| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative satisfies condition |
| Difficulty | Standard +0.3 Part (a) is a guided proof using quotient rule with standard trig derivatives. Part (b) applies product rule to a compound function using the result from (a). Part (c) requires setting derivative to zero and algebraic manipulation to reach the given form. This is a standard C3 multi-part question with clear scaffolding, slightly above average due to the algebraic manipulation in part (c) and the need to work with inverse trig functions, but all techniques are routine for this level. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07g Differentiation from first principles: for small positive integer powers of x1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(y=(\sin x)^{-1}\), then \(\frac{dy}{dx}=-1(\sin x)^{-2}\times \cos x\) | M1 A1 | M1: Use of chain rule \(\frac{dy}{dx}=-1(\sin x)^{-2}\times(\pm\cos x)\); A1: cao |
| \(\frac{dy}{dx}=\frac{-1}{\sin x}\times\frac{\cos x}{\sin x}=-\text{cosec}\,x\cot x\) | B1* (3) | Use of definitions of cosec and cot and conclusion; dependent on M1; need at least intermediate step \(\frac{dy}{dx}=\frac{-\cos x}{\sin x \sin x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx}=3e^{3x}\text{cosec}\,2x + e^{3x}\times -2\text{cosec}\,2x\cot 2x\) | M1 A1 A1 (3) | M1: product rule attempt; A1: one term correct; A1: both terms correct (need not simplify) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx}=e^{3x}\text{cosec}\,2x(3-2\cot 2x)=0\) | M1 | Puts \(\frac{dy}{dx}=0\) and factorises/cancels by \(e^{3x}\text{cosec}\,2x\) concluding \(a\pm b\cot 2x=0\) or \(\cot 2x=\pm\frac{a}{b}\) |
| \(\cot 2x=1.5\), \(\tan 2x=\frac{2}{3}\) so \(x=\frac{1}{2}\arctan\frac{2}{3}\) (or \(k=2/3\)) | A1 (2) | Draws correct conclusion \(\frac{1}{2}\arctan\frac{2}{3}\) or \(k=2/3\) |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $y=(\sin x)^{-1}$, then $\frac{dy}{dx}=-1(\sin x)^{-2}\times \cos x$ | M1 A1 | M1: Use of chain rule $\frac{dy}{dx}=-1(\sin x)^{-2}\times(\pm\cos x)$; A1: cao |
| $\frac{dy}{dx}=\frac{-1}{\sin x}\times\frac{\cos x}{\sin x}=-\text{cosec}\,x\cot x$ | B1* (3) | Use of definitions of cosec and cot and conclusion; dependent on M1; need at least intermediate step $\frac{dy}{dx}=\frac{-\cos x}{\sin x \sin x}$ |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=3e^{3x}\text{cosec}\,2x + e^{3x}\times -2\text{cosec}\,2x\cot 2x$ | M1 A1 A1 (3) | M1: product rule attempt; A1: one term correct; A1: both terms correct (need not simplify) |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=e^{3x}\text{cosec}\,2x(3-2\cot 2x)=0$ | M1 | Puts $\frac{dy}{dx}=0$ and factorises/cancels by $e^{3x}\text{cosec}\,2x$ concluding $a\pm b\cot 2x=0$ or $\cot 2x=\pm\frac{a}{b}$ |
| $\cot 2x=1.5$, $\tan 2x=\frac{2}{3}$ so $x=\frac{1}{2}\arctan\frac{2}{3}$ (or $k=2/3$) | A1 (2) | Draws correct conclusion $\frac{1}{2}\arctan\frac{2}{3}$ or $k=2/3$ |
---\begin{enumerate}
\item (a) By writing $\operatorname { cosec } x$ as $\frac { 1 } { \sin x }$, show that
\end{enumerate}
$$\frac { \mathrm { d } ( \operatorname { cosec } x ) } { \mathrm { d } x } = - \operatorname { cosec } x \cot x$$
Given that $y = \mathrm { e } ^ { 3 x } \operatorname { cosec } 2 x , 0 < x < \frac { \pi } { 2 }$,\\
(b) find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
The curve with equation $y = \mathrm { e } ^ { 3 x } \operatorname { cosec } 2 x , 0 < x < \frac { \pi } { 2 }$, has a single turning point.\\
(c) Show that the $x$-coordinate of this turning point is at $x = \frac { 1 } { 2 } \arctan k$ where the value\\
of the constant $k$ should be found. of the constant $k$ should be found.\\
\hfill \mbox{\textit{Edexcel C3 2014 Q3 [8]}}