Q.

Let⁡f(x)=[a]2−5[a]+4x3−6{a}2−5{a}+1x−(tan⁡x)×sgn⁡x be an even function for all x∈R . Then the sum of  all possible values of a is (where [ ] and {⋅} denote greatest  integer function and fractional part function, respectively)

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a

176

b

566

c

313

d

353

answer is D.

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Detailed Solution

fx=αx3-βx-tanx· Sgnx ,  where α=a 2-5a+4 ,β = 6a2-5a+1 since f is even function ⇒f-x=fx ⇒ -αx3+βx-tanx ·Sgnx=αx3-βx-tanx ·Sgnx ⇒ 2αx3-βx=0 for all x ∈R⇒α =0 and β=0   Identity in x therefore a2-5a+4=0  and 6a2-5a+1=0 ⇒          a=1 or a =4    and  a=13 or 12     ⇒  a = 1+13 ,1+12,4+13,4+12 ⇒  sum of the values of a =353
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