if the maximum value of 3sinx+cosx is √x.Then value of x.......
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(x) = sqrt(3) cos x + sin x
f'(x) = -sqrt(3) sin x + cos x = 0
- sqrt(3) sin x = - cos x
sqrt(3) sin x = cos x
divide both sides by cos x
sqrt(3) tan x = 1
tan x = 1/sqrt(3)
x = pi/6, 7pi/6
f''(x) =-sqrt(3) cos x - sin x
f''(pi/6) = -sqrt(3) sqrt(3)/2 - 1/2
f''(pi/6) = -3/2-1/2 =-2 < 0, so f has a relative maximum at x=pi/6
The maximum value is f(pi/6) = sqrt(3)sqrt(3)/2 + 1/2 = 2
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