2024 H2 Mathematics Paper 1 Question 8

Maclaurin Series

Answers

$k = 2.$

$1 - 2 x^{2} - 4 x^{3} + \dots$

$a = 1,$ $b = 4.$

(a)

\begin{aligned} y & = \cos (1 - e^{2 x}) \\ \frac{d y}{d x} & = 2 e^{2 x} \sin (1 - e^{2 x}) \\ \frac{d^{2} y}{d x^{2}} & = 4 e^{2 x} \sin (1 - e^{2 x}) - 4 e^{4 x} \cos (1 - e^{2 x}) \\ = 2 \frac{d y}{d x} - 4 y e^{4 x} \\ = 2 (\frac{d y}{d x} - 2 y e^{4 x}) ∎ \end{aligned}

(b)

\begin{aligned} \frac{d^{2} y}{d x^{2}} & = 2 (\frac{d y}{d x} - 2 y e^{4 x}) \\ \frac{d^{3} y}{d x^{3}} & = 2 (\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} e^{4 x} - 8 e^{4 x} y) \end{aligned}

When $x = 0,$

\begin{aligned} y & = \cos (1 - e^{2 (0)}) \\ = 1 \\ \frac{d y}{d x} & = 2 e^{2 (0)} \sin (1 - e^{2 (0)}) \\ = 0 \\ \frac{d^{2} y}{d x^{2}} & = 2 (0 - 2 (1) e^{4 (0)}) \\ = - 4 \\ \frac{d^{3} y}{d x^{3}} & = 2 (- 4 - 2 (0) e^{4 (0)} - 8 e^{4 (0)} (1)) \\ = - 24 \end{aligned}

\begin{aligned} Maclaurin expansion of \cos (1 - e^{2 x}) \\ = f (0) + x f^{'} (0) + \frac{x^{2}}{2!} f^{''} (0) + \frac{x^{3}}{3!} f^{'''} (0) + \dots \\ = 1 + x (0) + \frac{x^{2}}{2} (- 4) + \frac{x^{3}}{6} (- 24) + \dots \\ = 1 - 2 x^{2} - 4 x^{3} + \dots \end{aligned}

(c)

\begin{aligned} \frac{1}{\sqrt{a + b x^{2}}} \\ = {(a + b x^{2})}^{- \frac{1}{2}} \\ = a^{- \frac{1}{2}} {(1 + \frac{b x^{2}}{a})}^{- \frac{1}{2}} \\ = a^{- \frac{1}{2}} (1 - \frac{1}{2} (\frac{b x^{2}}{a}) + \dots) \\ = a^{- \frac{1}{2}} (1 - \frac{b x^{2}}{2 a} + \dots) \\ = a^{- \frac{1}{2}} - \frac{b x^{2}}{2 a^{\frac{3}{2}}} + \dots \end{aligned}

Comparing constant terms,

\begin{aligned} a^{- \frac{1}{2}} & = 1 \\ a & = 1 ∎ \end{aligned}

Comparing coefficient of $x^{2},$

\begin{aligned} - \frac{b}{2 a^{\frac{3}{2}}} & = - 2 \\ - \frac{b}{2} & = - 2 \\ b & = 4 ∎ \end{aligned}