| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | First-order integration |
| Difficulty | Moderate -0.3 This is a straightforward integration problem requiring two successive integrations with given boundary conditions. While it involves multiple parts (finding f(x), locating stationary points, and determining their nature), each step follows standard A-level procedures: integrate twice using the power rule, apply conditions to find constants, solve f'(x)=0, and use the second derivative test. The algebra is clean and no novel problem-solving insight is required, making it slightly easier than average. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f''(x) = \frac{12}{x^3}\) | ||
| \(f'(x) = -\frac{6}{x^2}\ (+c)\) | B1 | Correct integration |
| \(= 0\) when \(x=2 \rightarrow c = \frac{3}{2}\) | M1 A1 | Uses \(x=2\), \(f'(x)=0\) |
| \(f(x) = \frac{6}{x} + \frac{3x}{2}\ (+A)\) | B1\(^\checkmark\) B1\(^\checkmark\) | For each integral |
| \(= 10\) when \(x=2 \rightarrow A=4\) | A1 [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-\frac{6}{x^2} + \frac{3}{2} = 0 \rightarrow x = \pm 2\) | M1 | Sets their 2 term \(f'(x)\) to 0 |
| Other point is \((-2, -2)\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \(x=2\), \(f''(x) = 1.5\) Min | B1 | |
| At \(x=-2\), \(f''(x) = -1.5\) Max | B1 [2] |
## Question 9:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f''(x) = \frac{12}{x^3}$ | | |
| $f'(x) = -\frac{6}{x^2}\ (+c)$ | **B1** | Correct integration |
| $= 0$ when $x=2 \rightarrow c = \frac{3}{2}$ | **M1 A1** | Uses $x=2$, $f'(x)=0$ |
| $f(x) = \frac{6}{x} + \frac{3x}{2}\ (+A)$ | **B1$^\checkmark$ B1$^\checkmark$** | For each integral |
| $= 10$ when $x=2 \rightarrow A=4$ | **A1** [6] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-\frac{6}{x^2} + \frac{3}{2} = 0 \rightarrow x = \pm 2$ | **M1** | Sets their 2 term $f'(x)$ to 0 |
| Other point is $(-2, -2)$ | **A1** [2] | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $x=2$, $f''(x) = 1.5$ Min | **B1** | |
| At $x=-2$, $f''(x) = -1.5$ Max | **B1** [2] | |9 The curve $y = \mathrm { f } ( x )$ has a stationary point at $( 2,10 )$ and it is given that $\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 12 } { x ^ { 3 } }$.\\
(i) Find $\mathrm { f } ( x )$.\\
(ii) Find the coordinates of the other stationary point.\\
(iii) Find the nature of each of the stationary points.
\hfill \mbox{\textit{CAIE P1 2015 Q9 [10]}}