| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a standard C2 calculus question covering routine differentiation, solving a cubic equation (with two roots given), and finding area between curves. All techniques are straightforward applications with no novel problem-solving required, making it slightly easier than the average A-level question which would typically require more integration of concepts or less scaffolding. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(3x^2 - 6x - 22\); their \(y' = 0\) soi; \(3.89\); \(-1.89\) | M1, M1, A1, A1 | condone one incorrect term, but must be three terms; at least one term correct in their \(y'\); if A0A0, SC1 for \(\frac{3 \pm 5\sqrt{3}}{3}\) or \(1 \pm \sqrt{1/3}\) or better, or both decimal answers given to a different accuracy or from truncation; 3.886751346 and \(-1.886751346\); condone "\(y=\)" may be implied by use of eg quadratic formula, completing square, attempt to factorise |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^3 - 3x^2 - 22x + 24 = 6x + 24\); \(x^3 - 3x^2 - 28x [= 0]\) | M1, M1 | may be implied by \(x^3 - 3x^2 - 28x [= 0]\); may be implied by \(x^2 - 3x - 28[=0]\) |
| other point when \(x = 7\) isw | A1 | dependent on award of both M marks; ignore other values of x |
| Answer | Marks | Guidance |
|---|---|---|
| \(F[x] = \frac{x^4}{4} - \frac{3x^3}{3} - \frac{22x^2}{2} + 24x\); \(F[0] - F[-4]\); area of triangle \(= 48\); area required \(= 96\) from fully correct working | M1*, M1dep, B1, A1 | allow for three terms correct; condone \(+c\); on award of either of previous M1s; A0 for \(-96\), ignore units; no marks for 96 unsupported; alternative method M1 for \(\int((x^3 - 3x^2 - 22x + 24) - (6x + 24))dx\) may be implied by 2nd M1; M1* for \(F[x] = \frac{x^4}{4} - \frac{3x^3}{3} - \frac{28x^2}{2}\) condone one error in integration; M1dep for \(F[0] - F[-4]\) |
### Part (i)
$3x^2 - 6x - 22$; their $y' = 0$ soi; $3.89$; $-1.89$ | M1, M1, A1, A1 | condone one incorrect term, but must be three terms; at least one term correct in their $y'$; if A0A0, SC1 for $\frac{3 \pm 5\sqrt{3}}{3}$ or $1 \pm \sqrt{1/3}$ or better, or both decimal answers given to a different accuracy or from truncation; 3.886751346 and $-1.886751346$; condone "$y=$" may be implied by use of eg quadratic formula, completing square, attempt to factorise
### Part (ii)
$x^3 - 3x^2 - 22x + 24 = 6x + 24$; $x^3 - 3x^2 - 28x [= 0]$ | M1, M1 | may be implied by $x^3 - 3x^2 - 28x [= 0]$; may be implied by $x^2 - 3x - 28[=0]$
other point when $x = 7$ isw | A1 | dependent on award of both M marks; ignore other values of x
### Part (iii)
$F[x] = \frac{x^4}{4} - \frac{3x^3}{3} - \frac{22x^2}{2} + 24x$; $F[0] - F[-4]$; area of triangle $= 48$; area required $= 96$ from fully correct working | M1*, M1dep, B1, A1 | allow for three terms correct; condone $+c$; on award of either of previous M1s; A0 for $-96$, ignore units; no marks for 96 unsupported; alternative method M1 for $\int((x^3 - 3x^2 - 22x + 24) - (6x + 24))dx$ may be implied by 2nd M1; M1* for $F[x] = \frac{x^4}{4} - \frac{3x^3}{3} - \frac{28x^2}{2}$ condone one error in integration; M1dep for $F[0] - F[-4]$Fig. 9 shows a sketch of the curve $y = x^3 - 3x^2 - 22x + 24$ and the line $y = 6x + 24$.
\includegraphics{figure_9}
\begin{enumerate}[label=(\roman*)]
\item Differentiate $y = x^3 - 3x^2 - 22x + 24$ and hence find the $x$-coordinates of the turning points of the curve. Give your answers to 2 decimal places. [4]
\item You are given that the line and the curve intersect when $x = 0$ and when $x = -4$. Find algebraically the $x$-coordinate of the other point of intersection. [3]
\item Use calculus to find the area of the region bounded by the curve and the line $y = 6x + 24$ for $-4 \leq x \leq 0$, shown shaded on Fig. 9. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2013 Q9 [11]}}