| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Evaluate composite at point |
| Difficulty | Moderate -0.3 This is a straightforward composite and inverse functions question with standard techniques. Parts (a)-(c) involve routine procedures (finding range, inverse function, and evaluating a composition), while part (d) requires solving an equation but follows directly from substitution and basic algebra. Slightly easier than average due to the accessibility of all parts. |
| 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 | Mark | Guidance |
| \(f(x) > 3\) | B1 | e.g. \(y > 3\), range \(> 3\), \(f(x) \in (3,\infty)\), \(\{f(x): f(x)>3\}\), \(f > 3\). Not e.g. \(x > 3\), \(f(x) \geqslant 3\), \([3,\infty)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 3 + \sqrt{x-2} \Rightarrow x = \ldots\) | M1 | Sets \(y = 3 + \sqrt{x-2}\) and attempts to make \(x\) the subject. Look for correct order of operations: expression of form \((x=)(y\pm3)^2 \pm 2\) or \((y=)(x\pm3)^2\pm2\) |
| \(f^{-1}(x) = (x-3)^2 + 2\) | A1 | Also accept \(f^{-1}: x \to (x-3)^2+2\). Also accept \(f^{-1}(x) = x^2 - 6x + 11\) (simplified or unsimplified) |
| \(x > 3\) | B1ft | Follow through on their part (a). Allow \(x \in (\text{"3"},\infty)\) or \(\{x: x > \text{"3"}\}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(6) = 3 + \sqrt{6-2} = 5 \Rightarrow g(\text{"5"}) = \frac{15}{\text{"5"}-3} = \ldots\) | M1 | Substitutes \(x=6\) into f and substitutes result into g. Allow attempt to substitute \(x=6\) into \(gf(x) = \frac{15}{\sqrt{x-2}}\) condoning slips. Condone where \(\sqrt{x-2}\) leads to two answers e.g. \(\frac{15}{\sqrt{6-2}} \to \pm\frac{15}{2}\) |
| \(= \frac{15}{2}\) | A1 | \(\frac{15}{2}\) only oe isw once correct answer seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3 + \sqrt{a^2+2-2} = \frac{15}{a-3} \Rightarrow \text{"} a^2 - 9 = 15\text{"}\) | M1 | Attempts to form equation \(3+\sqrt{a^2+2-2} = \frac{15}{a-3}\) and proceeds to quadratic in \(a\) (usually \(a^2 = k\) or \(a^2 - k = 15\)). Condone arithmetical/sign slips. Alternatively forms \(a^2+2 = \left(\frac{15}{a-3}-3\right)^2+2 \Rightarrow (a+3)(a-3)=15 \Rightarrow a^2-9=15\). May be implied by correct exact answer. |
| \(a = 2\sqrt{6}\) | A1 | \((a=) 2\sqrt{6}\) or accept \(\sqrt{24}\). Must reject negative solution. \(\sqrt{6}\times\sqrt{4}\) is A0. isw \(\sqrt{24}\) followed by \(4\sqrt{6}\) but not followed by \(\pm\sqrt{24}\). Decimal answer alone or \(\pm\sqrt{24}\) scores A0. |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) > 3$ | B1 | e.g. $y > 3$, range $> 3$, $f(x) \in (3,\infty)$, $\{f(x): f(x)>3\}$, $f > 3$. **Not** e.g. $x > 3$, $f(x) \geqslant 3$, $[3,\infty)$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 3 + \sqrt{x-2} \Rightarrow x = \ldots$ | M1 | Sets $y = 3 + \sqrt{x-2}$ and attempts to make $x$ the subject. Look for correct order of operations: expression of form $(x=)(y\pm3)^2 \pm 2$ or $(y=)(x\pm3)^2\pm2$ |
| $f^{-1}(x) = (x-3)^2 + 2$ | A1 | Also accept $f^{-1}: x \to (x-3)^2+2$. Also accept $f^{-1}(x) = x^2 - 6x + 11$ (simplified or unsimplified) |
| $x > 3$ | B1ft | Follow through on their part (a). Allow $x \in (\text{"3"},\infty)$ or $\{x: x > \text{"3"}\}$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(6) = 3 + \sqrt{6-2} = 5 \Rightarrow g(\text{"5"}) = \frac{15}{\text{"5"}-3} = \ldots$ | M1 | Substitutes $x=6$ into f and substitutes result into g. Allow attempt to substitute $x=6$ into $gf(x) = \frac{15}{\sqrt{x-2}}$ condoning slips. Condone where $\sqrt{x-2}$ leads to two answers e.g. $\frac{15}{\sqrt{6-2}} \to \pm\frac{15}{2}$ |
| $= \frac{15}{2}$ | A1 | $\frac{15}{2}$ only oe isw once correct answer seen |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3 + \sqrt{a^2+2-2} = \frac{15}{a-3} \Rightarrow \text{"} a^2 - 9 = 15\text{"}$ | M1 | Attempts to form equation $3+\sqrt{a^2+2-2} = \frac{15}{a-3}$ and proceeds to quadratic in $a$ (usually $a^2 = k$ or $a^2 - k = 15$). Condone arithmetical/sign slips. Alternatively forms $a^2+2 = \left(\frac{15}{a-3}-3\right)^2+2 \Rightarrow (a+3)(a-3)=15 \Rightarrow a^2-9=15$. May be implied by correct exact answer. |
| $a = 2\sqrt{6}$ | A1 | $(a=) 2\sqrt{6}$ or accept $\sqrt{24}$. Must reject negative solution. $\sqrt{6}\times\sqrt{4}$ is A0. isw $\sqrt{24}$ followed by $4\sqrt{6}$ but not followed by $\pm\sqrt{24}$. Decimal answer alone or $\pm\sqrt{24}$ scores A0. |\begin{enumerate}
\item The function f is defined by
\end{enumerate}
$$f ( x ) = 3 + \sqrt { x - 2 } \quad x \in \mathbb { R } \quad x > 2$$
(a) State the range of f\\
(b) Find f-1
The function $g$ is defined by
$$g ( x ) = \frac { 15 } { x - 3 } \quad x \in \mathbb { R } \quad x \neq 3$$
(c) Find $g f ( 6 )$\\
(d) Find the exact value of the constant $a$ for which
$$\mathrm { f } \left( a ^ { 2 } + 2 \right) = \mathrm { g } ( a )$$
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q7 [8]}}