For the recurrence equation $$T(N)=aT(N/b)+f(N)$$, if $$af(N/b)=

For the recurrence equation $$T(N)=aT(N/b)+f(N)$$, if $$af(N/b)=Kf(N)$$ for some constant $$K>1$$, then $$T(N)=\Theta(f(N))$$. ~@[](2)

答案：FALSE