| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2003 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation with inverses |
| Difficulty | Moderate -0.3 This is a straightforward composite and inverse functions question requiring standard techniques: finding fg(x) by substitution, finding inverses by swapping and rearranging, solving a simple equation, and showing no real roots by discriminant or algebraic manipulation. All parts are routine P1 exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow x = 1\frac{1}{3}\) | M1, DM1, A1 [3] | Correct order - g first, then into f. Correct method of solution of \(f\mathrm{g}=7\). Co. (nb gf gets 0/3). (M1 for 6. M1 for g(x)=6. A1). |
| (or f(A)=7, A = 6, g(x) = 6, \(\rightarrow x = 1\frac{1}{3}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow g^{-1} = 2 - (4-x)\) | B1, M1, A1 [3] | Anywhere in the question. For changing the subject. Co – any correct answer. (A0 if f(y,).) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow\) No roots. | M1, M1, A1 [3] | Algebra leading to a quadratic. Quadratic=0 + use of \(b^2-4ac\). Correct deduction from correct quadratic. |
| (iv) | B1, B1, B1 [3] | Sketch of \(f\). Sketch of \(f^{-1}\). Evidence of symmetry about \(y=x\). |
**(i)** $f\mathrm{g}(x) = \mathrm{g}$ first, then $f$
$= \frac{8}{(2-x) - 5} = 7$
$\rightarrow x = 1\frac{1}{3}$ | M1, DM1, A1 [3] | Correct order - g first, then into f. Correct method of solution of $f\mathrm{g}=7$. Co. (nb gf gets 0/3). (M1 for 6. M1 for g(x)=6. A1).
(or f(A)=7, A = 6, g(x) = 6, $\rightarrow x = 1\frac{1}{3}$) |
**(ii)** $f^{-1} = \frac{1}{2}(x+5)$
Makes $y$ the subject $y = 4-(2-x)$
$\rightarrow g^{-1} = 2 - (4-x)$ | B1, M1, A1 [3] | Anywhere in the question. For changing the subject. Co – any correct answer. (A0 if f(y,).)
**(iii)** $2-4/x = \frac{1}{2}(x+5)$
$\rightarrow x^2+x+8=0$
Use of $b^2-4ac \rightarrow$ Negative value
$\rightarrow$ No roots. | M1, M1, A1 [3] | Algebra leading to a quadratic. Quadratic=0 + use of $b^2-4ac$. Correct deduction from correct quadratic.
**(iv)** | B1, B1, B1 [3] | Sketch of $f$. Sketch of $f^{-1}$. Evidence of symmetry about $y=x$.10 Functions $f$ and $g$ are defined by
$$\begin{aligned}
& \mathrm { f } : x \mapsto 2 x - 5 , \quad x \in \mathbb { R } , \\
& \mathrm {~g} : x \mapsto \frac { 4 } { 2 - x } , \quad x \in \mathbb { R } , \quad x \neq 2 .
\end{aligned}$$
(i) Find the value of $x$ for which $\mathrm { fg } ( x ) = 7$.\\
(ii) Express each of $\mathrm { f } ^ { - 1 } ( x )$ and $\mathrm { g } ^ { - 1 } ( x )$ in terms of $x$.\\
(iii) Show that the equation $\mathrm { f } ^ { - 1 } ( x ) = \mathrm { g } ^ { - 1 } ( x )$ has no real roots.\\
(iv) Sketch, on a single diagram, the graphs of $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$, making clear the relationship between these two graphs.
\hfill \mbox{\textit{CAIE P1 2003 Q10 [12]}}