When `f(x)=sin2x(1+cos2x)` `f(x)` has a maximum value for `x` is equal to (i)`-pi/2` (ii)`pi/2` (iii)`pi/3` (iv)`pi/6`

Prove that f(x)=sin x+3^(1/2) cos x has maximum value at x=(pi)/(6)

Let f(x)=sinx+2cos^2x , x in [pi/6,(2pi)/3] , then maximum value of f(x) is

Let f(x)=sin x+2cos^(2)x,x in[(pi)/(6),(2 pi)/(3)] then maximum value of f(x) is

If 0 <= x <= pi, then the solution of the equation 16^(sin^2) x + 16 ^(cos^2) x = 10 is given by x equal to (i) pi/6,pi/3 (ii) pi/3,pi/2 (iii) pi/6,pi/2 (iv) none of these

The minimum value of f(x)= "sin"x, [(-pi)/(2),(pi)/(2)] is

Let f(x) = |(2cos^2x, sin2x, -sinx), (sin2x, 2 sin^2x, cosx), (sinx, -cosx,0)|, the value of int_0^(pi//2){f(x) + f'(x)} dx, is (i)pi/2 (ii)pi (iii)(3pi)/2 (iv)2pi

The maximum value of f(x)=sin x+cos2x when in[0,2 pi] is

If f(x)=|cos x-sin x|, then f'((pi)/(2)) is equal to

Period of f(x)=sin, pi/2 x+2cos, pi/3 x-tan, pi/4 x is equal to

Let f(x)=sin^(-1)2x + cos^(-1)2x + sec^(-1)2x . Then the sum of the maximum and minimum values of f(x) is (a) pi (b) pi/2 (c) 2pi (d) (3pi)/2