| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of composite quotient/product |
| Difficulty | Standard +0.3 Part (a) is a guided proof of a standard derivative using quotient rule. Part (b) applies product rule with chain rule to a straightforward combination. Part (c) requires setting derivative to zero and solving a simple trigonometric equation, then substituting back. All techniques are standard C3 material with clear structure and no novel insight required, making this slightly easier than average. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07h Differentiation from first principles: for sin(x) and cos(x)1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -1(\cos x)^{-2}(-\sin x)\) | M1 | \(\frac{dy}{dx} = \pm\left((\cos x)^{-2}(\sin x)\right)\) |
| \(-1(\cos x)^{-2}(-\sin x)\) or \((\cos x)^{-2}(\sin x)\) | A1 | |
| \(\frac{dy}{dx} = \left\{\frac{\sin x}{\cos^2 x}\right\} = \left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) = \sec x \tan x\) | A1 AG | Convincing proof; must see both underlined steps |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = e^{2x},\ v = \sec 3x\); \(\frac{du}{dx} = 2e^{2x}\), \(\frac{dv}{dx} = 3\sec 3x \tan 3x\) | M1 | Either \(e^{2x} \to 2e^{2x}\) or \(\sec 3x \to 3\sec 3x\tan 3x\) (seen or implied) |
| Both \(e^{2x} \to 2e^{2x}\) and \(\sec 3x \to 3\sec 3x\tan 3x\) | A1 | |
| Applies \(vu' + uv'\) correctly for their \(u, u', v, v'\) | M1 | |
| \(\frac{dy}{dx} = 2e^{2x}\sec 3x + 3e^{2x}\sec 3x\tan 3x\) | A1 isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(\frac{dy}{dx} = 0\); \(e^{2x}\sec 3x(2 + 3\tan 3x) = 0\) | M1 | Sets \(\frac{dy}{dx} = 0\) and factorises/cancels \(e^{2x}\) from at least two terms |
| \(\{e^{2x} \neq 0,\ \sec 3x \neq 0\}\) so \(\tan 3x = -\frac{2}{3}\) | M1 | \(\tan 3x = \pm k;\ k \neq 0\) |
| \(3x = -0.58800 \Rightarrow x = \{a\} = -0.19600\ldots\) | A1 | Either awrt \(-0.196^c\) or awrt \(-11.2°\) |
| \(y = \{b\} = e^{2(-0.196)}\sec(3 \times -0.196) = 0.812\) | A1 cao |
## Question 7:
### Part (a): $y = \sec x = \frac{1}{\cos x} = (\cos x)^{-1}$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -1(\cos x)^{-2}(-\sin x)$ | M1 | $\frac{dy}{dx} = \pm\left((\cos x)^{-2}(\sin x)\right)$ |
| $-1(\cos x)^{-2}(-\sin x)$ or $(\cos x)^{-2}(\sin x)$ | A1 | |
| $\frac{dy}{dx} = \left\{\frac{\sin x}{\cos^2 x}\right\} = \left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) = \sec x \tan x$ | A1 AG | Convincing proof; must see both underlined steps |
### Part (b): $y = e^{2x}\sec 3x$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = e^{2x},\ v = \sec 3x$; $\frac{du}{dx} = 2e^{2x}$, $\frac{dv}{dx} = 3\sec 3x \tan 3x$ | M1 | Either $e^{2x} \to 2e^{2x}$ or $\sec 3x \to 3\sec 3x\tan 3x$ (seen or implied) |
| Both $e^{2x} \to 2e^{2x}$ and $\sec 3x \to 3\sec 3x\tan 3x$ | A1 | |
| Applies $vu' + uv'$ correctly for their $u, u', v, v'$ | M1 | |
| $\frac{dy}{dx} = 2e^{2x}\sec 3x + 3e^{2x}\sec 3x\tan 3x$ | A1 isw | |
### Part (c): Turning point
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\frac{dy}{dx} = 0$; $e^{2x}\sec 3x(2 + 3\tan 3x) = 0$ | M1 | Sets $\frac{dy}{dx} = 0$ and factorises/cancels $e^{2x}$ from at least two terms |
| $\{e^{2x} \neq 0,\ \sec 3x \neq 0\}$ so $\tan 3x = -\frac{2}{3}$ | M1 | $\tan 3x = \pm k;\ k \neq 0$ |
| $3x = -0.58800 \Rightarrow x = \{a\} = -0.19600\ldots$ | A1 | Either awrt $-0.196^c$ or awrt $-11.2°$ |
| $y = \{b\} = e^{2(-0.196)}\sec(3 \times -0.196) = 0.812$ | A1 cao | |
> **Note:** If there are EXTRA solutions for $x$ inside $-\frac{\pi}{6} < x < \frac{\pi}{6}$ or ANY extra solutions for $y$, withhold the final accuracy mark.
---\begin{enumerate}
\item (a) By writing $\sec x$ as $\frac { 1 } { \cos x }$, show that $\frac { \mathrm { d } ( \sec x ) } { \mathrm { d } x } = \sec x \tan x$.
\end{enumerate}
Given that $y = \mathrm { e } ^ { 2 x } \sec 3 x$,\\
(b) find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
The curve with equation $y = \mathrm { e } ^ { 2 x } \sec 3 x , - \frac { \pi } { 6 } < x < \frac { \pi } { 6 }$, has a minimum turning point at $( a , b )$.\\
(c) Find the values of the constants $a$ and $b$, giving your answers to 3 significant figures.
\hfill \mbox{\textit{Edexcel C3 2010 Q7 [11]}}