(i) To find the term independent of x, consider the general term in the expansion:
\(T_r = \binom{6}{r} (2x)^{6-r} \left( \frac{k}{x} \right)^r\)
The term independent of x occurs when the powers of x cancel out, i.e.,
\((6-r) - r = 0\)
\(6 - 2r = 0\)
\(r = 3\)
Substitute \(r = 3\) into the term:
\(T_3 = \binom{6}{3} (2x)^3 \left( \frac{k}{x} \right)^3\)
\(= 20 \cdot 8x^3 \cdot \frac{k^3}{x^3}\)
\(= 20 \cdot 8 \cdot k^3\)
\(= 160k^3\)
Given that this term is 540, we have:
\(160k^3 = 540\)
\(k^3 = \frac{540}{160}\)
\(k^3 = \frac{27}{8}\)
\(k = \frac{1}{2}\)
(ii) To find the coefficient of x2, consider the term:
\(T_r = \binom{6}{r} (2x)^{6-r} \left( \frac{k}{x} \right)^r\)
We need \((6-r) - r = 2\), so:
\(6 - 2r = 2\)
\(4 = 2r\)
\(r = 2\)
Substitute \(r = 2\) into the term:
\(T_2 = \binom{6}{2} (2x)^4 \left( \frac{k}{x} \right)^2\)
\(= 15 \cdot 16x^4 \cdot \frac{k^2}{x^2}\)
\(= 15 \cdot 16 \cdot k^2 \cdot x^2\)
Substitute \(k = \frac{1}{2}\):
\(= 15 \cdot 16 \cdot \left( \frac{1}{2} \right)^2\)
\(= 15 \cdot 16 \cdot \frac{1}{4}\)
\(= 240\)
