| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Standard +0.3 This is a standard C4 quotient rule question with routine follow-up parts. Part (i) requires applying the quotient rule to a trigonometric function—straightforward with careful algebra. Parts (ii)-(iv) involve standard techniques: finding tangent equations, locating stationary points by solving f'(x)=0, and reasoning about periodicity. All steps are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation |
| Answer | Marks |
|---|---|
| \(f'(x) = \frac{(-\sin x)\times(2-\sin x)-\cos x\times(-\cos x)}{(2-\sin x)^2}\) | M1 A1 |
| \(= \frac{-2\sin x+\sin^2 x+\cos^2 x}{(2-\sin x)^2}\) | |
| \(= \frac{1-2\sin x}{(2-\sin x)^2}\) | A1 |
| Answer | Marks |
|---|---|
| \(x=\pi,\ y=-\frac{1}{2},\ \text{grad}=\frac{1}{4}\) | B1 |
| \(\therefore y+\frac{1}{2}=\frac{1}{4}(x-\pi) \quad [x-4y-2-\pi=0]\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| SP: \(\frac{1-2\sin x}{(2-\sin x)^2}=0\) | from graph, min. and max. values at SP | |
| \(\sin x = \frac{1}{2}\) | M1 | |
| \(x=\frac{\pi}{6},\ \pi-\frac{\pi}{6}=\frac{\pi}{6},\ \frac{5\pi}{6}\) | A1 | |
| at SP, \(y=\frac{\pm\frac{\sqrt{3}}{2}}{\frac{3}{2}}=\pm\frac{1}{3}\sqrt{3}\) | M1 | |
| \(\therefore\) min. \(= -\frac{1}{3}\sqrt{3}\), max. \(= \frac{1}{3}\sqrt{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin x\) and \(\cos x\) both have period \(2\pi\); \(f(x)\) is a function of \(\sin x\) and \(\cos x\) \(\therefore\) also has period \(2\pi\); \(\therefore\) values of \(f(x)\) in interval \(0\leq x\leq 2\pi\) are just repeated | B2 | (12) |
# Question 8:
## Part (i)
| $f'(x) = \frac{(-\sin x)\times(2-\sin x)-\cos x\times(-\cos x)}{(2-\sin x)^2}$ | M1 A1 | |
|---|---|---|
| $= \frac{-2\sin x+\sin^2 x+\cos^2 x}{(2-\sin x)^2}$ | | |
| $= \frac{1-2\sin x}{(2-\sin x)^2}$ | A1 | |
## Part (ii)
| $x=\pi,\ y=-\frac{1}{2},\ \text{grad}=\frac{1}{4}$ | B1 | |
|---|---|---|
| $\therefore y+\frac{1}{2}=\frac{1}{4}(x-\pi) \quad [x-4y-2-\pi=0]$ | M1 A1 | |
## Part (iii)
| SP: $\frac{1-2\sin x}{(2-\sin x)^2}=0$ | | from graph, min. and max. values at SP |
|---|---|---|
| $\sin x = \frac{1}{2}$ | M1 | |
| $x=\frac{\pi}{6},\ \pi-\frac{\pi}{6}=\frac{\pi}{6},\ \frac{5\pi}{6}$ | A1 | |
| at SP, $y=\frac{\pm\frac{\sqrt{3}}{2}}{\frac{3}{2}}=\pm\frac{1}{3}\sqrt{3}$ | M1 | |
| $\therefore$ min. $= -\frac{1}{3}\sqrt{3}$, max. $= \frac{1}{3}\sqrt{3}$ | A1 | |
## Part (iv)
| $\sin x$ and $\cos x$ both have period $2\pi$; $f(x)$ is a function of $\sin x$ and $\cos x$ $\therefore$ also has period $2\pi$; $\therefore$ values of $f(x)$ in interval $0\leq x\leq 2\pi$ are just repeated | B2 | **(12)** |
|---|---|---|8.\\
\includegraphics[max width=\textwidth, alt={}, center]{a86f277c-a2ec-4ba0-ab08-575cad2a5e53-3_424_698_246_479}
The diagram shows the curve $y = \mathrm { f } ( x )$ in the interval $0 \leq x \leq 2 \pi$ where
$$\mathrm { f } ( x ) = \frac { \cos x } { 2 - \sin x } , \quad x \in \mathbb { R }$$
(i) Show that $\mathrm { f } ^ { \prime } ( x ) = \frac { 1 - 2 \sin x } { ( 2 - \sin x ) ^ { 2 } }$.\\
(ii) Find an equation for the tangent to the curve $y = \mathrm { f } ( x )$ at the point where $x = \pi$.\\
(iii) Find the minimum and maximum values of $\mathrm { f } ( x )$ in the interval $0 \leq x \leq 2 \pi$.\\
(iv) Explain why your answers to part (c) are the minimum and maximum values of $\mathrm { f } ( x )$ for all real values of $x$.
\hfill \mbox{\textit{OCR C4 Q8 [12]}}