| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2007 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.3 This is a straightforward application of the chain rule to find derivatives, followed by standard stationary point analysis. The algebra is simple (expanding or factorising a quadratic), and determining nature via second derivative is routine. Slightly easier than average due to clean numbers and no conceptual challenges. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = 3 × (2x-3)^2 × 2 - 6\); \(\frac{d^2y}{dx^2} = 12×(2x-3) × 2\) | B1, B1, B1∨ [3] | For \(3 × (2x-3)^2 - 6\); For \(x\) 2; Nb \(24x^2 - 72x + 48\) B2,1; √ from his \(dy/dx\). |
| (ii) s.p → \(\frac{dy}{dx} = 0\) → \((2x-3)^2 = 1\) → \(x = 2\) or \(x = 1\) | M1, DM1, A1 | Sets \(dy/dx\) to 0; Solution to give two values of \(x\). For both values correct. |
| If \(x = 2\), 2nd diff = +ve → MIN; If \(x = 1\), 2nd diff = −ve → MAX | M1, A1 [5] | Looks at sign of 2nd differential or other method with one \(x\) value. A1 for correct conclusions |
$y = (2x - 3)^3 - 6x$
**(i)** $\frac{dy}{dx} = 3 × (2x-3)^2 × 2 - 6$; $\frac{d^2y}{dx^2} = 12×(2x-3) × 2$ | B1, B1, B1∨ [3] | For $3 × (2x-3)^2 - 6$; For $x$ 2; Nb $24x^2 - 72x + 48$ B2,1; √ from his $dy/dx$.
**(ii)** s.p → $\frac{dy}{dx} = 0$ → $(2x-3)^2 = 1$ → $x = 2$ or $x = 1$ | M1, DM1, A1 | Sets $dy/dx$ to 0; Solution to give two values of $x$. For both values correct.
If $x = 2$, 2nd diff = +ve → MIN; If $x = 1$, 2nd diff = −ve → MAX | M1, A1 [5] | Looks at sign of 2nd differential or other method with one $x$ value. A1 for correct conclusions8 The equation of a curve is $y = ( 2 x - 3 ) ^ { 3 } - 6 x$.\\
(i) Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ in terms of $x$.\\
(ii) Find the $x$-coordinates of the two stationary points and determine the nature of each stationary point.
\hfill \mbox{\textit{CAIE P1 2007 Q8 [8]}}