| 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 calculus question covering routine differentiation, tangent equations, and stationary points. While it has multiple parts (6+ marks total), each step follows textbook procedures: differentiate a polynomial, find gradient at a point, verify tangent passes through a coordinate, solve dy/dx=0 for turning points, and sketch using given information. Slightly easier than average due to straightforward algebra and no novel problem-solving required. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| \(y' = 6x^2 - 18x + 12\) | M1 | condone one error |
| \(= 12\) | M1 | subst of \(x = 3\) in their \(y'\) |
| \(y = 7\) when \(x = 3\) | B1 | |
| tgt is \(y - 7 = 12(x - 3)\) | M1 | f.t. their \(y\) and \(y'\) |
| verifying \((-1, -41)\) on tgt | A1 | or B2 for showing line joining \((3, 7)\) and \((-1, -41)\) has gradient \(12\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y' = 0\) soi | M1 | Their \(y'\) |
| quadratic with 3 terms | M1 | Any valid attempt at solution |
| \(x = 1\) or \(2\) | A1 | or A1 for \((1, 3)\) and A1 for \((2, 2)\) marking to benefit of candidate |
| \(y = 3\) or \(2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| cubic curve correct orientation | G1 | |
| touching \(x\)-axis only at \((0.2, 0)\) | G1 | |
| max and min correct | G1 | |
| curve crossing \(y\) axis only at \(-2\) | f.t. |
# Question 1
## (i)
$y' = 6x^2 - 18x + 12$ | M1 | condone one error
$= 12$ | M1 | subst of $x = 3$ in their $y'$
$y = 7$ when $x = 3$ | B1 |
tgt is $y - 7 = 12(x - 3)$ | M1 | f.t. their $y$ and $y'$
verifying $(-1, -41)$ on tgt | A1 | or B2 for showing line joining $(3, 7)$ and $(-1, -41)$ has gradient $12$
## (ii)
$y' = 0$ soi | M1 | Their $y'$
quadratic with 3 terms | M1 | Any valid attempt at solution
$x = 1$ or $2$ | A1 | or A1 for $(1, 3)$ and A1 for $(2, 2)$ marking to benefit of candidate
$y = 3$ or $2$ | A1 |
## (iii)
cubic curve correct orientation | G1 |
touching $x$-axis only at $(0.2, 0)$ | G1 |
max and min correct | G1 |
curve crossing $y$ axis only at $-2$ | | f.t.1 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.
\hfill \mbox{\textit{OCR MEI C2 Q1 [12]}}