If f(x) is continuous function in `[0,2pi]` and f(0)=f(2`pi`), then prove that there exists a point `c in (0,pi)` such that `f(x)=f(x+pi).`

Does there exist a point c(in)(0,pi/2) such that f' (c)= 0 where f(x) = sin 2x ?

Let f(0)=0,f((pi)/(2))=1,f((3pi)/(2))=-1 be a continuos and twice differentiable function. Statement I |f''(x)|le1 for atleast one x in (0,(3pi)/(2)) because Statement II According to Rolle's theorem, if y=g(x) is continuos and differentiable, AAx in [a,b]andg(a)=g(b), then there exists atleast one such that g'(c)=0.

Can we find a point c^(in) (-pi/2,pi/2) such that f'© = 0 where f(x) = |sinx|

Let f(x) = [sinx] , is f continuous at x = pi/2

Prove that f(x)="cos"x is strictly decreasing (0,pi)

f(x) = cos x is strictly decreasing in (0, pi)

f(x) = cos x is strictly increasing in (pi,2pi)

f(x) = sin x is a strictly increasing in (0, pi/2)

A function f is defined by f(x)=int_(0)^(pi)cos t cos (x-t)dt, 0 lexle2pi then the minimum value of f(x) is

f(x) = [(1-sin^3x)/(3cos^2x) ,if x pi/2]] continuous function at x = (pi/2) find a and b.