| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a standard C2 calculus question involving fractional and negative powers. Part (a) requires solving a quadratic equation after substitution, (b) uses routine differentiation and solving, (c) applies the second derivative test, and (d) is a straightforward sketch. All techniques are core C2 material with no novel insights required, making it slightly easier than average. |
| Spec | 1.07i 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 |
|---|---|---|
| (a) \(3 - x^{\frac{1}{2}} - 2x^{-\frac{1}{2}} = 0\), \(3x^{\frac{1}{2}} - x - 2 = 0\) | M1 | |
| \(x - 3x^{\frac{1}{2}} + 2 = 0\), \((x^{\frac{1}{2}} - 1)(x^{\frac{1}{2}} - 2) = 0\) | M1 | |
| \(x^{\frac{1}{2}} = 1, 2\) | A1 | |
| \(x = 1, 4\) \(\therefore (1, 0), (4, 0)\) | A1 | |
| (b) \(\frac{dy}{dx} = -\frac{1}{2}x^{-\frac{1}{2}} + x^{-\frac{3}{2}}\) | M1 A1 | |
| for minimum, \(-\frac{1}{2}x^{-\frac{1}{2}} + x^{-\frac{3}{2}} = 0\) | M1 | |
| \(-\frac{1}{2}x^{-\frac{1}{2}}(x - 2) = 0\) | ||
| \(x = 2, y = 3 - \sqrt{2} - \frac{2}{\sqrt{2}}\) \(\therefore (2, 3 - 2\sqrt{2})\) | A2 | |
| (c) \(\frac{d^2y}{dx^2} = \frac{1}{4}x^{-\frac{3}{2}} - \frac{3}{2}x^{-\frac{5}{2}}\) | 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 | |
| (d) Sketch showing curve with maximum, x-intercepts at 1 and 4, y-intercept | B2 | (13) |
**(a)** $3 - x^{\frac{1}{2}} - 2x^{-\frac{1}{2}} = 0$, $3x^{\frac{1}{2}} - x - 2 = 0$ | M1 |
$x - 3x^{\frac{1}{2}} + 2 = 0$, $(x^{\frac{1}{2}} - 1)(x^{\frac{1}{2}} - 2) = 0$ | M1 |
$x^{\frac{1}{2}} = 1, 2$ | A1 |
$x = 1, 4$ $\therefore (1, 0), (4, 0)$ | A1 |
**(b)** $\frac{dy}{dx} = -\frac{1}{2}x^{-\frac{1}{2}} + x^{-\frac{3}{2}}$ | M1 A1 |
for minimum, $-\frac{1}{2}x^{-\frac{1}{2}} + x^{-\frac{3}{2}} = 0$ | M1 |
$-\frac{1}{2}x^{-\frac{1}{2}}(x - 2) = 0$ | |
$x = 2, y = 3 - \sqrt{2} - \frac{2}{\sqrt{2}}$ $\therefore (2, 3 - 2\sqrt{2})$ | A2 |
**(c)** $\frac{d^2y}{dx^2} = \frac{1}{4}x^{-\frac{3}{2}} - \frac{3}{2}x^{-\frac{5}{2}}$ | 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 |
**(d)** Sketch showing curve with maximum, x-intercepts at 1 and 4, y-intercept | B2 | (13)The curve $C$ has the equation
$$y = 3 - x^{\frac{1}{2}} - 2x^{-\frac{1}{2}}, \quad x > 0.$$
\begin{enumerate}[label=(\alph*)]
\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{Edexcel C2 Q8 [13]}}