Year 11 Methods — Paper 1 + Paper 2

Full Unit 1 review

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Pap Q 1 3 marks Symmetry of sin

\(\sin(\theta) = 0.46\) and \(0 < \theta < \dfrac{\pi}{2}\). Find:
(a) \(\sin(\pi + \theta)\) (1)
(b) \(\sin(-\theta)\) (1)
(c) \(\cos\!\left(\dfrac{3\pi}{2} + \theta\right)\) (1)

Approach

Use the unit-circle symmetry identities. \(\theta\) is in the first quadrant (top-right), so everything starts positive.
\(\sin(\pi + \theta)\) is in the third quadrant — sine is negative there.
\(\sin(-\theta)\) is just the reflection of \(\sin\theta\) through 0 — negative.
\(\cos\!\left(\dfrac{3\pi}{2} + \theta\right)\) shifts the cosine graph by \(\dfrac{3\pi}{2}\), which turns cosine into sine.

Key step

(a) \(\sin(\pi + \theta) = -\sin\theta\).
(b) \(\sin(-\theta) = -\sin\theta\).
(c) \(\cos\!\left(\dfrac{3\pi}{2} + \theta\right) = \sin\theta\).

Answer

(a) \(-0.46\) (b) \(-0.46\) (c) \(0.46\)

Watch out

Sketch the unit circle if you can't remember the identities. The signs come from which quadrant the angle lands in.

Handwritten solution

Pap Q 2 3 marks Exact values

Evaluate:
(a) \(\sin\!\left(-\dfrac{5\pi}{6}\right)\)
(b) \(\tan(120^{\circ})\)
(c) \(\tan\!\left(\dfrac{\pi}{4}\right) + \sin\!\left(\dfrac{\pi}{3}\right)\)

Approach

For each one: find the reference angle (the acute angle made with the x-axis), use the exact-value table, then attach the correct sign based on which quadrant the angle is in.
(a) \(-\dfrac{5\pi}{6}\) is in the third quadrant; reference angle \(\dfrac{\pi}{6}\); sine is negative.
(b) \(120°\) is in the second quadrant; reference angle \(60°\); tan is negative.
(c) Both angles are in the first quadrant; everything positive.

Key step

(a) \(-\sin\!\left(\dfrac{\pi}{6}\right) = -\dfrac{1}{2}\).
(b) \(-\tan(60°) = -\sqrt{3}\).
(c) \(1 + \dfrac{\sqrt{3}}{2} = \dfrac{2 + \sqrt{3}}{2}\).

Answer

(a) \(-\dfrac{1}{2}\) (b) \(-\sqrt{3}\) (c) \(\dfrac{2 + \sqrt{3}}{2}\)

Watch out

The sign depends only on the quadrant. A negative angle just means rotating clockwise.

Handwritten solution

Pap Q 3 4 marks Sketch a translated reciprocal-squared curve

Sketch \(f:[-4, 4] \to \mathbb{R},\ f(x) = \dfrac{1}{(x+1)^{2}} - 4\). Label endpoints, intercepts and asymptotes.

Approach

Start from the parent curve \(\dfrac{1}{x^{2}}\) (always positive, two branches going up either side of \(x = 0\)). Shift it 1 left (so asymptote is now \(x = -1\)) and 4 down (so the horizontal asymptote is \(y = -4\)).
Find the x-intercepts by setting \(f(x) = 0\). Find the y-intercept by plugging in \(x = 0\). Find the endpoints by plugging in \(x = -4\) and \(x = 4\).

Key step

x-intercepts: \(\dfrac{1}{(x+1)^{2}} = 4 \Rightarrow (x+1)^{2} = \dfrac{1}{4} \Rightarrow x = -\dfrac{1}{2}\) or \(x = -\dfrac{3}{2}\).
y-intercept: \(f(0) = 1 - 4 = -3\).
Endpoints: \(f(-4) = \dfrac{1}{9} - 4\); \(f(4) = \dfrac{1}{25} - 4\).

Answer

Curve has two branches around \(x = -1\); crosses x-axis at \(x = -\dfrac{3}{2}\) and \(x = -\dfrac{1}{2}\); approaches \(y = -4\) on both ends.

Watch out

Both branches go UP near the asymptote because \(\dfrac{1}{(\text{anything})^{2}}\) is always positive. Don't draw the right branch going down.

