June 2023 Paper 1 Q11
11 \(\quad y = \dfrac{1}{12}x^4 + 3x^2 + 4\)
Work out the positive value of \(x\) for which \(\quad \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 55\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) \dfrac{x^3}{3} + 6x\) | M1 | oe eg \(\dfrac{4}{12}x^3\) either term correct ignore LHS |
| \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} =\right) x^2 + 6\) or \(x^2 + 6 = 55\) | M1dep | oe eg \(x^2 = 49\) must come from fully correct first M |
| 7 | A1 | do not accept \(-7\) or \(\pm 7\) |
Additional guidance
| M1 dep (and hence A1) can only be gained if the first M mark is fully correct (no incorrect terms and no additional terms) eg an answer of \(\dfrac{1}{3}x^3 + 6x + 4\) would be allowed for M1 but as it’s not fully correct would lose the following 2 marks | M1 M0 A0 |