Option : (C)
f(x) = sin x + \(\sqrt{3}\)cos x
Differentiating f(x) with respect to x, we get
f'(x) = cos x - \(\sqrt{3}\)sin x
Differentiating f’(x) with respect to x, we get
f''(x) = - sin x - \(\sqrt{3}\)cos x
For maxima at x = c,
f’(c) = 0 and f’’(c) < 0
f’(x) = 0
⇒ tan x = \(\frac{1}{\sqrt 3}\)
or, x = \(\frac{\pi}{6}\) or \(\frac{7\pi}{6}\)
f"(\(\frac{\pi}{6}\)) = - 2 < 0 and
f" (\(\frac{7\pi}{6}\)) = 2 > 0
Hence,
x = \(\frac{\pi}{6}\) is a point of maxima for f(x).