Prove that f(x) = sin x + √3cos x has maximum value at x = π/6 . ​

Question

Prove that f(x) = sin x + \(\sqrt3\) cos x has maximum value at x = \(\frac{\pi}{6}\).

Answer 1

f(x) = sin x + √3 cos x 

f’(x) = cos x – √3 sin x 

Now, 

f’(x) = 0 

cos x – √3 sin x = 0 

cos x = √3 sin x 

cot x = √3 

x = \(\frac{\pi}{6}\) 

Differentiate f’’(x), we get 

f’’(x) = – sin x –√3 cos x

f’’(\(\frac{\pi}{6}\)) = - sin (\(\frac{\pi}{6}\)) - √3 cos(\(\frac{\pi}{6}\)) < 0

Hence, at x = \(\frac{\pi}{6}\) is the point of local maxima.