2026 HSC Mathematics Advanced — Intuition Education Predicted Paper

Answers and marking notes

(Intuition Education — not part of the examination paper.)

Section I — correct responses

Brief reasons:

1. Need $4 - x^2 > 0$ (strict — the surd is in a denominator): $-2 < x < 2$. Distractor A closes the endpoints; D forgets the radicand must be positive, not merely non-zero. C
2. $2x = 4 \Rightarrow x = 2$, $y = f(4) + 1 = 4$: the point is $(2, 4)$. Distractor A doubles instead of halving; C forgets the $+1$. B
3. An extreme high outlier drags the mean strongly; the median moves at most to the next data value. D
4. Amplitude $2$ about centre $3$: $[3-2,\ 3+2] = [1, 5]$; the $4$ affects only the period. A
5. $h'(1) = f'(g(1)) \cdot g'(1) = f'(2) \times 6 = 3 \times 6 = 18$. Distractor A uses $f'(1) \cdot g'(1) = 24$. B
6. $3 - x > 0 \Rightarrow x < 3$ (strict inequality — $\ln 0$ undefined). C
7. $E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 1.1$. Distractor D is $E(X^2) = 1.7$. B
8. $P(-1 < Z < 2) = 34\% + 47.5\% = 81.5\%$. C
9. $f(-x) = -x^3 + x = -f(x)$ for $f(x) = x^3 - x$; the others fail $f(-x) = -f(x)$. A
10. $f'$ changes sign from positive to negative at $x = 2$: local maximum of $f$. At $x = 3$, $f' < 0$ so $f$ is decreasing (C wrong); B confuses the zero of $f'$ with a zero of $f''$. D

Question 11

(a) $T_n = 6 + 5(n-1) = 5n + 1$. Setting $5n + 1 = 2026$ gives $n = 405$, a positive integer — so $2026$ is the $405$th term. (1 mark forming $T_n$, 1 mark solving and stating $n$ integer.) (b) $S_{405} = \dfrac{405}{2}(6 + 2026) = 405 \times 1016 = 411\,480$.

Question 12

(a) $0.1 + 0.3 + k + 0.2 = 1 \Rightarrow k = 0.4$. (b) $E(X) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7$. $E(X^2) = 0.1 + 1.2 + 3.6 + 3.2 = 8.1$, so $\text{Var}(X) = 8.1 - 2.7^2 = 0.81$ and $\sigma = 0.9$. (Marking note: the recurring flag — the standard deviation is the square root of the variance.)

Question 13

LHS $= \dfrac{(1 - \sin\theta) + (1 + \sin\theta)}{(1+\sin\theta)(1-\sin\theta)} = \dfrac{2}{1 - \sin^2\theta} = \dfrac{2}{\cos^2\theta} = 2\sec^2\theta$ = RHS. (Work one side only; 1 mark common denominator, 1 mark Pythagorean identity and completion.)

Question 14

(a) $x^2 + 2 \ge 2 > 0$ for all $x$, so the domain is all real $x$. (b) The denominator has minimum value $2$ (at $x = 0$), so $h$ has maximum value $\tfrac{3}{2}$. (c) As $|x| \to \infty$, $h(x) \to 0^+$ but never reaches $0$: range is $0 < h(x) \le \tfrac{3}{2}$, i.e. $\left(0, \tfrac{3}{2}\right]$.

Question 15

(a) $\dfrac{dy}{dx} = 2x\ln x + x^2 \cdot \dfrac{1}{x} = 2x \ln x + x$. (b) From (a), $\displaystyle\int (2x\ln x + x)\,dx = x^2 \ln x$, so $\displaystyle\int x \ln x\;dx = \tfrac{1}{2}\left(x^2 \ln x - \tfrac{x^2}{2}\right) + C = \tfrac{x^2 \ln x}{2} - \tfrac{x^2}{4} + C$. (The "hence" is compulsory in spirit: recognition, not integration by parts, which is not in this course.)