Handwritten solution

Pap Q 4 4 marks Distance between two points on a hyperbola

\(f:\mathbb{R}\setminus\{0\} \to \mathbb{R},\ f(x) = \dfrac{1}{x} - 2\). \(P\) is the x-intercept. Find the distance from \(P\) to the point on \(f\) where \(x = -2\).

Approach

Find \(P\) by setting \(f(x) = 0\). Find the second point by plugging in \(x = -2\). Then use the distance formula. Keep surds exact.

Key step

\(P\): \(\dfrac{1}{x} - 2 = 0 \Rightarrow x = \dfrac{1}{2}\), so \(P = \left(\dfrac{1}{2}, 0\right)\).
At \(x = -2\): \(f(-2) = -\dfrac{1}{2} - 2 = -\dfrac{5}{2}\), so the other point is \(Q = \left(-2, -\dfrac{5}{2}\right)\).
Distance \(= \sqrt{\left(\dfrac{1}{2} - (-2)\right)^{2} + \left(0 - \left(-\dfrac{5}{2}\right)\right)^{2}} = \sqrt{\left(\dfrac{5}{2}\right)^{2} + \left(\dfrac{5}{2}\right)^{2}} = \sqrt{\dfrac{50}{4}} = \dfrac{5\sqrt{2}}{2}\).

Answer

\(\dfrac{5\sqrt{2}}{2}\) units (equivalently \(\dfrac{\sqrt{50}}{2}\)).

Watch out

Simplify the surd: \(\sqrt{50} = 5\sqrt{2}\). 'Exact' means leaving the surd, not a decimal.

Handwritten solution

Pap Q 5 6 marks Projectile quadratic

A ball is thrown into the air. Starts at ground level, reaches \(16\,\text{m}\) at \(t = 2\), returns to ground.
(a) Show \(h(t) = -4t^{2} + 16t\). (3)
(b) Height at \(t = 1\). (1)
(c) When is the height \(12\,\text{m}\)? (2)

Approach

Use vertex form: \(h(t) = a(t - 2)^{2} + 16\). Then use the fact that the ball starts at ground level (\(t = 0,\ h = 0\)) to find \(a\). Expand into standard form to match the required answer.
For (c) solve \(-4t^{2} + 16t = 12\) — quadratic, two solutions because the ball passes 12 m on the way up AND on the way down.

Key step

(a) \(0 = a(0 - 2)^{2} + 16 \Rightarrow a = -4\). Expand: \(-4(t^{2} - 4t) + 16 = -4t^{2} + 16t\).
(b) \(h(1) = -4 + 16 = 12\,\text{m}\).
(c) \(-4t^{2} + 16t = 12 \Rightarrow t^{2} - 4t + 3 = 0 \Rightarrow (t-1)(t-3) = 0\), so \(t = 1\) or \(t = 3\) seconds.

Answer

(b) \(12\,\text{m}\). (c) \(t = 1\,\text{s}\) (going up) and \(t = 3\,\text{s}\) (coming down).

Watch out

Two answers in (c). One on the way up, one on the way down. Both valid.

Handwritten solution

Pap Q 6 6 marks Solve and sketch 1 − 2sin(2x)

(a) Solve \(1 - 2\sin(2x) = 0\) for \(x \in [-\pi, \pi]\). (3)
(b) Sketch \(y = 1 - 2\sin(2x)\) on \([-\pi, \pi]\), labelling intercepts and endpoints. (3)

Approach

For (a), rearrange to \(\sin(2x) = \dfrac{1}{2}\). The reference angle is \(\dfrac{\pi}{6}\). Solve over the WIDER range \(2x \in [-2\pi, 2\pi]\) first, find all answers, then divide by 2 at the end.
For (b), think of it as the sine graph with: amplitude 2, flipped upside down (because of the \(-2\)), squashed (period changes from \(2\pi\) to \(\pi\)), shifted up by 1. Range is \([-1, 3]\).

Key step

For \(\sin(2x) = \dfrac{1}{2}\) with \(2x \in [-2\pi, 2\pi]\):
\(2x = \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ -\dfrac{11\pi}{6},\ -\dfrac{7\pi}{6}\). Divide by 2:
\(x = -\dfrac{11\pi}{12},\ -\dfrac{7\pi}{12},\ \dfrac{\pi}{12},\ \dfrac{5\pi}{12}\).

Answer

Four solutions: \(x = -\dfrac{11\pi}{12},\ -\dfrac{7\pi}{12},\ \dfrac{\pi}{12},\ \dfrac{5\pi}{12}\).

Watch out

Don't divide your range by 2 at the start. Solve for \(2x\) first over the bigger range, then divide.

