2017 H2 Maths Paper 1 Q1 — Worked Solution

Solution

4 marks

Using the standard series from MF26,

e^{2x}

= 1 + (2x) + \frac{(2x)^{2}}{2!} + \frac{(2x)^{3}}{3!} + \ldots

= 1 + 2x + 2x^{2} + \frac{4}{3}x^{3} + \ldots

and

\ln(1 + ax)

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

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

Multiplying the two series and keeping terms up to $x^{3}$ , only products of powers of $x$ adding to $3$ or less are needed:

e^{2x}\ln(1 + ax)

= \left( 1 + 2x + 2x^{2} + \frac{4}{3}x^{3} \right)\left( ax - \frac{a^{2}}{2}x^{2} + \frac{a^{3}}{3}x^{3} \right) + \ldots

= ax + \left( 2a - \frac{a^{2}}{2} \right)x^{2} + \left( 2a - a^{2} + \frac{a^{3}}{3} \right)x^{3} + \ldots

Hence, for there to be no term in $x^{2}$ , the coefficient of $x^{2}$ must vanish:

2a - \frac{a^{2}}{2} = 0

\Rightarrow a\left( 2 - \frac{a}{2} \right) = 0

\Rightarrow a = 0\ \left( \text{rej. since a is non}\text{-}\text{zero} \right),\ \ a = 4

$\therefore$ $a = 4$ .