# f(x) = sin x+ √3cos x is maximum when x = A.π/3 B.π/4 C.π/6 D. 0 ​

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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).