Question 16

(a) $P(D) = 0.5(0.02) + 0.3(0.04) + 0.2(0.05) = 0.01 + 0.012 + 0.01 = 0.032$. (1 mark converting the ratio $5:3:2$ to $0.5, 0.3, 0.2$; 1 mark total probability.) (b) $P(C \mid D) = \dfrac{P(C \cap D)}{P(D)} = \dfrac{0.01}{0.032} = \dfrac{5}{16} = 0.3125$. (The conditioning event is the outcome — the reduced sample space is the defective items.)

Question 17

(a) $6\%$ p.a. quarterly $\Rightarrow r = 1.5\%$ per quarter, $n = 20$ periods: factor $23.124$. $FV = 2000 \times 23.124 = \$46\,248$. (b) $8\%$ p.a. quarterly $\Rightarrow r = 2\%$, $n = 24$: factor $30.422$. Deposit $= \dfrac{80\,000}{30.422} = \$2629.68$. (Divide by the factor — the unknown is the contribution.) (c) Total deposits $= 24 \times 2629.68 = \$63\,112.32$; interest $= 80\,000 - 63\,112.32 = \$16\,887.68$.

Question 18

$\dfrac{f(x+h) - f(x)}{h} = \dfrac{(x+h)^2 - 3(x+h) - x^2 + 3x}{h} = \dfrac{2xh + h^2 - 3h}{h} = 2x + h - 3$. As $h \to 0$, $f'(x) = 2x - 3$. (1 mark correct expansion of the difference quotient, 1 mark taking the limit.)

Question 19

(a) $\Delta L = 30 = 10\log_{10}\left(\tfrac{I_1}{I_2}\right) \Rightarrow \tfrac{I_1}{I_2} = 10^3 = 1000$ times more intense. (b) $\tfrac{I_P}{I_Q} = 10^{6/10} = 10^{0.6} \approx 3.98 \approx 4$.

Question 20

$(5 - k)^2 + k = 11 \Rightarrow 25 - 10k + k^2 + k = 11 \Rightarrow k^2 - 9k + 14 = 0 \Rightarrow (k-2)(k-7) = 0$, so $k = 2$ or $k = 7$. Since $0 < k < 5$, reject $k = 7$: $k = 2$. (1 mark substituting the point into the transformed equation — apply the translation to the equation, not the coordinate; 1 mark solving the quadratic; 1 mark the explicit rejection. Both roots satisfy the equation — the rejection must cite the condition.)

Question 21

(a) $h = 1$: $\displaystyle\int_0^4 f(x)\,dx \approx \frac{1}{2}\left[2 + 9.2 + 2(2.9 + 4.4 + 6.5)\right] = \frac{1}{2}(11.2 + 27.6) = 19.4$. (Five function values = four applications; the flagged error is confusing the two counts.) (b) Concave up $\Rightarrow$ each chord lies above the curve, so the estimate is greater than the exact value (an overestimate).

Question 22

(a) $x^2 - 4x = 2x - x^2 \Rightarrow 2x^2 - 6x = 0 \Rightarrow 2x(x - 3) = 0 \Rightarrow x = 0, 3$. (b) At $x = 1$: $2x - x^2 = 1$ and $x^2 - 4x = -3$, so $y = 2x - x^2$ lies above. (c) $\displaystyle A = \int_0^3 \left[(2x - x^2) - (x^2 - 4x)\right] dx = \int_0^3 (6x - 2x^2)\,dx = \left[3x^2 - \tfrac{2x^3}{3}\right]_0^3 = 27 - 18 = 9$ square units. (Subtract in the correct order; substitute limits with brackets.)

Question 23

(a) $z = \dfrac{514 - 505}{6} = 1.5$: $P(X < 514) = 0.9332$. (b) $z$-scores $-1$ and $1.5$: $P = 0.9332 - (1 - 0.8413) = 0.9332 - 0.1587 = 0.7745$. (1 mark both $z$-scores, 1 mark symmetry for the lower tail.) (c) $P(X > 514) = 0.0668$: expected number $= 2000 \times 0.0668 = 133.6 \approx 134$ bags. (d) $0.0808 = 1 - 0.9192 \Rightarrow z = -1.4$: mass $= 505 - 1.4 \times 6 = 496.6$ g. (Reverse use of the table with symmetry.)

