| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.3 This is a straightforward stationary points question requiring differentiation (including negative power), solving dy/dx=0, and using the second derivative test. While it involves multiple steps and the quotient rule rewritten as a negative power, it's a standard C1 exercise with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = 6x + 6x^{-2}\) | M1, A1 | Attempt to differentiate (one non-zero term correct) |
| \(6x + \frac{6}{x^2} = 0\) | M1 | Sets their \(\frac{dy}{dx} = 0\) |
| \(x = -1\) | A1 | Correct value for x - www |
| \(y = 7\) | A1 ft, 5 | Correct value of y for their value of x |
| (ii) \(\frac{d^2y}{dx^2} = 6 - 12x^{-3}\) | M1 | Correct method e.g. substitutes their x from (i) into their \(\frac{d^2y}{dx^2}\) (must involve x) and considers sign. |
| When \(x = -1\), \(\frac{d^2y}{dx^2} > 0\) so minimum pt | A1 ft, 2, 7 | ft from their \(\frac{dy}{dx}\) differentiated correctly and correct substitution of their value of x and consistent final conclusion |
**(i)** $\frac{dy}{dx} = 6x + 6x^{-2}$ | M1, A1 | Attempt to differentiate (one non-zero term correct) | Completely correct | NB – $x = -1$ (and therefore possibly $y = 7$) can be found from equating the incorrect differential $\frac{dy}{dx} = 6x + 6$ to 0. This could score M1A0 M1A0A1 ft
$6x + \frac{6}{x^2} = 0$ | M1 | Sets their $\frac{dy}{dx} = 0$ |
$x = -1$ | A1 | Correct value for x - www |
$y = 7$ | A1 ft, 5 | Correct value of y for their value of x | If more than one value of x found, allow A1 ft for one correct value of y
**(ii)** $\frac{d^2y}{dx^2} = 6 - 12x^{-3}$ | M1 | Correct method e.g. substitutes their x from (i) into their $\frac{d^2y}{dx^2}$ (must involve x) and considers sign. | Allow comparing signs of their $\frac{d\phi}{dx}$ either side of their "-1", comparing values of y to their "7" | SC $\frac{d^2y}{dx^2}$ = a constant correctly obtained from their $\frac{dy}{dx}$ and correct conclusion (ft) B1
When $x = -1$, $\frac{d^2y}{dx^2} > 0$ so minimum pt | A1 ft, 2, 7 | ft from their $\frac{dy}{dx}$ differentiated correctly and correct substitution of their value of x and consistent final conclusion | NB If second derivative evaluated, it must be correct (18 for x = -1). If more than one value of x used, max M1 A08 (i) Find the coordinates of the stationary point on the curve $y = 3 x ^ { 2 } - \frac { 6 } { x } - 2$.\\
(ii) Determine whether the stationary point is a maximum point or a minimum point.
\hfill \mbox{\textit{OCR C1 2011 Q8 [7]}}