| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Topic | Product & Quotient Rules |
| Type | Integration with differentiation context |
| Difficulty | Standard +0.3 Part (i) is a straightforward application of the quotient rule with chain rule to verify a given result (tan 3θ = sin 3θ/cos 3θ), requiring standard differentiation techniques. Part (ii) uses the result to integrate tan²3θ by rearranging to isolate it, then integrating—a standard 'hence' question requiring algebraic manipulation and basic integration. This is slightly easier than average as it's guided, routine calculus with no novel problem-solving required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.07q Product and quotient rules: differentiation1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
**Question 8:**
**(i)** $\tan 3\theta = \frac{\sin 3\theta}{\cos 3\theta}$ **M1** (OR $\frac{\text{d}(\tan 3\theta)}{\text{d}\theta} = 3\sec^2 3\theta$ M1; $3(1+\tan^2 3\theta)$ M1)
$\frac{\text{d}(\tan 3\theta)}{\text{d}\theta} = \frac{3\cos^2 3\theta + 3\sin^2 3\theta}{\cos^2 3\theta}$ **M1**
$= 3 + 3\tan^2 3\theta$ **A1** (Dep on M marks)
**(ii)** $3\int \tan^2 3\theta\,\text{d}\theta = \tan 3\theta - \int 3\,\text{d}\theta + c$ **M1** (Some attempt to show evidence for the use of (i))
$\int_{\pi/12}^{\pi/9} \tan^2 3\theta\,\text{d}\theta = \left[\frac{1}{3}\tan 3\theta - \theta\right]_{\pi/12}^{\pi/9}$ **M1** (Award for $a\tan 3\theta + b\theta$; $x$ appearing loses the A1)
[Correct integral] **A1**
$\left(\frac{1}{3}\tan\frac{\pi}{3} - \frac{\pi}{9}\right) - \left(\frac{1}{3}\tan\frac{\pi}{4} - \frac{\pi}{12}\right)$ **M1** (Award for sight only of substitution of correct limits)
$\frac{1}{3}(\sqrt{3}-1) - \frac{\pi}{36}$ **A1** (aef)
**Total: 7 marks**8 (i) Using the quotient rule, show that $\frac { \mathrm { d } } { \mathrm { d } \theta } ( \tan 3 \theta ) = 3 + 3 \tan ^ { 2 } 3 \theta$ for $- \frac { 1 } { 6 } \pi < \theta < \frac { 1 } { 6 } \pi$.\\
(ii) Hence find the value of $\int _ { \frac { 1 } { 12 } \pi } ^ { \frac { 1 } { 9 } \pi } \tan ^ { 2 } 3 \theta \mathrm {~d} \theta$, giving your answer in the simplest exact form.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2018 Q8 [7]}}