| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 This is a comprehensive composite/inverse functions question covering standard techniques: finding inverse of a rational function (routine algebra), determining range and whether a function has an inverse (monotonicity check), and composing functions. While multi-part with several marks, each component uses well-practiced A-level methods without requiring novel insight. Slightly easier than average due to straightforward algebraic manipulation throughout. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{2x+3}{x-2}\), \(xy - 2y = 2x + 3\), \(xy - 2x = 2y + 3\), \(x(y-2) = 2y + 3\) | M1 | Rearranges to make \(x\) the subject by isolating \(x\) terms; or swaps \(x\) and \(y\) and isolates \(y\) terms |
| \(x = \frac{2y+3}{y-2}\) | A1 | Obtains correct rearrangement and factorises; ACF PI by final correct answer |
| \(f^{-1}(x) = \frac{2x+3}{x-2},\ x \neq 2\) | R1 | Obtains \(f^{-1}(x)\) and states domain; must use fully correct notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(ff(x) = x\) | B1 | Obtains any valid expression in \(x\) for \(ff(x)\); can be left unsimplified; ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(g(4) = 6\) | B1 | Deduces the greatest value of \(g\) by evaluating \(g(4)\) |
| Vertex at \((1.25,\ -1.5625)\) | B1 | Obtains the minimum value of \(g\) |
| \(\{y : -1.5625 \leq y \leq 6\}\) | R1F | States the range using their finite greatest value and finite minimum value using set or interval notation; accept \([-1.5625, 6]\) in interval notation; for set notation use of none curly brackets or commas scores R0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(g(0) = 0 = g(2.5)\) | E1 | Demonstrates that \(g\) is a many to one function using an appropriate method; e.g. sketches the function, or evaluates \(g(x)\) at two points that give the same answer |
| \(g\) is many to one so it does not have an inverse | E1 | Deduces that \(g\) is many to one and states \(g\) has no inverse; or explains that \(g\) is not one to one and states \(g\) has no inverse |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(gf(x) = \frac{2\left(\frac{2x+3}{x-2}\right)^2 - 5\left(\frac{2x+3}{x-2}\right)}{2}\) | M1 (1.1a) | Substitutes \(f(x)\) into \(g(x)\) correctly |
| \(= \frac{2(2x+3)^2 - 5(2x+3)(x-2)}{2(x-2)^2}\) | A1 (1.1b) | Obtains common denominator of \(2(x-2)^2\) or \((x-2)^2\) correctly; fraction(s) must have fully correct structure |
| \(= \frac{2(4x^2+12x+9)-5(2x^2-x-6)}{2(x^2-4x+4)}\) | M1 (1.1a) | Expands at least two quadratics correctly |
| \(= \frac{48+29x-2x^2}{2x^2-8x+8}\) | R1 (2.1) | Completes rigorous argument; must have expanded all three quadratics correctly. Terms in numerator and denominator can be in any order. AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States \(g(x)=2\) or states \(2x^2-5x-4=0\) | M1 (3.1a) | PI by solving correct quadratic; PI by sight of \(\frac{5+\sqrt{57}}{4}\) or \(\frac{5-\sqrt{57}}{4}\) |
| \(x = \frac{5 \pm \sqrt{57}}{4}\) | ||
| \(a > 0\) since \(0 \leq x \leq 4\) | ||
| \(a = \frac{5+\sqrt{57}}{4}\) | R1 (2.4) | Determines exact value of \(a\) giving a clear reason for rejection of the negative root |
## Question 13(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{2x+3}{x-2}$, $xy - 2y = 2x + 3$, $xy - 2x = 2y + 3$, $x(y-2) = 2y + 3$ | M1 | Rearranges to make $x$ the subject by isolating $x$ terms; or swaps $x$ and $y$ and isolates $y$ terms |
| $x = \frac{2y+3}{y-2}$ | A1 | Obtains correct rearrangement and factorises; ACF PI by final correct answer |
