1. [Maximum mark: 3]
It is given that \(\theta\) is an acute angle in degrees such that \(\sin\theta=\frac{2}{3}\).
Find the exact value of \(\sin(\theta+60^{\circ})\).
▶️Answer/Explanation
Find \(\cos \theta\) using the Pythagorean identity:
\(\cos \theta = \sqrt{1 – \sin^2 \theta} = \sqrt{1 – \left(\frac{2}{3}\right)^2} = \frac{\sqrt{5}}{3}\).
Use the sine addition formula:
\(\sin(\theta + 60^\circ) = \sin \theta \cos 60^\circ + \cos \theta \sin 60^\circ\).
Substitute the known values:
\(\sin(\theta + 60^\circ) = \frac{2}{3} \cdot \frac{1}{2} + \frac{\sqrt{5}}{3} \cdot \frac{\sqrt{3}}{2} = \frac{1}{3} + \frac{\sqrt{15}}{6}\).
Final Answer: \(\frac{1}{3} + \frac{\sqrt{15}}{6}\).
2. [Maximum mark: 5]
A curve has equation \(y = 3 \tan \left(\frac{1}{2}x\right) \cos 2x\).
Find the gradient of the curve at the point for which \(x = \frac{1}{3}\pi\).
▶️Answer/Explanation
Differentiate \(y = 3 \tan \left(\frac{1}{2}x\right) \cos 2x\) using the product rule:
\(\frac{dy}{dx} = 3 \cdot \frac{1}{2} \sec^2 \left(\frac{1}{2}x\right) \cos 2x + 3 \tan \left(\frac{1}{2}x\right) (-2 \sin 2x)\).
Simplify the derivative:
\(\frac{dy}{dx} = \frac{3}{2} \sec^2 \left(\frac{1}{2}x\right) \cos 2x – 6 \tan \left(\frac{1}{2}x\right) \sin 2x\).
Substitute \(x = \frac{1}{3}\pi\):
\(\sec^2 \left(\frac{\pi}{6}\right) = \frac{4}{3}\), \(\cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2}\), \(\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\), \(\sin \left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\).
Calculate the gradient:
\(\frac{dy}{dx} = \frac{3}{2} \cdot \frac{4}{3} \cdot \left(-\frac{1}{2}\right) – 6 \cdot \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = -1 – 3 = -4\).
Final Answer: \(-4\).
3. [Maximum mark: 6]
(a) Find \(\int_{4}^{10} \frac{4}{2x – 5} \, dx\), giving your answer as \(\ln a\) where \(a\) is an integer. [4]
(b) Find the exact value of \(\int_{4}^{10} e^{2x – 5} \, dx\). [2]
▶️Answer/Explanation
(a)
Integrate to get \(2 \ln |2x – 5|\). Apply limits: \(2 \ln 15 – 2 \ln 3 = 2 \ln 5 = \ln 25\).
Answer: \(\ln 25\).
(b)
Integrate to get \(\frac{1}{2} e^{2x – 5}\). Apply limits: \(\frac{1}{2} e^{15} – \frac{1}{2} e^{3}\).
Answer: \(\frac{1}{2} (e^{15} – e^3)\).
4. [Maximum mark: 7]
(a) Sketch \(y = |3x – 5|\) and \(y = 2x + 7\) on the same diagram. [2]
(b) Solve \(|3x – 5| = 2x + 7\). [3]
(c) Hence solve \(|3^{y+1} – 5| = 2 \times 3^y + 7\) to 3 s.f. [2]
▶️Answer/Explanation
(a)
V-shaped graph for \(|3x – 5|\) (vertex at \(x = \frac{5}{3}\)) and straight line \(y = 2x + 7\).
(b)
Solve \(3x – 5 = 2x + 7 \Rightarrow x = 12\) and \(3x – 5 = -2x – 7 \Rightarrow x = -\frac{2}{5}\).
Answer: \(x = 12\) and \(x = -\frac{2}{5}\).
(c)
Let \(3^y = k\). From (b), \(k = 12 \Rightarrow y = \log_3 12 \approx 2.26\).
Answer: \(y \approx 2.26\).
5. [Maximum mark: 9]
(a) Given \(p(x) = 6x^3 + ax^2 + bx – 20\) has factor \((x + 2)\) and remainder \(-11\) when divided by \((x + 1)\), find \(a\) and \(b\). [5]
(b) Factorise \(p(x)\) and solve \(p(3x) = 0\). [4]
▶️Answer/Explanation
(a)
Using \(p(-2) = 0\) and \(p(-1) = -11\):
Solve \(4a – 2b = 68\) and \(a – b = 15\) to get \(a = 19\), \(b = 4\).
Answer: \(a = 19\), \(b = 4\).
(b)
Factorise: \(p(x) = (x + 2)(2x + 5)(3x – 2)\).
Solve \(p(3x) = 0 \Rightarrow x = -\frac{2}{3}\), \(-\frac{5}{6}\), \(\frac{2}{9}\).
Answer: \(x = -\frac{2}{3}\), \(-\frac{5}{6}\), \(\frac{2}{9}\).
6. [Maximum mark: 9]
(a) Show \(\csc \theta (3 \sin 2\theta + 4 \sin^3 \theta) \equiv 4 + 6 \cos \theta – 4 \cos^2 \theta\). [3]
(b) Solve \(\csc \theta (3 \sin 2\theta + 4 \sin^3 \theta) + 3 = 0\) for \(-\pi < \theta < 0\). [3]
(c) Find \(\int \csc \theta (3 \sin 2\theta + 4 \sin^3 \theta) \, d\theta\). [3]
▶️Answer/Explanation
(a)
Simplify LHS to \(6 \cos \theta + 4 \sin^2 \theta\), then use \(\sin^2 \theta = 1 – \cos^2 \theta\).
Answer: RHS proven.
(b)
Solve \(4 \cos^2 \theta – 6 \cos \theta – 7 = 0\) for \(\cos \theta \approx -0.770\), \(\theta \approx -2.45\).
Answer: \(\theta \approx -2.45\).
(c)
Integrate to get \(2\theta + 6 \sin \theta – \sin 2\theta + C\).
Answer: \(2\theta + 6 \sin \theta – \sin 2\theta + C\).
7. [Maximum mark: 11]
(a) For the curve \(e^{2x} – 18x + y^3 + y = 11\), find the exact \(x\)-coordinate of the stationary point \((p, q)\). [4]
(b) Show \(q = \sqrt[3]{2 + 18 \ln 3 – q}\). [2]
(c) Verify \(q \in (2.5, 3.0)\). [2]
(d) Use iteration to find \(q\) to 4 s.f. [3]
▶️Answer/Explanation
(a)
Differentiate implicitly and set \(\frac{dy}{dx} = 0\) to find \(p = \ln 3\).
Answer: \(p = \ln 3\).
(b)
Substitute \(x = p\) into the curve equation and rearrange.
Answer: As shown.
(c)
Evaluate \(q^3 + q – (2 + 18 \ln 3)\) at \(q = 2.5\) (≈ \(-0.18\)) and \(q = 3.0\) (≈ \(0.34\)).
Answer: Sign change confirms \(q \in (2.5, 3.0)\).
(d)
Iterate \(q_{n+1} = \sqrt[3]{2 + 18 \ln 3 – q_n}\) starting at \(q_0 = 2.5\):
Converges to \(q \approx 2.673\).
Answer: \(q \approx 2.673\).