| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary point then sketch curve |
| Difficulty | Moderate -0.3 This is a straightforward C1 stationary points question requiring standard techniques: solving a quadratic (via substitution u=x^{1/2}), differentiating using power rule, finding stationary points by setting dy/dx=0, and using second derivative test. All steps are routine with no novel problem-solving required, though the fractional/negative powers add minor complexity beyond the most basic examples. |
| 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 derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(3 - x^{\frac{1}{4}} - 2x^{-\frac{1}{4}} = 0\) | ||
| \(3x^{\frac{1}{4}} - x - 2 = 0\) | M1 | |
| \(x - 3x^{\frac{1}{4}} + 2 = 0, (x^{\frac{1}{4}} - 1)(x^{\frac{1}{4}} - 2) = 0\) | M1 | |
| \(x^{\frac{1}{4}} = 1, 2\) | A1 | |
| \(x = 1, 4 \therefore (1,0), (4, 0)\) | A1 | |
| (ii) \(\frac{dy}{dx} = -\frac{1}{4}x^{-\frac{3}{4}} + x^{-\frac{3}{4}}\) | M1 A1 | |
| for minimum, \(-\frac{1}{4}x^{-\frac{3}{4}} + x^{-\frac{3}{4}} = 0\) | M1 | |
| \(-\frac{1}{4}x^{-\frac{3}{4}}(x - 2) = 0\) | ||
| \(x = 2\) | A1 | |
| \(y = 3 - \sqrt{2} - \frac{2}{\sqrt{2}} \therefore (2, 3 - 2\sqrt{2})\) | A1 | |
| (iii) \(\frac{d^2y}{dx^2} = \frac{1}{4}x^{-\frac{7}{4}} - \frac{3}{2}x^{-\frac{7}{4}}\) | M1 | |
| when \(x = 2, \frac{d^2y}{dx^2} = \frac{1}{8\sqrt{2}} - \frac{3}{8\sqrt{2}} = -\frac{1}{4\sqrt{2}}, \frac{d^2y}{dx^2} < 0 \therefore\) maximum | A1 | |
| (iv) [Graph showing curve passing through origin, with maximum in first quadrant, crossing x-axis at two points] | B2 | (13) |
**(i)** $3 - x^{\frac{1}{4}} - 2x^{-\frac{1}{4}} = 0$ | |
$3x^{\frac{1}{4}} - x - 2 = 0$ | M1 |
$x - 3x^{\frac{1}{4}} + 2 = 0, (x^{\frac{1}{4}} - 1)(x^{\frac{1}{4}} - 2) = 0$ | M1 |
$x^{\frac{1}{4}} = 1, 2$ | A1 |
$x = 1, 4 \therefore (1,0), (4, 0)$ | A1 |
**(ii)** $\frac{dy}{dx} = -\frac{1}{4}x^{-\frac{3}{4}} + x^{-\frac{3}{4}}$ | M1 A1 |
for minimum, $-\frac{1}{4}x^{-\frac{3}{4}} + x^{-\frac{3}{4}} = 0$ | M1 |
$-\frac{1}{4}x^{-\frac{3}{4}}(x - 2) = 0$ | |
$x = 2$ | A1 |
$y = 3 - \sqrt{2} - \frac{2}{\sqrt{2}} \therefore (2, 3 - 2\sqrt{2})$ | A1 |
**(iii)** $\frac{d^2y}{dx^2} = \frac{1}{4}x^{-\frac{7}{4}} - \frac{3}{2}x^{-\frac{7}{4}}$ | M1 |
when $x = 2, \frac{d^2y}{dx^2} = \frac{1}{8\sqrt{2}} - \frac{3}{8\sqrt{2}} = -\frac{1}{4\sqrt{2}}, \frac{d^2y}{dx^2} < 0 \therefore$ maximum | A1 |
**(iv)** [Graph showing curve passing through origin, with maximum in first quadrant, crossing x-axis at two points] | B2 | (13) |
---
**Total (72)**The curve $C$ has the equation
$$y = 3 - x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}, \quad x > 0.$$
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the points where $C$ crosses the $x$-axis. [4]
\item Find the exact coordinates of the stationary point of $C$. [5]
\item Determine the nature of the stationary point. [2]
\item Sketch the curve $C$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 Q9 [13]}}