Sarthaks Test
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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

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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).

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