Handwritten solution

Pap Q 7 3 marks Discriminant of a family of quadratics

\(y = 2kx^{2} - kx + (1 - k)\), \(k \in \mathbb{R}\setminus\{0\}\).
(a) Show that the discriminant is \(\Delta = k(9k - 8)\). (2)
(b) For which \(k\) does the graph have two x-intercepts? (1)

Approach

Discriminant formula: \(\Delta = b^{2} - 4ac\). Here \(a = 2k\), \(b = -k\), \(c = 1 - k\).
Two x-intercepts means \(\Delta > 0\). Solve \(k(9k - 8) > 0\) by considering each factor's sign.

Key step

(a) \(\Delta = (-k)^{2} - 4(2k)(1 - k) = k^{2} - 8k + 8k^{2} = 9k^{2} - 8k = k(9k - 8)\).
(b) \(k(9k - 8) > 0\) when both factors are positive or both negative: \(k > \dfrac{8}{9}\) or \(k < 0\).

Answer

\(k > \dfrac{8}{9}\) or \(k < 0\) (and \(k \neq 0\)).

Watch out

Both factors positive OR both negative. A sign table or quick sketch helps.

Handwritten solution

Pap Q 8 7 marks Hyperbola, function notation, inverse

Graph of \(h\) shown: y-intercept \((0,\dfrac{5}{2})\), horizontal asymptote \(y = 3\), vertical asymptote \(x = 2\).
(a) Find \(h(x)\). (2)
(b) Write \(h\) in full function notation. (1)
(c) State the domain of \(h^{-1}\). (1)
(d) Write \(h^{-1}(x) = \dfrac{A}{x + B} + C\) and find \(A,\ B,\ C\). (3)

Approach

(a) The general hyperbola is \(h(x) = \dfrac{a}{x - 2} + 3\) (asymptotes built in). Use the y-intercept to find \(a\).
(b) Add the domain: everything except where the vertical asymptote is.
(c) Domain of the inverse = range of the original.
(d) Swap \(x\) and \(y\), then rearrange to solve for \(y\). Compare to the required form to read off \(A, B, C\).

Key step

(a) \(\dfrac{5}{2} = \dfrac{a}{-2} + 3 \Rightarrow a = 1\). So \(h(x) = \dfrac{1}{x - 2} + 3\).
(b) \(h:\mathbb{R}\setminus\{2\} \to \mathbb{R},\ h(x) = \dfrac{1}{x - 2} + 3\).
(c) Range of \(h\) is \(\mathbb{R}\setminus\{3\}\), so that's the domain of \(h^{-1}\).
(d) Swap: \(x = \dfrac{1}{y - 2} + 3 \Rightarrow y - 2 = \dfrac{1}{x - 3} \Rightarrow y = \dfrac{1}{x - 3} + 2\). So \(A = 1,\ B = -3,\ C = 2\).

Answer

\(h(x) = \dfrac{1}{x - 2} + 3\); \(h^{-1}(x) = \dfrac{1}{x - 3} + 2\); \(A = 1,\ B = -3,\ C = 2\).

Watch out

The asymptote of the original becomes the EXCLUDED value of the inverse's domain. Watch the sign of \(B\).

Handwritten solutions

Pap Q 9 4 marks Probability — with and without independence

\(\Pr(A) = \dfrac{1}{6},\ \Pr(B) = \dfrac{1}{4}\).
(a) Show that \(\Pr(A \cap B') = \dfrac{1}{15}\) when \(\Pr(A \cap B) = \dfrac{1}{10}\). (1)
(b) Calculate \(\Pr(A | B')\). (1)
(c) If \(A\) and \(B\) are independent, calculate \(\Pr(A \cap B')\). (2)

Approach

(a) Split \(\Pr(A)\) into 'A and B' plus 'A and not B'. So \(\Pr(A \cap B') = \Pr(A) - \Pr(A \cap B)\).
(b) Conditional formula: \(\Pr(A | B') = \dfrac{\Pr(A \cap B')}{\Pr(B')}\). And \(\Pr(B') = 1 - \dfrac{1}{4} = \dfrac{3}{4}\).
(c) For independence, multiply the probabilities: \(\Pr(A \cap B') = \Pr(A) \cdot \Pr(B')\).

