If `f: RR->RR` be defined as follows: `f(x)={1,\ if\ x in QQ, -1,\ if\ x !in QQ` Find: `f(1//2),f(pi),\ f(sqrt(2))`

Let RR and QQ be the set of real numbers and rational numbers respectively and f:RR to RR be defined as follows: f(x)={:{(5",", "when "x in QQ),(-5","," when " x !in QQ):} Find f(3),f(sqrt(3)),f(3.6),f(pi),f(e) and f(3.36),

Let, the function f:RRrarrRR be defined by, f(x)={:{(1 ,"when " x in QQ),(-1,"when "x !in QQ):}" Find" f(2),f(sqrt(2)),f(pi),f(2.23),f(e)

Let f: RR to RR be defined by f(x)=1 -|x|. Then the range of f is

Let, the function f:RRrarrRR be defined by, f(x)={:{(1 ,"when " x in QQ),(-1,"when "x !in QQ):}" find" pre-image of 1 and (-1)

Let, the function f:RRrarrRR be defined by, f(x)={:{(1 ,"when " x in QQ),(-1,"when "x !in QQ):}" find" range of f

Let f , g : RR to RR be defined as f( x) = 2x -|x| and g(x) = 2x+|x|. Find f(g).

If f,g : RR to RR are defined by f(x) = |x| +x and g(x) = |x|-x, find g o f and f o g.

If f, g : RR to RR are defined by f(x) = |x| - x , and g(x) = |x| - x , find g^(@) f and f^(@)g .