(a)     Using the Maclaurin series for \({(1 + x)^n}\), write down and simplify the Maclaurin series approximation for \({(1 - {x^2})^{ - \frac{1}{2}}}\) as far as the term in \({x^4}\)

(b)     Use your result to show that a series approximation for arccos x is

\[\arccos x \approx \frac{\pi }{2} - x - \frac{1}{6}{x^3} - \frac{3}{{40}}{x^5}.\]

(c)     Evaluate \(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} - \arccos ({x^2}) - {x^2}}}{{{x^6}}}\).

(d)     Use the series approximation for \(\arccos x\) to find an approximate value for

\[\int_0^{0.2} {\arccos \left( {\sqrt x } \right){\text{d}}x} ,\]

giving your answer to 5 decimal places. Does your answer give the actual value of the integral to 5 decimal places?

(a)     using or obtaining \({(1 + x)^n} = 1 + nx + \frac{{n(n - 1)}}{2}{x^2} + \ldots \)     (M1)

\({(1 - {n^2})^{ - \frac{1}{2}}} = 1 + ( - {x^2}) \times \left( { - \frac{1}{2}} \right) + \frac{{{{( - {x^2})}^2}}}{2} \times \left( { - \frac{1}{2}} \right) \times \left( { - \frac{3}{2}} \right) + \ldots \)     (A1)

\( = 1 + \frac{1}{2}{x^2} + \frac{3}{8}{x^4} + \ldots \)     A1

[3 marks]

 

(b)     integrating, and changing sign

\(\arccos x = - x - \frac{1}{6}{x^3} - \frac{3}{{40}}{x^5} + C + \ldots \)     M1A1

put x = 0,

\(\frac{\pi }{2} = C\)     M1

\(\left( {\arccos x \approx \frac{\pi }{2} - x - \frac{1}{6}{x^3} - \frac{3}{{40}}{x^5}} \right)\)     AG

[3 marks]

 

(c)     EITHER

using \(\arccos {x^2} \approx \frac{\pi }{2} - {x^2} - \frac{1}{6}{x^6} - \frac{3}{{40}}{x^{10}}\)     M1A1

\(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{\pi }{2} - \arccos {x^2} - {x^2}}}{{{x^6}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{x^6}}}{6} + {\text{higher powers}}}}{{{x^6}}}\)     M1A1

\( = \frac{1}{6}\)     A1

OR

using l’Hôpital’s Rule     M1

\({\text{limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^4}} }} \times 2x - 2x}}{{6{x^5}}}\)     M1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^4}} }} - 1}}{{3{x^4}}}\)     A1

\( = \mathop {\lim }\limits_{x \to 0} \frac{{ - \frac{1}{2} \times \frac{1}{{{{(1 - {x^4})}^{3/2}}}} \times - 4{x^3}}}{{12{x^3}}}\)     M1

\( = \frac{1}{6}\)     A1

[5 marks]

 

(d)     \(\int_0^{0.2} {\arccos \sqrt x {\text{d}}x \approx \int_0^{0.2} {\left( {\frac{\pi }{2} - {x^{\frac{1}{2}}} - \frac{1}{6}{x^{\frac{3}{2}}} - \frac{3}{{40}}{x^{\frac{5}{2}}}} \right){\text{d}}x} } \)     M1

\( = \left[ {\frac{\pi }{2}x - \frac{2}{3}{x^{\frac{3}{2}}} - \frac{1}{{15}}{x^{\frac{5}{2}}} - \frac{3}{{140}}{x^{\frac{7}{2}}}} \right]_0^{0.2}\)     (A1)

\( = \frac{\pi }{2} \times 0.2 - \frac{2}{3} \times {0.2^{\frac{3}{2}}} - \frac{1}{{15}} \times {0.2^{\frac{5}{2}}} - \frac{3}{{140}} \times {0.2^{\frac{7}{2}}}\)     (A1)

= 0.25326 (to 5 decimal places)     A1

Note: Accept integration of the series approximation using a GDC.

 

using a GDC, the actual value is 0.25325     A1

so the approximation is not correct to 5 decimal places     R1

[6 marks]

Total [17 marks]

Many candidates ignored the instruction in the question to use the series for \({(1 + x)^n}\) to deduce the series for \({(1 - {x^2})^{ - 1/2}}\) and attempted instead to obtain it by successive differentiation. It was decided at the standardisation meeting to award full credit for this method although in the event the algebra proved to be too difficult for many. Many candidates used l’Hopital’s Rule in (c) – this was much more difficult algebraically than using the series and it usually ended unsuccessfully. Candidates should realise that if a question on evaluating an indeterminate limit follows the determination of a Maclaurin series then it is likely that the series will be helpful in evaluating the limit. Part (d) caused problems for many candidates with algebraic errors being common. Many candidates failed to realise that the best way to find the exact value of the integral was to use the calculator.