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Question

Let f(x)=([a]25[a]+4)x3+(6{a}25{a}+1)x(tanx)×sgn x be an even function for all xR, then the sum of all possible values of a is (where [.],{.},sgn x 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 D 353
Let α=[a]25[a]+4,β=6{a}25{a}+1, then
f(x)=αx3+βx(tanx)×sgn x

f(x) is an even function
f(x)=f(x)
αx3βx(tanx)×(sgn x)=αx3+βx(tanx)×sgn x
2(αx2+β)x=0 xR
α=0,β=0
[a]25[a]+4=0 and 6{a}25{a}+1=0
([a]1)([a]4)=0 and (3{a}1)(2{a}1)=0
a=1+13,1+12,4+13,4+12
a=43,32,133,92

Sum of values of a
=353

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