| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find stationary points from derivative |
| Difficulty | Moderate -0.8 This is a straightforward multi-part integration question requiring standard power rule integration, finding a constant using a given point, differentiating to find the second derivative, and routine stationary point analysis. All techniques are basic AS-level calculus with no problem-solving insight needed—purely procedural application of standard methods. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = \frac{2}{3}x^3 - 2x^2 + (c)\) oe | B1B1 [4] | Attempt to integrate |
| \(\frac{2}{3} - \frac{16}{3} - 4 + c\) | M1 | Sub \(\left(4, \frac{2}{3}\right)\). Dependent on c present |
| \(c = -\frac{2}{3}\) | A1 | |
| (ii) \(\frac{1}{2}x^4 + \frac{1}{2}x^{-2}\) oe | B1B1 [2] | |
| (iii) \(x^{\frac{1}{2}} - x^{-\frac{1}{2}} = 0 \to \frac{x - 1}{\sqrt{x}} = 0\) | M1 | Equate to zero and attempt to solve |
| \(x = 1\) | A1 | |
| When \(x = 1, y = \frac{2}{3} - 2 - \frac{2}{3} = -2\) | M1A1 | Sub. their '1' into their 'y' |
| When \(x = 1, \frac{d^2y}{dx^2} = (1) > 0\) Hence minimum | B1 [5] | Everything correct on final line. Also dep on correct (ii). Accept other valid methods |
**(i)** $y = \frac{2}{3}x^3 - 2x^2 + (c)$ oe | B1B1 [4] | Attempt to integrate
$\frac{2}{3} - \frac{16}{3} - 4 + c$ | M1 | Sub $\left(4, \frac{2}{3}\right)$. Dependent on c present
$c = -\frac{2}{3}$ | A1 |
**(ii)** $\frac{1}{2}x^4 + \frac{1}{2}x^{-2}$ oe | B1B1 [2] |
**(iii)** $x^{\frac{1}{2}} - x^{-\frac{1}{2}} = 0 \to \frac{x - 1}{\sqrt{x}} = 0$ | M1 | Equate to zero and attempt to solve
$x = 1$ | A1 |
When $x = 1, y = \frac{2}{3} - 2 - \frac{2}{3} = -2$ | M1A1 | Sub. their '1' into their 'y'
When $x = 1, \frac{d^2y}{dx^2} = (1) > 0$ Hence minimum | B1 [5] | Everything correct on final line. Also dep on correct (ii). Accept other valid methods12 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { \frac { 1 } { 2 } } - x ^ { - \frac { 1 } { 2 } }$. The curve passes through the point $\left( 4 , \frac { 2 } { 3 } \right)$.\\
(i) Find the equation of the curve.\\
(ii) Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.\\
(iii) Find the coordinates of the stationary point and determine its nature.
\hfill \mbox{\textit{CAIE P1 2014 Q12 [11]}}