Let `f(x)=ax+b` and `g(x)=cx+d,a!=0,c!=0` .Assume `a=1,b=2` .If `(fog)(x)=(gof)(x)` for all `x` ,what can you say about c and d

Let f(x)=ax+b and g(x)=cx+d,a!=0 Assume a=1,b=2. If (fog)(x)=(gof)(x) for all x, what can you say about c and d?

Let f(x)=ax+b and g(x)=x+2,a,b in{1,2,3...,8} If (fog)(x)=(gof)(x) for all x then write the number of different possible values of a+b?

If f(x)=-1+|x-2|,0<=x<=4 and g(x)=2-|x|,-1<=x<=3 then find (fog)(x) and (gof)(x)

Let f(x)=x^2+x+1 and g(x)=sinx . Show that fog!=gof .

If f(x)=ax+b and g(x)=cx+d, then f(g(x))=g(f(x)) is equivalent to

Let f(x)=x^2 and g(x)=2^x . Then the solution set of the equation fog(x)=gof(x) is R (b) {0} (c) {0, 2} (d) none of these

Let f(x) = (ax +b)/(cx + d) (da-cb ne 0, c ne0) then f(x) has