All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
From MF15,
\mathrm{ln}(1+2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} + \ldots

\approx 2x - 2x^2 + \frac{8}{3}x^3

(ii)
ax(1+bx)^c = ax(1 + cbx + \frac{c(c)-1}{2!}(bx)^2 + \ldots)

Comparing coefficients,

a = 2

abc = -2 \Rightarrow bc = -1

2[\frac{c(c-1)}{2!}b^2] = \frac{8}{3}

\Rightarrow 2[\frac{c^2-c}{2c^2}] = \frac{8}{3}

c = -\frac{3}{5}

b = \frac{5}{3}

Coefficient of x^4 = a[\frac{c(c-1)(c-2)}{3!}(\frac{5}{3})^3] = -\frac{104}{27}

Back to 2015 A-level H2 Mathematics (9740) Paper 1 Suggested Solutions

KS Comments:

Hopefully you didn’t copy the formula wrongly.
The next parts, just need to be really careful. It is quite a lot of variables to juggle around here. For the last part, dont forget the a and b^2

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