| Date | May 2012 | Marks available | 3 | Reference code | 12M.2.sl.TZ1.6 |
| Level | SL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
Consider the expansion of \({\left( {2{x^3} + \frac{b}{x}} \right)^8} = 256{x^{24}} + 3072{x^{20}} + \ldots + k{x^0} + \ldots \) .
Find b.
Find k.
Markscheme
valid attempt to find term in \({x^{20}}\) (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
8\\
1
\end{array}} \right)({2^7})(b)\) , \({(2{x^3})^7}\left( {\frac{b}{x}} \right) = 3072\)
correct equation A1
e.g. \(\left( {\begin{array}{*{20}{c}}
8\\
1
\end{array}} \right)({2^7})(b) = 3072\)
\(b = 3\) A1 N2
[3 marks]
evidence of choosing correct term (M1)
e.g. 7th term, \(r = 6\)
correct expression A1
e.g. \(\left( {\begin{array}{*{20}{c}}
8\\
6
\end{array}} \right){(2{x^3})^2}{\left( {\frac{3}{x}} \right)^6}\)
\(k = 81648\) (accept \(81600\) ) A1 N2
[3 marks]
Examiners report
An unfamiliar presentation confused a number of candidates who attempted to set up an equation with the wrong term in part (a). Time and again, candidates omitted the binomial coefficient in their set up leading to an incorrect result. In part (b) it was common to see the constant term treated as the last term of the expansion rather than the 7th term.
An unfamiliar presentation confused a number of candidates who attempted to set up an equation with the wrong term in part (a). Time and again, candidates omitted the binomial coefficient in their set up leading to an incorrect result. In part (b) it was common to see the constant term treated as the last term of the expansion rather than the 7th term.