Key step

(a) \(\dfrac{1}{6} - \dfrac{1}{10} = \dfrac{5}{30} - \dfrac{3}{30} = \dfrac{2}{30} = \dfrac{1}{15}\). ✓
(b) \(\dfrac{1/15}{3/4} = \dfrac{1}{15} \times \dfrac{4}{3} = \dfrac{4}{45}\).
(c) Independent: \(\dfrac{1}{6} \times \dfrac{3}{4} = \dfrac{3}{24} = \dfrac{1}{8}\).

Answer

(b) \(\dfrac{4}{45}\) (c) \(\dfrac{1}{8}\)

Watch out

Parts (a) and (c) calculate the SAME probability — but only (c) assumes independence. They give different answers, which is the point.

Handwritten solution

Pap MC 1 1 mark Domain of f(x) = x − 5

\(f:D \to \mathbb{R},\ f(x) = x - 5\) has range \((-5, 4]\). Find \(D\).

Approach

The range tells us the smallest and largest values \(f\) reaches. Solve \(f(x) = -5\) and \(f(x) = 4\) to find the corresponding x-values. Match the brackets: round bracket on \(-5\) → round bracket on the x-value too; square bracket on \(4\) → square bracket on the x-value.

Key step

\(x - 5 = -5 \Rightarrow x = 0\) (round); \(x - 5 = 4 \Rightarrow x = 9\) (square). So \(D = (0, 9]\).

Answer

A — \((0, 9]\)

Watch out

Brackets carry over. Don't swap them.

Handwritten solution

Pap MC 2 1 mark Period and range of 3sin(2x/5) − 2

\(f:\mathbb{R} \to \mathbb{R},\ f(x) = 3\sin\!\left(\dfrac{2x}{5}\right) - 2\). State the period and range.

Approach

Period: take the coefficient of \(x\) inside the sine — here \(\dfrac{2}{5}\) — and divide \(2\pi\) by it.
Range: the amplitude (number out front) is 3, so sine oscillates between \(-3\) and \(+3\). Shift down by 2 gives a range of \([-5, 1]\).

Key step

Period \(= \dfrac{2\pi}{2/5} = 5\pi\). Range \(= [-3, 3] - 2 = [-5, 1]\).

Answer

B — \(5\pi\) and \([-5, 1]\)

Watch out

Don't forget to apply the \(-2\) shift when calculating the range.

Handwritten solution

Pap MC 3 1 mark Axis of symmetry

For \(y = ax^{2} - 2bx + c\), the axis of symmetry is:

Approach

Axis of symmetry formula: \(x = -\dfrac{\text{coefficient of }x}{2 \cdot \text{coefficient of }x^{2}}\). The coefficient of \(x\) here is \(-2b\), NOT \(2b\) — careful with the sign.

Key step

\(x = -\dfrac{-2b}{2a} = \dfrac{b}{a}\).

Answer

D — \(x = \dfrac{b}{a}\)

Watch out

Read the sign of the middle term carefully. The minus is part of the coefficient.

Handwritten solution

Pap MC 4 1 mark Period of tan(πx/4)

The period of \(y = \tan\!\left(\dfrac{\pi x}{4}\right)\) is:

Approach

The basic tan curve has period \(\pi\) (not \(2\pi\) like sin and cos). For \(\tan(kx)\), the period is \(\dfrac{\pi}{|k|}\). Here \(k = \dfrac{\pi}{4}\).

Key step

Period \(= \dfrac{\pi}{\pi/4} = \pi \cdot \dfrac{4}{\pi} = 4\).

Answer

C — \(4\)

Watch out

Tan has period \(\pi\), NOT \(2\pi\). Easy to mix up with sin/cos.

Handwritten solution

Pap MC 5 1 mark Transformed point on g(x) = ½f(x + 1)

\(g(x) = \dfrac{1}{2} f(x + 1)\). The point \((3, 2)\) lies on \(f\). What point lies on \(g\)?

Approach

Two transformations: the \((x + 1)\) inside means shift LEFT by 1 (opposite sign of what's written). The \(\dfrac{1}{2}\) outside means halve the y-value. So \((x, y) \to (x - 1,\ y/2)\).

Key step

\((3, 2) \to (3 - 1,\ 2/2) = (2, 1)\).

Answer

A — \((2, 1)\)

Watch out

Inside the function = horizontal change (opposite sign). Outside = vertical change (same sign).

Handwritten solution

Pap MC 6 1 mark Inverse of g(x) = (x² + 6)/2

\(g:[0, \infty) \to \mathbb{R},\ g(x) = \dfrac{x^{2} + 6}{2}\). Find \(g^{-1}\).

