| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Show derivative equals expression - algebraic/trigonometric identity proof |
| Difficulty | Standard +0.8 This question requires applying the product rule twice to expressions involving trigonometric functions, then simplifying to a specific form (part a), followed by finding the second derivative and manipulating it to show a given trigonometric identity (part b). While the techniques are standard A-level calculus, the algebraic manipulation is substantial and the need to work with both derivatives and trigonometric identities across 8 marks makes this moderately challenging, though not requiring novel insight. |
| Spec | 1.07f Convexity/concavity: points of inflection1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07p Points of inflection: using second derivative1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | Uses the product rule for either | |
| term | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| terms | AO1.1a | M1 |
| Differentiates both terms correctly | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| CAO | AO1.1b | A1 |
| (b) | d2y |
| Answer | Marks | Guidance |
|---|---|---|
| derivative and equates to zero | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| awarded in part (a) | AO1.1b | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| marks in (b) have been awarded. | AO1.1b | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| reasoning at the start. AG | AO2.1 | R1 |
| Total | 8 | |
| Q | Marking Instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Uses the product rule for either
term | AO1.1a | M1 | y 2xcos3x(3x2 4)sin3x
dy
2cos3x6xsin3x
dx
6xsin3x3(3x2 4)cos3x
= (9x2 – 10) cos 3x
Uses the product rule for both
terms | AO1.1a | M1
Differentiates both terms correctly | AO1.1b | A1
Rearranges to correct form
CAO | AO1.1b | A1
(b) | d2y
Finds from ‘their’ first
dx2
derivative and equates to zero | AO3.1a | M1 | d2y
18xcos3x3(9x2 10)sin3x
dx2
d2y
point of inflection 0
dx2
18xcos3x3(9x2 10)sin3x0
cos3x 3(9x2 10)
sin3x 18x
9x2 10
cot3x
6x
(AG)
Applies product rule correctly on
dy
‘their’
dx
FT only applies if both M1 marks
awarded in part (a) | AO1.1b | A1F
Arrives at a result using ‘their’
second derivative through correct
algebraic manipulation that is
correct for ‘their’ second derivative
FT only applies if both previous
marks in (b) have been awarded. | AO1.1b | A1F
Constructs a clearly explained
rigorous mathematical argument,
to show the required result
This must include a concluding
statement or an explanation of
reasoning at the start. AG | AO2.1 | R1
Total | 8
Q | Marking Instructions | AO | Marks | Typical SolutionA curve has equation $y = 2x \cos 3x + (3x^2 - 4) \sin 3x$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac{dy}{dx}$, giving your answer in the form $(mx^2 + n) \cos 3x$, where $m$ and $n$ are integers.
[4 marks]
\item Show that the $x$-coordinates of the points of inflection of the curve satisfy the equation
$$\cot 3x = \frac{9x^2 - 10}{6x}$$
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 Q8 [8]}}