| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a straightforward multi-part curve sketching question requiring standard techniques: identifying a root from factored form, finding stationary points via differentiation, computing area under a curve with integration, and finding the minimum of the gradient function. All parts follow routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a = 2\) | B1 | [1] |
| (ii) \(y = x^3 - 4x^2 + 4x\); \(\frac{dy}{dx} = 3x^2 - 8x + 4\); \((x-2)(3x-2) = 0\); \(b = \frac{2}{3}\) | B1, B2, 1, 0, \(\checkmark\), B1 | \(-1\) for eeoo; cao [4] |
| (iii) area \(= \int y\, dx = \left[\frac{x^4}{4} - \frac{4x^3}{3} + 2x^2\right]\); \(4 - \frac{32}{3} + 8\); \(\frac{4}{3}\) | B2, 1, 0, M1, A1 | \(-1\) for eeoo; Apply limits 0→2 — ft their \(a\) from (i); cao [4] |
| (iv) \(\frac{d^2y}{dx^2} = 6x - 8 = 0 \Rightarrow x = \frac{4}{3}\); When \(x = \frac{4}{3}\), \(\frac{dy}{dx}\) (or \(m\)) \(= -\frac{4}{3}\) | M1, A1, DM1, A1 | Attempt \(\frac{d^2y}{dx^2}\) and set = 0; cao [4] |
(i) $a = 2$ | B1 | [1]
(ii) $y = x^3 - 4x^2 + 4x$; $\frac{dy}{dx} = 3x^2 - 8x + 4$; $(x-2)(3x-2) = 0$; $b = \frac{2}{3}$ | B1, B2, 1, 0, $\checkmark$, B1 | $-1$ for eeoo; cao [4]
(iii) area $= \int y\, dx = \left[\frac{x^4}{4} - \frac{4x^3}{3} + 2x^2\right]$; $4 - \frac{32}{3} + 8$; $\frac{4}{3}$ | B2, 1, 0, M1, A1 | $-1$ for eeoo; Apply limits 0→2 — ft their $a$ from (i); cao [4]
(iv) $\frac{d^2y}{dx^2} = 6x - 8 = 0 \Rightarrow x = \frac{4}{3}$; When $x = \frac{4}{3}$, $\frac{dy}{dx}$ (or $m$) $= -\frac{4}{3}$ | M1, A1, DM1, A1 | Attempt $\frac{d^2y}{dx^2}$ and set = 0; cao [4]11\\
\includegraphics[max width=\textwidth, alt={}, center]{d3c76ceb-cff7-4155-9697-5c302a9d63a9-4_526_974_822_587}
The diagram shows the curve with equation $y = x ( x - 2 ) ^ { 2 }$. The minimum point on the curve has coordinates $( a , 0 )$ and the $x$-coordinate of the maximum point is $b$, where $a$ and $b$ are constants.\\
(i) State the value of $a$.\\
(ii) Find the value of $b$.\\
(iii) Find the area of the shaded region.\\
(iv) The gradient, $\frac { \mathrm { d } y } { \mathrm {~d} x }$, of the curve has a minimum value $m$. Find the value of $m$.
\hfill \mbox{\textit{CAIE P1 2012 Q11 [13]}}