| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find normal line equation at given point |
| Difficulty | Moderate -0.8 This is a straightforward C2 differentiation question requiring basic power rule application (rewriting 2/x as 2x^{-1}), substituting x=2 to verify a given gradient, then using the perpendicular gradient property to find the normal equation. All steps are routine with no problem-solving or insight required, making it easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x + 2x^{-1}\); \(\frac{dy}{dx} = 1 - 2x^{-2}\) | B1, M1, A1 | PI by sight of \(-2x^{-2}\); One term correct; OE |
| Total for 1(a)(i): 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 2\), \(\frac{dy}{dx} = 1 - \frac{2}{4} = \frac{1}{2}\) | A1 | CSO AG (be convinced) |
| Total for 1(a)(ii): 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 2\), \(y = 3\); gradient of normal \(= -2\); Equation normal: \(y - 3 = -2(x - 2)\) | B1, M1, M1, A1 | For \(y = 3\); \(m \times m' = -1\) used; \(y - "3" = m(x-2)\) OE; Award at 1st correct form |
| Total for 1(b): 4 marks | ||
| Total for Question 1: 8 marks |
**1(a)(i)**
| $y = x + 2x^{-1}$; $\frac{dy}{dx} = 1 - 2x^{-2}$ | B1, M1, A1 | PI by sight of $-2x^{-2}$; One term correct; OE |
| Total for 1(a)(i): 3 marks |
**1(a)(ii)**
| When $x = 2$, $\frac{dy}{dx} = 1 - \frac{2}{4} = \frac{1}{2}$ | A1 | CSO AG (be convinced) |
| Total for 1(a)(ii): 1 mark |
**1(b)**
| When $x = 2$, $y = 3$; gradient of normal $= -2$; Equation normal: $y - 3 = -2(x - 2)$ | B1, M1, M1, A1 | For $y = 3$; $m \times m' = -1$ used; $y - "3" = m(x-2)$ OE; Award at 1st correct form |
| Total for 1(b): 4 marks |
| **Total for Question 1: 8 marks** |
---1 A curve is defined for $x > 0$ by the equation $y = x + \frac { 2 } { x }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence show that the gradient of the curve at the point $P$ where $x = 2$ is $\frac { 1 } { 2 }$.
\end{enumerate}\item Find an equation of the normal to the curve at this point $P$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q1 [8]}}