| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard techniques: part (a)(i) involves solving simultaneous equations from composite function values (routine algebra), part (a)(ii) requires finding an inverse function using standard methods, and part (b) is a standard related rates calculus problem. All components are textbook exercises requiring competent execution of learned procedures but no novel insight or complex problem-solving. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \((a + b)^{\frac{1}{3}} = 2, (9a + b)^{\frac{1}{3}} = 16\) | B1B1 | Ignore 2nd soln (\(-9, 17\)) throughout |
| \(a + b = 8, 9a + b = 64\) | M1 | Cube etc. & attempt to solve |
| \(a = 7, b = 1\) | A1 | Correct answers without any working 0/4 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = (7y + 1)^{\frac{1}{3}}\) (y/x interchange as first or last step) | B1 | ft on from their a, b or in terms of a, b |
| \(x^3 = 7y + 1\) or \(x^3 = 7x + 1\) | B1 | ft on from their a, b or in terms of a, b |
| \(f^{-1}(x) = \frac{1}{7}(x^3 - 1)\) cao | B1 | A function of x required |
| Domain of \(f^{-1}\) is \(x \geq 1\) cao | B1 | Accept \(>\). Must be x [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \left[\frac{1}{3}(7x^2 + 1)^{-\frac{2}{3}}\right] \times [14x]\) | B1B1 | |
| When \(x = 3, \frac{dy}{dx} = \frac{1}{3} \times (64)^{-\frac{2}{3}} \times 42 \quad \left(= \frac{-8}{8}\right)\) | M1 | |
| \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = \frac{7}{8} \times 8\) | DM1 | Use chain rule |
| \(\frac{7}{7}\) | A1 | [5] |
### (a)
#### (i)
$(a + b)^{\frac{1}{3}} = 2, (9a + b)^{\frac{1}{3}} = 16$ | **B1B1** | Ignore 2nd soln ($-9, 17$) throughout
$a + b = 8, 9a + b = 64$ | **M1** | Cube etc. & attempt to solve
$a = 7, b = 1$ | **A1** | Correct answers without any working 0/4 [4]
#### (ii)
$x = (7y + 1)^{\frac{1}{3}}$ (y/x interchange as first or last step) | **B1** | ft on from their a, b or in terms of a, b
$x^3 = 7y + 1$ or $x^3 = 7x + 1$ | **B1** | ft on from their a, b or in terms of a, b
$f^{-1}(x) = \frac{1}{7}(x^3 - 1)$ cao | **B1** | A function of x required
Domain of $f^{-1}$ is $x \geq 1$ cao | **B1** | Accept $>$. Must be x [4]
### (b)
$\frac{dy}{dx} = \left[\frac{1}{3}(7x^2 + 1)^{-\frac{2}{3}}\right] \times [14x]$ | **B1B1** |
When $x = 3, \frac{dy}{dx} = \frac{1}{3} \times (64)^{-\frac{2}{3}} \times 42 \quad \left(= \frac{-8}{8}\right)$ | **M1** |
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = \frac{7}{8} \times 8$ | **DM1** | Use chain rule
$\frac{7}{7}$ | **A1** | [5]\begin{enumerate}[label=(\alph*)]
\item The functions $f$ and $g$ are defined for $x \geq 0$ by
$$f : x \mapsto (ax + b)^{\frac{1}{3}}, \text{ where } a \text{ and } b \text{ are positive constants,}$$
$$g : x \mapsto x^2.$$
Given that $fg(1) = 2$ and $gf(9) = 16$,
\begin{enumerate}[label=(\roman*)]
\item calculate the values of $a$ and $b$, [4]
\item obtain an expression for $f^{-1}(x)$ and state the domain of $f^{-1}$. [4]
\end{enumerate}
\item A point $P$ travels along the curve $y = (7x^2 + 1)^{\frac{1}{3}}$ in such a way that the $x$-coordinate of $P$ at time $t$ minutes is increasing at a constant rate of 8 units per minute. Find the rate of increase of the $y$-coordinate of $P$ at the instant when $P$ is at the point $(3, 4)$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2014 Q10 [13]}}