Let f(x)=([a]2−5[a]+4)x3+(6{a}2−5{a}+1)x−(tanx)×sgnx be an even function for all x∈R, then the sum of all possible values of a is (where [.],{.},sgnx represents greatest integer function , fractional part function and signum function respectively)
A
176
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B
536
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C
313
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D
353
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Solution
The correct option is D353 Let α=[a]2−5[a]+4,β=6{a}2−5{a}+1, then f(x)=αx3+βx−(tanx)×sgnx
∵f(x) is an even function ∴f(−x)=f(x) ⇒−αx3−βx−(−tanx)×(−sgnx)=αx3+βx−(tanx)×sgnx ⇒−2(αx2+β)x=0∀x∈R ⇒α=0,β=0 ⇒[a]2−5[a]+4=0and6{a}2−5{a}+1=0 ⇒([a]−1)([a]−4)=0and(3{a}−1)(2{a}−1)=0 ⇒a=1+13,1+12,4+13,4+12 ⇒a=43,32,133,92