| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - trigonometric functions |
| Difficulty | Moderate -0.3 Part (a) is a guided proof using the chain rule with explicit hints—straightforward application requiring d/dx(cos x) and the chain rule. Part (b) involves finding a stationary point by setting f'(x)=0, solving a trigonometric equation (2sec²x = 3sec x tan x leads to standard values), then substituting back. While it requires multiple steps and algebraic manipulation, the techniques are all standard C3 material with no novel insight required. Slightly easier than average due to the scaffolding in part (a). |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = -1 \times (\cos x)^{-2} \times -\sin x\) | M1 | For M1, condone 'dropping one minus sign' and/or poor use of brackets; candidates must use chain rule; clear use of quotient rule scores 0/2 |
| \(= \dfrac{\sin x}{\cos^2 x}\) | Must see 'a middle line' | |
| \(= \dfrac{\sin x}{\cos x} \times \dfrac{1}{\cos x}\) | ||
| \(= \tan x \sec x\) | A1 | AG, all correct and no errors seen; Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = 2\sec^2 x - 3\sec x \tan x\) | M1 | \(m\sec^2 x + n\sec x\tan x\) |
| \([\sec x](2\sec x - 3\tan x) [= 0]\) | m1 | \([\sec x](m\sec x + n\tan x)[=0]\) |
| \(\sin x = \dfrac{2}{3}\) | A1 | Finding any correct exact trig ratio |
| \(\cos x = \dfrac{\sqrt{5}}{3}\), \(\quad \tan x = \dfrac{2}{\sqrt{5}}\), \(\quad \sec x = \dfrac{3}{\sqrt{5}}\) | A1 | Finding a second correct exact trig ratio |
| \(y = 2 \times \dfrac{2}{\sqrt{5}} - 3 \times \dfrac{3}{\sqrt{5}}\) | M1 | For subst their exact values correctly into '\(y\)'; PI by correct final answer following previous 4 marks earned |
| \(y = -\sqrt{5}\) | A1 CSO | Must have used correct exact values throughout; if second M mark not earned, SC1 for AWRT \(-2.24\) or \(-\sqrt{5}\); Total: 6 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = -1 \times (\cos x)^{-2} \times -\sin x$ | M1 | For M1, condone 'dropping one minus sign' and/or poor use of brackets; candidates must use chain rule; clear use of quotient rule scores 0/2 |
| $= \dfrac{\sin x}{\cos^2 x}$ | | Must see 'a middle line' |
| $= \dfrac{\sin x}{\cos x} \times \dfrac{1}{\cos x}$ | | |
| $= \tan x \sec x$ | A1 | AG, all correct and no errors seen; **Total: 2** |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = 2\sec^2 x - 3\sec x \tan x$ | M1 | $m\sec^2 x + n\sec x\tan x$ |
| $[\sec x](2\sec x - 3\tan x) [= 0]$ | m1 | $[\sec x](m\sec x + n\tan x)[=0]$ |
| $\sin x = \dfrac{2}{3}$ | A1 | Finding **any** correct exact trig ratio |
| $\cos x = \dfrac{\sqrt{5}}{3}$, $\quad \tan x = \dfrac{2}{\sqrt{5}}$, $\quad \sec x = \dfrac{3}{\sqrt{5}}$ | A1 | Finding a second correct exact trig ratio |
| $y = 2 \times \dfrac{2}{\sqrt{5}} - 3 \times \dfrac{3}{\sqrt{5}}$ | M1 | For subst their exact values correctly into '$y$'; PI by correct final answer following previous 4 marks earned |
| $y = -\sqrt{5}$ | A1 CSO | Must have used correct exact values throughout; if second M mark not earned, **SC1** for AWRT $-2.24$ or $-\sqrt{5}$; **Total: 6** |7
\begin{enumerate}[label=(\alph*)]
\item By writing $\sec x = ( \cos x ) ^ { - 1 }$, use the chain rule to show that, if $y = \sec x$, then
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec x \tan x$$
\item The function f is defined by
$$\mathrm { f } ( x ) = 2 \tan x - 3 \sec x , \text { for } 0 < x < \frac { \pi } { 2 }$$
Find the value of the $y$-coordinate of the stationary point of the graph of $y = \mathrm { f } ( x )$, giving your answer in the form $p \sqrt { q }$, where $p$ and $q$ are integers.\\[0pt]
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2016 Q7 [8]}}