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).`

Suppose that f(x) is continuous in [0, 1] and f(0) = 0, f(1) = 0 . Prove f(c) = 1 - 2c^(2) for some c in (0, 1)

let f(x)=2+cosx for all real x Statement 1: For each real t, there exists a pointc in [t,t+pi] such that f'(c)=0 Because statement 2: f(t)=f(t+2pi) for each real t

If f(x)=x sin x then f'(pi/2) = ……

The function f(x) is a non - negative continous function .The area of the region bounded by the curve y=f(x) , X - axis , x=(pi)/(4), x= beta gt (pi)/(4) is beta sinbeta + (pi)/(4) cos beta + sqrt(2) beta then f ((pi)/(2)) ……

Prove that f(x)= 2x-|x| is a continuous function at x= 0.

Prove that the function f(x)= x^(2) is continuous at x=0.

Discuss the continuity of the function f given by f(x)= |x| " at " x=0 .

Show that the function f(x)= |sinx + cos x| is continuous at x= pi

If f:R rarr R is continuous function satisfying f(0) =1 and f(2x) -f(x) =x, AAx in R , then lim_(nrarroo) (f(x)-f((x)/(2^(n)))) is equal to

A function f: R rarr R satisfies the equation f(x+ y)= f(x).f(y) "for all" x, y in R, f(x) ne 0 . Suppose that the function is differentiable at x=0 and f'(0)=2, then prove that f'(x)= 2f(x)