Approach

Swap \(x\) and \(y\), then solve for \(y\). Take the POSITIVE square root because the original domain is \([0, \infty)\). The domain of the inverse is the range of the original — the smallest value \(g\) reaches is \(g(0) = 3\).

Key step

Swap: \(x = \dfrac{y^{2} + 6}{2} \Rightarrow y^{2} = 2x - 6 \Rightarrow y = \sqrt{2x - 6}\) (positive root).
Domain: \([3, \infty)\) (= range of \(g\)).

Answer

B — \(g^{-1}:[3, \infty) \to \mathbb{R},\ g^{-1}(x) = \sqrt{2x - 6}\)

Watch out

The restricted original domain means take only the positive square root. Domain of inverse = range of original.

Handwritten solution

Pap MC 7 1 mark Function with maximal domain R \ {5}

The maximal domain of \(f\) is \(\mathbb{R}\setminus\{5\}\). Which rule could \(f\) have?

Approach

\(\mathbb{R}\setminus\{5\}\) means everything except \(x = 5\). So plugging in \(x = 5\) must make the function undefined — usually a zero denominator.

Key step

Try each: \(\dfrac{x + 4}{x - 5}\) is undefined when \(x - 5 = 0\), i.e. \(x = 5\). ✓

Answer

B — \(f(x) = \dfrac{x + 4}{x - 5}\)

Watch out

Option (a) \(\dfrac{x^{2} - 5}{x - 1}\) is undefined at \(x = 1\), not 5.

Handwritten solution

Pap MC 8 1 mark Graph of the inverse

Part of \(f\) shown (decreasing curve in two pieces). Which graph is \(f^{-1}\)?

Approach

The graph of \(f^{-1}\) is the reflection of \(f\) in the line \(y = x\). Swap the x and y axes. Wherever the original curve goes, the inverse goes with x and y swapped.

Key step

Reflect in \(y = x\): each point \((x, y)\) becomes \((y, x)\).

Answer

C

Watch out

Reflection is in \(y = x\) — not the x-axis or y-axis.

Handwritten solution

Pap MC 9 1 mark Clock minute-hand height

Minute hand 10 cm, clock face radius 15 cm. At 12:45 the hand points horizontally LEFT and moves clockwise. \(h(t)\) is the height of the tip above the base of the clock face, \(t\) minutes after 12:45.

Approach

The centre of the clock sits 15 cm above the base (since the radius is 15). The tip's height above the centre is \(10\sin(\text{angle})\). A full rotation is 60 min, so the angular speed is \(\dfrac{2\pi}{60} = \dfrac{\pi}{30}\) rad/min.
At \(t = 0\) the hand points LEFT, so the tip is at clock-centre height: \(h(0) = 15\). The sine model that starts at 0 and goes UP first matches the clockwise motion (the tip rises towards 12 first).

Key step

\(h(t) = 15 + 10\sin\!\left(\dfrac{\pi t}{30}\right)\).

Answer

A — \(h(t) = 15 + 10\sin\!\left(\dfrac{\pi t}{30}\right)\)

Watch out

Direction of rotation determines whether you use \(+\sin\) or \(-\sin\). Check by plugging in \(t = 15\) (a quarter rotation later) and seeing whether the tip is at the right height.

Handwritten solution

Pap MC 10 1 mark Reflect then dilate g(x) = √(2x + 5)

\(g(x) = \sqrt{2x + 5}\). Reflect in the y-axis, then dilate from the y-axis by a factor of \(\dfrac{1}{2}\). Find \(f\).

Approach

Reflect in the y-axis: replace \(x\) with \(-x\).
Dilate from the y-axis by factor \(\dfrac{1}{2}\): this squashes the graph horizontally, which means inputs are scaled UP by 2. So replace \(x\) with \(2x\).

Key step

After reflection: \(\sqrt{-2x + 5}\). After dilation: replace \(x\) with \(2x\): \(\sqrt{-2(2x) + 5} = \sqrt{5 - 4x}\).

Answer

A — \(f(x) = \sqrt{5 - 4x}\)

Watch out

'Dilation from the y-axis by factor 1/2' means the graph gets HALF AS WIDE. Inside \(x\) becomes \(2x\), not \(\dfrac{x}{2}\).

Handwritten solution

Pap MC 11 1 mark Find Pr(A) given independence

\(A, B\) are independent. \(\Pr(B) = 2\Pr(A)\) and \(\Pr(A \cup B) = 0.52\). Find \(\Pr(A)\).

