| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent parallel to given line |
| Difficulty | Standard +0.3 This is a straightforward C2 differentiation question requiring standard techniques: differentiating x^(-3) and x, finding stationary points by setting dy/dx=0, and finding a normal using the negative reciprocal gradient. All steps are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{x^3} = x^{-3}\) | B1 | PI by its correct derivative |
| \(\frac{dy}{dx} = -3x^{-4} + 48\) | M1 | A power decreased by 1; could be \(+48\) or ft after B0 |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-3x^{-4} + 48 = 0\) | M1 | c's answer to (a) equated to 0 |
| \(x^{-4} = 16\) | A1F | To \(x^p = q\) but only ft on eqns of the form \(ax^{2k}+48=0\) where \(a\) and \(k\) are negative integers |
| \(x = \pm\frac{1}{2}\) | A1 | |
| Equations of tangents: \(y = 32\) and \(y = -32\) | A1F | Only ft if answer is of the form \(y = \pm k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x=1\), \(\frac{dy}{dx} = -3 + 48 = 45\) | M1 | Attempt to find value of \(\frac{dy}{dx}\) at \(x=1\) |
| Gradient of normal at \((1,49)\) is \(-\frac{1}{45}\) | m1 | Correct use of \(m \times m' = -1\) with c's value of \(\frac{dy}{dx}\) when \(x=1\) |
| Normal at \((1,49)\): \(y - 49 = -\frac{1}{45}(x-1)\) | A1 | CSO. Apply ISW after ACF; accept \(49.02\) or better in place of \(49\frac{1}{45}\) |
## Question 5:
**Part (a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{x^3} = x^{-3}$ | B1 | PI by its correct derivative |
| $\frac{dy}{dx} = -3x^{-4} + 48$ | M1 | A power decreased by 1; could be $+48$ or ft after B0 |
| | A1 | |
**Part (b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-3x^{-4} + 48 = 0$ | M1 | c's answer to (a) equated to 0 |
| $x^{-4} = 16$ | A1F | To $x^p = q$ but only ft on eqns of the form $ax^{2k}+48=0$ where $a$ and $k$ are **negative integers** |
| $x = \pm\frac{1}{2}$ | A1 | |
| Equations of tangents: $y = 32$ and $y = -32$ | A1F | Only ft if answer is of the form $y = \pm k$ |
**Part (c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x=1$, $\frac{dy}{dx} = -3 + 48 = 45$ | M1 | Attempt to find value of $\frac{dy}{dx}$ at $x=1$ |
| Gradient of normal at $(1,49)$ is $-\frac{1}{45}$ | m1 | Correct use of $m \times m' = -1$ with c's value of $\frac{dy}{dx}$ when $x=1$ |
| Normal at $(1,49)$: $y - 49 = -\frac{1}{45}(x-1)$ | A1 | CSO. Apply ISW after ACF; accept $49.02$ or better in place of $49\frac{1}{45}$ |5 A curve has equation $y = \frac { 1 } { x ^ { 3 } } + 48 x$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence find the equation of each of the two tangents to the curve that are parallel to the $x$-axis.
\item Find an equation of the normal to the curve at the point $( 1,49 )$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q5 [10]}}