2017 H2 Mathematics Paper 1 Question 1

Maclaurin Series

Answers

ax+(−a22+2a)x2+(a33−a2+2a)x3+…ax + \left( -\frac{a^2}{2} + 2a \right) x^2 + \left( \frac{a^3}{3} - a^2 + 2a \right)x^3 + \ldots
a=4{a=4}

Full solutions

e2xln⁡(1+ax)=(1+2x+4x22+8x33!+…)(ax−a2x22+a3x33+…)=(1+2x+2x2+…)(ax−a2x22+a3x33+…)=ax+(−a22+2a)x2+(a33−a2+2a)x3+…  ■\begin{align*} & \mathrm{e}^{2x} \ln (1+ax) \\ &= \left(1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{3!} + \ldots\right) \left( ax - \frac{a^2x^2}{2} + \frac{a^3x^3}{3} +\ldots \right) \\ &= \left(1 + 2 x + 2 x^2 + \ldots\right) \left( ax - \frac{a^2x^2}{2} + \frac{a^3x^3}{3} +\ldots \right) \\ &= ax + \left( -\frac{a^2}{2} + 2a \right) x^2 + \left( \frac{a^3}{3} - a^2 + 2a \right)x^3 + \ldots \; \blacksquare \end{align*}
If there is no term in x2,{x^2,}
−a22+2a=0-\frac{a^2}{2} + 2a = 0
Since a{a} is non-zero,
−a2+2=0−a+4=0a=4  ■\begin{align*} -\frac{a}{2} + 2 &= 0 \\ -a + 4 &= 0 \\ a &= 4 \; \blacksquare \end{align*}