| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Sketch function and inverse graphs |
| Difficulty | Moderate -0.3 This is a standard multi-part question on functions covering routine techniques: differentiation of a rational function, finding an inverse function algebraically, sketching inverse graphs using reflection, and solving a composite function equation. All parts are textbook exercises requiring no novel insight, though the multiple parts and careful domain work make it slightly more substantial than the most basic questions. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = -6(2x+3)^{-2} \times 2\) | B1, B1 | co (\(-\text{ve}\) power ok); B1 for \(\times 2\) |
| Always \(-\text{ve} \rightarrow\) Decreasing | B1\(\sqrt{}\) [3] | Answer given. Correct explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \dfrac{6}{2x+3}\) | M1 | Reasonable attempt in making \(x\) the subject (ok to interchange \(x, y\) first) |
| \(\rightarrow f^{-1}(x) = \dfrac{1}{2}\left(\dfrac{6}{x} - 3\right)\) | M1, A1 | Order of operations must be correct ie \(\div y\), \(-3\) then \(\div 2\); correct expression as \(f^{-1}(x)\); gets \(\frac{2}{3}\) for correct expression with \(y\) |
| Domain of \(f^{-1}\): \(0 < x \leq 2\) | B1 [4] | Could be independent of answer for \(f^{-1}\); condone \(<\) or \(\leq\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct graph for \(f^{-1}\) (curve, stops on axis) | B1 | Correct graph for \(f^{-1}\) (curve, stops on axis) |
| Symmetry about \(y = x\) shown | B1 [2] | Makes clear on graph, or in words by the line \(y = x\) marked |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(fg(x) = \dfrac{6}{x+3}\) | M1 | \(g\) first, then \(f\). Reverse (\(3 \div (2x+3)\)) M0 |
| \(= 1.5 \rightarrow x = 1\) | M1, A1 [3] | Not DM – so can get this if attempt ok; co |
| [or, using \(f^{-1}\), \(\rightarrow g(x) = \frac{1}{2}\), \(\Rightarrow x = 1\)] | [M1, M1, A1] |
## Question 11(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = -6(2x+3)^{-2} \times 2$ | B1, B1 | co ($-\text{ve}$ power ok); B1 for $\times 2$ |
| Always $-\text{ve} \rightarrow$ Decreasing | B1$\sqrt{}$ [3] | Answer given. Correct explanation |
## Question 11(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \dfrac{6}{2x+3}$ | M1 | Reasonable attempt in making $x$ the subject (ok to interchange $x, y$ first) |
| $\rightarrow f^{-1}(x) = \dfrac{1}{2}\left(\dfrac{6}{x} - 3\right)$ | M1, A1 | Order of operations must be correct ie $\div y$, $-3$ then $\div 2$; correct expression as $f^{-1}(x)$; gets $\frac{2}{3}$ for correct expression with $y$ |
| Domain of $f^{-1}$: $0 < x \leq 2$ | B1 [4] | Could be independent of answer for $f^{-1}$; condone $<$ or $\leq$ |
## Question 11(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct graph for $f^{-1}$ (curve, stops on axis) | B1 | Correct graph for $f^{-1}$ (curve, stops on axis) |
| Symmetry about $y = x$ shown | B1 [2] | Makes clear on graph, or in words by the line $y = x$ marked |
## Question 11(iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $fg(x) = \dfrac{6}{x+3}$ | M1 | $g$ first, then $f$. Reverse ($3 \div (2x+3)$) M0 |
| $= 1.5 \rightarrow x = 1$ | M1, A1 [3] | Not DM – so can get this if attempt ok; co |
| [or, using $f^{-1}$, $\rightarrow g(x) = \frac{1}{2}$, $\Rightarrow x = 1$] | [M1, M1, A1] | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{b24ed4c7-ab07-45f4-adf2-027734c36b62-4_862_892_932_628}
The diagram shows the graph of $y = \mathrm { f } ( x )$, where $\mathrm { f } : x \mapsto \frac { 6 } { 2 x + 3 }$ for $x \geqslant 0$.\\
(i) Find an expression, in terms of $x$, for $\mathrm { f } ^ { \prime } ( x )$ and explain how your answer shows that f is a decreasing function.\\
(ii) Find an expression, in terms of $x$, for $\mathrm { f } ^ { - 1 } ( x )$ and find the domain of $\mathrm { f } ^ { - 1 }$.\\
(iii) Copy the diagram and, on your copy, sketch the graph of $y = \mathrm { f } ^ { - 1 } ( x )$, making clear the relationship between the graphs.
The function g is defined by $\mathrm { g } : x \mapsto \frac { 1 } { 2 } x$ for $x \geqslant 0$.\\
(iv) Solve the equation $\operatorname { fg } ( x ) = \frac { 3 } { 2 }$.
\hfill \mbox{\textit{CAIE P1 2007 Q11 [12]}}