| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Moderate -0.3 This is a straightforward multi-part integration and differentiation question requiring standard techniques: integrating x^(-1/2), finding constants using boundary conditions, setting dy/dx=0, computing second derivatives, and basic coordinate geometry with normals. All parts are routine P1 exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = 4\sqrt{x} - x(+c)\) | B1 B1 | Ignore \(+c\). |
| Uses \((9, 5)\) in an integrated expression | M1 | Substitution of point after integration. |
| \(\rightarrow c = 2\) | A1 | co. |
| [4] | ||
| (ii) \(\frac{dy}{dx} = 0 \rightarrow x = 4, y = 6\) | M1 A1 | Attempt to solve \(dy/dx = 0\). \(x\) correct. |
| A1 | \(y\) correct. | |
| [3] | ||
| (iii) \(\frac{d^2y}{dx^2} = -x^{-\frac{3}{2}} \rightarrow -ve \rightarrow \text{Max}\) | B1 B1 √ | co. √ for correct deduction. |
| [2] | ||
| (iv) \(\frac{dy}{dx} = -\frac{1}{3}\) | M1 | Use of \(m_1m_2 = -1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = 3\) | A1 | Needs \(k = 3\) |
| Angle is \(\tan^{-1}3\) | ||
| \(k = 3\) | [2] |
$\frac{dy}{dx} = \frac{2}{\sqrt{x}} - 1$ and $P(9, 5)$
(i) $y = 4\sqrt{x} - x(+c)$ | B1 B1 | Ignore $+c$.
Uses $(9, 5)$ in an integrated expression | M1 | Substitution of point after integration.
$\rightarrow c = 2$ | A1 | co.
| [4] |
(ii) $\frac{dy}{dx} = 0 \rightarrow x = 4, y = 6$ | M1 A1 | Attempt to solve $dy/dx = 0$. $x$ correct.
| A1 | $y$ correct.
| [3] |
(iii) $\frac{d^2y}{dx^2} = -x^{-\frac{3}{2}} \rightarrow -ve \rightarrow \text{Max}$ | B1 B1 √ | co. √ for correct deduction.
| [2] |
(iv) $\frac{dy}{dx} = -\frac{1}{3}$ | M1 | Use of $m_1m_2 = -1$
Perpendicular $m = 3$
$\tan\theta = 3$ | A1 | Needs $k = 3$
Angle is $\tan^{-1}3$ |
$k = 3$ | [2] |9 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { \sqrt { } x } - 1$ and $P ( 9,5 )$ is a point on the curve.\\
(i) Find the equation of the curve.\\
(ii) Find the coordinates of the stationary point on the curve.\\
(iii) Find an expression for $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ and determine the nature of the stationary point.\\
(iv) The normal to the curve at $P$ makes an angle of $\tan ^ { - 1 } k$ with the positive $x$-axis. Find the value of $k$.
\hfill \mbox{\textit{CAIE P1 2011 Q9 [11]}}