Prove that f(x) = sin x + \(\sqrt3\) cos x has maximum value at x = \(

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

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

Answer»

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.