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).`

If f(x) is a continuous function in [0,pi] such that f(0)=f(x)=0, then the value of int_(0)^(pi//2) {f(2x)-f''(2x)}sin x cos x dx is equal to

A function f is defined by f(x)=int_0^pi costcos(x-t)dt, 0lexle2pi . Which of the following hold(s) good? (A) f(x) is continuous but not differentiable in (0,2pi) (B) There exists at least one c in (0,2pi) such that f\'(c)=0 (C) Maximum value of f is pi/2 (D) Minimum value of f is -pi/2

Let f:R->R be a twice differential function such that f(pi/4)=0,f((5pi)/4)=0 and f(3)=4 , then show that there exist a cin(0,2pi) such that f^(primeprime)(c)+sinc-cosc<0

If f and g are continuous functions on [ 0, pi] satisfying f(x) +f(pi-x) =1=g (x)+g(pi-x) then int_(0)^(pi) [f(x)+g(x)] dx is equal to

If f(x) is continuous in [0,pi] such that f(pi)=3 and int_0^(pi/2)(f(2x)+f^(primeprime)(2x))sin2x dx=7 then find f(0).

Prove that the function f(x)=cosx is strictly decreasing in (0,\ pi)

If f(x)|cos(2x)cos(2x)"sin"(2x)-cosxcosx-sinxsinxsinxcosx|,t h e n: a f^(prime)(x)=0 at exactly three point in (-pi,pi) b f^(prime)(x)=0 at more than three point in (-pi,pi) c f(x) attains its maximum at x=0 d f(x) attains its minimum at x=0

A point where function f(x) = [sin [x]] is not continuous in (0,2pi) [.] denotes the greatest integer le x, is

If int_(0)^(x)f(x)sint dt=" constant, " 0 lt x lt 2pi and f(pi)=2 , then the value of f(pi//2) is

If f'(x)=x|sinx|AAx""in(0,pi/2)andf(0)=(1)/(sqrt3), then f(x) will be