| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary point then sketch curve |
| Difficulty | Moderate -0.3 This is a standard C2 differentiation question requiring routine techniques: differentiating a polynomial, finding a tangent equation, locating stationary points by solving dy/dx=0, and sketching with given information. All steps are textbook exercises with no novel problem-solving, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| \((x+5)(x-2)(x+2)\) | 2 | M1 for \(a(x+5)(x-2)(x+2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \([(x+2)](x^2+3x-10)\) | M1 | for correct expansion of one pair of their brackets |
| \(x^3+3x^2-10x+2x^2+6x-20\) | M1 | for clear expansion of correct factors – accept given answer from \((x+5)(x^2-4)\) as first step |
| Answer | Marks | Guidance |
|---|---|---|
| \(y' = 3x^2+10x-4\) | M2 | M1 if one error |
| their \(3x^2+10x-4=0\) s.o.i. | M1 | or M1 for substitution of 0.4 if trying to obtain 0 |
| \(x=0.36\ldots\) from formula o.e. | A1 | and A1 for correct demonstration of sign change |
| \((-3.7, 12.6)\) | B1+1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((-1.8, 12.6)\) | B1+1 | accept \((-1.9, 12.6)\) or f.t.(\(\frac{1}{2}\) their max \(x\), their max \(y\)) |
## Question 2:
**Part i:**
$(x+5)(x-2)(x+2)$ | 2 | M1 for $a(x+5)(x-2)(x+2)$
**Part ii:**
$[(x+2)](x^2+3x-10)$ | M1 | for correct expansion of one pair of their brackets
$x^3+3x^2-10x+2x^2+6x-20$ | M1 | for clear expansion of correct factors – accept given answer from $(x+5)(x^2-4)$ as first step | 2
**Part iii:**
$y' = 3x^2+10x-4$ | M2 | M1 if one error
their $3x^2+10x-4=0$ s.o.i. | M1 | or M1 for substitution of 0.4 if trying to obtain 0
$x=0.36\ldots$ from formula o.e. | A1 | and A1 for correct demonstration of sign change
$(-3.7, 12.6)$ | B1+1 | | 6
**Part iv:**
$(-1.8, 12.6)$ | B1+1 | accept $(-1.9, 12.6)$ or f.t.($\frac{1}{2}$ their max $x$, their max $y$) | 2
---2 The equation of a cubic curve is $y = 2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and show that the tangent to the curve when $x = 3$ passes through the point $( - 1 , - 41 )$.\\
(ii) Use calculus to find the coordinates of the turning points of the curve. You need not distinguish between the maximum and minimum.\\
(iii) Sketch the curve, given that the only real root of $2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2 = 0$ is $x = 0.2$ correct to 1 decimal place.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b6ea89e3-a8a4-41a2-8ed5-eed6c2dfda7e-2_1017_935_285_638}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Fig. 11 shows a sketch of the cubic curve $y = \mathrm { f } ( x )$. The values of $x$ where it crosses the $x$-axis are - 5 , - 2 and 2 , and it crosses the $y$-axis at $( 0 , - 20 )$.\\
\hfill \mbox{\textit{OCR MEI C2 Q2 [12]}}