C1 January 2009 Q11
11. The curve \(C\) has equation\[y = 9 - 4x - \frac{8}{x}, \qquad x > 0.\]The point \(P\) on \(C\) has \(x\)-coordinate equal to 2.
The tangent at \(P\) meets the \(x\)-axis at \(A\) and the normal at \(P\) meets the \(x\)-axis at \(B\).
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) -4 + 8x^{-2}\) (4 or \(8x^{-2}\) for M1… sign can be wrong) | M1A1 |
| \(x = 2 \Rightarrow \quad m = -4 + 2 = -2\) | M1 |
| \(y = 9 - 8 - \dfrac{8}{2} = -3\) The first 4 marks could be earned in part (b) | B1 |
| Equation of tangent is: \(\quad y + 3 = -2(x - 2) \to \quad y = 1 - 2x\) (*) | M1 A1cso |
| (6) |
Notes
1st M1: for 4 or \(8x^{-2}\) (ignore the signs).
1st A1: for both terms correct (including signs).
2nd M1: for substituting \(x = 2\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (must be different from their \(y\))
B1: for \(y_P = -3\), but not if clearly found from the given equation of the tangent.
3rd M1: for attempt to find the equation of tangent at \(P\), follow through their \(m\) and \(y_P\).
Apply general principles for straight line equations (see end of scheme).
NO DIFFERENTIATION ATTEMPTED: Just assuming \(m = -2\) at this stage is M0
2nd A1cso: for correct work leading to printed answer (allow equivalents with \(2x\), \(y\), and 1 terms… such as \(2x + y - 1 = 0\)).
| Scheme | Marks |
|---|---|
| Gradient of normal \(= \tfrac{1}{2}\) | B1ft |
| Equation is: \(\dfrac{y + 3}{x - 2} = \dfrac{1}{2}\) or better equivalent, e.g. \(y = \dfrac{1}{2}x - 4\) | M1A1 |
| (3) |
Notes
B1ft: for correct use of the perpendicular gradient rule. Follow through their \(m\), but if \(m \ne -2\) there must be clear evidence that the \(m\) is thought to be the gradient of the tangent.
M1: for an attempt to find normal at \(P\) using their changed gradient and their \(y_P\).
Apply general principles for straight line equations (see end of scheme).
A1: for any correct form as specified above (correct answer only).
| Scheme | Marks |
|---|---|
| \((A:)\ \dfrac{1}{2}, \qquad (B:)\ 8\) | B1, B1 |
| Area of triangle is: \(\dfrac{1}{2}(x_B \pm x_A)\times y_P\) with values for all of \(x_B, x_A\) and \(y_P\) | M1 |
| \(\dfrac{1}{2}\left(8 - \dfrac{1}{2}\right)\times 3 = \quad \dfrac{45}{4}\) or 11.25 | A1 |
| (4) | |
| (13 marks) |
Notes
1st B1 for \(\dfrac{1}{2}\) and 2nd B1 for 8.
M1: for a full method for the area of triangle \(ABP\). Follow through their \(x_A, x_B\) and their \(y_P\), but the mark is to be awarded ‘generously’, condoning sign errors..
The final answer must be positive for A1, with negatives in the working condoned.
Determinant: Area \(= \dfrac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \dfrac{1}{2}\begin{vmatrix} 2 & -3 & 1 \\ 0.5 & 0 & 1 \\ 8 & 0 & 1 \end{vmatrix} = \ldots\) (Attempt to multiply out required for M1)
Alternative: \(AP = \sqrt{(2 - 0.5)^2 + (-3)^2}\), \(BP = \sqrt{(2 - 8)^2 + (-3)^2}\), Area \(= \dfrac{1}{2}AP\times BP = \ldots\) M1
Intersections with \(y\)-axis instead of \(x\)-axis: Only the M mark is available B0 B0 M1 A0.