Let `f(x)="max"{1+sin x ,1,1-cosx},x in [0,2pi],a n dg(x)=max{1,|x-1|}, x in Rdot` Then (a)`g(f(0))=1` (b) `g(f(1))=1` (c)`f(f(1))=1` (d) `f(g(0))=1sin1`

Let f(x)={1-|x|,|x|<=1 and 0,|x|<1 and g(x)=f(x-)+f(x+1) ,for all x in R .Then,the value of int_(-3)^(3)g(x)dx is

Let g^(prime)(x)>0a n df^(prime)(x)<0AAx in Rdot Then (f(x+1))>g(f(x-1)) f(g(x-1))>f(g(x+1)) g(f(x+1))

Let g(x)gt0andf'(x)lt0,AAx in R, then show g(f(x+1))ltg(f(x-1)) f(g(x+1))ltf(g(x-1))

If f(x) = (1)/(1 -x) x ne 1 and g(x) = (x-1)/(x) , x ne0 , then the value of g[f(x)] is :

If f(x)=[cosx-sinx -sin x cosc 1] and g(y)=[cosy sin y siny cos y], then [f(x)g(y)]^-1 is equal to (a) f(-x)g(-y) (b) g(-y)f(-x) (c) f(x^-1)g(y^-1) (d) g(y^-1)f(x^-1)

Let g(x)=2f((x)/(2))+f(1-x) and f'(x)<0 in 0<=x<=1 then g(x)

Let g(x)=1+x-[x] and f(x)=-1 if x 0, then f|g(x)]=1x>0

Sometimes functions are defined like f(x)=max{sinx,cosx} , then f(x) is splitted like f(x)={{:(cosx, x in (0,(pi)/(4)]),(sinx, x in ((pi)/(4),(pi)/(2)]):} etc. If f(x)=max{(1)/(2),sinx} , then f(x)=(1)/(2) is defined when x in

Let f(x)={x-1,-1<=x<0 and x^(2),0<=x<=1,g(x)=sin x and h(x)=f(|g(x)|)+[f(g(x)) Then