| Exam Board | OCR |
|---|---|
| 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 | Moderate -0.3 This is a straightforward multi-part C1 question requiring standard techniques: factorising a cubic to show one x-intercept, finding a normal equation via differentiation, and calculating a triangle area. All steps are routine with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x^3 - 5x^2 + 7x = 0\) | M1 | |
| \(x(x^2 - 5x + 7) = 0\) | M1 | |
| \(x = 0\) or \(x^2 - 5x + 7 = 0\) | ||
| \(b^2 - 4ac = (-5)^2 - 4(1)(7) = -3\) | M1 | |
| \(b^2 - 4ac < 0 \therefore\) no real roots | A1 | |
| \(\therefore\) only crosses x-axis at one point | A1 | |
| (ii) \(\frac{dy}{dx} = 3x^2 - 10x + 7\) | M1, A1 | |
| grad of tangent = \(27 - 30 + 7 = 4\) | M1 | |
| grad of normal = \(-\frac{1}{4}\) | A1 | |
| \(\therefore y - 3 = -\frac{1}{4}(x-3)\) | M1 | |
| \(4y - 12 = -x + 3\) | ||
| \(x + 4y = 15\) | A1 | |
| (iii) \(x = 0 \Rightarrow y = \frac{15}{4}\) | M1 | |
| \(y = 0 \Rightarrow x = 15\) | ||
| area = \(\frac{1}{2} \times \frac{15}{4} \times 15 = \frac{225}{8} = 28\frac{1}{8}\) | M1, A1 | (13) |
**(i)** $x^3 - 5x^2 + 7x = 0$ | M1 |
$x(x^2 - 5x + 7) = 0$ | M1 |
$x = 0$ or $x^2 - 5x + 7 = 0$ | |
$b^2 - 4ac = (-5)^2 - 4(1)(7) = -3$ | M1 |
$b^2 - 4ac < 0 \therefore$ no real roots | A1 |
$\therefore$ only crosses x-axis at one point | A1 |
**(ii)** $\frac{dy}{dx} = 3x^2 - 10x + 7$ | M1, A1 |
grad of tangent = $27 - 30 + 7 = 4$ | M1 |
grad of normal = $-\frac{1}{4}$ | A1 |
$\therefore y - 3 = -\frac{1}{4}(x-3)$ | M1 |
$4y - 12 = -x + 3$ | |
$x + 4y = 15$ | A1 |
**(iii)** $x = 0 \Rightarrow y = \frac{15}{4}$ | M1 |
$y = 0 \Rightarrow x = 15$ | |
area = $\frac{1}{2} \times \frac{15}{4} \times 15 = \frac{225}{8} = 28\frac{1}{8}$ | M1, A1 | **(13)** |
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**Total: (72)**A curve has the equation $y = x^3 - 5x^2 + 7x$.
\begin{enumerate}[label=(\roman*)]
\item Show that the curve only crosses the $x$-axis at one point. [4]
\end{enumerate}
The point $P$ on the curve has coordinates $(3, 3)$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find an equation for the normal to the curve at $P$, giving your answer in the form $ax + by = c$, where $a$, $b$ and $c$ are integers. [6]
\end{enumerate}
The normal to the curve at $P$ meets the coordinate axes at $Q$ and $R$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Show that triangle $OQR$, where $O$ is the origin, has area $28\frac{1}{8}$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 Q9 [13]}}