| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary points coordinates |
| Difficulty | Moderate -0.3 This is a standard multi-part differentiation question covering routine techniques: finding stationary points by setting dy/dx=0, finding intercepts, sketching, and finding a tangent equation. Part (iii) requires solving a cubic but helpfully factors as (x-1)²(x+2)=0. All techniques are textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks |
|---|---|
| \(y' = 3x^2 - 5\) | M1 |
| their \(y' = 0\) | M1 |
| \((1.3, 4.3)\) cao | A1 |
| \((-1.3, -4.3)\) cao | A1 |
| Answer | Marks |
|---|---|
| crosses axes at \((0, 0)\) and \((\pm\sqrt{5}, 0)\) | B1 |
| sketch of cubic with turning points in correct quadrants and of correct orientation and passing through origin | B1 |
| \(x\)-intercepts \(\pm\sqrt{5}\) marked | B1 |
| B1 |
| Answer | Marks |
|---|---|
| substitution of \(x = 1\) in \(f'(x) = 3x^2 - 5\) | M1 |
| \(-2\) | A1 |
| \(y - 4 = \text{(their } f'(1)) \times (x - 1)\) oe | M1* |
| \(-2x + 2 = x^3 - 5x\) and completion to given result www | M1dep* |
| use of Factor theorem in \(x^3 - 3x + 2\) with \(-1\) or \(\pm 2\) | M1 |
| \(x = -2\) obtained correctly | A1 |
# Question 1
## (i)
$y' = 3x^2 - 5$ | M1
their $y' = 0$ | M1
$(1.3, 4.3)$ cao | A1
$(-1.3, -4.3)$ cao | A1
**[4]**
**Guidance:**
- or A1 for $x = \pm\frac{5}{3}$ oe soi
- allow if not written as co-ordinates if pairing is clear
- ignore any work relating to second derivative
## (ii)
crosses axes at $(0, 0)$ and $(\pm\sqrt{5}, 0)$ | B1
sketch of cubic with turning points in correct quadrants and of correct orientation and passing through origin | B1
$x$-intercepts $\pm\sqrt{5}$ marked | B1
| B1
**[4]**
**Guidance:**
- condone $x$ and $y$ intercepts not written as co-ordinates; may be on graph
- $\pm(2.23 \text{ to } 2.24)$ implies $\pm\sqrt{5}$
- may be in decimal form ($\pm 2.2\ldots$)
- must meet the $x$-axis three times
- B0 eg if more than 1 point of inflection
- See examples in Appendix
## (iii)
substitution of $x = 1$ in $f'(x) = 3x^2 - 5$ | M1
$-2$ | A1
$y - 4 = \text{(their } f'(1)) \times (x - 1)$ oe | M1*
$-2x + 2 = x^3 - 5x$ and completion to given result www | M1dep*
use of Factor theorem in $x^3 - 3x + 2$ with $-1$ or $\pm 2$ | M1
$x = -2$ obtained correctly | A1
**[6]**
**Guidance:**
- or $-4 = -2 \times (1) + c$ or any other valid method; must be shown
- sight of $-2$ does not necessarily imply M1: check $f'(x) = 3x^2 - 5$ is correct in part (i)
- eg long division or comparing coefficients to find $(x-1)(x^2 + x - 2)$ or $(x+2)(x^2 - 2x + 1)$ is enough for M1 with both factors correct
- NB M0A0 for $x(x^2 - 3) = -2$ so $x = -2$ or $x^2 - 3 = -2$ oe1 (i) Use calculus to find, correct to 1 decimal place, the coordinates of the turning points of the curve $y = x ^ { 3 } - 5 x$. [You need not determine the nature of the turning points.]\\
(ii) Find the coordinates of the points where the curve $y = x ^ { 3 } - 5 x$ meets the axes and sketch the curve.\\
(iii) Find the equation of the tangent to the curve $y = x ^ { 3 } - 5 x$ at the point $( 1 , - 4 )$. Show that, where this tangent meets the curve again, the $x$-coordinate satisfies the equation
$$x ^ { 3 } - 3 x + 2 = 0$$
Hence find the $x$-coordinate of the point where this tangent meets the curve again.
\hfill \mbox{\textit{OCR MEI C2 Q1 [14]}}