| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent parallel to given line |
| Difficulty | Moderate -0.3 This is a straightforward tangent question requiring differentiation, point-gradient form, and solving a cubic equation. Part (a) is routine verification (5 marks for showing work). Part (b) requires finding where dy/dx = 3, giving a cubic that factors to reveal Q. All techniques are standard A-level with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x^n \to x^{n+1}\) | M1A1 | \(x^n \to x^{n-1}\) for any term including \(3 \to 0\). \(\frac{dy}{dx} = 3x^2 - 2 \times 2x - 1\). There is no need to see any simplification |
| \(\frac{dy}{dx} = 3x^2 - 2 \times 2x - 1 = (3)\) | M1 | Sub \(x=2\) into their \(f'(x)\) |
| Sub \(x=2\) | ||
| \(3 = \frac{y-1}{x-2}\) | dM1 | Uses their numerical gradient with \((2, 1)\) to find an equation of a tangent to \(y = f(x)\). It is dependent upon both M's. Accept their \(\frac{dy}{dx}\big |
| \(y = 3x - 5\) cso | A1* | Cso \(y = 3x - 5\). This is a given answer and all steps must be correct. Look for gradient = 3 having been achieved by differentiation. |
| (5 marks) | ||
| (b) | M1 | Sets their \(\frac{dy}{dx} = 3\) and proceeds to a 3TQ=0. Condone errors on \(\left(\frac{dy}{dx}\right)\) |
| At Q \(\frac{dy}{dx} = 3x^2 - 4x - 1 = 3\) | ||
| \(3x^2 - 4x - 4 = 0\) | ||
| \((3x + 2)(x - 2) = 0\) | dM1 | Factorises their 3TQ (usual rules) leading to a solution \(x= ...\) . It is dependent upon the previous M. |
| \(x = -\frac{2}{3}\) | A1 | \(x = -\frac{2}{3}\) |
| Sub \(x = -\frac{2}{3}\) into \(y = x^3 - 2x^2 - x + 3\) | dM1 | Sub their \(x = -\frac{2}{3}\) into \(y = x^3 - 2x^2 - x + 3\). It is dependent only upon the first M in (b) having been scored |
| \(y = \frac{67}{27}\) | A1 | Correct y coordinate \(y = \frac{67}{27}\) or equivalent |
| (5 marks) | ||
| (10 marks) |
(a) $x^n \to x^{n+1}$ | M1A1 | $x^n \to x^{n-1}$ for any term including $3 \to 0$. $\frac{dy}{dx} = 3x^2 - 2 \times 2x - 1$. There is no need to see any simplification
$\frac{dy}{dx} = 3x^2 - 2 \times 2x - 1 = (3)$ | M1 | Sub $x=2$ into their $f'(x)$
Sub $x=2$ | |
$3 = \frac{y-1}{x-2}$ | dM1 | Uses their numerical gradient with $(2, 1)$ to find an equation of a tangent to $y = f(x)$. It is dependent upon both M's. Accept their $\frac{dy}{dx}\big|_{x=2} = \frac{y-1}{x-2}$. Both signs must be correct
$y = 3x - 5$ cso | A1* | Cso $y = 3x - 5$. This is a given answer and all steps must be correct. Look for gradient = 3 having been achieved by differentiation.
| | **(5 marks)**
(b) | M1 | Sets their $\frac{dy}{dx} = 3$ and proceeds to a 3TQ=0. Condone errors on $\left(\frac{dy}{dx}\right)$
At Q $\frac{dy}{dx} = 3x^2 - 4x - 1 = 3$ | |
$3x^2 - 4x - 4 = 0$ | |
$(3x + 2)(x - 2) = 0$ | dM1 | Factorises their 3TQ (usual rules) leading to a solution $x= ...$ . It is dependent upon the previous M.
$x = -\frac{2}{3}$ | A1 | $x = -\frac{2}{3}$
Sub $x = -\frac{2}{3}$ into $y = x^3 - 2x^2 - x + 3$ | dM1 | Sub their $x = -\frac{2}{3}$ into $y = x^3 - 2x^2 - x + 3$. It is dependent only upon the first M in (b) having been scored
$y = \frac{67}{27}$ | A1 | Correct y coordinate $y = \frac{67}{27}$ or equivalent
| | **(5 marks)**
| | **(10 marks)**The curve $C$ has equation $y = x^3 - 2x^2 - x + 3$
The point $P$, which lies on $C$, has coordinates $(2, 1)$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the tangent to $C$ at the point $P$ is $y = 3x - 5$ [5]
\end{enumerate}
The point $Q$ also lies on $C$.
Given that the tangent to $C$ at $Q$ is parallel to the tangent to $C$ at $P$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the coordinates of the point $Q$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q10 [10]}}