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OCR MEI C3 (12th June 2015) Markscheme

1. (1.107, 4.094) [6]
2. (3(2x-1)(4/3))/8 + c [4]
3. u = 2x-1, (1/16)(64ln2 - 15) [5]
4. r=h, 1/20pi = 0.0159 [5]
5. gradient = ½ [6]
6. i) x = ½ [2]ii) x = sqrt(2)/2 [2]
7. i) f(f(x)) = x via substituting in. f-1(x) = f(x) = (1-x)/(1+x) [3]
ii) g(x) is even since g(-x) = g(x). Symmetrical about the y-axis. [3]
8. i) f'(x) = 1 - 4/x2, f''(x) = 8/x3which is less than zero when x=-1 and so Q is a maximum point. [7]
ii) Area was 3 take away integral which gave ½(15-8ln4) [6]
iii) f(x-1)-1=g(x) so g(x) = (x2-3x)/(x+1) [3]
iv) This is the negative of the previous answer, and yes it's negative, because it asked for the value. [2]
9. i) x = ln3 [2]
ii) x = ln2, y = -1 [4]
iii) The integral came out as negative, but this question asked for the area, so the given answer must be positive; 4 - 3ln3 [5]
iv) f-1(x) = ln(sqrt(x+1)+2), with the domain as x >= -1 and range as y >= ln2. The sketch of the graph was a reflection in the line y = x, but subject to the 'new' domain and range (i.e. not below ln2). [7]

Please quote me if I have made a mistake...

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