HSC Maths Extension 1 2024

(a)(i) [1 mark]

$$2\underline{a} - \underline{b} = 2(3\underline{i} + 2\underline{j}) - (-\underline{i} + 4\underline{j}) = (6 + 1)\underline{i} + (4 - 4)\underline{j} = 7\underline{i}.$$

(a)(ii) [1 mark]

The dot product is a scalar:

$$\underline{a} \cdot \underline{b} = (3)(-1) + (2)(4) = -3 + 8 = 5.$$

(b) [2 marks]

Factor: $x^2 - 8x - 9 = (x-9)(x+1)$. The upward parabola is at or below zero between its roots:

$$-1 \le x \le 9.$$

(c) [3 marks]

With $u = x - 1$, $x = u + 1$ and $du = dx$:

$$\int x\sqrt{x-1}\,dx = \int (u+1)u^{1/2}\,du = \int \left(u^{3/2} + u^{1/2}\right)du = \frac{2u^{5/2}}{5} + \frac{2u^{3/2}}{3} + C.$$

Returning to $x$:

$$= \frac{2(x-1)^{5/2}}{5} + \frac{2(x-1)^{3/2}}{3} + C.$$

(d) [2 marks]

Separate the variables (valid since $y > 0$):

$$\int \frac{1}{y}\,dy = \int x\,dx \implies \ln y = \frac{x^2}{2} + C \implies y = e^{x^2/2 + C} = A e^{x^2/2}.$$

(e) [1 mark]

With $f(x) = \arcsin(x^5)$ and inner derivative $5x^4$:

$$f'(x) = \frac{5x^4}{\sqrt{1 - (x^5)^2}} = \frac{5x^4}{\sqrt{1 - x^{10}}}.$$

(f) [2 marks]

Differentiate $V$ and use the chain rule:

$$\frac{dV}{dr} = 4\pi r^2, \qquad \frac{dr}{dt} = \frac{dr}{dV} \cdot \frac{dV}{dt} = \frac{1}{4\pi r^2} \times 10 = \frac{5}{2\pi r^2}\text{ cm s}^{-1}.$$

(g) [3 marks]

On $[0, \tfrac{\pi}{2}]$ the line $y = x$ lies above $y = \sin x$, so the area is the difference:

$$\text{Area} = \int_0^{\pi/2} (x - \sin x)\,dx = \left[\frac{x^2}{2} + \cos x\right]_0^{\pi/2} = \left(\frac{\pi^2}{8} + 0\right) - (0 + 1) = \frac{\pi^2}{8} - 1.$$

Marker's note. Keep vector notation tidy and remember the dot product is a scalar, not a vector. In (b) use the inequality sign correctly and state the interval with proper brackets. In (c) keep brackets around $(u+1)$ and convert back to $x$ at the end. In (g) take the difference of the two areas (line minus curve), not a volume, and handle the double negative from $-\cos x$ at $0$ carefully.

(a) [3 marks]

Perpendicular vectors have a zero dot product:

$$a^2(a + 5) + 2(a - 4) = 0 \implies a^3 + 5a^2 + 2a - 8 = 0.$$

Testing $a = 1$: $1 + 5 + 2 - 8 = 0$, so $(a - 1)$ is a factor. Dividing gives

$$(a - 1)(a^2 + 6a + 8) = (a - 1)(a + 2)(a + 4) = 0,$$

so $a = 1,\ -2,\ -4$.

