| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Moderate -0.8 This is a straightforward integration question requiring basic antiderivatives (x² and x⁻²), using a point to find the constant, then finding a normal line and stationary point. All techniques are routine P1/AS-level procedures with no problem-solving insight needed—easier than average A-level questions. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(2) = 4 - \frac{1}{2} = \frac{7}{2} \rightarrow\) gradient of normal \(= -\frac{2}{7}\) | B1M1 | |
| \(y - 6 = -\frac{2}{7}(x - 2)\) AEF | A1✓ | Ft from their \(f'(2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = x^2 + \frac{2}{x}\ (+c)\) | B1B1 | |
| \(6 = 4 + 1 + c \Rightarrow c = 1\) | M1A1 | Sub \((2, 6)\) — dependent on \(c\) being present |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2x - \frac{2}{x^2} = 0 \Rightarrow 2x^3 - 2 = 0\) | M1 | Put \(f'(x) = 0\) and attempt to solve |
| \(x = 1\) | A1 | Not necessary for last A mark as \(x > 0\) given |
| \(f''(x) = 2 + \frac{4}{x^3}\) or any valid method | M1 | |
| \(f''(1) = 6\ \text{OR} > 0\) hence minimum | A1 | Dependent on everything correct |
## Question 9:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(2) = 4 - \frac{1}{2} = \frac{7}{2} \rightarrow$ gradient of normal $= -\frac{2}{7}$ | **B1M1** | |
| $y - 6 = -\frac{2}{7}(x - 2)$ AEF | **A1**✓ | Ft from their $f'(2)$ |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = x^2 + \frac{2}{x}\ (+c)$ | **B1B1** | |
| $6 = 4 + 1 + c \Rightarrow c = 1$ | **M1A1** | Sub $(2, 6)$ — dependent on $c$ being present |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2x - \frac{2}{x^2} = 0 \Rightarrow 2x^3 - 2 = 0$ | **M1** | Put $f'(x) = 0$ and attempt to solve |
| $x = 1$ | **A1** | Not necessary for last A mark as $x > 0$ given |
| $f''(x) = 2 + \frac{4}{x^3}$ or any valid method | **M1** | |
| $f''(1) = 6\ \text{OR} > 0$ hence minimum | **A1** | Dependent on everything correct |
---9 The function f is defined for $x > 0$ and is such that $\mathrm { f } ^ { \prime } ( x ) = 2 x - \frac { 2 } { x ^ { 2 } }$. The curve $y = \mathrm { f } ( x )$ passes through the point $P ( 2,6 )$.\\
(i) Find the equation of the normal to the curve at $P$.\\
(ii) Find the equation of the curve.\\
(iii) Find the $x$-coordinate of the stationary point and state with a reason whether this point is a maximum or a minimum.
\hfill \mbox{\textit{CAIE P1 2014 Q9 [11]}}