Let α=[a]2−5[a]+4 and β=6{a}2−5{a}+1
∴f(x)=αx3−βx−(tanx)×sgn x
Also, as f(x) is an even function.∴f(−x)=f(x)
⇒−αx3+βx−(tanx)×sgn x=αx3−βx−(tanx)×sgn x
⇒2(−αx2−β)x=0 for all x∈R
⇒α=0 and β=0
∴[a]2−5[a]+4=0 and 6{a}2−5{a}+1=0
⇒([a]−1)([a]−4)=0 and (3{a}−1)(2{a}−1)=0
⇒[a]=1 or 4 and {a}=13 or 12
⇒a=1+13, 1+12, 4+13, 4+12
Therefore, sum of all possible values of a is 353∴3a=35