| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Complete the square |
| Difficulty | Moderate -0.8 This is a routine multi-part question on completing the square and function transformations. Part (a) is standard completing the square, parts (b)-(d) are direct applications of understanding vertex and domain/range for one-one functions, and part (e) is straightforward translation. All parts follow predictable patterns with no problem-solving required, making it easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\{-3(x-2)^2\}\ \ \{+14\}\) | B1 B1 | B1 for each correct term; condone \(a = 2\), \(b = 14\). |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([k =]\ 2\) | B1 | Allow \([x] \leqslant 2\). |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Range is] \([y] \leqslant -13\) | B1 | Allow \([\text{f}(x)] \leqslant -13\), \([\text{f}] \leqslant -13\) but NOT \(x \leqslant -13\). |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = -3(x-2)^2 + 14\) leading to \((x-2)^2 = \frac{14-y}{3}\) | M1 | Allow \(\frac{y-14}{-3}\). Allow 1 error in rearrangement if \(x\), \(y\) on opposite sides. |
| \(x = 2(\pm)\sqrt{\frac{14-y}{3}}\) | A1 | Allow \(\frac{y-14}{-3}\). |
| \([\text{f}^{-1}(x)] = 2 - \sqrt{\frac{14-x}{3}}\) | A1 | OE. Allow \(\frac{x-14}{-3}\). Must be \(x\) on RHS; must be negative square root only. |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = -3(y-2)^2 + 14\) leading to \((y-2)^2 = \frac{14-x}{3}\) | M1 | Allow \(\frac{x-14}{-3}\). Allow 1 error in rearrangement if \(x\), \(y\) on opposite sides. |
| \(= 2(\pm)\sqrt{\frac{14-x}{3}}\) | A1 | Allow \(\frac{x-14}{-3}\). |
| \([\text{f}^{-1}(x)] = 2 - \sqrt{\frac{14-x}{3}}\) | A1 | OE. Allow \(\frac{x-14}{-3}\). Must be \(x\) on RHS; must be negative square root only. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\text{g}(x) =]\ \{-3(x+3-2)^2\} + \{14+1\}\) | B2, 1, 0 | OR \(\{-3(x+3)^2\} + \{12(x+3)\} + \{3\}\) |
| \(g(x) = -3x^2 - 6x + 12\) | B1 | |
| Total: 3 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\{-3(x-2)^2\}\ \ \{+14\}$ | B1 B1 | B1 for each correct term; condone $a = 2$, $b = 14$. |
| **Total: 2** | | |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[k =]\ 2$ | B1 | Allow $[x] \leqslant 2$. |
| **Total: 1** | | |
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Range is] $[y] \leqslant -13$ | B1 | Allow $[\text{f}(x)] \leqslant -13$, $[\text{f}] \leqslant -13$ but NOT $x \leqslant -13$. |
| **Total: 1** | | |
## Question 8(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = -3(x-2)^2 + 14$ leading to $(x-2)^2 = \frac{14-y}{3}$ | M1 | Allow $\frac{y-14}{-3}$. Allow 1 error in rearrangement if $x$, $y$ on opposite sides. |
| $x = 2(\pm)\sqrt{\frac{14-y}{3}}$ | A1 | Allow $\frac{y-14}{-3}$. |
| $[\text{f}^{-1}(x)] = 2 - \sqrt{\frac{14-x}{3}}$ | A1 | OE. Allow $\frac{x-14}{-3}$. Must be $x$ on RHS; must be negative square root only. |
| **Total: 3** | | |
**Alternative method for 8(d):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = -3(y-2)^2 + 14$ leading to $(y-2)^2 = \frac{14-x}{3}$ | M1 | Allow $\frac{x-14}{-3}$. Allow 1 error in rearrangement if $x$, $y$ on opposite sides. |
| $= 2(\pm)\sqrt{\frac{14-x}{3}}$ | A1 | Allow $\frac{x-14}{-3}$. |
| $[\text{f}^{-1}(x)] = 2 - \sqrt{\frac{14-x}{3}}$ | A1 | OE. Allow $\frac{x-14}{-3}$. Must be $x$ on RHS; must be negative square root only. |
---
## Question 8(e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{g}(x) =]\ \{-3(x+3-2)^2\} + \{14+1\}$ | B2, 1, 0 | OR $\{-3(x+3)^2\} + \{12(x+3)\} + \{3\}$ |
| $g(x) = -3x^2 - 6x + 12$ | B1 | |
| **Total: 3** | | |
---8
\begin{enumerate}[label=(\alph*)]
\item Express $- 3 x ^ { 2 } + 12 x + 2$ in the form $- 3 ( x - a ) ^ { 2 } + b$, where $a$ and $b$ are constants.\\
The one-one function f is defined by $\mathrm { f } : x \mapsto - 3 x ^ { 2 } + 12 x + 2$ for $x \leqslant k$.
\item State the largest possible value of the constant $k$.\\
It is now given that $k = - 1$.
\item State the range of f.
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.\\
The result of translating the graph of $y = \mathrm { f } ( x )$ by $\binom { - 3 } { 1 }$ is the graph of $y = \mathrm { g } ( x )$.
\item Express $\mathrm { g } ( x )$ in the form $p x ^ { 2 } + q x + r$, where $p , q$ and $r$ are constants.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q8 [10]}}