C4 June 2014 Q2

Notes

In this question ignore part labelling and mark part (a) and part (b) together.

Note: Writing down \(\left\{(1 + kx)^{-4}\right\} = 1 + (-4)(kx) + \dfrac{(-4)(-4 - 1)}{2!}(kx)^2 + \ldots\)
gets all the method marks in Q2. i.e. (a) M1 and (b) M1M1

M1: Award M1 for

• either writing down \((-4)k = -6\) or \(4k = 6\)
• or expanding \((1 + kx)^{-4}\) to give \(1 + (-4)(kx)\)
• or writing down \((-4)kx = -6\) or \((-4k) = -6x\) or \(-4kx = -6x\)

A1: \(k = \dfrac{3}{2}\) or 1.5 or \(\dfrac{6}{4}\) from no incorrect sign errors.

Note: The M1 mark can be implied by a candidate writing down the correct value of \(k\).

Note: Award M1 for writing down \(4k = 6\) and then A1 for \(k = 1.5\) (or equivalent).

Note: Award M0 for \(4k = -6\) (if there is no evidence that \((1 + kx)^{-4}\) expands to give \(1 + (-4)(kx) + \ldots\))

Note: \(1 + (-4)(kx)\) leading to \((-4)k = 6\) leading to \(k = \dfrac{3}{2}\) is M1A0.

Notes

M1: For either \(\dfrac{(-4)(-4 - 1)}{2!}\) or \(\dfrac{(-4)(-5)}{2!}\) or 10 or \((k)^2\) or \((kx)^2\)

M1: Either \(\dfrac{(-4)(-4 - 1)}{2!}(k)^2\) or \(\dfrac{(-4)(-5)}{2!}(k)^2\) or \(\dfrac{(-4)(-5)}{2!}(kx)^2\) or \(\dfrac{(-4)(-5)}{2!}(\text{their } k)^2\) or \(10k^2\)

Note: Candidates are allowed to use 2 instead of 2!

A1: Uses \(k = 1.5\) to give \(A = \dfrac{45}{2}\) or 22.5

Note: \(A = \dfrac{90}{4}\) which has not been simplified is A0.

Note: Award A0 for \(A = \dfrac{45}{2}x^2\).

Note: Allow A1 for \(A = \dfrac{45}{2}x^2\) followed by \(A = \dfrac{45}{2}\)

Note: \(k = -1.5\) leading to \(A = \dfrac{45}{2}\) or 22.5 is A0.