Let `g(x) = f(f(x))` where `f(x) = { 1 + x ; 0 <=x<=2} and f(x) = {3 - x; 2 < x <= 3}` then the number of points of discontinuity of `g(x)` in [0,3] is :

Let F(x) = f(x) +g(x) , G(x) = f(x) - g( x) and H(x) =(f( x))/( g(x)) where f(x) = 1- 2 sin ^(2) x and g(x) =cos (2x) AA f: R to [-1,1] and g, R to [-1,1] Now answer the following If the solution of F(x) -G (x) =0 are x_1,x_2, x_3 --------- x_n Where x in [0,5 pi] then

Let f(x+y) = f(x) f(y) and f(x) = 1 + x g(x) G(x) where lim_(x->0) g(x) =a and lim_(x->o) G(x) = b. Then f'(x) is

Let g(x)= f(x) +f'(1-x) and f''(x) lt 0 ,0 le x le 1 Then

Let g(x)=f(x)+f(1-x) and f''(x)<0 , when x in (0,1) . Then f(x) is

Let g(x)=f(x)+f(1-x) and f''(x)<0 , when x in (0,1) . Then f(x) is

Let f(x+y)=f(x)f(y) and f(x) = 1 + (sin2x) g(x) where g(x) is continuous . Then f^(1) (x) equals

Let h(x)=f(x)-g(x) , where f(x)=sin^(4)pix and g(x)=log x . Let x_(0),x_(1),x_(2) , ....,x_(n+1_ be the roots of f(x)=g(x) in increasing order. In the above question, the value of n is