(i) The function is given by \(f(x) = a + b \cos 2x\). We have two conditions:
\(f(0) = a + b \cos 0 = -1\) which simplifies to \(a + b = -1\).
\(f\left(\frac{1}{2}\pi\right) = a + b \cos \pi = 7\) which simplifies to \(a - b = 7\).
Solving these two equations:
\(a + b = -1\)
\(a - b = 7\)
Add the equations: \(2a = 6\) so \(a = 3\).
Substitute \(a = 3\) into \(a + b = -1\):
\(3 + b = -1\) so \(b = -4\).
Thus, \(a = 3\) and \(b = -4\).
(ii) To find the x-coordinates where the curve intersects the x-axis, set \(f(x) = 0\):
\(3 - 4 \cos 2x = 0\)
\(\cos 2x = \frac{3}{4}\)
Solving for \(x\), we find:
\(2x = \cos^{-1}\left(\frac{3}{4}\right)\)
\(x = \frac{1}{2} \cos^{-1}\left(\frac{3}{4}\right)\)
Calculating gives \(x \approx 0.36\) and \(x \approx 2.78\).
(iii) The graph of \(y = f(x)\) is a cosine wave starting at \(y = -1\) when \(x = 0\) and reaching a maximum of \(y = 7\) at \(x = \frac{1}{2}\pi\), then returning to \(y = -1\) at \(x = \pi\). The graph oscillates once between \(x = 0\) and \(x = \pi\).
