| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Differentiate rational functions |
| Difficulty | Moderate -0.3 This is a standard C1/C2 differentiation question requiring quotient rule (or product rule after rewriting), finding stationary points, and determining their nature. While multi-part with several steps, each component is routine textbook material with no novel problem-solving required. Slightly easier than average due to straightforward algebraic manipulation and standard techniques. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 3x - 34 + \frac{75}{x}\) | B1 | Any correct equivalent 3 or 4 term polynomial |
| \(\frac{dy}{dx} = 3 - 75x^{-2} + \{0\}\ (x>0)\); accept \(\frac{dy}{dx} = \frac{3x^2-75}{x^2}\) | M1 A1 | M1: evidence of differentiation; \(x^n \to x^{n-1}\) at least once. A1: both terms correct and simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Put \(\frac{dy}{dx} = 3 - 75x^{-2} = 0\) | M1 | Puts \(\frac{dy}{dx}=0\) |
| \(x = 5\) | A1 | Ignore extra answer \(x=-5\) |
| Substitute to give \(y = -4\) | M1 A1 | Substitute into \(y\) to find \(y\); ignore extra answer \(-64\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Consider \(\frac{d^2y}{dx^2} = 150x^{-3} > 0\) | M1 | Considers second derivative (reducing by 1 power of \(\frac{dy}{dx}\)) and considers its sign, or considers gradient either side, or considers shape of curve |
| So minimum | A1 | Has correct second derivative, positive value for \(x\), states \(>0\) or equivalent and concludes "minimum" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x=2.5\), \(y=3.5\) | B1 | cao |
| Gradient of curve found by substituting 2.5 into \(\frac{dy}{dx}\) \((= -9)\) | M1 | Substitutes 2.5 into gradient function |
| Gradient of normal is \(-\frac{1}{m} \left(= \frac{1}{9}\right)\) | dM1 | Finds perpendicular gradient |
| Either: \(y - \text{"}3.5\text{"} = \text{"}\frac{1}{9}\text{"}(x-2.5)\) or: \(y=\text{"}\frac{1}{9}\text{"}x+c\) and \(\text{"3.5"} = \text{"}\frac{1}{9}\text{"}(2.5)+c \Rightarrow c=\text{"}3\frac{2}{9}\text{"}\) | dM1 | Equation of normal using normal gradient, \(x=2.5\) and their \(y\); depends on both previous M marks |
| \(x - 9y + 29 = 0\) or \(9y - x - 29 = 0\) or any multiple | A1 | Must have \(=0\) and integer coefficients |
## Question 13:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 3x - 34 + \frac{75}{x}$ | B1 | Any correct equivalent 3 or 4 term polynomial |
| $\frac{dy}{dx} = 3 - 75x^{-2} + \{0\}\ (x>0)$; accept $\frac{dy}{dx} = \frac{3x^2-75}{x^2}$ | M1 A1 | M1: evidence of differentiation; $x^n \to x^{n-1}$ at least once. A1: both terms correct and simplified |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Put $\frac{dy}{dx} = 3 - 75x^{-2} = 0$ | M1 | Puts $\frac{dy}{dx}=0$ |
| $x = 5$ | A1 | Ignore extra answer $x=-5$ |
| Substitute to give $y = -4$ | M1 A1 | Substitute into $y$ to find $y$; ignore extra answer $-64$ |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Consider $\frac{d^2y}{dx^2} = 150x^{-3} > 0$ | M1 | Considers second derivative (reducing by 1 power of $\frac{dy}{dx}$) and considers its sign, or considers gradient either side, or considers shape of curve |
| So minimum | A1 | Has correct second derivative, positive value for $x$, states $>0$ or equivalent and concludes "minimum" |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x=2.5$, $y=3.5$ | B1 | cao |
| Gradient of curve found by substituting 2.5 into $\frac{dy}{dx}$ $(= -9)$ | M1 | Substitutes 2.5 into gradient function |
| Gradient of normal is $-\frac{1}{m} \left(= \frac{1}{9}\right)$ | dM1 | Finds perpendicular gradient |
| Either: $y - \text{"}3.5\text{"} = \text{"}\frac{1}{9}\text{"}(x-2.5)$ or: $y=\text{"}\frac{1}{9}\text{"}x+c$ and $\text{"3.5"} = \text{"}\frac{1}{9}\text{"}(2.5)+c \Rightarrow c=\text{"}3\frac{2}{9}\text{"}$ | dM1 | Equation of normal using normal gradient, $x=2.5$ and their $y$; depends on both previous M marks |
| $x - 9y + 29 = 0$ or $9y - x - 29 = 0$ or any multiple | A1 | Must have $=0$ and integer coefficients |
---13. The curve $C$ has equation
$$y = \frac { ( x - 3 ) ( 3 x - 25 ) } { x } , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in a fully simplified form.
\item Hence find the coordinates of the turning point on the curve $C$.
\item Determine whether this turning point is a minimum or maximum, justifying your answer.
The point $P$, with $x$ coordinate $2 \frac { 1 } { 2 }$, lies on the curve $C$.
\item Find the equation of the normal at $P$, in the form $a x + b y + c = 0$, where $a$, b and $c$ are integers.\\
\includegraphics[max width=\textwidth, alt={}, center]{e878227b-d625-4ef2-ac49-a9dc05c5321a-35_90_72_2631_1873}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q13 [14]}}