June 2016 Paper 2 Q25
25 The curve \(\ y = 2x^3 - 5\ \) intersects the \(y\)-axis at \(C\).
The tangent to the curve at \(P(2, 11)\) intersects the \(y\)-axis at \(D\).
Not drawn accurately
Work out the length \(CD\). [6 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(y\)-coordinate of \(C\) is \(-5\) or \((0, -5)\) | B1 | May be on diagram or implied by final answer |
| \(3 \times 2x^2\) or \(6x^2\) | M1 | Differentiates correctly |
| \(6 \times 2 \times 2\) or 24 | M1 | Substitutes \(x = 2\) in their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) must be a function of \(x\) but cannot be \(2x^3 - 5\) |
| \(y - 11 =\) their \(24\,(x - 2)\) or \(y =\) their \(24x + c\) and substitutes (2, 11) or \(y = 24x - 37\) | M1dep | dep on 2nd M1 |
| \(y - 11 =\) their \(24\,(0 - 2)\) or \(y = 0 - 37\) or \(y = -37\) or (\(y\)-coordinate of \(D =\)) \(-37\) | M1dep | Substitutes \(x = 0\) into their linear equation dep on 3rd M1 May be seen on diagram 11 − 2 × their 24 scores 3rd M1 and 4th M1 |
| 32 | A1ft | ft B0 M4 with a positive length from their \(y\)-coordinate of \(C - -37\) |
Additional guidance
| \(-32\) | A0 |
| \((-5, 0)\) is B0 unless recovered | |
| 1st M1 Allow \(6x^2 + c\) (\(c\) an unknown constant) if \(c\) subsequently rejected | |
| Differentiates to \(6x^2 - 5\) 1st M0 gradient \(= 6 \times 2^2 - 5 = 19\) 2nd M1 Could continue and gain 3rd M1 and 4th M1 | |
| 3rd M1 Cannot be the equation of the normal at (2, 11) |