19N.3.AHL.TZ0.HCA_3

pestleMathematicsAAHLPaper 319N· ahl-5-19-maclaurin-series, ahl-5-13-limits-and-lhopitalssource ↗

The function $f$ is defined by $f\left( x \right) = {\text{arcsin}}\left( {2x} \right)$ , where $- \frac{1}{2} \leqslant x \leqslant \frac{1}{2}$ .

By finding a suitable number of derivatives of $f$ , find the first two non-zero terms in the Maclaurin series for $f$ .

[8]

Hence or otherwise, find $\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{\text{arcsin}}\left( {2x} \right) - 2x}}{{{{\left( {2x} \right)}^3}}}$ .

[3]

Markscheme / solution

$f\left( x \right) = {\text{arcsin}}\left( {2x} \right)$

$f'\left( x \right) = \frac{2}{{\sqrt {1 - 4{x^2}} }}$ M1A1

Note: Award M1A0 for $f'\left( x \right) = \frac{1}{{\sqrt {1 - 4{x^2}} }}$

$f''\left( x \right) = \frac{{8x}}{{{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}}}}$ A1

EITHER

$f'''\left( x \right) = \frac{{8{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}} - 8x\left( {\frac{3}{2}\left( { - 8x} \right){{\left( {1 - 4{x^2}} \right)}^{\frac{1}{2}}}} \right)}}{{{{\left( {1 - 4{x^2}} \right)}^3}}}\,\,\,\left( { = \frac{{8{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}} + 96{x^2}{{\left( {1 - 4{x^2}} \right)}^{\frac{1}{2}}}}}{{{{\left( {1 - 4{x^2}} \right)}^3}}}} \right)$ A1

$f'''\left( x \right) = 8{\left( {1 - 4{x^2}} \right)^{ - \frac{3}{2}}} + 8x\left( { - \frac{3}{2}{{\left( {1 - 4{x^2}} \right)}^{ - \frac{5}{2}}}} \right)\left( { - 8x} \right)\,\,\,\left( { = 8{{\left( {1 - 4{x^2}} \right)}^{ - \frac{3}{2}}} + 96{x^2}{{\left( {1 - 4{x^2}} \right)}^{ - \frac{5}{2}}}} \right)$ A1

THEN

substitute $x = 0$ into $f$ or any of its derivatives (M1)

$f\left( 0 \right) = 0$ , $f'\left( 0 \right) = 2$ and $f''\left( 0 \right) = 0$ A1

$f'''\left( 0 \right) = 8$

the Maclaurin series is

$f\left( x \right) = 2x + \frac{{8{x^3}}}{6} + \ldots \,\left( { = 2x + \frac{{4{x^3}}}{3} + \ldots } \right)$ (M1)A1

[8 marks]

METHOD 1

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{\text{arcsin}}\left( {2x} \right) - 2x}}{{{{\left( {2x} \right)}^3}}} = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{2x + \frac{{4{x^3}}}{3} + \ldots - 2x}}{{8{x^3}}}$ M1

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\frac{4}{3} + \ldots {\text{ terms with }}x}}{8}$ (M1)

$= \frac{1}{6}$ A1

Note: Condone the omission of +… in their working.

METHOD 2

$\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{\text{arcsin}}\left( {2x} \right) - 2x}}{{{{\left( {2x} \right)}^3}}} = \frac{0}{0}$ indeterminate form, using L’Hôpital’s rule

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\frac{2}{{\sqrt {1 - 4{x^2}} }} - 2}}{{24{x^2}}}$ M1

$= \frac{0}{0}$ indeterminate form, using L’Hôpital’s rule again

$= \mathop {{\text{lim}}}\limits_{x \to 0} \frac{{\frac{{8x}}{{{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}}}}}}{{48x}}\left( { = \mathop {{\text{lim}}}\limits_{x \to 0} \frac{1}{{6{{\left( {1 - 4{x^2}} \right)}^{\frac{3}{2}}}}}} \right)$ M1

Note: Award M1 only if their previous expression is in indeterminate form.

$= \frac{1}{6}$ A1

Note: Award FT for use of their derivatives from part (a).

[3 marks]

Examiners’ report

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