| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.8 This is a straightforward differentiation and stationary points question using basic power rule (rewriting 8/x as 8x^{-1}). Finding stationary points by solving dy/dx = 0 gives a simple equation, and using the second derivative test is routine. Below average difficulty as it requires only standard techniques with no problem-solving insight. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = -\frac{8}{x^2} + 2\) cao | B1B1 | |
| \(\frac{d^2y}{dx^2} = \frac{16}{x^3}\) cao | B1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-\frac{8}{x^2} + 2 = 0 \rightarrow 2x^2 - 8 = 0\) | M1 | Set \(= 0\) and rearrange to quadratic form |
| \(x = \pm 2\) | A1 | |
| \(y = \pm 8\) | A1 | If A0A0 scored, SCA1 for just \((2,8)\) |
| \(\frac{d^2y}{dx^2} > 0\) when \(x=2\) hence MINIMUM | B1\(\checkmark\) | Ft for "correct" conclusion if \(\frac{d^2y}{dx^2}\) incorrect or any valid method inc. a good sketch |
| \(\frac{d^2y}{dx^2} < 0\) when \(x=-2\) hence MAXIMUM | B1\(\checkmark\) [5] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -\frac{8}{x^2} + 2$ cao | B1B1 | |
| $\frac{d^2y}{dx^2} = \frac{16}{x^3}$ cao | B1 [3] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-\frac{8}{x^2} + 2 = 0 \rightarrow 2x^2 - 8 = 0$ | M1 | Set $= 0$ and rearrange to quadratic form |
| $x = \pm 2$ | A1 | |
| $y = \pm 8$ | A1 | If A0A0 scored, SCA1 for just $(2,8)$ |
| $\frac{d^2y}{dx^2} > 0$ when $x=2$ hence MINIMUM | B1$\checkmark$ | Ft for "correct" conclusion if $\frac{d^2y}{dx^2}$ incorrect or any valid method inc. a good sketch |
| $\frac{d^2y}{dx^2} < 0$ when $x=-2$ hence MAXIMUM | B1$\checkmark$ [5] | |
---5 A curve has equation $y = \frac { 8 } { x } + 2 x$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.\\
(ii) Find the coordinates of the stationary points and state, with a reason, the nature of each stationary point.
\hfill \mbox{\textit{CAIE P1 2015 Q5 [8]}}