| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring routine application of polynomial differentiation, finding a gradient from a linear equation, and using the tangent condition. All steps are standard textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part nature. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 6x^2 + 2kx + 5\) | M1 A1 | M1: \(x^n \to x^{n-1}\) for one term including \(6 \to 0\); A1: Correct derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of given line is \(\frac{17}{2}\) | B1 | Uses or states \(\frac{17}{2}\) or equivalent e.g. 8.5; must be stated or used in (b), not just seen as part of \(y = \frac{17}{2}x + \frac{1}{2}\) |
| \(\left(\frac{dy}{dx}\right)_{x=-2} = 6(-2)^2 + 2k(-2) + 5\) | M1 | Substitutes \(x = -2\) into their derivative (not the curve) |
| \(``24 - 4k + 5" = ``\frac{17}{2}" \Rightarrow k = \frac{41}{8}\) | dM1 A1 | dM1: Puts their expression \(=\) their \(\frac{17}{2}\), solves for \(k\) (dependent on previous M); A1: \(\frac{41}{8}\) or \(5\frac{1}{8}\) or \(5.125\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = -16 + 4k - 10 + 6 = ``k" - 20 = \frac{1}{2}\) | M1 A1 | M1: Substitutes \(x = -2\) and their numerical \(k\) into \(y = \ldots\); A1: \(y = \frac{1}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y - ``\frac{1}{2}" = ``\frac{17}{2}"(x - {}^{-}2) \Rightarrow -17x + 2y - 35 = 0\) or \(y = ``\frac{17}{2}"x + c \Rightarrow c = \ldots \Rightarrow -17x + 2y - 35 = 0\) or \(2y - 17x = 1 + 34 \Rightarrow -17x + 2y - 35 = 0\) | M1 A1 | M1: Correct attempt at linear equation with their 8.5 gradient (not normal gradient) using \(x = -2\) and their \(\frac{1}{2}\); A1: cao (allow any integer multiple) |
## Question 11:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 6x^2 + 2kx + 5$ | M1 A1 | M1: $x^n \to x^{n-1}$ for one term including $6 \to 0$; A1: Correct derivative |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of given line is $\frac{17}{2}$ | B1 | Uses or states $\frac{17}{2}$ or equivalent e.g. 8.5; must be stated or used in (b), not just seen as part of $y = \frac{17}{2}x + \frac{1}{2}$ |
| $\left(\frac{dy}{dx}\right)_{x=-2} = 6(-2)^2 + 2k(-2) + 5$ | M1 | Substitutes $x = -2$ into their derivative (not the curve) |
| $``24 - 4k + 5" = ``\frac{17}{2}" \Rightarrow k = \frac{41}{8}$ | dM1 A1 | dM1: Puts their expression $=$ their $\frac{17}{2}$, solves for $k$ (dependent on previous M); A1: $\frac{41}{8}$ or $5\frac{1}{8}$ or $5.125$ |
**Part (c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = -16 + 4k - 10 + 6 = ``k" - 20 = \frac{1}{2}$ | M1 A1 | M1: Substitutes $x = -2$ and their numerical $k$ into $y = \ldots$; A1: $y = \frac{1}{2}$ |
**Part (d):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y - ``\frac{1}{2}" = ``\frac{17}{2}"(x - {}^{-}2) \Rightarrow -17x + 2y - 35 = 0$ **or** $y = ``\frac{17}{2}"x + c \Rightarrow c = \ldots \Rightarrow -17x + 2y - 35 = 0$ **or** $2y - 17x = 1 + 34 \Rightarrow -17x + 2y - 35 = 0$ | M1 A1 | M1: Correct attempt at linear equation with their 8.5 gradient (not normal gradient) using $x = -2$ and their $\frac{1}{2}$; A1: cao (allow any integer multiple) |11. The curve $C$ has equation $y = 2 x ^ { 3 } + k x ^ { 2 } + 5 x + 6$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$
The point $P$, where $x = - 2$, lies on $C$.
The tangent to $C$ at the point $P$ is parallel to the line with equation $2 y - 17 x - 1 = 0$\\
Find
\item the value of $k$,
\item the value of the $y$ coordinate of $P$,
\item the equation of the tangent to $C$ at $P$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\begin{center}
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\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2016 Q11 [10]}}