2015 H2 Maths Paper 1 Q6 — Worked Solution

Part (i)

8 marks

Since $\ln(1 + u) = u - \frac{u^{2}}{2} + \frac{u^{3}}{3} - \ldots$ for $- 1 < u \leq 1$ , we substitute $u = 2x$ :

\ln(1 + 2x)

= 2x - \frac{(2x)^{2}}{2} + \frac{(2x)^{3}}{3} - \ldots

= 2x - 2x^{2} + \frac{8}{3}x^{3} - \ldots

$\therefore$ the first three non-zero terms are $2x - 2x^{2} + \frac{8}{3}x^{3}$ .

Part (ii)

Approach

We expand $ax(1 + bx)^{c}$ by the binomial series for a non-integer index and match coefficients of $x,x^{2},x^{3}$ with the series found in (i), one equation at a time, rather than trying to solve all three simultaneously.

Using the binomial series for $(1 + bx)^{c}$ :

(1 + bx)^{c} = 1 + c(bx) + \frac{c(c - 1)}{2}(bx)^{2} + \frac{c(c - 1)(c - 2)}{6}(bx)^{3} + \ldots

Multiplying by $ax$ :

ax(1 + bx)^{c} = ax + \text{abc}\, x^{2} + \frac{ab^{2}c(c - 1)}{2}x^{3} + \frac{ab^{3}c(c - 1)(c - 2)}{6}x^{4} + \ldots

Comparing the coefficient of $x$ with $2x$ :

a = 2.

Comparing the coefficient of $x^{2}$ with $- 2x^{2}$ :

\text{abc} = - 2.

Substituting $a = 2$ :

bc = - 1.

Comparing the coefficient of $x^{3}$ with $\frac{8}{3}x^{3}$ :

\frac{ab^{2}c(c - 1)}{2} = \frac{8}{3}

\Rightarrow ab^{2}c(c - 1) = \frac{16}{3}.

Substituting $a = 2$ and $b = - \frac{1}{c}$ (from $bc = - 1$ ):

2\left( \frac{1}{c^{2}} \right)c(c - 1) = \frac{16}{3}

Simplifying the left side:

\frac{2(c - 1)}{c} = \frac{16}{3}

6(c - 1) = 16c

6c - 6 = 16c

- 10c = 6

c = - \frac{3}{5}.

Using $b = - \frac{1}{c}$ :

b = \frac{5}{3}.

$\therefore a = 2$ , $b = \frac{5}{3}$ , $c = - \frac{3}{5}$ .

To check your answer: substituting

a = 2,\ b = \frac{5}{3},\ c = - \frac{3}{5}

back into

\frac{ab^{2}c(c - 1)}{2}

gives

\frac{8}{3}

, matching the coefficient of

x^{3}

found in (i). ✓

Now we find the coefficient of $x^{4}$ , using the fourth term of the binomial expansion above:

\text{coefficient of x}^{4} = \frac{ab^{3}c(c - 1)(c - 2)}{6}.

Substituting the values found:

= \frac{2\left( \frac{5}{3} \right)^{3}\left( - \frac{3}{5} \right)\left( - \frac{8}{5} \right)\left( - \frac{13}{5} \right)}{6}

= - \frac{104}{27}.

$\therefore$ the coefficient of $x^{4}$ is $- \frac{104}{27}$ .

2015 H2 Maths Paper 1, Question 6

Part (i)

Part (ii)

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