June 2024 Paper 1 Q4

Notes

Note: Throughout the question allow use of \(\delta x\) for \(h\) or any other letter e.g. \(\alpha\) if used consistently.
If \(\delta x\) is used then you can condone e.g. \(\delta^2 x\) for \(\delta x^2\) as well as condoning e.g. poorly formed \(\delta\)’s

M1: Begins the process by writing down the gradient of the chord and attempts to expand the correct squared bracket – you can condone “poor” squaring e.g. \((x+h)^2 = x^2 + h^2\) but the \(-x^2\) must be present.

A1: Reaches a correct fraction o.e. with the \(x^2\) terms cancelled out and with no algebraic errors, e.g. \(\dfrac{\cancel{x^2} + 2xh + h^2 - \cancel{x^2}}{h}\), \(2x + h\) is correct.

A1*: Completes the process by applying a limiting argument and deduces that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\) with no errors seen. They must have = \(2x\) and not just \(\lim\limits_{h \to 0} 2x\) to complete the proof.

\(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) or an equivalent e.g. \(\mathrm{f}^{\prime}(x) =\) or “Gradient =” must be evident somewhere in their working or final line. If \(\mathrm{f}^{\prime}(x)\) is used then there is no requirement to see \(\mathrm{f}(x)\) defined first. Condone e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} \rightarrow 2x\) or \(\mathrm{f}^{\prime}(x) \rightarrow 2x\).

Condone missing brackets to allow e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \lim\limits_{h \to 0} \dfrac{2xh + h^2}{h} = \lim\limits_{h \to 0} 2x + h = 2x\)

Do not allow \(h = 0\) if there is never a reference to \(h \to 0\).

e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \lim\limits_{h \to 0} \dfrac{2xh + h^2}{h} = \lim\limits_{h \to 0} 2x + 0 = 2x\) is acceptable

but e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2xh + h^2}{h} = 2x + 0 = 2x\) is not unless \(h \to 0\) is seen.

The \(h \to 0\) does not need to be present throughout the proof e.g. appear on every line but must appear at least once.

They must reach \(2x + h\) at the end and not \(\dfrac{2xh + h^2}{h}\) (without the \(h\)’s cancelled) to complete the limiting argument.