Let `g(x)=1+x-[x] and f(x)={-1,x < 00, x=0 f, x > 0.` Then for all `x,f(g(x))` is equal to (where [.] represents the greatest integer function). (a) `x` (b) `1` (c) `f(x)` (d) `g(x)`

Find the domain of f(x)=sqrt(([x]-1))+sqrt((4-[x])) (where [ ] represents the greatest integer function).

if f(x) ={{:((1-|x|)/(1+x),xne-1),(1, x=-1):} then f([2x]) , where [.] represents the greatest integer function , is

Let g(x)=1=x-[x] and f(x)={-1, x < 0 , 0, x=0 and 1, x > 0, then for all x, f(g(x)) is equal to (i) x (ii) 1 (iii) f(x) (iv) g(x)

The greatest value of f(x)=cos(x e^([x])+7x^2-3x),x in [-1,oo], is (where [.] represents the greatest integer function). -1 (b) 1 (c) 0 (d) none of these

Let f(x)=sec^(-1)[1+cos^(2)x], where [.] denotes the greatest integer function. Then the range of f(x) is

Let f(x)=(x^(2)+2)/([x]),1 le x le3 , where [.] is the greatest integer function. Then the least value of f(x) is

If f(x) = [x] , 0<= {x} < 0.5 and f(x) = [x]+1 , 0.5<{x}<1 then prove that f (x) = -f(-x) (where[.] and{.} represent the greatest integer function and the fractional part function, respectively).

if f(x) =ax +b and g(x) = cx +d , then f[g(x) ] - g[f(x) ] is equivalent to

Let f(x) = ([x]+1)/({x}+1) for f: [0, (5)/(2) ) to ((1)/(2) , 3] , where [*] represents the greatest integer function and {*} represents the fractional part of x. Draw the graph of y= f(x) . Prove that y=f(x) is bijective. Also find the range of the function.

Given f(x)=4-(1/2-x)^(2/3),g(x)={("tan"[x])/x ,x!=0 1,x=0 h(x)={x},k(x)=5^((log)_2(x+3)) Then in [0,1], lagranges mean value theorem is not applicable to (where [.] and {.} represents the greatest integer functions and fractional part functions, respectively). f (b) g (c) k (d) h