| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find constant from definite integral |
| Difficulty | Moderate -0.5 This is a straightforward integration problem requiring basic techniques: integrate the derivative to find y (with constant of integration), use the point (0,11) to find the constant, then use the stationary point condition (dy/dx=0 at x=2) and the point (2,3) to form two simultaneous equations. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{1}{3}ax^3 + \frac{1}{2}bx^2 - 4x\ (+c)\) | B1 | |
| \(11 = 0+0+0+c\) | M1 | Sub \(x=0\), \(y=11\) into integrated expression. \(c\) must be present |
| \(y = \frac{1}{3}ax^3 + \frac{1}{2}bx^2 - 4x + 11\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(4a + 2b - 4 = 0\) | M1 | Sub \(x=2\), \(\frac{dy}{dx}=0\) |
| \(\frac{1}{3}(8a) + 2b - 8 + 11 = 3\) | M1 | Sub \(x=2\), \(y=3\) into integrated expression. Allow if 11 missing |
| Solve simultaneous equations | DM1 | Dep. on both M marks |
| \(a=3,\ b=-4\) | A1A1 | Allow if no working seen for simultaneous equations |
## Question 8(i):
| $y = \frac{1}{3}ax^3 + \frac{1}{2}bx^2 - 4x\ (+c)$ | B1 | |
|---|---|---|
| $11 = 0+0+0+c$ | M1 | Sub $x=0$, $y=11$ into integrated expression. $c$ must be present |
| $y = \frac{1}{3}ax^3 + \frac{1}{2}bx^2 - 4x + 11$ | A1 | |
## Question 8(ii):
| $4a + 2b - 4 = 0$ | M1 | Sub $x=2$, $\frac{dy}{dx}=0$ |
|---|---|---|
| $\frac{1}{3}(8a) + 2b - 8 + 11 = 3$ | M1 | Sub $x=2$, $y=3$ into integrated expression. Allow if 11 missing |
| Solve simultaneous equations | DM1 | Dep. on both M marks |
| $a=3,\ b=-4$ | A1A1 | Allow if no working seen for simultaneous equations |8 A curve passes through $( 0,11 )$ and has an equation for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = a x ^ { 2 } + b x - 4$, where $a$ and $b$ are constants.\\
(i) Find the equation of the curve in terms of $a$ and $b$.\\
(ii) It is now given that the curve has a stationary point at $( 2,3 )$. Find the values of $a$ and $b$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q8 [8]}}