Let f:R→R be defined by f(x)=ax+b∀x∈R when a,b∈R and a≠1 If (fofofofof )(x) = 32x +93, then value of b is
(fof)(x)=f(f(x))=af(x)+b =a(ax+b)+b=a2x+ab+b ⇒ (fofof)(x)=(fof)(f(x)) =a2f(x)+ab+b =a3x+(a2+a+1)b ⇒ (fofofof)(x)=a3f(x)+(a2+a+1)b =a4x+(a3+a2+a+1)b =a4x+a4−1a−1b ⇒ (fofofofof)(x)=a4f(x)+a4−1a−1b
=a5x+a4b+a4−1a−1b =a5x+a5−1a−1b∴ a5=32⇒a=2
Thus, 31b = 93 ⇒b=3