| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find intersection points |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on inverse functions requiring standard techniques: stating domain/range (simple reflection), finding the inverse algebraically (routine manipulation of y = 1 + √x), and solving f(x) = f⁻¹(x) using the symmetry property about y = x. The final part guides students by mentioning graphs, making it easier than if they had to discover the approach independently. Slightly above average due to the multi-step nature and the final algebraic manipulation, but all techniques are standard C3 material. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Domain of \(f^{-1}\) is \(x \geq 1\) Range is \(x \geq 0\) | B1, B1 | For the correct set, in any notation; Ditto |
| (ii) If \(y = 1 + \sqrt{x}\), then \(x = (y - 1)^2\) Hence \(f^{-1}(x) = (x - 1)^2\) | M1, A1 | For changing the subject, or equivalent; For correct expression in terms of \(x\) |
| (iii) The graphs intersect on the line \(y = x\) Hence \(x\) satisfies \(x = (x - 1)^2\) i.e. \(x^2 - 3x + 1 = 0 \Rightarrow x = \frac{3 \pm \sqrt{5}}{2}\) So \(x = \frac{1}{2}(3 + \sqrt{5})\) as x must be greater than 1 | B1, B1, M1, A1 | For stating or using this fact; For either \(x = f(x)\) or \(x = f^{-1}(x)\); For solving the relevant quadratic equation; For showing the given answer fully |
**(i)** Domain of $f^{-1}$ is $x \geq 1$ Range is $x \geq 0$ | B1, B1 | For the correct set, in any notation; Ditto
**(ii)** If $y = 1 + \sqrt{x}$, then $x = (y - 1)^2$ Hence $f^{-1}(x) = (x - 1)^2$ | M1, A1 | For changing the subject, or equivalent; For correct expression in terms of $x$
**(iii)** The graphs intersect on the line $y = x$ Hence $x$ satisfies $x = (x - 1)^2$ i.e. $x^2 - 3x + 1 = 0 \Rightarrow x = \frac{3 \pm \sqrt{5}}{2}$ So $x = \frac{1}{2}(3 + \sqrt{5})$ as x must be greater than 1 | B1, B1, M1, A1 | For stating or using this fact; For either $x = f(x)$ or $x = f^{-1}(x)$; For solving the relevant quadratic equation; For showing the given answer fully
**Total for Question 6: 8 marks**
---6 The function f is defined by
$$\mathrm { f } : x \mapsto 1 + \sqrt { } x \quad \text { for } x \geqslant 0$$
(i) State the domain and range of the inverse function $\mathrm { f } ^ { - 1 }$.\\
(ii) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.\\
(iii) By considering the graphs of $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$, show that the solution to the equation
$$\mathrm { f } ( x ) = \mathrm { f } ^ { - 1 } ( x )$$
is $x = \frac { 1 } { 2 } ( 3 + \sqrt { } 5 )$.
\hfill \mbox{\textit{OCR C3 Q6 [8]}}