`f:R rarr R`, defined by `f(x)=sinx and g:R rarrR`, defined by `g(x)=x^(2), (f_(o)g)(x)=`

If f:R rarr R defined by f(x)=sin x and g:R rarr R defined by g(x)=x^(2), then (fog)(x)=(a)x^(2)+sin x (b) x^(2)sin x(c)sin^(2)x(d)sin x^(2)

If f: R rarr R be defined by f(x) =e^(x) and g:R rarr R be defined by g(x) = x^(2) the mapping gof : R rarr R be defined by (gof)(x) =g[f(x)] forall x in R then

If f : R rarr R is defined by f(x)=x^2+2 and g : R-{1} rarr R is defined by g(x)=x/(x-1) , find (fog)(x) and (gof)(x) .

Let f: R rarr R be defined by f(x) = 2x - 3 and g: R rarr R be defined by g(x) = (x+3)/(2) , show that fogʻ= I_(R)

Let f:R rarr R is defined by f(x)=3x-2 and g:R rarr R is defined by g(x)=(x+2)/3 . Show that f.g=g.f .

If f : R rarr R be defined by f(x)=2x-3 and g : R rarr R be defined by g(x)= (x+3)/2 , show that fog=gof=I_R

If f , R rarr R defined by f (x) = 3x - 6 and g : R rarr R defined by g(x) = 3x + k if fog = gof then the value of k is ........... .

Let f : R rarr R be defined by f(x) = 3x - 2 ad g : R rarr R be defined by g(x) = (x+2)/(3) . Shwo that fog = I_R = gof

Let f:R rarr R be defined as f(x)=2x-1 and g:R-{1}rarr R be defined as g(x)=(x-(1)/(2))/(x-1) .Then the composition function f(g(x)) is :

Let f : R rarr R be defined by f(x) = x^(2) + 3x + 1 and g : R rarr R is defined by g(x) = 2x - 3, Find, (i) fog (ii) gof