| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| 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 C1 differentiation question requiring standard techniques: differentiate a polynomial, find gradient at a point, verify a tangent equation, then solve dy/dx = 3 to find another point with the same gradient. All steps are routine with no problem-solving insight needed, making it slightly easier than average but not trivial due to the multi-part nature. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(x^n \to x^{n-1} \quad \frac{dy}{dx} = 3x^2 - 2 \times 2x - 1\) | M1 A1 |
| Sub \(x=2\) | M1 | Sub \(x=2\) into their \(f'(x)\). |
| \(\frac{dy}{dx} = 3 \times 2^2 - 2 \times 4 - 1 = (3)\) | ||
| \(3 = \frac{y-1}{x-2}\) | 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 \(\left |
| \(y = 3x - 5\) 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. |
| (b) | At \(Q\) \(\frac{dy}{dx} = 3x^2 - 4x - 1 = 3\) | M1 |
| \(3x^2 - 4x - 4 = 0\) | ||
| \((3x + 2)(x - 2) = 0\) | dM1 | Factorises their 3TQ (usual rules) leading to a solution \(x = ...\). It is dependent upon the previous M. Award also for use of formula/completion of square as long as the previous M has been awarded. |
| \(x = -\frac{2}{3}\) | 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. |
(a) | $x^n \to x^{n-1} \quad \frac{dy}{dx} = 3x^2 - 2 \times 2x - 1$ | M1 A1 | $x^n \to x^{n-1}$ for any term including $3 \to 0$. $\left(\frac{dy}{dx}\right) = 3x^2 - 2 \times 2x - 1$. There is no need to see any simplification.
| Sub $x=2$ | M1 | Sub $x=2$ into their $f'(x)$.
| $\frac{dy}{dx} = 3 \times 2^2 - 2 \times 4 - 1 = (3)$ | | |
| $3 = \frac{y-1}{x-2}$ | 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 $\left|\frac{dy}{dx}\right| = 3 \times 2^2 - 2 \times 4 - 1$. 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.
(b) | At $Q$ $\frac{dy}{dx} = 3x^2 - 4x - 1 = 3$ | M1 | Sets their $\frac{dy}{dx} = 3$ and proceeds to a 3TQ = 0. Condone errors on $\left(\frac{dy}{dx}\right)$.
| $3x^2 - 4x - 4 = 0$ | | |
| $(3x + 2)(x - 2) = 0$ | dM1 | Factorises their 3TQ (usual rules) leading to a solution $x = ...$. It is dependent upon the previous M. Award also for use of formula/completion of square as long as the previous M has been awarded.
| $x = -\frac{2}{3}$ | 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.10. The curve $C$ has equation $y = x ^ { 3 } - 2 x ^ { 2 } - x + 3$
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 $y = 3 x - 5$
The point $Q$ also lies on $C$.\\
Given that the tangent to $C$ at $Q$ is parallel to the tangent to $C$ at $P$,
\item find the coordinates of the point $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q10 [10]}}