Show that $f'\left( 0 \right) = 0$.

By differentiating the above equation twice, show that

$$\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0$$

where ${f^{\left( 3 \right)}}\left( x \right)$ and ${f^{\left( 4 \right)}}\left( x \right)$ denote the 3rd and 4th derivative of $f\left( x \right)$ respectively.

Hence show that the Maclaurin series for $f\left( x \right)$ up to and including the term in ${x^4}$ is ${x^2} + \frac{1}{3}{x^4}$.

Use this series approximation for $f\left( x \right)$ with $x = \frac{1}{2}$ to find an approximate value for ${\pi ^2}$.

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

Note: Award M1 for an attempt at chain rule differentiation.
Award M0A0 for $f'\left( x \right) = 2\,{\text{arcsin}}\,\left( x \right)$.

$f'\left( 0 \right) = 0$     AG

[2 marks]

differentiating gives $\left( {1 - {x^2}} \right){f^{\left( 3 \right)}}\left( x \right) - 2xf''\left( x \right) - f'\left( x \right) - xf''\left( x \right)\left( { = 0} \right)$      M1A1

differentiating again gives $\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 2x{f^{\left( 3 \right)}}\left( x \right) - 3f''\left( x \right) - 3x{f^{\left( 3 \right)}}\left( x \right) - f''\left( x \right)\left( { = 0} \right)$     M1A1

Note: Award M1 for an attempt at product rule differentiation of at least one product in each of the above two lines.
Do not penalise candidates who use poor notation.

$\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0$      AG

[4 marks]

attempting to find one of $f''\left( 0 \right)$, ${f^{\left( 3 \right)}}\left( 0 \right)$ or ${f^{\left( 4 \right)}}\left( 0 \right)$ by substituting $x = 0$ into relevant differential equation(s)       (M1)

Note: Condone $f''\left( 0 \right)$ found by calculating $\frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{2\,{\text{arcsin}}\,\left( x \right)}}{{\sqrt {1 - {x^2}} }}} \right)$ at $x = 0$.

$\left( {f\left( 0 \right) = 0,\,f'\left( 0 \right) = 0} \right)$

$f''\left( 0 \right) = 2$ and ${f^{\left( 4 \right)}}\left( 0 \right) - 4f''\left( 0 \right) = 0 \Rightarrow {f^{\left( 4 \right)}}\left( 0 \right) = 8$      A1

${f^{\left( 3 \right)}}\left( 0 \right) = 0$ and so $\frac{2}{{2{\text{!}}}}{x^2} + \frac{8}{{4{\text{!}}}}{x^4}$     A1

Note: Only award the above A1, for correct first differentiation in part (b) leading to ${f^{\left( 3 \right)}}\left( 0 \right) = 0$ stated or ${f^{\left( 3 \right)}}\left( 0 \right) = 0$ seen from use of the general Maclaurin series.
Special Case: Award (M1)A0A1 if ${f^{\left( 4 \right)}}\left( 0 \right) = 8$ is stated without justification or found by working backwards from the general Maclaurin series.

so the Maclaurin series for $f\left( x \right)$ up to and including the term in ${x^4}$ is ${x^2} + \frac{1}{3}{x^4}$     AG

[3 marks]

substituting $x = \frac{1}{2}$ into ${x^2} + \frac{1}{3}{x^4}$      M1

the series approximation gives a value of $\frac{{13}}{{48}}$

so ${\pi ^2} \simeq \frac{{13}}{{48}} \times 36$

$\simeq 9.75\,\,\left( { \simeq \frac{{39}}{4}} \right)$     A1

Note: Accept 9.76.

[2 marks]