Approach

For independent events, \(\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)\). Let \(p = \Pr(A)\), so \(\Pr(B) = 2p\) and \(\Pr(A \cap B) = 2p^{2}\). Now sub into the addition rule:
\(\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)\). Solve the resulting quadratic in \(p\).

Key step

\(0.52 = p + 2p - 2p^{2} \Rightarrow 2p^{2} - 3p + 0.52 = 0\). Solve (CAS): \(p = 0.2\).

Answer

B — \(0.2\)

Watch out

The quadratic has two solutions; pick the one with \(p < 1\) (probabilities can't exceed 1).

Handwritten solution

Pap MC 12 1 mark Two real solutions of x² + 2x − k = 0

For what values of \(k\) does \(x^{2} + 2x - k = 0\) have two real solutions?

Approach

'Two real solutions' means discriminant \(\Delta > 0\). Here \(a = 1,\ b = 2,\ c = -k\).
So \(\Delta = b^{2} - 4ac = 4 - 4(-k) = 4 + 4k\).

Key step

\(4 + 4k > 0 \Rightarrow k > -1\). So \(k \in (-1, \infty)\).

Answer

B — \((-1, \infty)\)

Watch out

\(c = -k\), NOT \(+k\), because of the minus sign in the equation. That flips the sign in the discriminant.

Handwritten solution

Pap MC 13 1 mark Transformed point through h(x) = f(x/2) + 5

\(f\) passes through \((-2, 7)\). \(h(x) = f(x/2) + 5\). Through what point does \(h\) pass?

Approach

The \(x/2\) inside scales x-inputs by 2 (graph gets twice as wide). The \(+5\) outside shifts everything up by 5. So \((x, y) \to (2x,\ y + 5)\).

Key step

\((-2, 7) \to (2 \cdot -2,\ 7 + 5) = (-4, 12)\).

Answer

C — \((-4, 12)\)

Watch out

\(x/2\) inside means x-values DOUBLE (graph stretches). Easy to think it halves instead.

Handwritten solution

Pap MC 14 1 mark sin(x) + cos(y) with quadrant info

\(\cos x = \dfrac{3}{5}\); \(\sin^{2} y = \dfrac{25}{169}\); \(x, y \in \left[\dfrac{3\pi}{2}, 2\pi\right]\). Find \(\sin x + \cos y\).

Approach

Both angles are in the fourth quadrant (\(\dfrac{3\pi}{2}\) to \(2\pi\)). In Q4: cos is positive, sin is negative.
Use the Pythagorean identity \(\sin^{2}\theta + \cos^{2}\theta = 1\) to find the missing function, then attach the sign based on the quadrant.

Key step

\(\sin x = -\sqrt{1 - \dfrac{9}{25}} = -\dfrac{4}{5}\) (negative in Q4).
\(\cos y = +\sqrt{1 - \dfrac{25}{169}} = +\dfrac{12}{13}\) (positive in Q4).
Sum: \(-\dfrac{4}{5} + \dfrac{12}{13} = \dfrac{-52 + 60}{65} = \dfrac{8}{65}\).

Answer

A — \(\dfrac{8}{65}\)

Watch out

Quadrant determines the SIGN. Always check before stamping a plus or minus on a square root.

Handwritten solution

Pap MC 15 1 mark Two marbles — same colour

Box: 6 red and 4 blue marbles. Two drawn without replacement. Probability they are the same colour?

Approach

Either both red, or both blue. Add the two probabilities.
\(\Pr(RR) = \dfrac{6}{10} \times \dfrac{5}{9}\) (after one red, only 5 reds and 9 total left).
\(\Pr(BB) = \dfrac{4}{10} \times \dfrac{3}{9}\).

Key step

\(\Pr(RR) + \Pr(BB) = \dfrac{30}{90} + \dfrac{12}{90} = \dfrac{42}{90} = \dfrac{7}{15}\).

Answer

C — \(\dfrac{7}{15}\)

Watch out

Without replacement — the denominator drops by 1 for the second draw.

Handwritten solution

Pap ER 1 19 marks Mission: helicopter, parachute, piecewise path

Helicopter: \(h(x) = -\dfrac{10000}{x - 100}\). Jump happens at height \(200\,\text{m}\). Parachute path: \(p(x) = (x - a)^{2} + b\) (from jump point). At \((100, 100)\) Ethan pulls his chute. After that, path is \(q(x) = p^{-1}(x)\).