Question 24

(a) $1200 = 800e^{5k} \Rightarrow e^{5k} = \tfrac{3}{2} \Rightarrow k = \tfrac{1}{5}\ln\tfrac{3}{2} \approx 0.0811$. (b) $N(10) = 800e^{10k} = 800\left(e^{5k}\right)^2 = 800 \times \left(\tfrac{3}{2}\right)^2 = 1800$ exactly. (c) $N'(t) = kN(t)$, so $N'(10) = 0.0811 \times 1800 \approx 146$ bacteria per hour. ("Rate" means the derivative — the perennial flag.) (d) $800e^{kt} = 3000 \Rightarrow e^{kt} = 3.75 \Rightarrow t = \dfrac{\ln 3.75}{k} \approx \dfrac{1.3218}{0.0811} \approx 16.3$ hours.

Question 25

(a) The back-bearing from $Q$ to $P$ is $040° + 180° = 220°$; the bearing of $R$ from $Q$ is $150°$; hence $\angle PQR = 220° - 150° = 70°$. (b) $PR^2 = 9^2 + 7^2 - 2(9)(7)\cos 70° = 130 - 126\cos 70° \approx 86.91$, so $PR \approx 9.32$ km. (Cosine rule — no right angle is available.) (c) Sine rule: $\dfrac{\sin \angle QPR}{7} = \dfrac{\sin 70°}{9.32} \Rightarrow \sin\angle QPR \approx 0.7055 \Rightarrow \angle QPR \approx 45°$ (acute, since the largest angle is at $Q$, opposite the longest side). Bearing of $R$ from $P$ $= 040° + 45° = 085°$. (Convert the internal angle to a bearing — the flagged step.)

Question 26

(a) For each additional hour of revision, the predicted test score increases by $4.2$ marks. (Context and value both required.) (b) $\hat{y} = 4.2(6) + 38 = 63.2$. (c) $x = 20$ is far outside the observed data range (\$1$ to $8$ hours) — extrapolation; the line predicts $4.2(20) + 38 = 122$, an impossible score out of $100$. (d) The gradient measures the rate of change of predicted score, not the strength of the association; the correlation is measured by $r = 0.87$. A steep line can have weak correlation and vice versa.

Question 27

(a) $\displaystyle\int_0^4 k(4 - x)\,dx = k\left[4x - \tfrac{x^2}{2}\right]_0^4 = 8k = 1 \Rightarrow k = \tfrac{1}{8}$. (b) $F(x) = \displaystyle\int_0^x \tfrac{4 - t}{8}\,dt = \tfrac{1}{8}\left(4x - \tfrac{x^2}{2}\right) = \dfrac{8x - x^2}{16}$ (and $F(4) = 1$, as required). (c) $F(m) = \tfrac{1}{2}$: $8m - m^2 = 8 \Rightarrow m^2 - 8m + 8 = 0 \Rightarrow m = 4 \pm 2\sqrt{2}$. Since $0 \le m \le 4$, the median is $m = 4 - 2\sqrt{2}$. (Equate the CDF to $0.5$ — pdf/CDF confusion is the flagged error.) (d) $f$ is decreasing on $[0, 4]$, so its maximum is at the left endpoint: the mode is $x = 0$. (Do not differentiate — the monotone-density endpoint trap.) (e) $P(X > 2) = 1 - F(2) = 1 - \tfrac{12}{16} = \tfrac{1}{4}$.

