Standard +0.3 This is a straightforward differentiation problem requiring students to find dy/dx, set it equal to 8 (the gradient of the given line), and solve the resulting equation. While it involves differentiating x^(-1) and solving a cubic equation, these are standard C1 techniques with no conceptual difficulty beyond routine application.
Equate their \(\frac{dy}{dx}\) to 8 (or their gradient of line, if clear)
Note: If equated to \(+1/8\) then M0 but the next M1 and its dependencies are available
*M1
Use a correct substitution to obtain a 3 term quadratic or factorise into 2 brackets each containing \(x^2\)
If no substitution stated and treated as quadratic (e.g. quadratic formula), no more marks until square rooting seen. SC: If spotted after first five marks- (3, 12) B1, (-3, -12) B1. Justifies exactly two solutions B3
DM1
Correct method to solve 3 term quadratic – dependent on previous M1
A1
No extras
DM1
Attempt to find \(x\) by square rooting – accept one value
A1
No extras
[10]
More Additional Guidance for Q10
If curve equated to line and before differentiating: First four marks B1 M1 A1 B1 available as main scheme. Then M0 for equating as this not been explicitly done. Allow the M1 for the substitution. DM1 for quadratic as main scheme (dependent on a correct substitution). A0 for the 9 (as follows wrong working). DM1 for square rooting (dependent on a correct substitution). A0 for the co-ordinates (as follows wrong working). Max mark 7/10
$\frac{dy}{dx} = x^2 - 9x^{-2}$, Gradient of line $= 8$, $x^2 - 9x^{-2} = 8$, $x^4 - 8x^2 - 9 = 0$, $k^2 - 8k - 9 = 0$, $(k - 9)(k + 1) = 0$, $k = 9$ (don't need $k = -1$), $x = 3, -3$, $y = 12, -12$
| B1 | $x^2$ from differentiating first term | |
| M1 | $kx^{-2}$ | |
| A1 | $-9x^{-2}$ (no + c) | |
| B1 | | |
| M1 | Equate their $\frac{dy}{dx}$ to 8 (or their gradient of line, if clear) | Note: If equated to $+1/8$ then M0 but the next M1 and its dependencies are available |
| *M1 | Use a correct substitution to obtain a 3 term quadratic or factorise into 2 brackets each containing $x^2$ | If no substitution stated and treated as quadratic (e.g. quadratic formula), no more marks until square rooting seen. SC: If spotted after first five marks- (3, 12) B1, (-3, -12) B1. Justifies exactly two solutions B3 |
| DM1 | Correct method to solve 3 term quadratic – dependent on previous M1 | |
| A1 | No extras | |
| DM1 | Attempt to find $x$ by square rooting – accept one value | |
| A1 | No extras | |
| [10] | | |
**More Additional Guidance for Q10**
If curve equated to line and before differentiating: First four marks B1 M1 A1 B1 available as main scheme. Then M0 for equating as this not been explicitly done. Allow the M1 for the substitution. DM1 for quadratic as main scheme (dependent on a correct substitution). A0 for the 9 (as follows wrong working). DM1 for square rooting (dependent on a correct substitution). A0 for the co-ordinates (as follows wrong working). Max mark 7/10
Find the coordinates of the points on the curve $y = \frac{1}{3}x^3 + \frac{9}{x}$ at which the tangent is parallel to the line $y = 8x + 3$. [10]
\hfill \mbox{\textit{OCR C1 2013 Q10 [10]}}