Given:
f(x)=ax−3
g(x)=xb+c,x∈R (fog)−1(x)=(2x−7)31
Now, let
h(x)=(f∘g)(x)
⇒h−1(x)=(2x−7)31
Let
y=(2x−7)31
⇒y3=(2x−7)
⇒x=2y3+7
So, inverse of h−1(x) is 2x3+7 i.e.,
h(x)=fog(x)=2x3+7
Also,
fog(x)=a(xb+c)−3
⇒axb+ac−3=2x3+7
On comparing, we get
a=2,b=3,c=5
So,
f(x)=2x−3g(x)=x3+5
Now,
fog(ac)=fog(10)=f(1005)=2007
(gof)(b)=gof(3)=g(3)=32
Therefore,
fog(ac)+(gof)(b)=2007+32=2039