(a) Height at \(x = 30\). (1) (b) Show \(x = 50\) when \(h = 200\). (1) (c) Jump coordinate. (1)
(d) Domain of \(h\). (1) (e) Sketch \(h\). (2)
(f) Find \(b\). (1) (g) Domain of \(p\). (1) (h) Write \(p\) in full function notation. (1) (i) Sketch \(p\). (2)
(j) Find \(p^{-1}(x)\). (3) (k) Sketch \(q\). (2)
(l) Does Ethan land on the headquarters? (1) (m) Full path as piecewise function. (2)

Approach

(a)–(c) Just substitute into \(h\). Jump point comes out at \((50, 200)\).
(d)–(e) Helicopter goes from take-off \(x=0\) to jump \(x=50\), so domain \([0, 50]\).
(f) The parachute path is \(f(x) = a(x-h)^2 + k\) where \((h,k)=(50,200)\) is the jump point. Use the point \((100,100)\) to find \(a\).
(g) Parachute is active from jump \(x=50\) to chute-pull \(x=100\), so domain \([50,100]\).
(j) Inverse: swap \(x\) and \(y\), solve for \(y\). You get \(\pm\) — pick the sign that matches the descending path \((100,100)\to(200,50)\).
(l) Test \(p(200)\). If it equals \(50\), he lands on the HQ roof.
(m) Stitch three pieces together. Use \([0, 50)\), \([50, 100)\), \([100, 200]\) so domains don't overlap.

Key step

(a) \(h(30) = -\dfrac{10000}{-70} \approx 142.86\,\text{m}\).
(b) \(200 = -\dfrac{10000}{x-100} \Rightarrow x-100 = -50 \Rightarrow x = 50\). ✓
(c) Jump at \((50, 200)\).
(f) \(100 = a(100-50)^2 + 200 \Rightarrow -100 = 2500a \Rightarrow a = -\dfrac{1}{25}\).
(j) Swap and solve: \(x = -\dfrac{1}{25}(y-50)^2 + 200 \Rightarrow y = 50 + 5\sqrt{200-x}\) (positive root for descent).

Answer

Jump coord: \((50, 200)\). \(a = -\dfrac{1}{25}\). \(f^{-1}(x) = 50 + 5\sqrt{200-x}\). YES — \(p(200) = 50\), he lands on the roof.
Piecewise: helicopter \(-\dfrac{10000}{x-100}\) on \([0,50)\); parachute \(-\dfrac{1}{25}(x-50)^2 + 200\) on \([50,100)\); chute \(50 + 5\sqrt{200-x}\) on \([100,200]\).

Watch out

In (j) you MUST show the \(\pm\) before picking the positive root — it's worth a mark. For the piecewise function in (m), adjust the brackets so the domains don't overlap.

Handwritten solutions

Pap ER 2 13 marks Fan blades — diving Ethan

Fan diameter \(2\,\text{m}\), spinning at 30 rpm. Blade 1: \(B_{1}(t) = \cos(at) + \dfrac{11}{10}\) (height in metres).
(a) Diameter of the shaft (\(>\) 2 m). (1)
(b) Time for one rotation. (1)
(c) Show \(a = \pi\). (1)
(d) Sketch \(B_{1}(t)\) for one full cycle. (3)
(e) Sketch Blade 2 \(B_{2}(t)\) on same axes (starts at lowest point). (2)
(f) Equation for \(B_{2}(t)\). (2)
(g) Safe dive time — both blades at least \(0.5\,\text{m}\) up. (3)

Approach

(b) Period = time for one rotation = \(\dfrac{60}{30} = 2\,\text{s}\).
(c) Period \(= \dfrac{2\pi}{a}\), so \(a = \dfrac{2\pi}{2} = \pi\).
(e)/(f) Blade 2 is opposite Blade 1 (they're 180° apart on the same fan). So shift the curve by half a period (1 second), giving \(B_{2}(t) = \cos(\pi(t - 1)) + \dfrac{11}{10} = -\cos(\pi t) + \dfrac{11}{10}\).
(g) 'Both blades above 0.5 m' means the minimum of the two curves is above 0.5. Solve to find the times when one of them dips to exactly 0.5, then measure the gap between those crossings.

Key step

(b) \(T = 2\,\text{s}\). (c) \(a = \pi\).
(f) \(B_{2}(t) = -\cos(\pi t) + \dfrac{11}{10}\).
(g) Use CAS to solve \(B_{1}(t) = 0.5\) and \(B_{2}(t) = 0.5\), then find the safe window length.

