| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Standard +0.3 This is a standard C3 composite/inverse functions question requiring routine techniques: finding an inverse by rearranging (cube root), verifying a composite by substitution, solving a rational equation (set numerator to zero), and finding a stationary point using quotient rule. All steps are textbook exercises with no novel insight required, making it slightly easier than the average A-level question. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 1 - 2y^3 \Rightarrow y = \left(\frac{1-x}{2}\right)^{1/3}\) or \(\sqrt[3]{\frac{1-x}{2}}\) | M1 A1 (2) | Ignore domain |
| Answer | Marks | Guidance |
|---|---|---|
| \(gf(x) = \frac{3}{1-2x^3} - 4\) | M1 A1 | |
| \(= \frac{3 - 4(1-2x^3)}{1-2x^3}\) | M1 | |
| \(= \frac{8x^3 - 1}{1-2x^3}\) ✻ | A1 (4) | cso |
| \(gf: x \mapsto \frac{8x^3 - 1}{1-2x^3}\) | Ignore domain |
| Answer | Marks | Guidance |
|---|---|---|
| \(8x^3 - 1 = 0\) | M1 | Attempting solution of numerator \(= 0\) |
| \(x = \frac{1}{2}\) | A1 (2) | Correct answer and no additional answers |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = \frac{(1-2x^3) \times 24x^2 + (8x^3 - 1) \times 6x^2}{(1-2x^3)^2}\) | M1 A1 |
| \(= \frac{18x^2}{(1-2x^3)^2}\) | A1 |
| Solving their numerator \(= 0\) and substituting to find \(y\) | M1 |
| \(x = 0,\ y = -1\) | A1 (5) |
## Question 8:
**Part (a):**
$x = 1 - 2y^3 \Rightarrow y = \left(\frac{1-x}{2}\right)^{1/3}$ or $\sqrt[3]{\frac{1-x}{2}}$ | M1 A1 (2) | Ignore domain
$f^{-1}: x \mapsto \left(\frac{1-x}{2}\right)^{1/3}$
---
**Part (b):**
$gf(x) = \frac{3}{1-2x^3} - 4$ | M1 A1 |
$= \frac{3 - 4(1-2x^3)}{1-2x^3}$ | M1 |
$= \frac{8x^3 - 1}{1-2x^3}$ ✻ | A1 (4) | cso
$gf: x \mapsto \frac{8x^3 - 1}{1-2x^3}$ | | Ignore domain
---
**Part (c):**
$8x^3 - 1 = 0$ | M1 | Attempting solution of numerator $= 0$
$x = \frac{1}{2}$ | A1 (2) | Correct answer and no additional answers
---
**Part (d):**
$\frac{dy}{dx} = \frac{(1-2x^3) \times 24x^2 + (8x^3 - 1) \times 6x^2}{(1-2x^3)^2}$ | M1 A1 |
$= \frac{18x^2}{(1-2x^3)^2}$ | A1 |
Solving their numerator $= 0$ and substituting to find $y$ | M1 |
$x = 0,\ y = -1$ | A1 (5) |
**Total: [13]**\begin{enumerate}
\item The functions $f$ and $g$ are defined by
\end{enumerate}
$$\begin{aligned}
& \mathrm { f } : x \mapsto 1 - 2 x ^ { 3 } , x \in \mathbb { R } \\
& \mathrm {~g} : x \mapsto \frac { 3 } { x } - 4 , x > 0 , x \in \mathbb { R }
\end{aligned}$$
(a) Find the inverse function $\mathrm { f } ^ { - 1 }$.\\
(b) Show that the composite function gf is
$$\text { gf } : x \mapsto \frac { 8 x ^ { 3 } - 1 } { 1 - 2 x ^ { 3 } }$$
(c) Solve $\operatorname { gf } ( x ) = 0$.\\
(d) Use calculus to find the coordinates of the stationary point on the graph of $y = \operatorname { gf } ( x )$.
\hfill \mbox{\textit{Edexcel C3 2008 Q8 [13]}}