f(x)=(x2−1)|x2−ax+2|+cos|x| Notice that cos(−x)=cosx=cos|x| which means cos|x| is differentiable everywhere in x∈R ⇒f(x) can be non differentiable where |x2−ax+2|=0 ⇒x2−ax+2=0
↗x=α=2
↘x=β
⇒4−2a+2=0⇒a=3 ⇒(x2−3x+2)=0⇒x=1,2 β=1 but f(x) is differentiable at x=1. so it should be bonus. distance of ( α,β ) from line 12x+5y+10=0 ⇒
