| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Determine if inverse exists |
| Difficulty | Standard +0.8 This question requires finding the domain (routine), identifying a discontinuity to explain why the intermediate value theorem fails (moderate insight), and using calculus to find turning points via quotient rule to determine injectivity for inverse existence (multi-step analysis). The combination of conceptual understanding about continuity, careful differentiation, and reasoning about one-to-one functions makes this above average difficulty. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.07a Derivative as gradient: of tangent to curve1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Domain excludes negatives or excludes 3 | M1 (AO1.1a) | Condone \(x > 0\) |
| \(x \geq 0\) and \(x \neq 3\) | A1 (AO2.2a) | No extras; condone \(x > 0\) |
| \(\{x: x \geq 0, x \neq 3\}\) | R1 (AO2.5) | Correct domain correctly stated in set notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h(x)\) has a discontinuity/asymptote at \(x = 3\) or in interval \((1,4)\) | M1 (AO2.4) | States discontinuity at \(x=3\); OE |
| \(h(x)\) is not continuous at \(x=3\); change of sign between \(x=1\) and \(4\) does not imply a root | A1 (AO2.3) | Explains discontinuity is at \(x=3\) and this is in the interval \((1,4)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h(x) = \frac{\sqrt{x}}{x-3}\); \(h'(x) = \frac{\frac{1}{2}x^{-\frac{1}{2}}(x-3) - x^{\frac{1}{2}}}{(x-3)^2}\) | M1 (AO3.1a) | Selects appropriate method; reaches \(h'(x)\) of form \(ax^{-\frac{1}{2}}(x-3)^{-1} + bx^{\frac{1}{2}}(x-3)^{-2}\); OE |
| Correct \(h'(x)\) | A1 (AO1.1b) | ACF |
| \(h'(x) = 0 \Rightarrow \frac{\frac{1}{2}x^{-\frac{1}{2}}(x-3) - x^{\frac{1}{2}}}{(x-3)^2} = 0\) | M1 (AO1.1a) | Equates \(h'(x)\) to 0 |
| \(\frac{1}{2}x^{-\frac{1}{2}}(x-3) - x^{\frac{1}{2}} = 0 \Rightarrow \frac{(x-3)}{2\sqrt{x}} - \sqrt{x} = 0 \Rightarrow x - 3 - 2x = 0 \Rightarrow x = -3\) | A1 (AO1.1b) | Obtains \(x = -3\) |
| \(x = -3\) not in domain of \(h\); no turning points | E1 (AO2.4) | Continuous function with no turning points is one-to-one therefore inverse exists; condone omission of 'continuous' |
| \(h(x) > 0\) for \(x > 3\); \(h(x) < 0\) for \(x < 3\); hence one-to-one and has an inverse | R1 (AO2.1) | Must explain \(x=-3\) not in domain; must explain no turning points; considers sign of \(h(x)\) either side of \(x=3\) |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Domain excludes negatives or excludes 3 | M1 (AO1.1a) | Condone $x > 0$ |
| $x \geq 0$ and $x \neq 3$ | A1 (AO2.2a) | No extras; condone $x > 0$ |
| $\{x: x \geq 0, x \neq 3\}$ | R1 (AO2.5) | Correct domain correctly stated in set notation |
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## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h(x)$ has a discontinuity/asymptote at $x = 3$ or in interval $(1,4)$ | M1 (AO2.4) | States discontinuity at $x=3$; OE |
| $h(x)$ is not continuous at $x=3$; change of sign between $x=1$ and $4$ does not imply a root | A1 (AO2.3) | Explains discontinuity is at $x=3$ and this is in the interval $(1,4)$ |
---
## Question 10(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h(x) = \frac{\sqrt{x}}{x-3}$; $h'(x) = \frac{\frac{1}{2}x^{-\frac{1}{2}}(x-3) - x^{\frac{1}{2}}}{(x-3)^2}$ | M1 (AO3.1a) | Selects appropriate method; reaches $h'(x)$ of form $ax^{-\frac{1}{2}}(x-3)^{-1} + bx^{\frac{1}{2}}(x-3)^{-2}$; OE |
| Correct $h'(x)$ | A1 (AO1.1b) | ACF |
| $h'(x) = 0 \Rightarrow \frac{\frac{1}{2}x^{-\frac{1}{2}}(x-3) - x^{\frac{1}{2}}}{(x-3)^2} = 0$ | M1 (AO1.1a) | Equates $h'(x)$ to 0 |
| $\frac{1}{2}x^{-\frac{1}{2}}(x-3) - x^{\frac{1}{2}} = 0 \Rightarrow \frac{(x-3)}{2\sqrt{x}} - \sqrt{x} = 0 \Rightarrow x - 3 - 2x = 0 \Rightarrow x = -3$ | A1 (AO1.1b) | Obtains $x = -3$ |
| $x = -3$ not in domain of $h$; no turning points | E1 (AO2.4) | Continuous function with no turning points is one-to-one therefore inverse exists; condone omission of 'continuous' |
| $h(x) > 0$ for $x > 3$; $h(x) < 0$ for $x < 3$; hence one-to-one and has an inverse | R1 (AO2.1) | Must explain $x=-3$ not in domain; must explain no turning points; considers sign of $h(x)$ either side of $x=3$ |
---10 The function h is defined by
$$\mathrm { h } ( x ) = \frac { \sqrt { x } } { x - 3 }$$
where $h$ has its maximum possible domain.\\
10
\begin{enumerate}[label=(\alph*)]
\item Find the domain of h .\\
Give your answer using set notation.
10
\item Alice correctly calculates
$$h ( 1 ) = - 0.5 \text { and } h ( 4 ) = 2$$
She then argues that since there is a change of sign there must be a value of $x$ in the interval $1 < x < 4$ that gives $\mathrm { h } ( x ) = 0$
Explain the error in Alice's argument.\\[0pt]
[2 marks]\\
10
\item By considering any turning points of h , determine whether h has an inverse function.
Fully justify your answer.\\[0pt]
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2021 Q10 [11]}}