If the function `g:(-oo,oo)->(-pi/2,pi/2)` is given by `g(u)=2tan^-1(e^u)-pi/2.` Then, `g` is

Let the function g,(-oo,oo)rarr(-pi/2,pi/2) be given by g(u)=2tan^(-1)(e^u)-pi/2, Then ,g is

Let the function g: (-oo,oo)rarr (-pi //2,pi//2) be given by g(u) = 2 tan^(-1) (e^u)-pi/2 Then g is

Let the function g:(-oo, oo) to (-(pi)/(2), (pi)/(2)) be given by g(u)=2tan^(-1)(e^(u))-(pi)/(2) . Then g is :

If g(x)=2tan^-1(e^x)-pi/2 is Even /odd

f: (0,oo) to (-pi/2,pi/2)" be defined as, "f(x)=tan^(-1) (log_(e)x) . The above function can be classified as :

f: (0,oo) to (-pi/2,pi/2)" be defined as, "f(x)=tan^(-1) (log_(e)x) . The above function can be classified as :

Consider the function f:[0,pi/2]rarrR given by f(x)=sinx and g:[0,pi/2]rarrR given by g(x)=cosx . Show that f and g are one-one functions.

Consider the function f:[0,pi/2]rarrR given by f(x)=sinx and g:[0,pi/2]rarrR given by g(x)=cosx . Is f+g one-one?Why?

If 1+2 x is a function having (-(pi)/(2), (pi)/(2)) as domain and (-oo, oo) as codomain, then it is

f:(-(pi)/(2),(pi)/(2))rarr(-oo,oo) defined by f(x)=tan x is