$$x = Ut, \qquad y = kh - \tfrac12 g t^2,$$

and for the particle from $B$,

$$x = V(t - T), \qquad y = h - \tfrac12 g (t - T)^2.$$

Landing means $y = 0$ for each, so $kh = \tfrac12 g t^2$ and $h = \tfrac12 g (t - T)^2$. Dividing,

$$k = \frac{t^2}{(t - T)^2} \implies \frac{t}{t - T} = \sqrt{k} \quad (t > T > 0).$$

Landing in the same place means $Ut = V(t - T)$, so

$$\frac{V}{U} = \frac{t}{t - T} = \sqrt{k},$$

which depends only on $k$.

(c) [3 marks]

The hands are perpendicular when $\overrightarrow{OA} \cdot \overrightarrow{OB} = 0$:

$$2\sin\frac{\pi t}{360}\sin\frac{\pi t}{30} + 2\cos\frac{\pi t}{360}\cos\frac{\pi t}{30} = 0.$$

By the cosine compound-angle formula this is $2\cos\!\left(\dfrac{\pi t}{30} - \dfrac{\pi t}{360}\right) = 2\cos\dfrac{11\pi t}{360} = 0$, so

$$\frac{11\pi t}{360} = \frac{\pi}{2},\ \frac{3\pi}{2},\ \ldots$$

The first time: $\dfrac{11\pi t}{360} = \dfrac{\pi}{2}$ gives $t = \dfrac{360}{2 \times 11} \approx 16.364$ minutes. The second time: $\dfrac{11\pi t}{360} = \dfrac{3\pi}{2}$ gives $t = \dfrac{3 \times 360}{2 \times 11} \approx 49.091$ minutes.

(d) [3 marks]

Differentiate $f(x) = \cos^{-1}(\sin x)$ by the chain rule:

$$f'(x) = -\frac{\cos x}{\sqrt{1 - \sin^2 x}} = -\frac{\cos x}{\sqrt{\cos^2 x}} = -\frac{\cos x}{|\cos x|}.$$

For $0 < x < \tfrac{\pi}{2}$, $\cos x > 0$, so $f'(x) = -1$. For $\tfrac{\pi}{2} < x < \pi$, $\cos x < 0$, so $f'(x) = 1$. At $x = \tfrac{\pi}{2}$ the denominator is zero, so $f'(x)$ is undefined there. Thus

$$f'(x) = -1 \text{ for } x \in \left(0, \tfrac{\pi}{2}\right), \qquad f'(x) = 1 \text{ for } x \in \left(\tfrac{\pi}{2}, \pi\right).$$

(e) [3 marks]

By the relations between roots and coefficients of $x^3 + bx^2 + cx + (b + c - 1)$:

$$\tan\alpha + \tan\beta + \tan\gamma = -b, \quad \sum \tan\alpha\tan\beta = c, \quad \tan\alpha\tan\beta\tan\gamma = -(b + c - 1) = 1 - b - c.$$

Using the tangent expansion for $\tan(\alpha + \beta + \gamma)$,

$$\tan(\alpha+\beta+\gamma) = \frac{(\tan\alpha + \tan\beta + \tan\gamma) - \tan\alpha\tan\beta\tan\gamma}{1 - (\tan\alpha\tan\beta + \tan\alpha\tan\gamma + \tan\beta\tan\gamma)} = \frac{-b - (1 - b - c)}{1 - c} = \frac{c - 1}{1 - c} = -1.$$

So $\alpha + \beta + \gamma = \ldots, -\tfrac{\pi}{4}, \tfrac{3\pi}{4}, \ldots$, and the smallest positive value is

$$\alpha + \beta + \gamma = \frac{3\pi}{4}.$$

Marker's note. In (a) build six factor-pairs each multiplying to $60$ and apply the pigeonhole principle; listing a few choices is not a proof. In (b) recognise the angle of projection is zero, that the two particles have different flight times, and equate horizontal displacements and the vertical landing conditions, keeping $U$, $V$, $T$, $t$ clearly distinct. In (c) set $\overrightarrow{OA}\cdot\overrightarrow{OB} = 0$ and convert to a single cosine via the compound-angle formula. In (d) note the denominator is $|\cos x|$, not $\cos x$, so $f'(x) = \pm 1$, and that it is undefined at $x = \tfrac{\pi}{2}$. In (e) use the root-coefficient relations and the tangent expansion, then choose the smallest positive angle.