f(x) = sin x + \(\sqrt{3}\) cos x is maximum when x =A. \(\frac{\p

1.	f(x) = sin x + \(\sqrt{3}\) cos x is maximum when x =A. \(\frac{\pi}{3}\)B. \(\frac{\pi}{4}\)C. \(\frac{\pi}{6}\)D. 0
Answer» 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).

f(x) = sin x + \(\sqrt{3}\) cos x is maximum when x =A. \(\frac{\pi}{3}\)B. \(\frac{\pi}{4}\)C. \(\frac{\pi}{6}\)D. 0

Answer»

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).