2015 A-level H2 Mathematics (9740) Paper 1 Question 6 Suggested Solutions

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

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

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

    pingbacks / trackbacks

    Leave a Comment

    five × 2 =

    Contact Us

    CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

    Not readable? Change text. captcha txt

    Start typing and press Enter to search