2008 H2 Mathematics Paper 1 Question 6

Maclaurin Series

Answers

(a)

2+34θ2{2 + \frac{3}{4} \theta^2}

(b)

f(0)=1{f(0)=1}
f′(0)=4{f'(0)=4}
f′′(0)=16{f''(0)=16}
f(x)≈1+4x+8x2{f(x) \approx 1 + 4x + 8x^2}

Full solutions

(a)

AC2=12+32−2(1)(3)cos⁡θ=10−6cos⁡θ≈10−6(1−θ22)AC≈(4+3θ2)12  ■\begin{align*} AC^2 &= 1^2 + 3^2 - 2(1)(3) \cos \theta \\ &= 10 - 6 \cos \theta \\ &\approx 10 - 6 \left( 1 - \frac{\theta^2}{2} \right) \\ AC &\approx \left( 4 + 3 \theta^2 \right)^{\frac{1}{2}}\; \blacksquare \\ \end{align*}
AC≈(4+3θ2)12=412(1+34θ2)12≈2(1+38θ2)AC≈2+34θ2  ■\begin{align*} AC &\approx \left( 4 + 3 \theta^2 \right)^{\frac{1}{2}} \\ &= 4^{\frac{1}{2}} \left( 1 + \frac{3}{4} \theta^2 \right)^{\frac{1}{2}} \\ &\approx 2 \left( 1 + \frac{3}{8} \theta^2 \right) \\ AC &\approx 2 + \frac{3}{4} \theta^2 \; \blacksquare \end{align*}

(b)

f′(x)=2sec⁡2(2x+14π)f′′(x)=2(2)(2)sec⁡(2x+14π)sec⁡(2x+14π)tan⁡(2x+14π)=8sec⁡2(2x+14π)tan⁡(2x+14π)\begin{align*} f'(x) &= 2 \sec^2 ( 2x + {\textstyle \frac{1}{4}} \pi ) \\ f''(x) &= 2(2)(2) \sec ( 2x + {\textstyle \frac{1}{4}} \pi ) \sec ( 2x + {\textstyle \frac{1}{4}} \pi ) \tan ( 2x + {\textstyle \frac{1}{4}} \pi ) \\ &= 8 \sec^2 ( 2x + {\textstyle \frac{1}{4}} \pi ) \tan ( 2x + {\textstyle \frac{1}{4}} \pi )\\ \end{align*}
f(0)=tan⁡14π=1  ■f′(0)=2sec⁡214π=2(1÷12)=4  ■f′′(0)=8sec⁡214πtan⁡14π=8(1÷12)(1)=16  ■\begin{align*} f(0) &= \tan {\textstyle \frac{1}{4}} \pi \\ &= 1 \; \blacksquare \\ f'(0) &= 2 \sec^2 {\textstyle \frac{1}{4}} \pi \\ &= 2 \left( 1 \div \frac{1}{2} \right) \\ &= 4 \; \blacksquare \\ f''(0) &= 8 \sec^2 {\textstyle \frac{1}{4}} \pi \tan {\textstyle \frac{1}{4}} \pi \\ &= 8\left( 1 \div \frac{1}{2} \right) (1) \\ &= 16 \; \blacksquare \\ \end{align*}
Maclaurin series for f(x):{f(x):}
f(x)=1+(4)x+162x2+…=1+4x+8x2+…  ■\begin{align*} f(x) &= 1 + (4)x + \frac{16}{2}x^2 + \ldots \\ &= 1 + 4x + 8x^2 + \ldots \; \blacksquare \end{align*}