2011 H2 Mathematics Paper 1 Question 4

Maclaurin Series

Answers

(i)

1−3x2+4x4{1 - 3 x^2 + 4 x^4}
(iia)
a−a3+45a5{a - a^3 + \frac{4}{5}a^5}
∫0π4g(x)  dx≈0.540{\int_0^{\frac{\pi}{4}} g(x) \; \mathrm{d}x \approx 0.540}
(iib)
0.475{0.475}
The approximation is not very good as 14π{\frac{1}{4}\pi} is not sufficiently small enough for the terms x5{x^5} and above that was omitted in the Maclaurin series in (i) to be negligible

Full solutions

(i)

cos⁡6x=(1−12x2+124x4+…)6=1+6(−12x2+124x4)+6(5)2(−12x2+124x4)2+…=1−3x2+14x4+15(14x4+…)+…=1−3x2+4x4+…  ■\begin{align*} & \cos^6 x \\ &= \left( 1 - \frac{1}{2} x^2 + \frac{1}{24} x^4 + \ldots \right)^6 \\ &= 1 + 6 \left( - \frac{1}{2} x^2 + \frac{1}{24} x^4 \right) + \frac{6(5)}{2} \left( - \frac{1}{2} x^2 + \frac{1}{24} x^4 \right)^2 + \ldots \\ &= 1 - 3x^2 + \frac{1}{4}x^4 + 15 \left( \frac{1}{4} x^4 + \ldots \right) + \ldots \\ &= 1 - 3 x^2 + 4 x^4 + \ldots \; \blacksquare \end{align*}
(iia)
∫0ag(x)  dx≈∫0a(1−3x2+4x4)  dx=[x−x3+45x5]0a=a−a3+45a5  ■\begin{align*} & \int_0^a g(x) \; \mathrm{d} x \\ &\approx \int_0^a \left( 1 - 3 x^2 + 4 x^4 \right) \; \mathrm{d} x \\ &= \left[ x - x^3 + \frac{4}{5}x^5 \right]_0^a \\ &= a - a^3 + \frac{4}{5}a^5 \; \blacksquare \end{align*}
When a=14π,{a=\frac{1}{4}\pi,}
∫0π4g(x)  dx≈π4−(π4)3+45(π4)5≈0.540 (3 sf)  ■\begin{align*} & \int_0^\frac{\pi}{4} g(x) \; \mathrm{d} x \\ &\approx \frac{\pi}{4} - \left( \frac{\pi}{4} \right)^3 + \frac{4}{5} \left( \frac{\pi}{4} \right)^5 \\ &\approx 0.540 \textrm{ (3 sf)} \; \blacksquare \end{align*}
(iib)
Using a GC,
∫0π4g(x)  dx≈0.475 (3 sf)  ■\int_0^{\frac{\pi}{4}} g(x) \; \mathrm{d}x \approx 0.475 \textrm{ (3 sf)} \; \blacksquare