Let `f(x)=1/(x-[x] )g(x)=sqrt(In{x}-In[x]) h(x)=log_((0.5+x))((x^2+2x-3)/(4x^2-4x-3)),` where `{x},[x]` represent fractional function and greatest integer function then

lim_(x rarr2^(+))((2{x}-4)/([x]-3)) is equal to,where {.} and [.] represents fractional part of x and greatest integer function

f:(2,3)rarr(0,1) defined by f(x)=x-[x], where [.] represents the greatest integer function.

The domain of the function f(x)=log_(e){sgn(9-x^(2))}+sqrt([x]^(3)-4[x]) (where [.] represents the greatest integer function is

f:(2,3)->(0,1) defined by f(x)=x-[x] ,where [dot] represents the greatest integer function.

The domain of the function f(x)=log_e {sgn(9-x^2)}+sqrt([x]^3-4[x]) (where [] represents the greatest integer function is

The domain of the function f(x)=log_e {sgn(9-x^2)}+sqrt([x]^3-4[x]) (where [] represents the greatest integer function is

Let f(x) = log _({x}) [x] g (x) =log _({x})-{x} h (x) = log _([x ]) {x} where [], {} denotes the greatest integer function and fractional part function respectively. Domine of h (x) is :

Let f(x)=|(x+(1)/(2))[x]| when -2<=x<=2| .where [.] represents greatest integer function.Then