| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a standard C3 inverse function question with routine completing the square, straightforward transformations, and algebraic manipulation to find the inverse. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02v Inverse and composite functions: graphs and conditions for existence1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(= 2[x^2 + 2x] + 2 = 2[(x+1)^2 - 1] + 2\) | M1 | |
| \(= 2(x+1)^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| Translation by 1 unit in negative \(x\) direction; stretch by scale factor 2 in \(y\) direction (either first) | B3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(\frac{y}{2} = (x+1)^2\); \(x + 1 = \pm\sqrt{\frac{y}{2}},\quad x = -1 \pm \sqrt{\frac{y}{2}}\) | M1 | |
| \(\therefore f^{-1}(x) = -1 + \sqrt{\frac{x}{2}},\ x \in \mathbb{R},\ x \geq 0\) | A2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| Sketch of \(y = f(x)\) | B2 | |
| \(y = f^{-1}(x)\) is reflection of \(y = f(x)\) in line \(y = x\) | B1 | (11) |
# Question 8:
## Part (i)
| Working/Answer | Mark | Notes |
|---|---|---|
| $= 2[x^2 + 2x] + 2 = 2[(x+1)^2 - 1] + 2$ | M1 | |
| $= 2(x+1)^2$ | A1 | |
## Part (ii)
| Working/Answer | Mark | Notes |
|---|---|---|
| Translation by 1 unit in negative $x$ direction; stretch by scale factor 2 in $y$ direction (either first) | B3 | |
## Part (iii)
| Working/Answer | Mark | Notes |
|---|---|---|
| $\frac{y}{2} = (x+1)^2$; $x + 1 = \pm\sqrt{\frac{y}{2}},\quad x = -1 \pm \sqrt{\frac{y}{2}}$ | M1 | |
| $\therefore f^{-1}(x) = -1 + \sqrt{\frac{x}{2}},\ x \in \mathbb{R},\ x \geq 0$ | A2 | |
## Part (iv)
| Working/Answer | Mark | Notes |
|---|---|---|
| Sketch of $y = f(x)$ | B2 | |
| $y = f^{-1}(x)$ is reflection of $y = f(x)$ in line $y = x$ | B1 | **(11)** |
---8.
$$\mathrm { f } ( x ) \equiv 2 x ^ { 2 } + 4 x + 2 , \quad x \in \mathbb { R } , \quad x \geq - 1$$
(i) Express $\mathrm { f } ( x )$ in the form $a ( x + b ) ^ { 2 } + c$.\\
(ii) Describe fully two transformations that would map the graph of $y = x ^ { 2 } , x \geq 0$ onto the graph of $y = \mathrm { f } ( x )$.\\
(iii) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$ and state its domain.\\
(iv) Sketch the graphs of $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$ on the same diagram and state the relationship between them.\\
\hfill \mbox{\textit{OCR C3 Q8 [11]}}