| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Differentiate rational functions |
| Difficulty | Standard +0.8 This is a substantial multi-part calculus question requiring domain analysis, quotient rule differentiation, finding and verifying a point of inflection through second derivative analysis, and interpreting concavity. Part (c) is particularly demanding as it requires computing f''(x) from an already complex f'(x), solving f''(x)=0, and proving uniqueness—this involves algebraic manipulation of nested radicals and rational expressions. The 7-mark allocation for part (c) confirms this is above-average difficulty, though it remains within standard A-level techniques without requiring novel mathematical insight. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.07f Convexity/concavity: points of inflection1.07p Points of inflection: using second derivative1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | Deduces that the lower bound of x | |
| is 1 | AO2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| States the domain in a correct form | AO2.5 | A1 |
| Answer | Marks |
|---|---|
| 6(b) | Differentiates using quotient rule |
| Answer | Marks | Guidance |
|---|---|---|
| ( 2x−2 ) | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| unsimplified form | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| show the correct form. AG | AO2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(c) | States that point of inflection | |
| requires second derivative to be 0 | AO2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Forms an equation x OE | AO1.1a | M1 |
| Solves their equation | AO1.1a | M1 |
| Obtains solution x = 4 | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (2x – 2)1/2 stating x ≠ 1. | AO2.4 | E1 |
| Tests either side of ‘their’ x = 4 | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| mark not awarded | AO2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(d) | Deduces values of x for convex | |
| section of graph | AO2.2a | B1 |
| Total | 13 | |
| Q | Marking Instructions | AO |
Question 6:
--- 6(a) ---
6(a) | Deduces that the lower bound of x
is 1 | AO2.2a | M1 | {x : x >1}
∈ℝ
States the domain in a correct form | AO2.5 | A1
--- 6(b) ---
6(b) | Differentiates using quotient rule
(condone correct use of product
rule)
Must have
( 2x−2 )1 2 −kx ( 2x−2 )−1 2
( )=
f ' x OE
( 2x−2 ) | AO1.1a | M1 | ( 2x−2 )1 2 − 1 x ( 2x−2 )−1 2 ×2
( )= 2
f ' x
( 2x−2 )
2x−2−x
=
( 2x−2 )3 2
x−2
=
( 2x−2 )3 2
Obtains correct derivative in
unsimplified form | AO1.1b | A1
Completes algebraic manipulation,
with all previous working correct, to
show the correct form. AG | AO2.1 | R1
--- 6(c) ---
6(c) | States that point of inflection
requires second derivative to be 0 | AO2.4 | E1 | ( )=0
For point of inflection f '' x
(2x−2)2 3 − 3 (x−2)(2x−2)1 2×2
f''(x)= 2
(2x−2)3
( 2x−2 )3 2 −3 ( x−2 )( 2x−2 )1 2 =0
( 2x−2 )1 2 ( 2x−2 )−3 ( x−2 ) =0
( 2x−2 )1 2 ( 4−x )=0
x=1orx=4
x≠1because of domain
1
f '' ( 3 )= >0
32
− 2
f '' ( 5 )= <0
256
Therefore point of inflection at x=4
f′′( )=0
Forms an equation x OE | AO1.1a | M1
Solves their equation | AO1.1a | M1
Obtains solution x = 4 | AO1.1b | A1
Gives a valid reason for rejecting
x = 1, or cancels factor of
(2x – 2)1/2 stating x ≠ 1. | AO2.4 | E1
Tests either side of ‘their’ x = 4 | AO1.1a | M1
Completes rigorous argument to
conclude they have one point of
inflection
Do not award this mark if 2nd E1
mark not awarded | AO2.1 | R1
--- 6(d) ---
6(d) | Deduces values of x for convex
section of graph | AO2.2a | B1 | 1< x < 4
Total | 13
Q | Marking Instructions | AO | Marks | Typical SolutionA function $f$ is defined by $f(x) = \frac{x}{\sqrt{2x - 2}}$
\begin{enumerate}[label=(\alph*)]
\item State the maximum possible domain of $f$.
[2 marks]
\item Use the quotient rule to show that $f'(x) = \frac{x - 2}{(2x - 2)^{\frac{3}{2}}}$.
[3 marks]
\item Show that the graph of $y = f(x)$ has exactly one point of inflection.
[7 marks]
\item Write down the values of $x$ for which the graph of $y = f(x)$ is convex.
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2018 Q6 [13]}}