| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2007 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Determine if inverse exists |
| Difficulty | Moderate -0.3 This is a standard P1 question on inverse functions requiring completing the square, identifying range from vertex form, explaining one-to-one requirement, and finding inverse by restricting domain. All parts follow routine procedures with no novel problem-solving, making it slightly easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(f(x) = 2(x - 2)^2 + 3\) | 3 × B1 [3] | B1 for each of 2, -2 and 3. – no need to equate with \(a\), \(b\) and \(c\). |
| (ii) Range is \(\geq 3\) | B1∨ [1] | Condone >3. √ for \(x \geq c\). |
| (iii) Not 1:1 (2 x-values for 1 y-value) or curve is quadratic – or has a minimum value | B1 [1] | |
| (iv) \(A = 2\) | B1∨ [1] | co - √ for \(x\) value from (i) |
| (v) \(y = 2(x - 2)^2 + 3\) → \(\frac{y-3}{2} = (x-2)^2\) → \(x = 2 + \sqrt{\frac{y-3}{2}}\) → \(g^{-1}(x) = 2 - \sqrt{\frac{x-3}{2}}\) | M1, M1, A1 | attempt to make \(x\) the subject – or y if \(x\) and \(y\) interchanged at start; order correct +3, +2, √, ±2; co – must be \(f(x)\) |
| Range of \(g^{-1} \leq 2\) | B1∨ [4] | condone \(\leq\) or \(>\). Nb if "+" root taken, answer will be \(\leq 2\), but could be \(\leq 2\) if returning to the original function. |
$f(x) = 2x^2 - 8x + 11$
**(i)** $f(x) = 2(x - 2)^2 + 3$ | 3 × B1 [3] | B1 for each of 2, -2 and 3. – no need to equate with $a$, $b$ and $c$.
**(ii)** Range is $\geq 3$ | B1∨ [1] | Condone >3. √ for $x \geq c$.
**(iii)** Not 1:1 (2 x-values for 1 y-value) or curve is quadratic – or has a minimum value | B1 [1] |
**(iv)** $A = 2$ | B1∨ [1] | co - √ for $x$ value from (i)
**(v)** $y = 2(x - 2)^2 + 3$ → $\frac{y-3}{2} = (x-2)^2$ → $x = 2 + \sqrt{\frac{y-3}{2}}$ → $g^{-1}(x) = 2 - \sqrt{\frac{x-3}{2}}$ | M1, M1, A1 | attempt to make $x$ the subject – or y if $x$ and $y$ interchanged at start; order correct +3, +2, √, ±2; co – must be $f(x)$
Range of $g^{-1} \leq 2$ | B1∨ [4] | condone $\leq$ or $>$. Nb if "+" root taken, answer will be $\leq 2$, but could be $\leq 2$ if returning to the original function.11 The function f is defined by $\mathrm { f } : x \mapsto 2 x ^ { 2 } - 8 x + 11$ for $x \in \mathbb { R }$.\\
(i) Express $\mathrm { f } ( x )$ in the form $a ( x + b ) ^ { 2 } + c$, where $a$, $b$ and $c$ are constants.\\
(ii) State the range of f .\\
(iii) Explain why f does not have an inverse.
The function g is defined by $\mathrm { g } : x \mapsto 2 x ^ { 2 } - 8 x + 11$ for $x \leqslant A$, where $A$ is a constant.\\
(iv) State the largest value of $A$ for which g has an inverse.\\
(v) When $A$ has this value, obtain an expression, in terms of $x$, for $\mathrm { g } ^ { - 1 } ( x )$ and state the range of $\mathrm { g } ^ { - 1 }$.
\hfill \mbox{\textit{CAIE P1 2007 Q11 [10]}}