| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find normal to curve or parallel tangent |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic C2 skills: factorizing a quadratic, finding a gradient by differentiation to get the normal equation, and integrating a quadratic between roots. All steps are routine textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(2(4)^2 - 11(4) + 12 = 32 - 44 + 12 = 0\) ✓ | B1 | Verification shown |
| \(2x^2 - 11x + 12 = (x-4)(2x-3) = 0\) | M1 | Factorising |
| Other intersection: \(x = \frac{3}{2}\), point \(\left(\frac{3}{2}, 0\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 4x - 11\); at \(x = 4\): gradient \(= 5\) | M1 A1 | Differentiation and evaluation |
| Normal gradient \(= -\frac{1}{5}\) | M1 | |
| Normal: \(y - 0 = -\frac{1}{5}(x - 4) \Rightarrow y = -\frac{1}{5}x + \frac{4}{5}\) | A1 | |
| \(x\)-intercept: \(x = 4\); \(y\)-intercept: \(y = \frac{4}{5}\) | M1 | Finding intercepts |
| Area \(= \frac{1}{2}(4)(\frac{4}{5}) = \frac{8}{5} = 1.6\) units² | A1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{\frac{3}{2}}^{4}(2x^2 - 11x + 12)\,dx\) | M1 | Correct limits and integral |
| \(= \left[\frac{2x^3}{3} - \frac{11x^2}{2} + 12x\right]_{\frac{3}{2}}^{4}\) | A1 | Correct integration |
| \(= \left(\frac{128}{3} - 88 + 48\right) - \left(\frac{9}{4} - \frac{99}{8} + 18\right) = \frac{125}{24} \approx 5.21\) units² | A1 | cao (note: take modulus if needed) |
## Question 12(i):
$2(4)^2 - 11(4) + 12 = 32 - 44 + 12 = 0$ ✓ | B1 | Verification shown
$2x^2 - 11x + 12 = (x-4)(2x-3) = 0$ | M1 | Factorising
Other intersection: $x = \frac{3}{2}$, point $\left(\frac{3}{2}, 0\right)$ | A1 |
## Question 12(ii):
$\frac{dy}{dx} = 4x - 11$; at $x = 4$: gradient $= 5$ | M1 A1 | Differentiation and evaluation
Normal gradient $= -\frac{1}{5}$ | M1 |
Normal: $y - 0 = -\frac{1}{5}(x - 4) \Rightarrow y = -\frac{1}{5}x + \frac{4}{5}$ | A1 |
$x$-intercept: $x = 4$; $y$-intercept: $y = \frac{4}{5}$ | M1 | Finding intercepts
Area $= \frac{1}{2}(4)(\frac{4}{5}) = \frac{8}{5} = 1.6$ units² | A1 | Shown
## Question 12(iii):
$\int_{\frac{3}{2}}^{4}(2x^2 - 11x + 12)\,dx$ | M1 | Correct limits and integral
$= \left[\frac{2x^3}{3} - \frac{11x^2}{2} + 12x\right]_{\frac{3}{2}}^{4}$ | A1 | Correct integration
$= \left(\frac{128}{3} - 88 + 48\right) - \left(\frac{9}{4} - \frac{99}{8} + 18\right) = \frac{125}{24} \approx 5.21$ units² | A1 | cao (note: take modulus if needed)
---12 Fig. 12 is a sketch of the curve $y = 2 x ^ { 2 } - 11 x + 12$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b4c0b4b0-f13c-49a9-9f98-f86f28d1f577-5_478_951_333_792}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
(i) Show that the curve intersects the $x$-axis at $( 4,0 )$ and find the coordinates of the other point of intersection of the curve and the $x$-axis.\\
(ii) Find the equation of the normal to the curve at the point $( 4,0 )$.
Show also that the area of the triangle bounded by this normal and the axes is 1.6 units $^ { 2 }$.\\
(iii) Find the area of the region bounded by the curve and the $x$-axis.
\hfill \mbox{\textit{OCR MEI C2 2007 Q12 [12]}}