| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Determine nature of stationary points |
| Difficulty | Moderate -0.3 This is a straightforward C2 calculus question requiring standard differentiation of powers (including fractional powers), finding stationary points by solving f'(x)=0, and using the second derivative test. All techniques are routine for this level with no problem-solving insight needed, making it slightly easier than average but not trivial due to the fractional power and multi-step nature. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f'(x) = -1 + 2x^{-1}\) | M1 A1 | |
| \(f''(x) = -\frac{2}{3}x^{-\frac{4}{3}}\) | A1 | |
| (b) for TP, \(-1 + 2x^{-1} = 0\) | M1 | |
| \(x^{\frac{1}{3}} = 2\) | M1 | |
| \(x = 8\) | A1 | |
| \(\therefore (8, 6)\) | A1 | |
| (c) \(f''(8) = -\frac{1}{24}, f''(x) < 0 \therefore\) maximum | M1 A1 |
**(a)** $f'(x) = -1 + 2x^{-1}$ | M1 A1 |
$f''(x) = -\frac{2}{3}x^{-\frac{4}{3}}$ | A1 |
**(b)** for TP, $-1 + 2x^{-1} = 0$ | M1 |
$x^{\frac{1}{3}} = 2$ | M1 |
$x = 8$ | A1 |
$\therefore (8, 6)$ | A1 |
**(c)** $f''(8) = -\frac{1}{24}, f''(x) < 0 \therefore$ maximum | M1 A1 | | (9 marks)$$f(x) = 2 - x + 3x^{\frac{1}{2}}, \quad x > 0.$$
\begin{enumerate}[label=(\alph*)]
\item Find $f'(x)$ and $f''(x)$. [3]
\item Find the coordinates of the turning point of the curve $y = f(x)$. [4]
\item Determine whether the turning point is a maximum or minimum point. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q6 [9]}}