AQA C3 June 2014 Unofficial Mark Scheme

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can someone explain how to do question 2ci please?

The intersection of the two lines satisfy x = 2ln(2e - x) so x = alpha is a solution to this equation. You can't solve this equation using elementary methods (i.e. rearranging/taking exponentials of both sides etc.) so you have to use a numerical method (parts (c)(ii) (c)(iii)). (c)(i) asks you to locate the root, alpha,in some given interval. Rearrange x=2ln(2e - x) to f(x) = 0. Find f(1) and f(3) and note that their signs are different. This means there is a solution to f(x)=0 in the interval 1 < alpha < 3.

It's the same process in many of these exam questions so just get some practice with it as these tend to be the 'giveaway' marks in the exam.

Reply 301

Teenage-Oddball

For 4 (c) (i), I put:
1. A stretch in the direction of the x-axis by scale factor 1/2.
2. Translation through the vector [-2 | 0]

But the mark scheme only allows:
1. Translation through the vector [-1 | 0]
2. A stretch in the direction of the x-axis by scale factor 1/2.

I've found that it could be either or but the mark scheme only allows the latter, so am I mistaken? If so, can someone please enlighten me as to why the answer in the mark scheme is the only possibility?