Show that

(i)     \(\frac{{{\text{d}}{f_n}(x)}}{{{\text{d}}x}} = n{g_n}(x)\);

(ii)     \(\frac{{{\text{d}}{g_n}(x)}}{{{\text{d}}x}} = (n + 1){f_{n + 2}}(x) - n{f_n}(x)\).

(i)     Use these results to show that the Maclaurin series for the function \({f_5}(x)\) up to and including the term in \({x^4}\) is \(1 + \frac{5}{2}{x^2} + \frac{{85}}{{24}}{x^4}\).

(ii)     By considering the general form of its higher derivatives explain briefly why all coefficients in the Maclaurin series for the function \({f_5}(x)\) are either positive or zero.

(iii)     Hence show that \({\sec ^5}(0.1) > 1.02535\).

(i)     \(\frac{{{\text{d}}{f_n}(x)}}{{{\text{d}}x}} = n{\sec ^{n - 1}}(x)\sec (x)\tan (x)\)     M1A1

\( = n{g_n}(x)\)    AG

(ii)     \(\frac{{{\text{d}}{g_n}(x)}}{{{\text{d}}x}} = \frac{{{\text{d}}{f_n}(x)}}{{{\text{d}}x}}\tan (x) + {f_n}(x){\sec ^2}(x)\)     M1

\(n{g_n}(x)\tan (x) + {f_{n + 2}}(x)\) or equivalent     A1

\(n{f_n}(x){\tan ^2}(x) + {f_{n + 2}}(x)\) or equivalent     A1

\( = (n + 1){f_{n + 2}}(x) - n{f_n}(x)\)    AG

Note:     Award M1A1 for the correct differentiation of a product and A1 for an intermediate result clearly leading to the AG.

[5 marks]

(i)     \({f_5}(0) = 1\)     A1

\(\frac{{{\text{d}}{f_5}}}{{{\text{d}}x}}(0) = 5{g_5}(0) = 0\)    A1

\(\frac{{{{\text{d}}^2}{f_5}}}{{{\text{d}}{x^2}}}(0) = 5\left( {6{f_7}(0) - 5{f_5}(0)} \right) = 5\)    A1

\(\frac{{{{\text{d}}^3}{f_5}}}{{{\text{d}}{x^3}}} = 30\frac{{{\text{d}}{f_7}}}{{{\text{d}}x}} - 25\frac{{{\text{d}}{f_5}}}{{{\text{d}}x}}\)    M1

hence \(\frac{{{{\text{d}}^3}{f_5}}}{{{\text{d}}{x^3}}}(0) = 30 \times 0 - 25 \times 0 = 0\)     A1

\(\frac{{{{\text{d}}^4}{f_5}}}{{{\text{d}}{x^4}}} = 30\frac{{{{\text{d}}^2}{f_7}}}{{{\text{d}}{x^2}}} - 25\frac{{{{\text{d}}^2}{f_5}}}{{{\text{d}}{x^2}}} = 210(8{f_9} - 7{f_7}) - 25\frac{{{{\text{d}}^2}{f_5}}}{{{\text{d}}{x^2}}}\)    M1A1

hence \(\frac{{{{\text{d}}^4}{f_5}}}{{{\text{d}}{x^4}}}(0) = 210 - 125 = 85\)     A1

hence \({f_5}(x) \approx 1 + \frac{5}{2}{x^2} + \frac{{85}}{{24}}{x^4}\)     AG

(ii)     each derivative of \({f_m}(x)\) is a sum of terms of the form \({\text{A}}\,{\sec ^p}(x)\,\,{\tan ^q}(x)\)     A1

where \(A \geqslant 0\)     A1

when \(x = 0\) is substituted the result is the sum of positive and/or zero terms     R1

(iii)     since the full series represents \({f_5}(x)\), the truncated series is a lower bound (or some equivalent statement)     R1

hence \({\sec ^5}(0.1) > 1 + \frac{5}{2}{0.1^2} + \frac{{85}}{{24}}{0.1^4}\)     M1

\( = 1.025354\)    A1

\( > 1.02535\)    AG

[14 marks]

Part (a) was generally answered, albeit either with an excess of algebraic manipulation or with too little – candidates need to realise that when an answer is given in the question, they need to convincingly reach that answer.

In part (b)(i), the results of part (a) were well used for up to the quadratic term. The obtaining of the cubic term, and more so the quartic term, was often not convincing. In part (ii), poor communication let down many candidates. In answering part (iii), many candidates failed to realise that in order to prove the stated inequality, they needed to actually write down the number 1.025354…, which is clearly greater than 1.02535.