(a) To find the coefficient of \(x^2\) in the expansion of \((4 + 2x)(2 - ax)^5\), consider the terms that contribute to \(x^2\):
- From \((4)(2 - ax)^5\), the term contributing to \(x^2\) is \(-80ax\).
- From \((2x)(2 - ax)^5\), the term contributing to \(x^2\) is \(80a^2x^2\).
The coefficient of \(x^2\) is given by:
\(320a^2 - 160a = -15\)
Rearranging gives:
\(320a^2 - 160a + 15 = 0\)
Solving this quadratic equation using the quadratic formula:
\(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
where \(a = 320\), \(b = -160\), \(c = 15\).
\(a = \frac{160 \pm \sqrt{(-160)^2 - 4 \times 320 \times 15}}{2 \times 320}\)
\(a = \frac{160 \pm \sqrt{25600 - 19200}}{640}\)
\(a = \frac{160 \pm \sqrt{6400}}{640}\)
\(a = \frac{160 \pm 80}{640}\)
\(a = \frac{240}{640} \text{ or } a = \frac{80}{640}\)
\(a = \frac{3}{8} \text{ or } a = \frac{1}{8}\)
(b) Given that there is only one value of \(a\) for which the coefficient of \(x^2\) is \(k\), we set the discriminant of the quadratic equation to zero:
\(160^2 - 4 \times 320 \times (15 - k) = 0\)
Solving for \(k\):
\(25600 - 1280(15 - k) = 0\)
\(25600 - 19200 + 1280k = 0\)
\(6400 + 1280k = 0\)
\(k = -\frac{6400}{1280}\)
\(k = -20\)
Substituting \(k = -20\) back into the quadratic equation gives:
\(320a^2 - 160a + 20 = 0\)
Solving this gives \(a = \frac{1}{4}\).
