| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a straightforward application of differentiation rules (power rule for the rational function rewritten as (x-3)^{-1}) followed by standard stationary point analysis. While it requires multiple steps (first derivative, second derivative, solving a quadratic, classifying points), each step is routine and the question follows a very standard template for A-level calculus problems. Slightly above average difficulty due to the algebraic manipulation needed when solving for stationary points. |
| 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 derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{-1}{(x-3)^2} + 1\) | B1 | oe |
| \(\frac{d^2y}{dx^2} = \frac{2}{(x-3)^3}\) | B1 | oe |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x-3)^2 = 1 \Rightarrow x - 3 = \pm 1\) | M1 | Set \(\frac{dy}{dx} = 0\) & reasonable attempt to solve |
| \(x = 4, 2\) | A1 | |
| \(y = 5, 1\) | A1 | |
| When \(x = 4\): \(\frac{d^2y}{dx^2} > 0\ (= 2) \Rightarrow\) min | M1 | Investigate signs of \(f''\) at a point or other method |
| When \(x = 2\): \(\frac{d^2y}{dx^2} < 0\ (= -2) \Rightarrow\) max | A1 | |
| [5] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{-1}{(x-3)^2} + 1$ | B1 | oe |
| $\frac{d^2y}{dx^2} = \frac{2}{(x-3)^3}$ | B1 | oe |
| **[2]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x-3)^2 = 1 \Rightarrow x - 3 = \pm 1$ | M1 | Set $\frac{dy}{dx} = 0$ & reasonable attempt to solve |
| $x = 4, 2$ | A1 | |
| $y = 5, 1$ | A1 | |
| When $x = 4$: $\frac{d^2y}{dx^2} > 0\ (= 2) \Rightarrow$ min | M1 | Investigate signs of $f''$ at a point or other method |
| When $x = 2$: $\frac{d^2y}{dx^2} < 0\ (= -2) \Rightarrow$ max | A1 | |
| **[5]** | | |
---5 A curve has equation $y = \frac { 1 } { x - 3 } + x$.\\
(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 maximum point $A$ and the minimum point $B$ on the curve.
\hfill \mbox{\textit{CAIE P1 2010 Q5 [7]}}