| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find stationary points of standard polynomial |
| Difficulty | Moderate -0.5 This is a straightforward curve sketching question on a cubic polynomial y = x(x-2)². Parts (a-d) involve: identifying the minimum point from a diagram, finding the maximum using differentiation (dy/dx = 3x² - 8x + 4 = 0), calculating area under curve via integration, and finding gradient at minimum. All are standard P1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| \(a = 2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x^3 - 4x^2 + 4x\) | B1 | |
| \(\frac{dy}{dx} = 3x^2 - 8x + 4\) | B1 | FT B1 for \(3x^2\), B1 for \(-8x + 4\) |
| \((x-2)(3x-2) = 0 \rightarrow b = \frac{2}{3}\) | B1 | Dependent on method seen for solving quadratic equation |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= \int y\, dx = \left[\frac{x^4}{4} - \frac{4x^3}{3} + 2x^2\right]\) | B2 | B1 for \(\frac{x^4}{4}\), B1 for \(\frac{4x^3}{3} + 2x^2\) |
| \(4 - \frac{32}{3} + 8\) | M1 | Apply limits \(0 \rightarrow 2\) |
| \(4\frac{1}{3}\) | A1 | Unsupported answer receives 0 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = 6x - 8 = 0,\ x = \frac{4}{3}\) | M1*A1 | Attempt 2nd derivative and set \(= 0\) |
| When \(x = \frac{4}{3}\), \(\frac{dy}{dx} = -\frac{4}{3}\) (or \(m\)) \(= -\frac{4}{3}\) | DM1A1 |
# Question 12:
## Part 12(a):
| $a = 2$ | B1 | |
## Part 12(b):
| $y = x^3 - 4x^2 + 4x$ | B1 | |
| $\frac{dy}{dx} = 3x^2 - 8x + 4$ | B1 | FT B1 for $3x^2$, B1 for $-8x + 4$ |
| $(x-2)(3x-2) = 0 \rightarrow b = \frac{2}{3}$ | B1 | Dependent on method seen for solving quadratic equation |
## Part 12(c):
| Area $= \int y\, dx = \left[\frac{x^4}{4} - \frac{4x^3}{3} + 2x^2\right]$ | B2 | B1 for $\frac{x^4}{4}$, B1 for $\frac{4x^3}{3} + 2x^2$ |
| $4 - \frac{32}{3} + 8$ | M1 | Apply limits $0 \rightarrow 2$ |
| $4\frac{1}{3}$ | A1 | Unsupported answer receives 0 marks |
## Part 12(d):
| $\frac{d^2y}{dx^2} = 6x - 8 = 0,\ x = \frac{4}{3}$ | M1*A1 | Attempt 2nd derivative and set $= 0$ |
| When $x = \frac{4}{3}$, $\frac{dy}{dx} = -\frac{4}{3}$ (or $m$) $= -\frac{4}{3}$ | DM1A1 | |12\\
\includegraphics[max width=\textwidth, alt={}, center]{9803d51b-215e-4d03-884f-a67fb7ed6442-20_524_972_274_548}
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.\\
(a) State the value of $a$.\\
(b) Calculate the value of $b$.\\
(c) Find the area of the shaded region.\\
(d) The gradient, $\frac { \mathrm { d } y } { \mathrm {~d} x }$, of the curve has a minimum value $m$.
Calculate the value of $m$.\\
\hfill \mbox{\textit{CAIE P1 2020 Q12 [13]}}