| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Standard +0.8 This requires applying the quotient rule to a trigonometric function, then performing substantial algebraic manipulation to simplify the result into the target form involving sin(2θ). The algebraic simplification requires recognizing double angle identities and factoring techniques, going beyond routine differentiation. However, it's a 'show that' question with a clear target, making it more accessible than open-ended problems. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{(2\sin\theta + 2\cos\theta)3\cos\theta - 3\sin\theta(2\cos\theta - 2\sin\theta)}{(2\sin\theta + 2\cos\theta)^2}\) | M1, A1 | 1.1b, 1.1b |
| Expands and uses \(\sin^2\theta + \cos^2\theta = 1\) at least once in numerator or denominator, or uses \(2\sin\theta\cos\theta = \sin 2\theta \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{\ldots}{\ldots C\sin\theta\cos\theta}\) | M1 | 3.1a |
| Expands and uses \(\sin^2\theta + \cos^2\theta = 1\) in numerator AND denominator AND uses \(2\sin\theta\cos\theta = \sin 2\theta \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{P}{Q + R\sin 2\theta}\) | M1 | 2.1 |
| \(\frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{3}{2 + 2\sin 2\theta} = \frac{\frac{3}{2}}{1+\sin 2\theta}\) | A1 | 1.1b |
# Question 5:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{(2\sin\theta + 2\cos\theta)3\cos\theta - 3\sin\theta(2\cos\theta - 2\sin\theta)}{(2\sin\theta + 2\cos\theta)^2}$ | M1, A1 | 1.1b, 1.1b |
| Expands and uses $\sin^2\theta + \cos^2\theta = 1$ at least once in numerator or denominator, or uses $2\sin\theta\cos\theta = \sin 2\theta \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{\ldots}{\ldots C\sin\theta\cos\theta}$ | M1 | 3.1a |
| Expands and uses $\sin^2\theta + \cos^2\theta = 1$ in numerator AND denominator AND uses $2\sin\theta\cos\theta = \sin 2\theta \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{P}{Q + R\sin 2\theta}$ | M1 | 2.1 |
| $\frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{3}{2 + 2\sin 2\theta} = \frac{\frac{3}{2}}{1+\sin 2\theta}$ | A1 | 1.1b |
**Notes:** M1 for choosing quotient, product rule or implicit differentiation and applying to given function. Condone $\frac{\mathrm{d}y}{\mathrm{d}x}$ notation and tolerate slips on coefficients; condone $\frac{\mathrm{d}(\sin\theta)}{\mathrm{d}\theta} = \pm\cos\theta$ and $\frac{\mathrm{d}(\cos\theta)}{\mathrm{d}\theta} = \pm\sin\theta$. A1: correct expression involving $\frac{\mathrm{d}y}{\mathrm{d}\theta}$. Final A1: fully correct proof with $A = \frac{3}{2}$ stated; allow $\frac{\frac{3}{2}}{1+\sin 2\theta}$. Allow recovery from missing brackets. This is not a given answer.\begin{enumerate}
\item Given that
\end{enumerate}
$$y = \frac { 3 \sin \theta } { 2 \sin \theta + 2 \cos \theta } \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$
show that
$$\frac { d y } { d \theta } = \frac { A } { 1 + \sin 2 \theta } \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$
where $A$ is a rational constant to be found.
\hfill \mbox{\textit{Edexcel Paper 1 2018 Q5 [5]}}