21M.2.AHL.TZ2.9

$1-t+t^2$               A1

Note: Accept $1, -t$ and $t^2$.

 

[1 mark]

$\text{sec} x=\frac{1}{1-{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^4}{4!}}\left(-\dots \right)} \left(={\left(1-\frac{x^2}{2!}+\left(\frac{x^4}{4!}\left(-\dots \right)\right)\right)}^{-1}\right)$                (M1)

$t=\cos x-1$  or  $\text{sec} x=1-\left(\cos x-1\right)+{\left(\cos x-1\right)}^2$               (M1)

$=1-\left(-\frac{x^2}{2!}+\frac{x^4}{4!}\left(-\dots \right)\right)+{\left(-\frac{x^2}{2!}+\frac{x^4}{4!}\left(-\dots \right)\right)}^2$               A1

$=1+\frac{x^2}{2}-\frac{x^4}{24}+\frac{x^4}{4}$               A1

so the Maclaurin series for $\text{sec} x$ up to and including the term in $x^4$ is $1+\frac{x^2}{2}+\frac{5x^4}{24}$               AG

Note: Condone the absence of ‘…’ 

 

[4 marks]

$\text{arctan} 2x=2x-\frac{{\left(2x\right)}^3}{3}+\dots$

$\underset{x\to 0}{\lim }\left(\frac{x\text{\hspace{0.05em}arctan} 2x}{\text{sec} x-1}\right)=\underset{x\to 0}{\lim }\left(\frac{x\left(2x-{\displaystyle \frac{{\left(2x\right)}^3}{3}}+\dots \right)}{\left(1+{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{5x^4}{24}}\right)-1}\right)$                      M1

$=\underset{x\to 0}{\lim }\left(\frac{2x^2-{\displaystyle \frac{8x^4}{3}}+\dots }{{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{5x^4}{24}}}\right)$              A1

$=\underset{x\to 0}{\lim }\left(\frac{2x^2\left(1-{\displaystyle \frac{4x^2}{3}}\right)}{{\displaystyle \frac{x^2}{2}\left(1+\frac{5x^2}{12}\right)}}\right)$

$=4$              A1

 

Note: Condone missing ‘lim’ and errors in higher derivatives.
Do not award M1 unless $x$ is replaced by $2x$ in $\text{arctan}$.

 

[3 marks]

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