The constant term in the expansion of \({\left( {\frac{x}{a} + \frac{{{a^2}}}{x}} \right)^6}\) , where \(a \in \mathbb{R}\) is \(1280\). Find \(a\) .

evidence of binomial expansion     (M1)

eg selecting correct term,\({\left( {\frac{x}{a}} \right)^6}{\left( {\frac{{{a^2}}}{x}} \right)^0} + \left( \begin{array}{l}
6\\
1
\end{array} \right){\left( {\frac{x}{a}} \right)^5}{\left( {\frac{{{a^2}}}{x}} \right)^1} +  \ldots \)

evidence of identifying constant term in expansion for power \(6\)     (A1)

eg   \(r = 3\) , 4^th term

evidence of correct term (may be seen in equation)     A2

eg   \(20\frac{{{a^6}}}{{{a^3}}}\) , \(\left( \begin{array}{l}
6\\
3
\end{array} \right){\left( {\frac{x}{a}} \right)^3}{\left( {\frac{{{a^2}}}{x}} \right)^3}\)

attempt to set up their equation     (M1)

eg   \(\left( \begin{array}{l}
6\\
3
\end{array} \right){\left( {\frac{x}{a}} \right)^3}{\left( {\frac{{{a^2}}}{x}} \right)^3} = 1280\), \({a^3} = 1280\)

correct equation in one variable \(a\)     (A1)

eg   \(20{a^3} = 1280\) , \({a^3} = 64\)

\(a = 4\)     A1     N4

[7 marks]

Many candidates struggled with this question. Some had difficulty with the binomial expansion, while others did not understand that the constant term had no \(x\) , while still others were unable to simplify a ratio of exponentials with a common base. Some candidates found \(r = 3\) using algebraic methods while others found it by writing out the first several terms. In some cases, candidates just set the entire expansion equal to 1280.