| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a straightforward C1 differentiation question with standard techniques: factorising a cubic to show one root, finding a derivative to get the normal equation, and calculating a triangle area. Part (a) requires basic factorisation, part (b) is routine differentiation and normal line work, and part (c) involves simple coordinate geometry. Slightly above average difficulty due to multiple parts and the need to work carefully through calculations, but all techniques are standard C1 material with no novel problem-solving required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.03a 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 |
|---|---|---|
| (a) \(x^3 - 5x^2 + 7x = 0\) | M1 | |
| \(x(x^2 - 5x + 7) = 0\) | ||
| \(x = 0\) or \(x^2 - 5x + 7 = 0\) | ||
| \(b^2 - 4ac = (-5)^2 - (4 \times 1 \times 7) = -3\) | M1 | |
| \(b^2 - 4ac < 0 \therefore\) no real roots | A1 | |
| \(\therefore\) only crosses \(x\)-axis at one point | A1 | |
| (b) \(\frac{dy}{dx} = 3x^2 - 10x + 7\) | M1 A1 | |
| grad of tangent \(= 27 - 30 + 7 = 4\) | ||
| grad of normal \(= -\frac{1}{4} = -\frac{1}{4}\) | M1 A1 | |
| \(\therefore y - 3 = -\frac{1}{4}(x-3)\) | M1 | |
| \(4y - 12 = -x + 3\) | ||
| \(x + 4y = 15\) | A1 | |
| (c) \(x = 0 \Rightarrow y = \frac{15}{4}\) | ||
| \(y = 0 \Rightarrow x = 15\) | M1 | |
| area \(= \frac{1}{2} \times \frac{15}{4} \times 15 = \frac{225}{8} = 28\frac{1}{8}\) | M1 A1 | (13) |
(a) $x^3 - 5x^2 + 7x = 0$ | M1 |
$x(x^2 - 5x + 7) = 0$ | |
$x = 0$ or $x^2 - 5x + 7 = 0$ | |
$b^2 - 4ac = (-5)^2 - (4 \times 1 \times 7) = -3$ | M1 |
$b^2 - 4ac < 0 \therefore$ no real roots | A1 |
$\therefore$ only crosses $x$-axis at one point | A1 |
(b) $\frac{dy}{dx} = 3x^2 - 10x + 7$ | M1 A1 |
grad of tangent $= 27 - 30 + 7 = 4$ | |
grad of normal $= -\frac{1}{4} = -\frac{1}{4}$ | M1 A1 |
$\therefore y - 3 = -\frac{1}{4}(x-3)$ | M1 |
$4y - 12 = -x + 3$ | |
$x + 4y = 15$ | A1 |
(c) $x = 0 \Rightarrow y = \frac{15}{4}$ | |
$y = 0 \Rightarrow x = 15$ | M1 |
area $= \frac{1}{2} \times \frac{15}{4} \times 15 = \frac{225}{8} = 28\frac{1}{8}$ | M1 A1 | (13)9. A curve has the equation $y = x ^ { 3 } - 5 x ^ { 2 } + 7 x$.
\begin{enumerate}[label=(\alph*)]
\item Show that the curve only crosses the $x$-axis at one point.
The point $P$ on the curve has coordinates $( 3,3 )$.
\item Find an equation for the normal to the curve at $P$, giving your answer in the form $a x + b y = c$, where $a , b$ and $c$ are integers.
The normal to the curve at $P$ meets the coordinate axes at $Q$ and $R$.
\item Show that triangle $O Q R$, where $O$ is the origin, has area $28 \frac { 1 } { 8 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q9 [13]}}