2009 H2 Mathematics Paper 1 Question 7

Maclaurin Series

Answers

(i)

f(0)=e{f(0)=\mathrm{e}}
f′(0)=0{f'(0)=0}
f′′(0)=−e{f''(0)=-\mathrm{e}}
f(x)=e−e2x2+…{f(x)=\mathrm{e} - \frac{\mathrm{e}}{2} x^2 + \ldots}

(ii)

a=1e,  b=12e{a=\frac{1}{\mathrm{e}}, \; b = \frac{1}{2\mathrm{e}}}

Full solutions

(i)

Let y=f(x)=ecos⁡x{y=f(x)=\mathrm{e}^{\cos x}}
dydx=−sin⁡x  ecos⁡x=−ysin⁡x\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= - \sin x \; \mathrm{e}^{\cos x} \\ &= -y \sin x \end{align*}
Differentiating implicitly w.r.t. x,{x,}
d2ydx2=−dydxsin⁡x−ycos⁡x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = - \frac{\mathrm{d}y}{\mathrm{d}x} \sin x - y \cos x
f(0)=ecos⁡0=e  ■f′(0)=−esin⁡0=0  ■f′′(0)=−0sin⁡0−ecos⁡0=−e  ■\begin{align*} f(0) &= \mathrm{e}^{\cos 0} \\ &= \mathrm{e} \; \blacksquare \\ f'(0) &= -\mathrm{e} \sin 0 \\ &= 0 \; \blacksquare \\ f''(0) &= -0\sin 0 - \mathrm{e} \cos 0 \\ &= -\mathrm{e} \; \blacksquare \\ \end{align*}
Hence the Maclaurin series
f(x)=e+0x+02!x2+…=e−e2x2+…  ■\begin{align*} f(x) &= \mathrm{e} + 0 x + \frac{0}{2!}x^2 + \ldots \\ &= \mathrm{e} - \frac{\mathrm{e}}{2} x^2 + \ldots \; \blacksquare \end{align*}

(ii)

1a+bx2=(a+bx2)−1=a−1(1+bx2a)−1=1a(1−bx2a+…)=1a−ba2x2+…\begin{align*} & \frac{1}{a+bx^2} \\ & = \left( a+bx^2 \right)^{-1} \\ & = a^{-1} \left( 1 + \frac{bx^2}{a} \right)^{-1} \\ &= \frac{1}{a} \left( 1 - \frac{bx^2}{a} + \ldots \right) \\ &= \frac{1}{a} - \frac{b}{a^2}x^2 + \ldots \end{align*}
Since the first two non-zero terms are equal,
1a=ea=1e  ■−ba2=−e2b=12e  ■\begin{align*} \frac{1}{a} &= \mathrm{e} \\ a &= \frac{1}{\mathrm{e}} \; \blacksquare \\ -\frac{b}{a^2} &= -\frac{\mathrm{e}}{2} \\ b &= \frac{1}{2\mathrm{e}} \; \blacksquare \end{align*}