(b) [3 marks]

Rotating about the $x$-axis, $V = \pi \int y^2\,dx$:

$$V = \pi \int_1^2 (x^3)^2\,dx = \pi \int_1^2 x^6\,dx = \pi \left[\frac{x^7}{7}\right]_1^2 = \frac{\pi}{7}(128 - 1) = \frac{127\pi}{7}.$$

(c) [3 marks]

Model the count as binomial with $n = 100$, $p = 0.31$, so $\mu = np = 31$ and

$$\sigma = \sqrt{np(1-p)} = \sqrt{100 \times 0.31 \times 0.69} = \sqrt{21.39} \approx 4.625.$$

At least $35\%$ of $100$ is at least $35$:

$$z = \frac{35 - 31}{4.625} \approx 0.86.$$

From the table $P(Z \le 0.86) = 0.8051$, so

$$P(X \ge 35) = 1 - 0.8051 = 0.1949 \approx 19.5\%.$$

(d) [3 marks]

Let $P(n)$ be the statement that $2^{3n} + 13$ is divisible by $7$.

Base case $n = 1$: $2^3 + 13 = 8 + 13 = 21 = 7 \times 3$, divisible by $7$. So $P(1)$ holds.

Inductive step. Assume $P(k)$: $2^{3k} + 13 = 7M$ for some integer $M$, so $2^{3k} = 7M - 13$. Then

This is $7$ times an integer, so $P(k+1)$ holds. By induction $P(n)$ is true for all integers $n \ge 1$.

(e) [3 marks]. Since $\tfrac{1}{|x-5|}$ is undefined at $x = 5$, consider two cases (and exclude $x = 5$).

Case $x > 5$: $|x - 5| = x - 5 > 0$, so $\tfrac{1}{x-5} \ge \tfrac16$ gives $6 \ge x - 5$, i.e. $x \le 11$. Combined with $x > 5$: $5 < x \le 11$.

Case $x < 5$: $|x - 5| = 5 - x > 0$, so $\tfrac{1}{5-x} \ge \tfrac16$ gives $6 \ge 5 - x$, i.e. $x \ge -1$. Combined with $x < 5$: $-1 \le x < 5$.

So the solution is $-1 \le x \le 11$, $x \ne 5$.

Marker's note. In (a) form the cubic from the dot product, then factor it fully by the factor theorem or polynomial division. In (b) use $V = \pi \int y^2\,dx$, not the area formula. In (c) use $\sigma = \sqrt{np(1-p)}$ and read the table correctly. In (d) set out every step of the induction and rearrange the assumption before substituting. In (e) respect the sign of the denominator in each case and exclude the value that makes it zero.

(a)(i) [1 mark]

The line $P = 2000$ is a horizontal asymptote of the logistic family (where $\tfrac{dP}{dt} = 0$). The point $S$ lies below it and $T$ lies on it, so the increasing solution through $S$ rises towards $P = 2000$ but never reaches it. Hence it cannot pass through $T$.

(a)(ii) [1 mark]

Sketch a smooth, increasing, concave-up-then-concave-down (S-shaped) curve starting at $S$, with an inflection near $P = 1000$, levelling off and following the asymptote $P = 2000$ from below.

(a)(iii) [2 marks]

The growth rate $\tfrac{dP}{dt}$ is largest where $\tfrac{d}{dP}\!\left(\tfrac{dP}{dt}\right) = 0$:

$$\frac{d}{dP}\big(2000P - P^2\big) = 2000 - 2P = 0 \implies P = 1000.$$

The direction field confirms the slope is greatest at $P = 1000$ (half the carrying capacity), so the rate of growth is largest when $P = 1000$.

(b)(i) [2 marks]

Write the sum of fourth powers using the squared identity:

$$\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\cos^2 x \sin^2 x = 1 - \tfrac12(2\cos x \sin x)^2 = 1 - \tfrac12 \sin^2 2x.$$

Now replace $\sin^2 2x = 1 - \cos^2 2x$:

$$= 1 - \tfrac12(1 - \cos^2 2x) = \tfrac12 + \tfrac12 \cos^2 2x = \frac{1 + \cos^2 2x}{2}.$$

(b)(ii) [3 marks]

Using part (i), then $\cos^2 2x = \tfrac12(1 + \cos 4x)$:

$$\int_0^{\pi/4} \frac{1 + \cos^2 2x}{2}\,dx = \frac12 \int_0^{\pi/4} \left(1 + \tfrac12(1 + \cos 4x)\right)dx = \frac12 \int_0^{\pi/4} \left(\tfrac32 + \tfrac12 \cos 4x\right)dx.$$

