June 2024 Paper 1 Q6
6 Given that \(\mathrm{f}(x) = 2x^2 + 3\), show from first principles that \(\mathrm{f}^{\prime}(x) = 4x\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{f}(x+h) - \mathrm{f}(x)}{h} = \dfrac{2(x+h)^2 + 3 - (2x^2+3)}{h}\) | M1 | 2.1 |
| \(= \dfrac{2x^2 + 4xh + 2h^2 + 3 - (2x^2+3)}{h}\) | M1 | 2.1 |
| \(= 4x + 2h\) | A1 | 2.1 |
| \(\mathrm{f}^{\prime}(x) = \lim\limits_{h \to 0}(4x + 2h) = 4x\) | A1 | 2.1 |
| [4] |
Notes
M1: Uses the given function in the formula. Allow a slip eg missing brackets
M1: Attempt to simplify the numerator
A1: Correct expression without \(h\) in the denominator from fully correct working
A1: AG Correct use of limit as \(h \to 0\) with their expression.