Year 11 Methods — Unit 1 Paper 1

Tech-free · 60 min · 40 marks · 9 short-answer
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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.
Solution page 2
Handwritten solution
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.
Solution page 2
Handwritten solution
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.
Solution page 3
Handwritten solution
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.
Solution page 4
Handwritten solution
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.
Solution page 5
Handwritten solution
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.
Solution page 6
Handwritten solution
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.
Solution page 7
Handwritten solution
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 \(B\).
Solution page 8Solution page 9
Handwritten solutions
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.
Solution page 10
Handwritten solution
Year 11 Methods — Unit 1 Paper 1
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Jump Cover Q1: Symmetry of sin Q2: Exact values Q3: Sketch a translated reciprocal... Q4: Distance between two points on... Q5: Projectile quadratic Q6: Solve and sketch 1 − 2sin(2x) Q7: Discriminant of a family of qu... Q8: Hyperbola, function notation, ... Q9: Probability — with and without...