Q.
Letf(x)=[a]2−5[a]+4x3−6{a}2−5{a}+1x−(tanx)×sgnx 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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