2021 H2 Mathematics Paper 1 Question 7

Maclaurin Series

Answers

(b)

1+x+12x2+13x3+…{1 + x + \frac{1}{2} x^2 + \frac{1}{3} x^3 + \ldots}

Full solutions

(a)

y=esin⁡−1xln⁡y=sin⁡−1x\begin{align*} y &= \mathrm{e}^{\sin^{-1} x} \\ \ln y &= \sin^{-1}x \end{align*}
Differentiating w.r.t. x,{x,}
1ydydx=11−x21−x2dydx=y(1−x2)(dydx)2=y2\begin{gather*} \frac{1}{y} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1-x^2}} \\ \sqrt{1-x^2} \frac{\mathrm{d}y}{\mathrm{d}x} = y \\ (1-x^2) \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 = y^2 \end{gather*}
Differentiating w.r.t. x,{x,}
2(1−x2)dydxd2ydx2−2x(dydx)2=2ydydx 2(1-x^2)\frac{\mathrm{d}y}{\mathrm{d}x} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2x \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 = 2y \frac{\mathrm{d}y}{\mathrm{d}x}
(1−x2)d2ydx2=xdydx+y  ■ (1-x^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = x \frac{\mathrm{d}y}{\mathrm{d}x} + y \; \blacksquare

(b)

Differentiating w.r.t. x,{x,}
(1−x2)d3ydx3−2xd2ydx2=dydx+xd2ydx2+dydx (1-x^2) \frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 2x \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}y}{\mathrm{d}x} + x \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \frac{\mathrm{d}y}{\mathrm{d}x}
When x=0,{x=0,}
y=esin⁡−10=1\begin{align*} y &= \mathrm{e}^{\sin^{-1} 0} \\ &= 1 \\ \end{align*}
11dydx=11−02dydx=1\begin{align*} \frac{1}{1} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{\sqrt{1-0^2}} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ \end{align*}
(1−02)d2ydx2=0(1)+1d2ydx2=1\begin{align*} (1-0^2) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= 0 (1) + 1 \\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= 1 \\ \end{align*}
(1−03)d3ydx3−0=1+0+1d3ydx3=2\begin{align*} (1-0^3) \frac{\mathrm{d}^3y}{\mathrm{d}x^3} - 0 &= 1 + 0 + 1 \\ \frac{\mathrm{d}^3y}{\mathrm{d}x^3} &= 2 \end{align*}
Maclaurin expansion for esin⁡−1x:{\mathrm{e}^{\sin^{-1}x}:}
esin⁡−1x=1+x−12!x2+13!x3+…=1+x+12x2+13x3+…  ■\begin{align*} \mathrm{e}^{\sin^{-1}x} &= 1 + x - \frac{1}{2!} x^2 + \frac{1}{3!} x^3 + \ldots \\ &= 1 + x + \frac{1}{2} x^2 + \frac{1}{3} x^3 + \ldots \; \blacksquare \end{align*}