A function `f` is defined by `f(x)=\ int_0^picostcos(x-t)dt ,\ 0lt=xlt=2pi` then find its minimum value.

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

A function fis defined by f(x)=int_(0)^( pi)cos t cos(x-t)dt,0<=x<=2 pi then which of the following.hold(s) good?

F(x)= int_(0)^(x)(cost)/((1+t^(2)))dt0lt=xlt=2pi . Then

Let f(x) = int_(0)^(x)(t-1)(t-2)^(2) dt , then find a point of minimum.

Let f(x) be a differentiable function satisfying f(x)=int_(0)^(x)e^((2tx-t^(2)))cos(x-t)dt , then find the value of f''(0) .

If f(x)=int_0^x (sint)/(t)dt,xgt0, then

A function f(x) satisfies f(x)=sin x+int_(0)^(x)f'(t)(2sin t-sin^(2)t)dt is

If f(x) =int_(0)^(x) sin^(4)t dt , then f(x+2pi) is equal to

The relation f is defined by f(x)={x^2,""0lt=xlt=3 3x ,""""3lt=xlt=10 The relating g is defined by g(x)={x^2,""0lt=xlt=3 3x ,""""2lt=xlt=10 Show that f is a function and g is not a function.