Integrating:

$$= \frac12 \left[\frac{3x}{2} + \frac{\sin 4x}{8}\right]_0^{\pi/4} = \frac12 \left(\frac{3\pi}{8} + \frac{\sin \pi}{8} - 0\right) = \frac12 \cdot \frac{3\pi}{8} = \frac{3\pi}{16}.$$

(c) [4 marks]

Let $\underline{x} = \begin{pmatrix} u \\ v \end{pmatrix}$. The projection onto $\underline{a}$ is $\dfrac{\underline{a} \cdot \underline{x}}{\underline{a} \cdot \underline{a}}\,\underline{a} = k\underline{a}$, so

$$k = \frac{u + 3v}{10} \implies u + 3v = 10k. \quad (1)$$

The projection onto $\underline{b}$ is $\dfrac{\underline{b} \cdot \underline{x}}{\underline{b} \cdot \underline{b}}\,\underline{b} = p\underline{b}$, so

$$p = \frac{2u - v}{5} \implies 2u - v = 5p. \quad (2)$$

Solving $(1)$ and $(2)$: from $3 \times (2)$, $6u - 3v = 15p$; adding to $(1)$ gives $7u = 10k + 15p$, so $u = \tfrac{10k + 15p}{7}$. Then $v = 2u - 5p = \tfrac{20k - 5p}{7}$. Hence

$$\underline{x} = \frac{1}{7}\begin{pmatrix} 10k + 15p \\ 20k - 5p \end{pmatrix} = \frac{5}{7}\begin{pmatrix} 2k + 3p \\ 4k - p \end{pmatrix}.$$

(d) [3 marks]

With $u = e^x + 2e^{-x}$, $\dfrac{du}{dx} = e^x - 2e^{-x}$, so $du = (e^x - 2e^{-x})\,dx$. Also

$$u^2 = e^{2x} + 4 + 4e^{-2x}.$$

Factor the denominator using $4 + 8e^{2x} + e^{4x} = e^{2x}(4e^{-2x} + 8 + e^{2x}) = e^{2x}\big[(4e^{-2x} + 4 + e^{2x}) + 4\big] = e^{2x}(u^2 + 4)$. The numerator is $e^{3x} - 2e^x = e^{2x}(e^x - 2e^{-x})$, so

$$\int \frac{e^{2x}(e^x - 2e^{-x})}{e^{2x}(u^2 + 4)}\,dx = \int \frac{du}{u^2 + 4} = \frac12 \tan^{-1}\!\frac{u}{2} + C = \frac12 \tan^{-1}\!\left(\frac{e^x + 2e^{-x}}{2}\right) + C.$$

Marker's note. In (a)(i) state the asymptote at $P = 2000$ and note $S$ and $T$ are on opposite sides of it. In (a)(iii) differentiate with respect to $P$ (not $t$) and use $P = -b/2a$ to get half the carrying capacity. In (b)(i) square the binomial carefully; in (b)(ii) integrate $\cos 4x$ correctly and evaluate $\sin \pi = 0$. In (c) write the unknown vector with its own components $u, v$, then solve the simultaneous equations. In (d) expand $u^2$ as an exponential and factor to reach the standard inverse-tangent integral, then rewrite in terms of $e^x$.

(a) [4 marks]. Separate the variables: $\dfrac{dy}{dx} = e^x e^y$, so

At the origin $x = y = 0$: $-1 = 1 + C$, so $C = -2$ and $e^{-y} = 2 - e^x$. Then

Domain: need $2 - e^x > 0$, so $e^x < 2$, i.e. $x < \ln 2$. Domain is $(-\infty, \ln 2)$.