| $f^{-1}(x) = \frac{2x+3}{x-2},\ x \neq 2$ | R1 | Obtains $f^{-1}(x)$ and states domain; must use fully correct notation |
---
## Question 13(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $ff(x) = x$ | B1 | Obtains any valid expression in $x$ for $ff(x)$; can be left unsimplified; ISW |
---
## Question 13(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $g(4) = 6$ | B1 | Deduces the greatest value of $g$ by evaluating $g(4)$ |
| Vertex at $(1.25,\ -1.5625)$ | B1 | Obtains the minimum value of $g$ |
| $\{y : -1.5625 \leq y \leq 6\}$ | R1F | States the range using their finite greatest value and finite minimum value using set or interval notation; accept $[-1.5625, 6]$ in interval notation; for set notation use of none curly brackets or commas scores R0 |
---
## Question 13(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $g(0) = 0 = g(2.5)$ | E1 | Demonstrates that $g$ is a many to one function using an appropriate method; e.g. sketches the function, or evaluates $g(x)$ at two points that give the same answer |
| $g$ is many to one so it does not have an inverse | E1 | Deduces that $g$ is many to one and states $g$ has no inverse; or explains that $g$ is not one to one and states $g$ has no inverse |
## Question 13(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $gf(x) = \frac{2\left(\frac{2x+3}{x-2}\right)^2 - 5\left(\frac{2x+3}{x-2}\right)}{2}$ | M1 (1.1a) | Substitutes $f(x)$ into $g(x)$ correctly |
| $= \frac{2(2x+3)^2 - 5(2x+3)(x-2)}{2(x-2)^2}$ | A1 (1.1b) | Obtains common denominator of $2(x-2)^2$ or $(x-2)^2$ correctly; fraction(s) must have fully correct structure |
| $= \frac{2(4x^2+12x+9)-5(2x^2-x-6)}{2(x^2-4x+4)}$ | M1 (1.1a) | Expands at least two quadratics correctly |
| $= \frac{48+29x-2x^2}{2x^2-8x+8}$ | R1 (2.1) | Completes rigorous argument; must have expanded all three quadratics correctly. Terms in numerator and denominator can be in any order. AG |
---
## Question 13(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $g(x)=2$ or states $2x^2-5x-4=0$ | M1 (3.1a) | PI by solving correct quadratic; PI by sight of $\frac{5+\sqrt{57}}{4}$ or $\frac{5-\sqrt{57}}{4}$ |
| $x = \frac{5 \pm \sqrt{57}}{4}$ | | |
| $a > 0$ since $0 \leq x \leq 4$ | | |
| $a = \frac{5+\sqrt{57}}{4}$ | R1 (2.4) | Determines exact value of $a$ giving a clear reason for rejection of the negative root |
---13 The function f is defined by
$$\mathrm { f } ( x ) = \frac { 2 x + 3 } { x - 2 } \quad x \in \mathbb { R } , x \neq 2$$
13
\begin{enumerate}[label=(\alph*)]
\item (i) Find f-1\\
13 (a) (ii) Write down an expression for $\mathrm { ff } ( x )$.\\
13
\item The function g is defined by
$$g ( x ) = \frac { 2 x ^ { 2 } - 5 x } { 2 } \quad x \in \mathbb { R } , 0 \leq x \leq 4$$
13 (b) (i) Find the range of g .\\
13 (b) (ii) Determine whether g has an inverse.\\
Fully justify your answer.
\item Show that
$$g f ( x ) = \frac { 48 + 29 x - 2 x ^ { 2 } } { 2 x ^ { 2 } - 8 x + 8 }$$
13
\item It can be shown that fg is given by
$$f g ( x ) = \frac { 4 x ^ { 2 } - 10 x + 6 } { 2 x ^ { 2 } - 5 x - 4 }$$
with domain $\{ x \in \mathbb { R } : 0 \leq x \leq 4 , x \neq a \}$\\
Find the value of $a$.\\
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2020 Q13 [15]}}