Let a non constant polynomial f satisfies the relation f(f(f(x)))+(1−p)f(x)=3∀x∈R, where p is some real number. If leading coefficient of f(x) is negative and f(0)=4 then ∫−11 f−1(x)dx is equal to

Degree is 1⇒f(x) must be linear  Let f(x)=ax+4⇒f(f(f(x)))=a3x+4a2+4a+4⇒a3x+4a2+4a+4+(1−p)(ax+4)=3, ∀x∈R⇒a3+(1−p)a=0⇒a2=p−1 4a2+4a+4+4(1-p)=3-4(1-p) +4a+4+4(1-p)=34a=-1⇒a=-14f(x)=-14x+4f-1(x)=4(4-x)