Answer

Safe dive interval \(\approx 0.41\) seconds per half-rotation (from solving \(B_{1}=0.5\) at \(t \approx 0.7048\) and \(B_{2}=0.5\) at \(t \approx 0.2951\); window \(\approx 0.4097\,\text{s}\)).

Watch out

'Both blades above 0.5 m' means take the MINIMUM of the two heights. The safe window is symmetric within each half-rotation, so you only need to solve once.

Handwritten solutions

Pap ER 3 7 marks Safe code probability

Ethan must guess a 3-digit code (digits 0–9, repeats allowed). Y = correct, N = incorrect.
(a) Complete tree diagram with all probabilities. (3)
(b) Probability he guesses correctly. (1)
(c) Probability he guesses the code INCORRECTLY (i.e. doesn't open the safe). (1)
(d) Probability of at least two digits correct. (1)
(e) Probability all three correct GIVEN at least two are correct. (1)

Approach

\(\Pr(Y) = \dfrac{1}{10}\), \(\Pr(N) = \dfrac{9}{10}\) for each digit. Digits are independent (with replacement).
IMPORTANT for (c): 'Guess the code INCORRECTLY' means the safe doesn't open — i.e. the code is wrong, which happens when ANY digit is wrong. So it's the COMPLEMENT of all three correct, not all three wrong.
For 'at least two correct': add the probability of all three right and the probability of exactly two right. Exactly two right has three orderings (YYN, YNY, NYY).

Key step

(b) All three correct: \(\left(\dfrac{1}{10}\right)^{3} = \dfrac{1}{1000}\).
(c) Incorrect code = 1 - P(all 3 correct) = \(1 - \dfrac{1}{1000} = \dfrac{999}{1000}\).
(d) At least 2 correct = \(\dfrac{1}{1000} + 3 \cdot \left(\dfrac{1}{10}\right)^{2} \cdot \dfrac{9}{10} = \dfrac{1 + 27}{1000} = \dfrac{28}{1000} = \dfrac{7}{250}\).
(e) Conditional: \(\dfrac{\Pr(\text{all 3})}{\Pr(\geq 2)} = \dfrac{1/1000}{28/1000} = \dfrac{1}{28}\).

Answer

(b) \(\dfrac{1}{1000}\) (c) \(\dfrac{999}{1000}\) (d) \(\dfrac{7}{250}\) (e) \(\dfrac{1}{28}\)

Watch out

(c) is the trap — 'incorrect code' means safe doesn't open, so just \(1 - \dfrac{1}{1000}\). It is NOT \(\left(\dfrac{9}{10}\right)^{3}\), which would be all three digits wrong.

Handwritten solutions

Pap ER 4 5 marks Power-saw cuts — maximise area

Total cut = \(30\,\text{cm}\). Three straight cuts: one horizontal of length \(x\) plus two vertical of length \(y\) each.
(a) Express \(y\) in terms of \(x\). (1)
(b) Area function \(A(x)\). (1)
(c) Domain of \(A\), given shortest cut is \(5\,\text{cm}\). (1)
(d) Maximum area. (1)
(e) \(x\) and \(y\) at the maximum. (1)

Approach

Total cut length: \(x + 2y = 30\), so \(y = \dfrac{30 - x}{2}\).
The cut traces out a rectangle of width \(x\) and height \(y\): \(A = x \cdot y = \dfrac{x(30-x)}{2}\). This is a downward parabola in \(x\).
Domain: both \(x \geq 5\) and \(y \geq 5\). \(y \geq 5\) gives \(\dfrac{30-x}{2} \geq 5 \Rightarrow x \leq 20\). So \(x \in [5, 20]\).
Maximum at the vertex of the parabola: \(x = 15\). Then \(y = \dfrac{30 - 15}{2} = 7.5\).

Key step

(a) \(y = \dfrac{30-x}{2}\). (b) \(A(x) = \dfrac{x(30-x)}{2}\). (c) \(x \in [5, 20]\).
(d) \(A(15) = \dfrac{15 \cdot 15}{2} = \dfrac{225}{2} = 112.5\,\text{cm}^{2}\). (e) \(x = 15\,\text{cm},\ y = 7.5\,\text{cm}\).

Answer

Maximum area = \(\dfrac{225}{2} = 112.5\,\text{cm}^{2}\) at \(x = 15\,\text{cm},\ y = 7.5\,\text{cm}\).

Watch out

There are TWO \(y\)-cuts but only ONE \(x\)-cut, so total length is \(x + 2y = 30\) (not \(2x + y\)). Domain restriction applies to BOTH \(x\) and \(y\).

Handwritten solution