C1 June 2007 Q10
10. The curve \(C\) has equation \(y = x^2(x - 6) + \dfrac{4}{x}\), \(x > 0\).
The points \(P\) and \(Q\) lie on \(C\) and have \(x\)-coordinates 1 and 2 respectively.
| Scheme | Marks |
|---|---|
| \(x = 1:\ y = -5 + 4 = \underline{-1},\qquad x = 2:\ y = -16 + 2 = \underline{-14}\) (can be given in (b) or (c)) | 1st B1 for −1 2nd B1 for −14 |
| \(PQ = \sqrt{(2 - 1)^2 + \left(-14 - (-1)\right)^2} = \sqrt{170} \qquad (*)\) | M1 A1cso |
| (4) |
Notes
M1: for attempting \(PQ\) or \(PQ^2\) using their \(P\) and their \(Q\). Usual rules about quoting formulae.
We must see attempt at \(1^2 + \left(y_P - y_Q\right)^2\) for M1. \(PQ^2 = \sqrt{\ldots}\) etc could be M1A0.
A1cso: for proceeding to the correct answer with no incorrect working seen.
| Scheme | Marks |
|---|---|
| \(y = x^3 - 6x^2 + 4x^{-1}\) | M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 12x - 4x^{-2}\) | M1 A1 |
| \(x = 1:\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 3 - 12 - 4 = -13\) M: Evaluate at one of the points | M1 |
| \(x = 2:\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 12 - 24 - 1 = -13 \qquad \therefore\) Parallel A: Both correct + conclusion | A1 |
| (5) |
Notes
1st M1: for multiplying by \(x^2\), the \(x^3\) or \(-6x^2\) must be correct.
2nd M1: for some correct differentiation, at least one term must be correct as printed.
1st A1: for a fully correct derivative.
These 3 marks can be awarded anywhere when first seen.3rd M1: for attempting to substitute \(x = 1\) or \(x = 2\) in their derivative. Substituting in \(y\) is M0.
2nd A1: for -13 from both substitutions and a brief comment.
The −13 must come from their derivative.
| Scheme | Marks |
|---|---|
| Finding gradient of normal \(\left(m = \dfrac{1}{13}\right)\) | M1 |
| \(y - -1 = \dfrac{1}{13}(x - 1)\) | M1 A1ft |
| \(\underline{x - 13y - 14 = 0}\) o.e. | A1cso |
| (4) | |
| (13 marks) |
Notes
1st M1: for use of the perpendicular gradient rule. Follow through their −13.
2nd M1: for full method to find the equation of the normal or tangent at \(P\). If formula is quoted allow slips in substitution, otherwise a correct substitution is required.
1st A1ft: for a correct expression. Follow through their −1 and their changed gradient.
2nd A1cso: for a correct equation with = 0 and integer coefficients.
This mark is dependent upon the −13 coming from their derivative in (b) hence cso.
Tangent can get M0M1A0A0, changed gradient can get M0M1A1A0orM1M1A1A0.
Condone confusion over terminology of tangent and normal, mark gradient and equation.
MR: Allow for \(-\dfrac{4}{x}\) or \((x+6)\) but not omitting \(4x^{-1}\) or treating it as \(4x\).