| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Identify transformation from equations |
| Difficulty | Moderate -0.5 This is a straightforward C1 question testing basic function transformations and differentiation. Part (i) is routine sketching, (ii) requires identifying a simple horizontal translation, (iii) is standard differentiation of x^(-1), and (iv) combines these in a predictable way. All parts are textbook exercises requiring recall and direct application rather than problem-solving, making it slightly easier than average. |
| Spec | 1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02w Graph transformations: simple transformations of f(x)1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (i) [Sketch of reciprocal function with correct 1st quadrant branch] | B1 B1 | For correct 1st quadrant branch; For both branches correct and nothing else |
| 2 | ||
| (ii) Translation of 2 units in the negative \(x\)-direction | B1 B1 B1 | For translation parallel to the \(x\)-axis; For correct magnitude; For correct direction |
| (iii) [Sketch of translated curve] | B1 √ B1 | For correct sketch of new curve; For some indication of location, e.g. \(\frac{1}{2}\) at \(y\)-intersection or \(-2\) at asymptote |
| 5 | ||
| (iii) Derivative is \(-x^{-2}\) | M1 A1 | For correct power \(-2\) in answer; For correct coefficient \(-1\) |
| 2 | ||
| (iv) Gradient of \(y = \frac{1}{x}\) at \(x = 2\) is required. This is \(2^{-2}\), which is \(-\frac{1}{4}\) | B1 M1 A1 | For correctly using the translation; For substituting \(x = 2\) in their (iii); For correct answer |
| 3 |
(i) [Sketch of reciprocal function with correct 1st quadrant branch] | B1 B1 | For correct 1st quadrant branch; For both branches correct and nothing else
| | 2 |
(ii) Translation of 2 units in the negative $x$-direction | B1 B1 B1 | For translation parallel to the $x$-axis; For correct magnitude; For correct direction
(iii) [Sketch of translated curve] | B1 √ B1 | For correct sketch of new curve; For some indication of location, e.g. $\frac{1}{2}$ at $y$-intersection or $-2$ at asymptote
| | 5 |
(iii) Derivative is $-x^{-2}$ | M1 A1 | For correct power $-2$ in answer; For correct coefficient $-1$
| | 2 |
(iv) Gradient of $y = \frac{1}{x}$ at $x = 2$ is required. This is $2^{-2}$, which is $-\frac{1}{4}$ | B1 M1 A1 | For correctly using the translation; For substituting $x = 2$ in their (iii); For correct answer
| | 3 |
---6 (i) Sketch the graph of $y = \frac { 1 } { x }$, where $x \neq 0$, showing the parts of the graph corresponding to both positive and negative values of $x$.\\
(ii) Describe fully the geometrical transformation that transforms the curve $y = \frac { 1 } { x }$ to the curve $y = \frac { 1 } { x + 2 }$. Hence sketch the curve $y = \frac { 1 } { x + 2 }$.\\
(iii) Differentiate $\frac { 1 } { x }$ with respect to $x$.\\
(iv) Use parts (ii) and (iii) to find the gradient of the curve $y = \frac { 1 } { x + 2 }$ at the point where it crosses the $y$-axis.
\hfill \mbox{\textit{OCR C1 Q6 [12]}}