Question 28

(a) Product rule: $f'(x) = e^{-x} + x(-e^{-x}) = (1 - x)e^{-x}$. (b) $f'(x) = 0 \Rightarrow x = 1$ (as $e^{-x} \ne 0$): stationary point $\left(1, e^{-1}\right)$. For $x < 1$, $f' > 0$; for $x > 1$, $f' < 0$: a local maximum. (Reject $e^{-x} = 0$; justify the nature.) (c) $f''(x) = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}$. $f''(2) = 0$, and $f''$ changes sign at $x = 2$ (negative before, positive after) — concavity changes, so $\left(2, 2e^{-2}\right)$ is a point of inflection. ($f'' = 0$ alone is not sufficient.) (d) Curve through the origin, rising to the maximum $\left(1, e^{-1}\right)$, inflection at $x = 2$, then decreasing and approaching the asymptote $y = 0$ from above as $x \to \infty$; unbounded below as $x \to -\infty$.

Question 29

(a) Centre $= \tfrac{9 + 3}{2} = 6 = A$; amplitude $= \tfrac{9 - 3}{2} = 3 = B$. (Check: period $= \tfrac{2\pi}{\pi/6} = 12$ h, maximum $9$ at $t = 0$, minimum $3$ at $t = 6$. ✓) (b) One full cosine cycle from $(0, 9)$ down to $(6, 3)$ and back to $(12, 9)$, passing through $(3, 6)$ and $(9, 6)$; axes labelled, only the stated domain $[0, 12]$ drawn. (c) $6 + 3\cos\tfrac{\pi t}{6} \ge 7.5 \Rightarrow \cos\tfrac{\pi t}{6} \ge \tfrac{1}{2}$. Within one cycle, $\tfrac{\pi t}{6} \in \left[0, \tfrac{\pi}{3}\right] \cup \left[\tfrac{5\pi}{3}, 2\pi\right]$, i.e. $t \in [0, 2] \cup [10, 12]$ — a total of $\mathbf{4}$ hours per cycle. (Adjust the domain before solving; take both quadrant solutions.)

Question 30

(a) $A_1 = 400\,000(1.005) - M$; $A_2 = A_1(1.005) - M = 400\,000(1.005)^2 - M(1.005) - M = 400\,000(1.005)^2 - M(1 + 1.005)$. (b) Iterating, $A_n = 400\,000(1.005)^n - M\left(1 + 1.005 + \cdots + (1.005)^{n-1}\right)$. The geometric series (with $n$ terms) sums to $\dfrac{(1.005)^n - 1}{0.005} = 200\left((1.005)^n - 1\right)$, giving the stated form. (1 mark building the series with the correct number of terms — the perennial flag; 1 mark summing.) (c) $A_{300} = 0$: $M = \dfrac{400\,000(1.005)^{300} \times 0.005}{(1.005)^{300} - 1}$. With $(1.005)^{300} \approx 4.46497$: $M = \dfrac{2000 \times 4.46497}{3.46497} \approx \$2577.20$. (d) With $M = 3000$: $A_n = 400\,000(1.005)^n - 600\,000\left((1.005)^n - 1\right) = 600\,000 - 200\,000(1.005)^n$. Then $A_n < 200\,000 \Rightarrow (1.005)^n > 2 \Rightarrow n > \dfrac{\ln 2}{\ln 1.005} \approx 138.98$, so $n = 139$ months. ($n$ must be a whole number of months, rounded up.)

Question 31

(a) $h = \dfrac{432}{x^2}$. Base and top: area $2x^2$, cost $2 \times 2x^2 = 4x^2$ cents. Sides: area $4xh = \dfrac{1728}{x}$, cost $1 \times \dfrac{1728}{x}$ cents. Total $C = 4x^2 + \dfrac{1728}{x}$. (b) $C'(x) = 8x - \dfrac{1728}{x^2} = 0 \Rightarrow x^3 = 216 \Rightarrow x = 6$. $C''(x) = 8 + \dfrac{3456}{x^3} > 0$ for $x > 0$, so $x = 6$ gives a minimum. (Verify the nature — the four-times-flagged omission.) (c) $C(6) = 4(36) + \dfrac{1728}{6} = 144 + 288 = 432$ cents $= \$4.32$. (Answer the quantity asked — the cost, in dollars.)