2020 H2 Mathematics Paper 1 Question 3

Maclaurin Series

Answers

(i)

f′′(x)=−91+sin⁡3x{f''(x)=\frac{-9}{1+\sin 3x}}

(ii)

3x−92x2+92x3+…{3 x - \frac{9}{2} x^2 + \frac{9}{2} x^3 + \ldots}

Full solutions

(i)

f(x)=ln⁡(1+sin⁡3x)f′(x)=3cos⁡3x1+sin⁡3xf′′(x)=(1+sin⁡3x)(−9sin⁡3x)−(3cos⁡3x)(3cos⁡3x)(1+sin⁡3x)2=−9sin⁡3x−9sin⁡23x−9cos⁡23x(1+sin⁡3x)2=−9sin⁡3x−9(1+sin⁡3x)2=−9(1+sin⁡3x)(1+sin⁡3x)2=−91+sin⁡3x  ■\begin{align*} f(x) &= \ln (1+\sin 3x) \\ f'(x) &= \frac{3 \cos 3x}{1+\sin 3x} \\ f''(x) &= \frac{(1+\sin 3x)(-9 \sin 3x) - (3 \cos 3x)(3 \cos 3x)}{(1+\sin 3x)^2} \\ &= \frac{-9\sin 3x - 9 \sin^2 3x - 9 \cos^2 3x}{(1+\sin 3x)^2} \\ &= \frac{-9\sin 3x - 9 }{(1+\sin 3x)^2} \\ &= \frac{-9(1+\sin 3x) }{(1+\sin 3x)^2} \\ &= \frac{-9}{1+\sin 3x} \; \blacksquare \end{align*}

(ii)

f′′′(x)=9(−1)(3cos⁡3x)(1+sin⁡3x)2=27cos⁡3x(1+sin⁡3x)2\begin{align*} f'''(x) &= \frac{9(-1)(3 \cos 3x)}{(1+\sin 3x)^2} \\ &= \frac{27 \cos 3x}{(1+\sin 3x)^2} \\ \end{align*}
f(0)=ln⁡(1+sin⁡3(0))=0f′(0)=3cos⁡3(0)1+sin⁡3(0)=3f′′(0)=−91+sin⁡3(0)=−9f′′′(0)=27cos⁡3(0)1+sin⁡3(0)=27\begin{align*} f(0) &= \ln(1 + \sin 3(0)) \\ &= 0 \\ f'(0) &= \frac{3 \cos 3(0)}{1+ \sin 3(0)} \\ &= 3 \\ f''(0) &= \frac{-9}{1+ \sin 3(0)} \\ &= -9 \\ f'''(0) &= \frac{27 \cos 3(0)}{1+ \sin 3(0)} \\ &= 27 \end{align*}
Maclaurin expansion for f(x):{f(x):}
f(x)=0+3x−92!x2+273!x3+…=3x−92x2+92x3+…  ■\begin{align*} f(x) &= 0 + 3x - \frac{9}{2!} x^2 + \frac{27}{3!} x^3 + \ldots \\ &= 3 x - \frac{9}{2} x^2 + \frac{9}{2} x^3 + \ldots \; \blacksquare \end{align*}