| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find minimum domain for inverse |
| Difficulty | Standard +0.3 This is a straightforward composite and inverse functions question requiring standard techniques: identifying the vertex of a parabola for part (i), finding an inverse by swapping and rearranging for part (ii), and solving ff(x) by working backwards through two function applications in part (iii). All steps are routine A-level procedures with no novel insight 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 |
|---|---|---|
| 10(i) | Carry out a correct method for finding a vector equation for AB | M1 |
| Obtain r = 2i + j +3k +λ(2i – 2k), or equivalent | A1 | |
| Equate pair(s) of components AB and l and solve for λ or µ | M1(dep*) | |
| Obtain correct answer for λ or µ | A1 | |
| Verify that all three component equations are not satisfied | A1 | |
| Total: | 5 |
| Answer | Marks |
|---|---|
| 10(ii) | State or imply a direction vector for AP has components |
| (2 + t, 5 + 2t, – 3 – 2t) | B1 |
| Answer | Marks |
|---|---|
| product of their moduli | M1 |
| Answer | Marks |
|---|---|
| moduli in terms of t, and obtain an equation in terms of t | M1 |
| Obtain the given equation correctly | A1 |
| Solve the quadratic and use a root to find a position vector for P | M1 |
| Answer | Marks |
|---|---|
| 3 | A1 |
| Total: | 6 |
Question 10:
--- 10(i) ---
10(i) | Carry out a correct method for finding a vector equation for AB | M1
Obtain r = 2i + j +3k +λ(2i – 2k), or equivalent | A1
Equate pair(s) of components AB and l and solve for λ or µ | M1(dep*)
Obtain correct answer for λ or µ | A1
Verify that all three component equations are not satisfied | A1
Total: | 5
--- 10(ii) ---
10(ii) | State or imply a direction vector for AP has components
(2 + t, 5 + 2t, – 3 – 2t) | B1
(cid:74)(cid:74)(cid:74)(cid:71) (cid:74)(cid:74)(cid:74)(cid:71)
State or imply that cos 120° equals the scalar product of AP and AB divided by the
product of their moduli | M1
Carry out the correct processes for finding the scalar product and the product of the
moduli in terms of t, and obtain an equation in terms of t | M1
Obtain the given equation correctly | A1
Solve the quadratic and use a root to find a position vector for P | M1
2
Obtain position vector 2i + 2j + 4k from t = – 2, having rejected the root t = −
3 | A1
Total: | 6The one-one function f is defined by $\mathrm{f}(x) = (x - 2)^2 + 2$ for $x \geqslant c$, where $c$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item State the smallest possible value of $c$. [1]
\end{enumerate}
In parts (ii) and (iii) the value of $c$ is 4.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find an expression for $\mathrm{f}^{-1}(x)$ and state the domain of $\mathrm{f}^{-1}$. [3]
\item Solve the equation $\mathrm{f f}(x) = 51$, giving your answer in the form $a + \sqrt{b}$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2018 Q10 [9]}}