Answer:
Option A
Explanation:
We have ,
f(x)={cos2xfor−∞<x<0e3xfor0≤x<3x2−4x+3,for3≤x≤6log(15x−89)x−6,forx>6
∵limx→3−e3x=e9 and limx→3+x2−4x+3=9−12+3=0
Clearly , limx→3−f(x)≠limx→3+f(x)
∴ f(x) is discontinuous at x=3
∴ a=3,
Now, limx→3x2−9x3−5x2+9x−9=limx→3(x−3)(x+3)(x−3)(x2−2x+3)
= 3+3(3)2−2(3)+3=22=1