June 2022 Paper 2 Q12
12 Work out the gradient of the curve \(\qquad y = \dfrac{12x^3 - 8x + 3}{4x^2}\)
at the point where \(\quad x = -1\)
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3x\) or \(-2x^{-1}\) or \(0.75x^{-2}\) | M1 | oe must have powers of \(x\) simplified eg \(\dfrac{12x}{4}\) or \(-\dfrac{2}{x}\) or \(\dfrac{3}{4x^2}\) |
| \(3x\) and \(-2x^{-1}\) and \(0.75x^{-2}\) | M1dep | oe must have powers of \(x\) simplified eg \(\dfrac{12x}{4}\) and \(-\dfrac{2}{x}\) and \(\dfrac{3}{4x^2}\) |
| Any one of \(3x\) and \(3(x^0)\) or \(-2x^{-1}\) and \(2x^{-2}\) or \(0.75x^{-2}\) and \(-1.5x^{-3}\) | M1 | oe eg \(\dfrac{12x}{4}\) and \(\dfrac{12}{4}x^{1 - 1}\) or \(-\dfrac{2}{x}\) and \(\dfrac{2}{x^2}\) or \(-\dfrac{2}{x}\) and \(- -2x^{-2}\) or \(\dfrac{3}{4x^2}\) and \(-\dfrac{3}{2x^3}\) implies 1st M1 for the derivatives \(x\) may be \((-1)\) |
| At least two of \(3x\) and \(3(x^0)\) or \(-2x^{-1}\) and \(2x^{-2}\) or \(0.75x^{-2}\) and \(-1.5x^{-3}\) | M1dep | oe dep on 3rd M1 for the derivatives \(x\) may be \((-1)\) |
| All three terms and their derivatives correct and 6.5 | A1 | oe eg all three terms and their derivatives correct and \(\dfrac{13}{2}\) for the derivatives \(x\) may be \((-1)\) SC3 104 |
Additional guidance
| Up to M4 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts | |
| \(\dfrac{3}{4x^2}\) seen but subsequently incorrectly simplified eg \(12x^{-2}\) (subsequent marks may be scored) | (1st) M1 |
| Correct answer after correct use of quotient rule or product rule | M4A1 |
| Incorrect answer after use of quotient rule or product rule | Zero |
| Condone \(y = 3 + 2x^{-2} \ldots\) etc | |
| All three terms and their derivatives correct and 6.5 in working but different answer eg \(y = 6.5x \ldots\) | M4A0 |
| SC3 is for multiplying the numerator by \(4x^{-2}\) with no subsequent errors |