| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Moderate -0.3 This is a straightforward composite function question requiring standard techniques: substitution to find gf(x), domain transformation using f's range, identifying range from the quadratic form, and finding the inverse. While multi-part, each step follows routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| \((2x+3-1)^2\) or \((2x+3)^2 - 2(2x+3)+1\) seen | M1 | Substitution |
| Simplified to eg \(4(x+1)^2\) or \(4x^2+8x+4\) or \((2x+2)^2\) | A1 | Mark the final answer; ignore superfluous work on eg finding roots |
| Domain is \(-1 < x < 0\) | B1 [3] | From \(2x+3>1\) |
| Answer | Marks |
|---|---|
| \(0 < \text{gf}(x) < 4\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Factorise their \(\text{gf}(x)\) to obtain perfect square or complete the square | M1 | |
| \(y = 4(x+1)^2\) or \((2x+2)^2\) oe | A1 | FT |
| \((x+1) = (\pm)\sqrt{\frac{y}{4}}\) oe | M1 | |
| \([(\text{gf})^{-1}(x) =]\ \sqrt{\frac{x}{4}}-1\) or \(\frac{\sqrt{x}}{2}-1\) oe | A1 | A1 for \((\text{gf})^{-1}(x) = \sqrt{\frac{x}{4}}-1\) or \(\frac{\sqrt{x}}{2}-1\) oe; or \(g^{-1}(x)=\sqrt{x}+1\) or \(f^{-1}(x)=\frac{1}{2}(x-3)\) for M1; M1 for their \(f^{-1}\)(their \(\sqrt{x}+1\)) |
| Domain is \(0 < x < 4\) | B1 [5] | FT their (b); \(x\) and \(y\) may be interchanged for the first 3 marks but not for the final A1 |
## Question 9:
### Part (a):
$(2x+3-1)^2$ or $(2x+3)^2 - 2(2x+3)+1$ seen | **M1** | Substitution
Simplified to eg $4(x+1)^2$ or $4x^2+8x+4$ or $(2x+2)^2$ | **A1** | Mark the final answer; ignore superfluous work on eg finding roots
Domain is $-1 < x < 0$ | **B1 [3]** | From $2x+3>1$
### Part (b):
$0 < \text{gf}(x) < 4$ | **B1 [1]** |
### Part (c):
Factorise their $\text{gf}(x)$ to obtain perfect square or complete the square | **M1** |
$y = 4(x+1)^2$ or $(2x+2)^2$ oe | **A1** | FT
$(x+1) = (\pm)\sqrt{\frac{y}{4}}$ oe | **M1** |
$[(\text{gf})^{-1}(x) =]\ \sqrt{\frac{x}{4}}-1$ or $\frac{\sqrt{x}}{2}-1$ oe | **A1** | **A1** for $(\text{gf})^{-1}(x) = \sqrt{\frac{x}{4}}-1$ or $\frac{\sqrt{x}}{2}-1$ oe; or $g^{-1}(x)=\sqrt{x}+1$ or $f^{-1}(x)=\frac{1}{2}(x-3)$ for **M1**; **M1** for their $f^{-1}$(their $\sqrt{x}+1$)
Domain is $0 < x < 4$ | **B1 [5]** | FT their (b); $x$ and $y$ may be interchanged for the first 3 marks but not for the final **A1**
---9 You are given that\\
$\mathrm { f } ( x ) = 2 x + 3 \quad$ for $x < 0 \quad$ and\\
$\mathrm { g } ( x ) = x ^ { 2 } - 2 x + 1$ for $x > 1$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { gf } ( x )$, stating the domain.
\item State the range of $\mathrm { gf } ( x )$.
\item Find (gf) ${ } ^ { - 1 } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2019 Q9 [9]}}