| Date | November 2008 | Marks available | 12 | Reference code | 08N.3ca.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
| Command term | Determine, Find, Show that, State, and Hence | Question number | 3 | Adapted from | N/A |
Question
The function \(f(x) = \frac{{1 + ax}}{{1 + bx}}\) can be expanded as a power series in x, within its radius of convergence R, in the form \(f(x) \equiv 1 + \sum\limits_{n = 1}^\infty {{c_n}{x^n}} \) .
(a) (i) Show that \({c_n} = {( - b)^{n - 1}}(a - b)\).
(ii) State the value of R.
(b) Determine the values of a and b for which the expansion of f(x) agrees with that of \({{\text{e}}^x}\) up to and including the term in \({x^2}\) .
(c) Hence find a rational approximation to \({{\text{e}}^{\frac{1}{3}}}\) .
Markscheme
(a) (i) \(f(x) = (1 + ax){(1 + bx)^{ - 1}}\)
\( = (1 + ax)(1 - bx + ...{( - 1)^n}{b^n}{x^n} + …\) M1A1
it follows that
\({c_n} = {( - 1)^n}{b^n} + {( - 1)^{n - 1}}a{b^{n - 1}}\) M1A1
\( = {( - b)^{n - 1}}(a - b)\) AG
(ii) \({\text{R}} = \frac{1}{{\left| b \right|}}\) A1
[5 marks]
(b) to agree up to quadratic terms requires
\(1 = - b + a,{\text{ }}\frac{1}{2} = {b^2} - ab\) M1A1A1
from which \(a = - b = \frac{1}{2}\) A1
[4 marks]
(c) \({{\text{e}}^x} \approx \frac{{1 + 0.5x}}{{1 - 0.5x}}\) A1
putting \(x = \frac{1}{3}\) M1
\({{\text{e}}^{\frac{1}{3}}} \approx \frac{{\left( {1 + \frac{1}{6}} \right)}}{{\left( {1 - \frac{1}{6}} \right)}} = \frac{7}{5}\) A1
[3 marks]
Total [12 marks]
Examiners report
Most candidates failed to realize that the first step was to write f(x) as \((1 + ax){(1 + bx)^{ - 1}}\) . Given the displayed answer to part(a), many candidates successfully tackled part(b). Few understood the meaning of the ‘hence’ in part(c).
