| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Differentiate after index conversion |
| Difficulty | Moderate -0.8 This is a straightforward C2 differentiation question requiring basic algebraic manipulation (dividing through by x to convert to index form), standard differentiation of powers, finding a normal line, and second derivative analysis. All steps are routine applications of learned techniques with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{x^3 + \sqrt{x}}{x} = x^2 + x^{-\frac{1}{2}}\) | B1, B1 | Award B1 for each correct term; \(p=2\), \(q=-\frac{1}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 2x - \frac{1}{2}x^{-\frac{3}{2}}\) | M1 A1 | M1 for attempt to differentiate; A1 for both terms correct |
| Answer | Marks | Guidance |
|---|---|---|
| At \(x=1\): \(y = 1 + 1 = 2\) | B1 | |
| \(\frac{dy}{dx} = 2 - \frac{1}{2} = \frac{3}{2}\) | M1 | Substituting \(x=1\) |
| Gradient of normal \(= -\frac{2}{3}\) | M1 | Using \(m_1 m_2 = -1\) |
| \(y - 2 = -\frac{2}{3}(x-1)\) i.e. \(3y + 2x = 8\) | A1 |
| Answer | Marks |
|---|---|
| \(\frac{d^2y}{dx^2} = 2 + \frac{3}{4}x^{-\frac{5}{2}}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| For \(x > 0\), \(\frac{d^2y}{dx^2} = 2 + \frac{3}{4}x^{-\frac{5}{2}} > 0\) always | M1 | Must state positive for all \(x>0\) |
| Since \(\frac{d^2y}{dx^2} > 0\) for all \(x > 0\), any stationary point would be a minimum, so no maximum points | A1 |
# Question 6:
## Part (a):
$\frac{x^3 + \sqrt{x}}{x} = x^2 + x^{-\frac{1}{2}}$ | B1, B1 | Award B1 for each correct term; $p=2$, $q=-\frac{1}{2}$
## Part (b)(i):
$\frac{dy}{dx} = 2x - \frac{1}{2}x^{-\frac{3}{2}}$ | M1 A1 | M1 for attempt to differentiate; A1 for both terms correct
## Part (b)(ii):
At $x=1$: $y = 1 + 1 = 2$ | B1 |
$\frac{dy}{dx} = 2 - \frac{1}{2} = \frac{3}{2}$ | M1 | Substituting $x=1$
Gradient of normal $= -\frac{2}{3}$ | M1 | Using $m_1 m_2 = -1$
$y - 2 = -\frac{2}{3}(x-1)$ i.e. $3y + 2x = 8$ | A1 |
## Part (c)(i):
$\frac{d^2y}{dx^2} = 2 + \frac{3}{4}x^{-\frac{5}{2}}$ | M1 A1 |
## Part (c)(ii):
For $x > 0$, $\frac{d^2y}{dx^2} = 2 + \frac{3}{4}x^{-\frac{5}{2}} > 0$ always | M1 | Must state positive for all $x>0$
Since $\frac{d^2y}{dx^2} > 0$ for all $x > 0$, any stationary point would be a minimum, so no maximum points | A1 |
---6 A curve $C$ has the equation
$$y = \frac { x ^ { 3 } + \sqrt { x } } { x } , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { x ^ { 3 } + \sqrt { x } } { x }$ in the form $x ^ { p } + x ^ { q }$.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Find an equation of the normal to the curve $C$ at the point on the curve where $x = 1$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Hence deduce that the curve $C$ has no maximum points.\\
\includegraphics[max width=\textwidth, alt={}, center]{f9a7a4dd-f7fd-4135-8872-2c1270d46a14-7_1463_1707_1244_153}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q6 [13]}}