`Off(x)=int_0^x("cos"(sint)+"cos"(cost)dt ,t h e nf(x+pi)i s` (a)`f(x)+f(pi)` (b) `f(x)+2(pi)` (c)`f(x)+f(pi/2)` (d) `f(x)+2f(pi/2)`

If f(x)=int_0^x("cos"(sint)+"cos"(cost)dt ,t h e nf(x+pi)i s f(x)+f(pi) (b) f(x)+2(pi) f(x)+f(pi/2) (d) f(x)+2f(pi/2)

Let f(x)=1/x^2 int_0^x (4t^2-2f'(t))dt then find f'(4)

f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt The range of f(x) is

If f(x)=cosx-int_0^x(x-t)f(t)dt ,t h e nf^(primeprime)(x)+f(x) is equal to (a) -cosx (b) -sinx (c) int_0^x(x-t)f(t)dt (d) 0

A function f: R->R satisfies sinxcosy(f(2x+2y)-f(2x-2y)=cosxsiny(f(2x+2y)+f(2x-2y))dot If f^(prime)(0)=1/2,t h e n (a) f''(x)=f(x)=0 (b) 4f''(x)+f(x)=0 (c) f''(x)+f(x)=0 (d) 4f''(x)-f(x)=0

f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt The value of int_(0)^(pi//2) f(x)dx is

f(x)=tanx" at "x=pi/2

If f(x)=sin^(-1)cosx , then the value of f(10)+f^(prime)(10) is (a) 11-(7pi)/2 (b) (7pi)/2-11 (c) (5pi)/2-11 (d) none of these

If f(x)=log((1+x)/(1-x)),t h e n (a) f(x_1)f(x)=f(x_1+x_2) (b) f(x+2)-2f(x+1)+f(x)=0 (c) f(x)+f(x+1)=f(x^2+x) (d) f(x_1)+f(x_2)=f((x_1+x_2)/(1+x_1x_2))

If f(x)=cos[pi^2]x +cos[-pi^2]x , where [x] stands for the greatest integer function, then (a) f(pi/2)=-1 (b) f(pi)=1 (c) f(-pi)=0 (d) f(pi/4)=1