| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.3 This is a straightforward simultaneous equations problem using basic differentiation. Students must use two conditions (point on curve, gradient at point) to find two unknowns. While it requires multiple steps, the techniques are standard Core 1/2 material with no conceptual challenges—slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
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| Q16 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 3ax^2 + 2bx + 2\) | B1 | States or uses this derivative |
| Sub \(x=1, y=4\) into \(y = ax^3 + bx^2 + 2x - 5\) giving \(a + b + 2 - 5 = 4\), i.e. \(a + b = 7\) | M1 | Attempts to substitute \(x=1, y=4\) in \(y = f(x)\). Also scored by substituting \(x=1\) into \(ax^3 + bx^2 + 2x - 5 = 12x - 8 \Rightarrow a + b + 2 - 5 = 12 - 8\) |
| Sub \(x=1, \frac{dy}{dx} = 12\) into \(\frac{dy}{dx} = 3ax^2 + 2bx + 2\) giving \(3a + 2b + 2 = 12\), i.e. \(3a + 2b = 10\) | M1 | Attempts to substitute \(x=1, \frac{dy}{dx}=12\) in their \(\frac{dy}{dx} = 3ax^2 + 2bx + 2\) |
| Solves simultaneously: \(a + b = 7\), \(3a + 2b = 10 \Rightarrow a = -4, b = 11\) | dM1 | Solves simultaneously to find both \(a\) and \(b\). Both M marks must have been awarded. Allow from a graphical calculator. Sight of both values is sufficient. |
| \(a = -4, b = 11\) | A1 |
## Question 16:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 3ax^2 + 2bx + 2$ | B1 | States or uses this derivative |
| Sub $x=1, y=4$ into $y = ax^3 + bx^2 + 2x - 5$ giving $a + b + 2 - 5 = 4$, i.e. $a + b = 7$ | M1 | Attempts to substitute $x=1, y=4$ in $y = f(x)$. Also scored by substituting $x=1$ into $ax^3 + bx^2 + 2x - 5 = 12x - 8 \Rightarrow a + b + 2 - 5 = 12 - 8$ |
| Sub $x=1, \frac{dy}{dx} = 12$ into $\frac{dy}{dx} = 3ax^2 + 2bx + 2$ giving $3a + 2b + 2 = 12$, i.e. $3a + 2b = 10$ | M1 | Attempts to substitute $x=1, \frac{dy}{dx}=12$ in their $\frac{dy}{dx} = 3ax^2 + 2bx + 2$ |
| Solves simultaneously: $a + b = 7$, $3a + 2b = 10 \Rightarrow a = -4, b = 11$ | dM1 | Solves simultaneously to find both $a$ and $b$. Both M marks must have been awarded. Allow from a graphical calculator. Sight of both values is sufficient. |
| $a = -4, b = 11$ | A1 | |
**Total: 5 marks**\begin{enumerate}
\item $\mathrm { f } ( x ) = a x ^ { 3 } + b x ^ { 2 } + 2 x - 5$, where $a$ and $b$ are constants The point $P ( 1,4 )$ lies on the curve with equation $y = \mathrm { f } ( x )$.
\end{enumerate}
The tangent to $y = \mathrm { f } ( x )$ at the point $P$ has equation $y = 12 x - 8$
Calculate the value of $a$ and the value of $b$.\\
(5)\\
VILIV SIMI NI III IM I ON OC\\
VILV SIHI NI JAHMMION OC\\
VALV SIHI NI JIIIM ION OC
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\hfill \mbox{\textit{Edexcel C12 2017 Q16 [5]}}