Range: as $x \to -\infty$, $e^x \to 0$ so $y \to -\ln 2$; as $x \to \ln 2^-$, $2 - e^x \to 0^+$ so $y \to \infty$. Since $\tfrac{dy}{dx} = e^{x+y} > 0$ the function increases throughout, so the range is $(-\ln 2, \infty)$.

(b) [3 marks]. The function is defined and differentiable for all reals, so it has an inverse if $f'(x)$ keeps one sign. Differentiating,

The denominator is positive, so the sign is set by the numerator $(k+1) + (1-k)x^2$.

Case $f'(x) \ge 0$ for all $x$: need $1 - k \ge 0$ and $k + 1 \ge 0$, i.e. $-1 \le k \le 1$.

Case $f'(x) \le 0$ for all $x$: need $1 - k \le 0$ and $k + 1 \le 0$, i.e. $k \ge 1$ and $k \le -1$, which is impossible.

So $f$ has an inverse when $-1 \le k \le 1$.

(c)(i) [1 mark]

Both $\tan^{-1}(3x)$ and $\tan^{-1}(10x)$ are strictly increasing with range $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$, so their sum is strictly increasing with range $(-\pi, \pi)$. A horizontal line $y = \theta$ with $-\pi < \theta < \pi$ meets a strictly increasing curve that spans this range exactly once, so the equation has exactly one solution.

(c)(ii) [2 marks]

Take the tangent of both sides and use the compound-angle formula (with $\tan\tfrac{3\pi}{4} = -1$):

$$\frac{3x + 10x}{1 - (3x)(10x)} = -1 \implies 13x = -(1 - 30x^2) \implies 30x^2 - 13x - 1 = 0.$$

Factoring, $(2x - 1)(15x + 1) = 0$, so $x = \tfrac12$ or $x = -\tfrac{1}{15}$. Only $x = \tfrac12$ gives a positive sum equal to $\tfrac{3\pi}{4}$ (the negative root gives a negative sum), so

$$x = \tfrac12.$$

(d) [4 marks]

Work with $D^2$ to avoid the square root:

$$D^2 = V^2t^2\cos^2\theta + \left(Vt\sin\theta - \tfrac12 gt^2\right)^2 = V^2t^2 - Vgt^3\sin\theta + \tfrac14 g^2 t^4.$$

Differentiate:

$$\frac{d(D^2)}{dt} = 2V^2 t - 3Vg t^2 \sin\theta + g^2 t^3 = t\big(2V^2 - 3Vg t\sin\theta + g^2 t^2\big).$$

For $t > 0$ the sign is set by the quadratic in $t$, $g^2 t^2 - 3Vg\sin\theta\, t + 2V^2$. Its leading coefficient $g^2 > 0$, so it is positive for all $t$ when its discriminant is negative:

$$\Delta = 9V^2 g^2 \sin^2\theta - 4g^2(2V^2) = V^2 g^2(9\sin^2\theta - 8) < 0.$$

Since $V^2 g^2 > 0$, this requires $9\sin^2\theta < 8$, i.e. $\sin^2\theta < \tfrac{8}{9}$. For a projection angle $0 < \theta < \tfrac{\pi}{2}$ where $\sin\theta > 0$, this is $\sin\theta < \tfrac{\sqrt{8}}{3}$, i.e. $\theta < \sin^{-1}\!\left(\tfrac{\sqrt{8}}{3}\right)$. Then $\tfrac{d(D^2)}{dt} > 0$ for all $t > 0$, so $D(t)$ is increasing for all $t > 0$.

Marker's note. In (a) integrate $e^{-y}$ carefully, include the constant, and use exact form $\ln 2$; mind the bracket-versus-parenthesis distinction in the interval notation. In (b) differentiate the quotient correctly and treat the numerator as a quadratic in $x$ whose sign must be fixed. In (c)(i) argue monotonicity and range; in (c)(ii) take the tangent of both sides, build the quadratic, and reject the invalid root with justification. In (d) use $D^2$ for an easier derivative and show the discriminant is negative.