| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on a standard P1 topic. Part (i) requires routine differentiation of a quotient/chain rule, part (ii) is direct interpretation of the derivative's sign, and part (iii) involves standard algebraic manipulation to find an inverse function plus identifying domain/range from the original function's range/domain. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.07l Derivative of ln(x): and related functions1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(f'(x) = \frac{-5}{(1-3x)^2} \times -3\) | B1 B1 | B1 without "\(\times -3\)". B1 for "×−3", even if first B mark is incorrect |
| [2] | ||
| (ii) \(15 > 0\) and \((1-3x)^2 > 0\), \(f'(x) > 0\) → increasing | B1 | \(\checkmark\) providing \(()^2\) in denominator |
| [1] | ||
| (iii) \(y = \frac{5}{1-3x} \rightarrow 3x y = 1 - \frac{5}{y}\) → \(f^{-1}(x) = \frac{x-5}{3x}\) or \(\frac{1}{3} - \frac{5}{3x}\) | M1 A1 | Attempt to make \(x\) the subject. Must be in terms of \(x\) |
| Range is \(\geq 1\). Domain is \(-2.5 \leq x < 0\) | B1 B1 B1 | must be \(\geq\) condone \(<\) |
| [5] |
$f(x) = \frac{5}{1-3x}$, $x \geq 1$
(i) $f'(x) = \frac{-5}{(1-3x)^2} \times -3$ | B1 B1 | B1 without "$\times -3$". B1 for "×−3", even if first B mark is incorrect
[2] |
(ii) $15 > 0$ and $(1-3x)^2 > 0$, $f'(x) > 0$ → increasing | B1 | $\checkmark$ providing $()^2$ in denominator
[1] |
(iii) $y = \frac{5}{1-3x} \rightarrow 3x y = 1 - \frac{5}{y}$ → $f^{-1}(x) = \frac{x-5}{3x}$ or $\frac{1}{3} - \frac{5}{3x}$ | M1 A1 | Attempt to make $x$ the subject. Must be in terms of $x$
Range is $\geq 1$. Domain is $-2.5 \leq x < 0$ | B1 B1 B1 | must be $\geq$ condone $<$
[5] |
---9 A function f is defined by $\mathrm { f } ( x ) = \frac { 5 } { 1 - 3 x }$, for $x \geqslant 1$.\\
(i) Find an expression for $\mathrm { f } ^ { \prime } ( x )$.\\
(ii) Determine, with a reason, whether $f$ is an increasing function, a decreasing function or neither.\\
(iii) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$, and state the domain and range of $\mathrm { f } ^ { - 1 }$.
\hfill \mbox{\textit{CAIE P1 2013 Q9 [8]}}