| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 This is a straightforward two-part question on functions. Part (i) involves substituting values and solving simultaneous equations (routine algebra). Part (ii) requires finding an inverse function by completing the square and solving for x, then determining the range of f as the domain of f^(-1) - all standard techniques for P1 level with no novel insight required. Slightly above average due to the algebraic manipulation with parameters, but still a textbook exercise. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02v Inverse and composite functions: graphs and conditions for existence1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2a + 4b = 8\) | M1 | Substitute in \(-2\) and \(-3\) |
| \(2a^2 + 3a + 4b = 14\) | A1 | |
| \(2a^2 + 3a + (8 - 2a) = 14 \rightarrow (a+2)(2a-3) = 0\) | M1 | Sub linear into quadratic & attempt solution |
| \(a = -2\) or \(3/2\) | A1 | If A0A0 scored allow SCA1 for either \((-2, 3)\) or \((3/2, 5/4)\) |
| \(b = 3\) or \(5/4\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \left(x - \dfrac{1}{2}\right)^2 - \dfrac{13}{4}\) Attempt completing the square | M1A1 | Allow with \(x/y\) transposed |
| \(x - \dfrac{1}{2} = (\pm)\sqrt{y + \dfrac{13}{4}}\) oe | DM1 | Allow with \(x/y\) transposed |
| \(f^{-1}(x) = \dfrac{1}{2} - \sqrt{x + \dfrac{13}{4}}\) oe | A1 | Allow \(y = \ldots\) Must be a function of \(x\) |
| Domain of \(f^{-1}\) is \((x) \geqslant -13/4\) | B1\(\checkmark\) [5] | Allow \(>\), \(-13/4 \leqslant x \leqslant \infty\), \(\left[-\dfrac{13}{4}, \infty\right)\) etc |
## Question 8:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2a + 4b = 8$ | M1 | Substitute in $-2$ and $-3$ |
| $2a^2 + 3a + 4b = 14$ | A1 | |
| $2a^2 + 3a + (8 - 2a) = 14 \rightarrow (a+2)(2a-3) = 0$ | M1 | Sub linear into quadratic & attempt solution |
| $a = -2$ or $3/2$ | A1 | If A0A0 scored allow SCA1 for either $(-2, 3)$ or $(3/2, 5/4)$ |
| $b = 3$ or $5/4$ | A1 [5] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \left(x - \dfrac{1}{2}\right)^2 - \dfrac{13}{4}$ Attempt completing the square | M1A1 | Allow with $x/y$ transposed |
| $x - \dfrac{1}{2} = (\pm)\sqrt{y + \dfrac{13}{4}}$ oe | DM1 | Allow with $x/y$ transposed |
| $f^{-1}(x) = \dfrac{1}{2} - \sqrt{x + \dfrac{13}{4}}$ oe | A1 | Allow $y = \ldots$ Must be a function of $x$ |
| Domain of $f^{-1}$ is $(x) \geqslant -13/4$ | B1$\checkmark$ [5] | Allow $>$, $-13/4 \leqslant x \leqslant \infty$, $\left[-\dfrac{13}{4}, \infty\right)$ etc |
---8 The function f is such that $\mathrm { f } ( x ) = a ^ { 2 } x ^ { 2 } - a x + 3 b$ for $x \leqslant \frac { 1 } { 2 a }$, where $a$ and $b$ are constants.\\
(i) For the case where $\mathrm { f } ( - 2 ) = 4 a ^ { 2 } - b + 8$ and $\mathrm { f } ( - 3 ) = 7 a ^ { 2 } - b + 14$, find the possible values of $a$ and $b$.\\
(ii) For the case where $a = 1$ and $b = - 1$, find an expression for $\mathrm { f } ^ { - 1 } ( x )$ and give the domain of $\mathrm { f } ^ { - 1 }$.
\hfill \mbox{\textit{CAIE P1 2016 Q8 [10]}}