June 2023 Paper 1 Q17
17 The curve \(\quad y = ax^3 + bx^2 + 7 \quad\) has a stationary point at \((-2, 11)\)
Work out the values of \(a\) and \(b\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 3ax^2 + 2bx\) | M1 | either term correct ignore LHS |
| \(3a(-2)^2 + 2b(-2) = 0\) or \(12a - 4b = 0\) or \(3a - b = 0\) or \(3a(-2)^2 + 2(1 + 2a)(-2) = 0\) (this would come from doing the third M mark before the differentiation) | M1dep | when \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) oe can follow through from an error in first M mark (as long as there are at least two terms in \(x^n\)) |
| \(11 = a(-2)^3 + b(-2)^2 + 7\) or \(-8a + 4b = 4\) or \(-2a + b = 1\) or \(11 = a(-2)^3 + 3a(4) + 7\) | M1 | oe |
| \(a = 1\) or \(b = 3\) | A1 | dependent on M3 awarded |
| \(a = 1\) and \(b = 3\) | A1 |
Additional guidance
| Correct answer could come from incorrect working (follow usual guidance) | |
| \(a = 1\) and \(b = 3\) not from incorrect working will be full marks | M3 A2 |
| Only one correct answer without working is 0 marks | M0 A0 |